Solucionário Fundamentos Circuitos Elétricos-Sadiku 5ed

W

ith its objective to present circuit analysis in a manner that is clearer, more interesting, and easier to understand than other texts, Fundamentals of Electric Circuits by Charles Alexander and Matthew Sadiku has become the student choice for introductory electric circuits courses.

FiFth Edition

FiFt h Edition

Building on the success of the previous editions, the fifth edition features the latest updates and advances in the field, while continuing to present material with an unmatched pedagogy and communication style.

Fundamentals of

Pedagogical Features

Matched Example Problems and Extended Examples. Each illustrative example is immediately followed by a practice problem and answer to test understanding of the preceding example. one extended example per chapter shows an example problem worked using a detailed outline of the six-step method so students can see how to practice this technique. Students follow the example step-by-step to solve the practice problem without having to flip pages or search the end of the book for answers.

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Comprehensive Coverage of Material. not only is Fundamentals the most comprehensive text in terms of material, but it is also self-contained in regards to mathematics and theory, which means that when students have questions regarding the mathematics or theory they are using to solve problems, they can find answers to their questions in the text itself. they will not need to seek out other references.

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Computer tools. PSpice® for Windows is used throughout the text with discussions and examples at the end of each appropriate chapter. MAtLAB® is also used in the book as a computational tool.

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new to the fifth edition is the addition of 120 national instruments Multisim™ circuit files. Solutions for almost all of the problems solved using PSpice are also available to the instructor in Multisim.

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We continue to make available KCidE for Circuits (a Knowledge Capturing integrated design Environment for Circuits).

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An icon is used to identify homework problems that either should be solved or are more easily solved using PSpice, Multisim, and/or KCidE. Likewise, we use another icon to identify problems that should be solved or are more easily solved using MAtLAB.

Teaching Resources McGraw-hill Connect® Engineering is a web-based assignment and assessment platform that gives students the means to better connect with their coursework, with their instructors, and with the important concepts that they will need to know for success now and in the future. Contact your McGraw-hill sales representative or visit www. connect.mcgraw-hill.com for more details.

Electric Circuits INSTRUCTOR SOLUTIONS MANUAL

MD DALIM 1167970 10/30/11 CYAN MAG YELO BLACK

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Fundamentals of

Problem-Solving Methodology. A six-step method for solving circuits problems is introduced in Chapter 1 and used consistently throughout the book to help students develop a systems approach to problem solving that leads to better understanding and fewer mistakes in mathematics and theory.

Electric Circuits

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the text also features a website of student and instructor resources. Check it out at www.mhhe.com/alexander.

Alexander Sadiku

Charles K. Alexander | Matthew n. o. Sadiku

Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = –103.84 mC (b) q = 1. 24x1018 x [-1.602x10-19 C] = –198.65 mC (c) q = 2.46x1019 x [-1.602x10-19 C] = –3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = –26.08 C

Chapter 1, Solution 2

(a) (b) (c) (d) (e)

i = dq/dt = 3 mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200 cos 120 t pA i =dq/dt =  e 4t (80 cos 50 t  1000 sin 50 t )  A

Chapter 1, Solution 3 (a) q(t)   i(t)dt  q(0)  (3t  1) C (b) q(t)   (2t  s) dt  q(v)  (t 2  5t) mC

(c) q(t)   20 cos 10t   / 6   q(0)  (2sin(10t   / 6)  1)  C

10e -30t q(t)   10e sin 40t  q(0)  ( 30 sin 40t - 40 cos t) (d) 900  1600   e - 30t (0.16cos40 t  0.12 sin 40t) C -30t

Chapter 1, Solution 4 q = it = 7.4 x 20 = 148 C

Chapter 1, Solution 5 10

1 t 2 10 q   idt   tdt   25 C 2 4 0 0

Chapter 1, Solution 6 (a) At t = 1ms, i 

dq 30   15 A dt 2

(b) At t = 6ms, i 

dq  0A dt

(c) At t = 10ms, i 

dq  30   –7.5 A dt 4

Chapter 1, Solution 7 25A, dq  i  - 25A, dt   25A,

0t2 2t6 6t8

which is sketched below:


q   idt 

10  1  10  1  15 μC 2


(a) q   idt   10 dt  10 C 0

3 5 1  q   idt  10  1  10    5 1 0 (b) 2    15  7.5  5  22.5C 5

(c) q   idt  10  10  10  30 C 0

Chapter 1, Solution 10 q = it = 10x103x15x10-6 = 150 mC

Chapter 1, Solution 11 q= it = 90 x10-3 x 12 x 60 x 60 = 3.888 kC E = pt = ivt = qv = 3888 x1.5 = 5.832 kJ

Chapter 1, Solution 12 For 0 < t < 6s, assuming q(0) = 0, t

t





0

0

q (t )  idt  q (0 )  3tdt  0  1.5t 2

At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s, t

t





6

6

q (t )  idt  q (6 )  18 dt  54  18 t  54

At t=10, q(10) = 180 – 54 = 126 For 10
t





10

10

q (t )  idt  q (10 )  ( 12)dt  126  12t  246

At t=15, q(15) = -12x15 + 246 = 66 For 15


q (t )  0 dt  q (15) 66 15

Thus,  1.5t 2 C, 0 < t < 6s   18 t  54 C, 6 < t < 10s q (t )   12t  246 C, 10 < t < 15s  66 C, 15 < t < 20s 

The plot of the charge is shown below.

140

120

100

q(t)

80

60

40

20

0 0

5

10 t

15

20

Chapter 1, Solution 13 (a) i = [dq/dt] = 20πcos(4πt) mA p = vi = 60πcos2(4πt) mW At t=0.3s,

p = vi = 60πcos2(4π0.3) mW = 123.37 mW

(b) W = W = 30π[0.6+(1/(8π))[sin(8π0.6)–sin(0)]] = 58.76 mJ








(a) q   idt   0.02 1 - e -0.5t dt  0.02 t  2e -0.5t 0

(b)



1 0





 0.02 1  2e -0.5  2 = 4.261 mC

p(t) = v(t)i(t) p(1) = 10cos(2)x0.02(1–e–0.5) = (–4.161)(0.007869) = –32.74 mW

Chapter 1, Solution 15  0.006 2t q   idt   0.006e dt  e 0 (a) 2 2



2

- 2t



0

 0.003 e  1  2.945 mC

(b)

-4

10di  0.012e - 2t (10)  0.12e - 2t V this leads to p(t) = v(t)i(t) = dt (-0.12e-2t)(0.006e-2t) = –720e–4t µW v

3

(c) w   pdt  -0.72 e 0

 720 -4t 6 dt  e 10 = –180 µJ 4 0 3

- 4t

Chapter 1, Solution 16 (a)  30t mA, 0 < t <2 i (t )   120-30t mA, 2 < t<4 5 V, 0 < t <2 v(t )    -5 V, 2 < t<4  150t mW, 0 < t <2 p(t )   -600+150t mW, 2 < t<4 which is sketched below. p(mW) 300

1

2

-300

(b) From the graph of p, 4

W   pdt  0 J 0

4

t (s)


 p=0

 -205 + 60 + 45 + 30 + p 3 = 0

p 3 = 205 – 135 = 70 W Thus element 3 receives 70 W.

Chapter 1, Solution 18 p 1 = 30(-10) = -300 W p 2 = 10(10) = 100 W p 3 = 20(14) = 280 W p 4 = 8(-4) = -32 W p 5 = 12(-4) = -48 W

Chapter 1, Solution 19 I = 8 –2 = 6 A Calculating the power absorbed by each element means we need to find vi for each element. p 8 amp source = –8x9 = –72 W p element with 9 volts across it = 2x9 = 18 W p element with 3 bolts across it = 3x6 = 18 W p 6 volt source = 6x6 = 36 W One check we can use is that the sum of the power absorbed must equal zero which is what it does.

Chapter 1, Solution 20 p 30 volt source = 30x(–6) = –180 W p 12 volt element = 12x6 = 72 W p 28 volt e.ement with 2 amps flowing through it = 28x2 = 56 W p 28 volt element with 1 amp flowing through it = 28x1 = 28 W p the 5Io dependent source = 5x2x(–3) = –30 W Since the total power absorbed by all the elements in the circuit must equal zero, or 0 = –180+72+56+28–30+p into the element with Vo or p into the element with Vo = 180–72–56–28+30 = 54 W Since p into the element with Vo = V o x3 = 54 W or V o = 18 V.


p 60   0.5 A v 120 q = it = 0.5x24x60x60 = 43.2 kC N e  qx6.24 x1018  2.696 x1023 electrons p  vi

 

i

Chapter 1, Solution 22 q = it = 40x103x1.7x10–3 = 68 C

Chapter 1, Solution 23 W = pt = 1.8x(15/60) x30 kWh = 13.5kWh C = 10cents x13.5 = $1.35

Chapter 1, Solution 24 W = pt = 60 x24 Wh = 0.96 kWh = 1.44 kWh C = 8.2 centsx0.96 = 11.808 cents


Cost  1.5 kW 

3.5 hr  30  8.2 cents/kWh = 21.52 cents 60


0.8A  h  80 mA 10h (b) p = vi = 6  0.08 = 0.48 W (c) w = pt = 0.48  10 Wh = 0.0048 kWh (a) i 


(a) Let T  4h  4  3600 T

q   idt   3dt  3T  3  4  3600  43.2 kC 0

T T 0. 5t   ( b) W   pdt   vidt   ( 3) 10  dt 0 0 3600   43600

 0. 25t 2    310t  3600  0   475.2 kJ

( c)

 340  3600  0. 25  16  3600

W  475.2 kWs, (J  Ws) 475.2 kWh  9 cent  1.188 cents Cost  3600


(a) i 

P 60  V 120 = 500 mA

(b) W  pt  60  365  24 Wh  525.6 kWh Cost  $0.095  525.6

= $49.93

Chapter 1, Solution 29 (20  40  15  45)  30  hr  1.8 kW  hr 60  60   2.4  0.9  3.3 kWh Cost  12 cents  3.3  39.6 cents w  pt  1. 2kW

Chapter 1, Solution 30 Monthly charge = $6 First 250 kWh @ $0.02/kWh = $5 Remaining 2,436–250 kWh = 2,186 kWh @ $0.07/kWh= $153.02 Total = $164.02

Chapter 1, Solution 31 Total energy consumed = 365(120x4 + 60x8) W Cost = $0.12x365x960/1000 = $42.05

Chapter 1, Solution 32 i = 20 µA q = 15 C t = q/i = 15/(20x10-6) = 750x103 hrs


i

dq  q   idt  2000  3  10  3  6 C dt

Chapter 1, Solution 34 (a)

Energy =

 pt

= 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2

= 10 kWh (b)

Average power = 10,000/24 = 416.7 W


energy = (5x5 + 4x5 + 3x5 + 8x5 + 4x10)/60 = 2.333 MWhr

Chapter 1, Solution 36 160A  h 4A 40 160Ah 160, 000h ( b) t    6,667 days 0.001A 24h / day

(a)

i

Chapter 1, Solution 37 W = pt = vit = 12x 40x 60x60 = 1.728 MJ

Chapter 1, Solution 38 P = 10 hp = 7460 W W = pt = 7460  30  60 J = 13.43  106 J

Chapter 1, Solution 39 W = pt = 600x4 = 2.4 kWh C = 10cents x2.4 = 24 cents

Chapter 2, Solution 1. Design a problem, complete with a solution, to help students to better understand Ohm’s Law. Use at least two resistors and one voltage source. Hint, you could use both resistors at once or one at a time, it is up to you. Be creative. Although there is no correct way to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem The voltage across a 5-k resistor is 16 V. Find the current through the resistor. Solution

v = iR

i = v/R = (16/5) mA = 3.2 mA

Chapter 2, Solution 2 p = v2/R 

R = v2/p = 14400/60 = 240 ohms

Chapter 2, Solution 3 For silicon,   6.4 x102 -m. A   r 2 . Hence,

R

L A



L  r2

 

r2 

 L 6.4 x102 x 4 x102   0.033953 R  x 240

r = 184.3 mm

Chapter 2, Solution 4 (a) (b)

i = 40/100 = 400 mA i = 40/250 = 160 mA

Chapter 2, Solution 5 n = 9;

l = 7; b = n + l – 1 = 15

Chapter 2, Solution 6 n = 12;

l = 8;

b = n + l –1 = 19

Chapter 2, Solution 7 6 branches and 4 nodes

Chapter 2, Solution 8. Design a problem, complete with a solution, to help other students to better understand Kirchhoff’s Current Law. Design the problem by specifying values of i a , i b , and i c , shown in Fig. 2.72, and asking them to solve for values of i 1 , i 2 , and i 3 . Be careful specify realistic currents. Although there is no correct way to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Use KCL to obtain currents i 1, i 2, and i 3 in the circuit shown in Fig. 2.72.

Solution 12 A a i1 b

8A i2

i3 12 A

c At node a, At node c, At node d,

9A d

8 = 12 + i 1 9 = 8 + i2 9 = 12 + i 3

i 1 = - 4A i 2 = 1A i 3 = -3A

Chapter 2, Solution 9 At A, 1+6–i 1 = 0 or i 1 = 1+6 = 7 A At B, –6+i 2 +7 = 0 or i 2 = 6–7 = –1 A At C, 2+i 3 –7 = 0 or i 3 = 7–2 = 5 A


–8A

1

4A i2

i1

3

–6A At node 1,

–8–i 1 –6 = 0 or i 1 = –8–6 = –14 A

At node 2,

–(–8)+i 1 +i 2 –4 = 0 or i 2 = –8–i 1 +4 = –8+14+4 = 10 A

Chapter 2, Solution 11 V1  1  5  0 5  2  V2  0

   

V1  6 V V2  3 V

Chapter 2, Solution 12 + 30v –

loop 2 – 50v + + 40v -

+ 20v –

loop 1

+ v1 –

+ v2 –

loop 3

+ v3 –

For loop 1,

–40 –50 +20 + v 1 = 0 or v 1 = 40+50–20 = 70 V

For loop 2,

–20 +30 –v 2 = 0 or v 2 = 30–20 = 10 V

For loop 3,

–v 1 +v 2 +v 3 = 0 or v 3 = 70–10 = 60 V

Chapter 2, Solution 13 2A

I2

7A

1

I4

2

3

4 4A

I1 3A

At node 2, 3  7  I2  0

 

I 2  10 A

At node 1, I1  I 2  2

 

I 1  2  I 2  12 A

At node 4, 2  I4  4

 

I 4  2  4  2 A

At node 3, 7  I4  I3

 

I3  7  2  5 A

Hence, I 1  12 A,

I 2  10 A,

I 3  5 A,

I 4  2 A

I3


+ 3V -

+ I3

4V

+

-

V3 -

+

I4 2V -

+ - V4

I2 +

V2 +

+ 5V

I1 -

For mesh 1, V4  2  5  0

 

V4  7V

For mesh 2, 4  V3  V4  0

 

V3  4  7  11V

 

V1  V3  3  8V

 

V2  V1  2  6V

For mesh 3, 3  V1  V3  0

For mesh 4, V1  V2  2  0

Thus, V1  8V ,

V1

V2  6V ,

V3  11V ,

V4  7V

Chapter 2, Solution 15 Calculate v and i x in the circuit of Fig. 2.79. 12  + v

10 V

+ _

+ 16 V – –

ix +

+

4V

_

_

Figure 2.79 For Prob. 2.15.

Solution

For loop 1, –10 + v +4 = 0, v = 6 V For loop 2, –4 + 16 + 3i x =0, i x =

–4 A

3 ix


Determine V o in the circuit in Fig. 2.80. 16 

14 

 + 10 V

+ _

Vo

+ _

25 V

_

 Figure 2.80 For Prob. 2.16.

Solution

Apply KVL, –10 + (16+14)I + 25 = 0 or 30I = 10–25 = – or I = –15/30 = –500 mA Also, –10 + 16I + V o = 0 or V o = 10 – 16(–0.5) = 10+8 = 18 V

Chapter 2, Solution 17 Applying KVL around the entire outside loop we get, –24 + v 1 + 10 + 12 = 0 or v 1 = 2V Applying KVL around the loop containing v 2 , the 10-volt source, and the 12-volt source we get, v 2 + 10 + 12 = 0 or v 2 = –22V Applying KVL around the loop containing v 3 and the 10-volt source we get, –v 3 + 10 = 0 or v 3 = 10V


Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32

I = 4A

-V ab + 5I + 8 = 0

V ab = 28V

Chapter 2, Solution 19 Applying KVL around the loop, we obtain –(–8) – 12 + 10 + 3i = 0 Power dissipated by the resistor: p 3 = i2R = 4(3) = 12W Power supplied by the sources: p 12V = 12 ((–2)) = –24W p 10V = 10 (–(–2)) = 20W p 8V = (–8)(–2) = 16W

i = –2A

Chapter 2, Solution 20 Determine i o in the circuit of Fig. 2.84. io

54V

22 

+ 

+ –

Figure 2.84 For Prob. 2.20 Solution Applying KVL around the loop, –54 + 22i o + 5i o = 0

i o = 4A

5i o

Chapter 2, Solution 21 Applying KVL, -15 + (1+5+2)I + 2 V x = 0 But V x = 5I, -15 +8I + 10I =0, I = 5/6 V x = 5I = 25/6 = 4.167 V

Chapter 2, Solution 22 Find V o in the circuit in Fig. 2.86 and the power absorbed by the dependent source. 10  + 10 

V1

Vo  25A

2V o

Figure 2.86 For Prob. 2.22 Solution At the node, KCL requires that [–V o /10]+[–25]+[–2V o ] = 0 or 2.1V o = –25 or V o = –11.905 V The current through the controlled source is i = 2V 0 = –23.81 A and the voltage across it is V 1 = (10+10) i 0 (where i 0 = –V 0 /10) = 20(11.905/10) = 23.81 V. Hence, p dependent source = V 1 (–i) = 23.81x(–(–23.81)) = 566.9 W Checking, (25–23.81)2(10+10) + (23.81)(–25) + 566.9 = 28.322 – 595.2 + 566.9 = 0.022 which is equal zero since we are using four places of accuracy!

Chapter 2, Solution 23 8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3 The circuit is reduced to that shown below. 1

ix +

20A

vx

–

2

3

Applying current division, i x = [2/(2+1+3)]20 = 6.667 and v x = 1x6.667 = 6.667 V ix 

2 (6 A)  2 A, 2  1 3

v x  1i x  2V

The current through the 1.2-  resistor is 0.5i x = 3.333 A. The voltage across the 12-  resistor is 3.333 x 4.8 = 16V. Hence the power absorbed by the 12-ohm resistor is equal to (16)2/12 = 21.33 W


I0 =

Vs R1  R2

V0   I 0 R3 R4  = 

R 3R 4 Vs  R1  R 2 R 3  R 4

V0  R3 R4  Vs  R1  R2  R3  R4 

(b)

If R 1 = R 2 = R 3 = R 4 = R, V0  R      10 VS 2R 2 4

 = 40

Chapter 2, Solution 25 V 0 = 5 x 10-3 x 10 x 103 = 50V Using current division, I 20 

5 (0.01x50)  0.1 A 5  20

V 20 = 20 x 0.1 kV = 2 kV p 20 = I 20 V 20 = 0.2 kW

Chapter 2, Problem 26. For the circuit in Fig. 2.90, i o = 3 A. Calculate i x and the total power absorbed by the entire circuit.

ix

10 

io

8

4

2

16 

Figure 2.90 For Prob. 2.26. Solution If i 16 = i o = 3A, then v = 16x3 = 48 V and i 8 = 48/8 = 6A; i 4 = 48/4 = 12A; and i 2 = 48/2 = 24A. Thus, i x = i 8 + i 4 + i 2 + i 16 = 6 + 12 + 24 + 3 = 45 A p = (45)210 + (6)28 + (12)24 + (24)22 + (3)216 = 20,250 + 288 + 576 +1152 + 144 = 20250 + 2106 = 22.356 kW.

Chapter 2, Problem 27. Calculate I o in the circuit of Fig. 2.91. 8

10V

+ 

Io

3

6

Figure 2.91 For Prob. 2.27. Solution The 3-ohm resistor is in parallel with the c-ohm resistor and can be replaced by a [(3x6)/(3+6)] = 2-ohm resistor. Therefore, I o = 10/(8+2) = 1 A.

Chapter 2, Solution 28 Design a problem, using Fig. 2.92, to help other students better understand series and parallel circuits. Although there is no correct way to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find v 1 , v 2 , and v 3 in the circuit in Fig. 2.92.

Solution We first combine the two resistors in parallel 15 10  6 

We now apply voltage division, v1 =

14 (40)  28 V 14  6

v2 = v3 = Hence,

6 (40)  12 V 14  6

v 1 = 28 V, v 2 = 12 V, v s = 12 V

Chapter 2, Solution 29 All resistors in Fig. 2.93 are 5  each. Find R eq .

R eq

Figure 2.93 For Prob. 2.29. Solution

R eq = 5 + 5||[5+5||(5+5)] = 5 + 5||[5+(5x10/(5+10))] = 5+5||(5+3.333) = 5 + 41.66/13.333 = 8.125 Ω

Chapter 2, Problem 30. Find R eq for the circuit in Fig. 2.94.

25 

180  60 

R eq

60 

Figure 2.94 For Prob. 2.30. Solution We start by combining the 180-ohm resistor with the 60-ohm resistor which in turn is in parallel with the 60-ohm resistor or = [60(180+60)/(60+180+60)] = 48. Thus, R eq = 25+48 = 73 Ω.


Req  3  2 // 4 //1  3 

1  3.5714 1/ 2  1/ 4  1 i 1 = 200/3.5714 = 56 A

v 1 = 0.5714xi 1 = 32 V and i 2 = 32/4 = 8 A i 4 = 32/1 = 32 A; i 5 = 32/2 = 16 A; and i 3 = 32+16 = 48 A


Find i 1 through i 4 in the circuit in Fig. 2.96. 60 

i4

i2

200  50 

40  i3

i1

16 A

Figure 2.96 For Prob. 2.32. Solution We first combine resistors in parallel. 40 60 

40x 60 50x 200  24  and 50 200   40  100 250

Using current division principle, 24 40 i1  i 2  (16)  6A, i 3  i 4  (16)  10A 24  40 64

i1 

200 50 (6)  –4.8 A and i 2  (6)  –1.2 A 250 250

i3 

60 40 (10)  –6 A and i 4  (10)  –4 A 100 100

Chapter 2, Solution 33 Combining the conductance leads to the equivalent circuit below i + v -

9A

6S 3S 

1S

4S

i

4S

6x3  2S and 2S + 2S = 4S 9

Using current division,

i

1 1 1 2

(9)  6 A, v = 3(1) = 3 V

9A

+ v -

1S

2S

Chapter 2, Solution 34 160//(60 + 80 + 20)= 80 , 160//(28+80 + 52) = 80  R eq = 20+80 = 100 Ω I = 200/100 = 2 A or p = VI = 200x2 = 400 W.


i 70  200V

+

a

-

+ V1 i1 -

30  Io +

20  i2

b

Vo 5  -

Combining the resistors that are in parallel, 70 30 

i=

70x30  21 , 100

20 5 

20 x5 4  25

200 8 A 21  4

v 1 = 21i = 168 V, v o = 4i = 32 V v v i 1 = 1  2.4 A, i 2 = o  1.6 A 70 20 At node a, KCL must be satisfied i1 = i2 + Io

2.4 = 1.6 + I o

I o = 0.8 A

Hence, v o = 32 V and I o = 800 mA

Chapter 2, Solution 36 20//(30+50) = 16, 24 + 16 = 40, 60//20 = 15 R eq = 80+(15+25)40 = 80+20 = 100 Ω i = 20/100 = 0.2 A If i 1 is the current through the 24- resistor and i o is the current through the 50- resistor, using current division gives i 1 = [40/(40+40)]0.2 = 0.1 and i o = [20/(20+80)]0.1 = 0.02 A or v o = 30i o = 30x0.02 = 600 mV.


Applying KVL, -20 + 10 + 10I – 30 = 0, I = 4

10  RI

  R

10  2.5  I

Chapter 2, Solution 38 20//80 = 80x20/100 = 16, 6//12 = 6x12/18 = 4 The circuit is reduced to that shown below. 2.5 

4 60  15 

16 

R eq (4 + 16)//60 = 20x60/80 = 15 R eq = 2.5+15||15 = 2.5+7.5 = 10 Ω and i o = 35/10 = 3.5 A.

Chapter 2, Solution 39 (a) We note that the top 2k-ohm resistor is actually in parallel with the first 1k-ohm resistor. This can be replaced (2/3)k-ohm resistor. This is now in series with the second 2k-ohm resistor which produces a 2.667k-ohm resistor which is now in parallel with the second 1k-ohm resistor. This now leads to,

R eq = [(1x2.667)/3.667]k = 727.3 Ω. (b) We note that the two 12k-ohm resistors are in parallel producing a 6k-ohm resistor. This is in series with the 6k-ohm resistor which results in a 12k-ohm resistor which is in parallel with the 4k-ohm resistor producing, R eq = [(4x12)/16]k = 3 kΩ.

Chapter 2, Solution 40 Req = 8  4 (2  6 3)  8  2  10 

I=

15 15   1.5 A R eq 10

Chapter 2, Solution 41 Let R 0 = combination of three 12 resistors in parallel

1 1 1 1    R o 12 12 12

Ro = 4

R eq  30  60 (10  R 0  R )  30  60 (14  R )

50  30 

60(14  R ) 74  R

or R = 16 

74 + R = 42 + 3R

Chapter 2, Solution 42 5x 20  4 25

(a)

R ab = 5 (8  20 30)  5 (8  12) 

(b)

R ab = 2  4 (5  3) 8  5 10 4  2  4 4  5 2.857  2  2  1.8181  5.818 

Chapter 2, Solution 43 5x 20 400   4  8  12  25 50

(a)

R ab = 5 20  10 40 

(b)

1 1   1 60 20 30       60 20 30  R ab = 80 (10  10) 

1



60  10 6

80  20  16  100

Chapter 2, Solution 44 For the circuits in Fig. 2.108, obtain the equivalent resistance at terminals a-b.

5

20 

a 2

3

b

Figure 2.108 For Prob. 2.44 Solution First we note that the 5 Ω and 20 Ω resistors are in parallel and can be replaced by a 4 Ω [(5x20)/(5+20)] resistor which in now in series with the 2 Ω resistor and can be replaced by a 6 Ω resistor in parallel with the 3 Ω resistor thus, R ab = [(6x3)/(6+3)] = 2 Ω.


(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8

Rab  5  50  4.8  59.8 

(b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus, Rab  5  12.8  15  32.5

Chapter 2, Solution 46 R eq = 12 + 5||20 + [1/((1/15)+(1/15)+(1/15))] + 5 + 24||8 = 12 + 4 + 5 + 5 + 6 = 32 Ω I = 80/32 = 2.5 A

Chapter 2, Solution 47 5 20 

6 3

5x 20  4 25

6x3  2 9 10 

8 a

b

4

R ab = 10 + 4 + 2 + 8 = 24 

2


(b)

R 1 R 2  R 2 R 3  R 3 R 1 100  100  100   30 R3 10 R a = R b = R c = 30 

Ra =

30x 20  30x50  20x 50 3100   103.3 30 30 3100 3100 Rb   155, R c   62 20 50 R a = 103.3 , R b = 155 , R c = 62  Ra 


R1 =

RaRc 12 *12   4 Ra  Rb  Rc 36

R1 = R2 = R3 = 4 

(b)

60x30  18 60  30  10 60 x10 R2   6 100 30x10 R3   3 100 R1 

R 1 = 18, R 2 = 6, R 3 = 3

2.50 Design a problem to help other students better understand wye-delta transformations using Fig. 2.114. Although there is no correct way to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem What value of R in the circuit of Fig. 2.114 would cause the current source to deliver 800 mW to the resistors. Solution Using R  = 3R Y = 3R, we obtain the equivalent circuit shown below:

R 30mA

3R 3R

3R

3R R 

30mA R

3RxR 3  R 4 4R

3 3Rx R 3  3 3 2 =R 3R  R  R   3R R  3 4  2 4 3R  R 2 P = I2R R = 889 

800 x 10-3 = (30 x 10-3)2 R

3R

3R/2


(a)

30 30  15 and 30 20  30x 20 /(50)  12 R ab = 15 (12  12)  15x 24 /(39)  9.231 

a

a 30 

30 

30  30 

b

12 

20  15 

12 

20  b

(b)

Converting the T-subnetwork into its equivalent  network gives R a'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70  R b'c' = 350/(10) = 35, Ra'c' = 350/(20) = 17.5 

Also

30 70  30x 70 /(100)  21 and 35/(15) = 35x15/(50) = 10.5 R ab = 25 + 17.5 (21  10.5)  25  17.5 31.5 R ab = 36.25  30 

30  25 

10 

20 

a

a 5

b

15 

25 

a’

17.5 

b

70 

b’

35 

c’

15 

c’

Chapter 2, Solution 52 Converting the wye-subnetwork to delta-subnetwork, we obtain the circuit below. 9

3 9

3

9

3

3

3 3

6

3//1 = 3x1/4 = 0.75, 2//1 =2x1/3 = 0.6667. Combining these resistances leads to the circuit below. 3 2.25 

3

2.25

9

3

2

We now convert the wye-subnetwork to the delta-subnetwork. R a = [(2.25x3+2.25x3+2.25x2.25)/3] = 6.188 Ω R b = R c = 18.562/2.25 = 8.25 Ω This leads to the circuit below. 3 6.188 8.25 3

9

8.25 2

R = 9||6.188+8.25||2 = 3.667+1.6098 = 5.277 R eq = 3+3+8.25||5.277 = 9.218 Ω.


Converting one  to T yields the equivalent circuit below: 30 

a’

20 

a

60 

b’

b

4

20 

c’

5

80 

40x10 10 x 50 40x 50  4, R b 'n   5, R c 'n   20 40  10  50 100 100 R ab = 20 + 80 + 20 + (30  4) (60  5)  120  34 65

R a'n =

R ab = 142.32  (b) We combine the resistor in series and in parallel. 30 (30  30) 

30 x 60  20 90

We convert the balanced  s to Ts as shown below: a

30 

30 

a 10 

30  30 

20 

10 

30 

b 30 

10  10 

b R ab = 10 + (10  10) (10  20  10)  10  20  20 40 R ab = 33.33 

10 

10 

20 

Chapter 2, Solution 54 (a) Rab  50  100 / /(150  100  150 )  50  100 / /400  130  (b) Rab  60  100 / /(150  100  150 )  60  100 / /400  140 

Chapter 2, Solution 55 We convert the T to  .

I0 a 24 V +

-

I0 20  40 

60 

10  20 

50 

a 140 

24 V +

35 

-

70 

b R eq

b

R eq R R  R 2 R 3  R 3 R 1 20 x 40  40 x10  10 x 20 1400    35 R ab = 1 2 R3 40 40 R ac = 1400/(10) = 140, R bc = 1400/(20) = 70 70 70  35 and 140 160  140x60/(200) = 42 R eq = 35 (35  42)  24.0625 I 0 = 24/(R ab ) = 997.4mA

60  70 

Chapter 2, Solution 56 We need to find R eq and apply voltage division. We first tranform the Y network to  . 30 

+ 100 V

16 

15 

35 

12 

30  16 

10  20 

-

+ 100 V

a

35 

R eq

R eq

15x10  10x12  12x15 450   37.5 12 12 R ac = 450/(10) = 45, R bc = 450/(15) = 30 R ab =

Combining the resistors in parallel, 30||20 = (600/50) = 12 , 37.5||30 = (37.5x30/67.5) = 16.667  35||45 = (35x45/80) = 19.688  R eq = 19.688||(12 + 16.667) = 11.672 By voltage division, v =

11.672 100 = 42.18 V 11.672  16

37.5  30  45 

c

b 20 

Chapter 2, Solution 57 4 a

2 27  1

18 

b d 10 

36 

c e

7

14 

28 

f 6 x12  12 x8  8x 6 216   18  12 12 R ac = 216/(8) = 27, R bc = 36  4 x 2  2 x8  8x 4 56 R de =  7 8 8 R ef = 56/(4) = 14, R df = 56/(2) = 28 

R ab =

Combining resistors in parallel, 280 36 x 7  7.368, 36 7   5.868 38 43 27 x 3 27 3   2 .7  30

10 28 

4

4

18  5.868 

7.568 

R cn

1.829  3.977 

0.5964 

14  7.568 

18x 2.7 18x 2.7   1.829  18  2.7  5.867 26.567 18x 5.868   3.977  26.567 5.868x 2.7   0.5904  26.567

R an  R bn

2.7 

14 

R eq  4  1.829  (3.977  7.368) (0.5964  14)

 5.829  11.346 14.5964  12.21  i = 20/(R eq ) = 1.64 A

Chapter 2, Solution 58 The resistance of the bulb is (120)2/60 = 240 40 

2A

+ 90 V - 0.5 A VS

240 

+

-

1.5 A

+ 120 V

80 

-

Once the 160 and 80 resistors are in parallel, they have the same voltage 120V. Hence the current through the 40 resistor is equal to 2 amps. 40(0.5 + 1.5) = 80 volts. Thus v s = 80 + 120 = 200 V.

Chapter 2, Solution 59 Three light bulbs are connected in series to a 120-V source as shown in Fig. 2.123. Find the current I through each of the bulbs. Each bulb is rated at 120 volts. How much power is each bulb absorbing? Do they generate much light?

Figure 2.123 For Prob. 2.59. Solution Using p = v2/R, we can calculate the resistance of each bulb. R 30W = (120)2/30 = 14,400/30 = 480 Ω R 40W = (120)2/40 = 14,400/40 = 360 Ω R 50W = (120)2/50 = 14,400/50 = 288 Ω The total resistance of the three bulbs in series is 480+360+288 = 1128 Ω. The current flowing through each bulb is 120/1128 = 0.10638 A. p 30 = (0.10638)2480 = 0.011317x480 = 5.432 W. p 40 = (0.10638)2360 = 0.011317x360 = 4.074 W. p 50 = (0.10638)2288 = 0.011317x288 = 3.259 W. Clearly these values are well below the rated powers of each light bulb so we would not expect very much light from any of them. To work properly, they need to be connected in parallel.

Chapter 2, Solution 60 If the three bulbs of Prob. 2.59 are connected in parallel to the 120-V source, calculate the current through each bulb. Solution Using p = v2/R, we can calculate the resistance of each bulb. R 30W = (120)2/30 = 14,400/30 = 480 Ω R 40W = (120)2/40 = 14,400/40 = 360 Ω R 50W = (120)2/50 = 14,400/50 = 288 Ω The current flowing through each bulb is 120/R. i 30 = 120/480 = 250 mA. i 40 = 120/360 = 333.3 mA. i 30 = 120/288 = 416.7 mA. Unlike the light bulbs in 2.59, the lights will glow brightly!

Chapter 2, Solution 61 There are three possibilities, but they must also satisfy the current range of 1.2 + 0.06 = 1.26 and 1.2 – 0.06 = 1.14. (a)

Use R 1 and R 2 : R = R 1 R 2  80 90  42.35 p = i2R = 70W i2 = 70/42.35 = 1.6529 or i = 1.2857 (which is outside our range) cost = $0.60 + $0.90 = $1.50

(b)

Use R 1 and R 3 : R = R 1 R 3  80 100  44.44  i2 = 70/44.44 = 1.5752 or i = 1.2551 (which is within our range), cost = $1.35

(c)

Use R 2 and R 3 : R = R 2 R 3  90 100  47.37 i2 = 70/47.37 = 1.4777 or i = 1.2156 (which is within our range), cost = $1.65

Note that cases (b) and (c) satisfy the current range criteria and (b) is the cheaper of the two, hence the correct choice is: R 1 and R 3

Chapter 2, Solution 62 p A = 110x8 = 880 W,

p B = 110x2 = 220 W

Energy cost = $0.06 x 365 x10 x (880 + 220)/1000 = $240.90

Chapter 2, Solution 63 Use eq. (2.61), Im 2 x10 3 x100 Rn = Rm   0.04 I  Im 5  2 x10 3 I n = I - I m = 4.998 A p = I 2n R  (4.998) 2 (0.04)  0.9992  1 W

Chapter 2, Solution 64 When R x = 0, i x  10A When R x is maximum, i x = 1A

i.e., R x = 110 - R = 99  R x = 99  Thus, R = 11 ,

R=

110  11  10

R  Rx 

110  110  1

Chapter 2, Solution 65 Rn 

Vfs 50  Rm   1 k  4 k I fs 10mA

Chapter 2, Solution 66 20 k/V = sensitivity =

1 I fs

1 k / V  50 A 20 V The intended resistance R m = fs  10(20k / V)  200k I fs V 50 V R n  fs  R m   200 k  800 k (a) i fs 50 A i.e., I fs =

(b)

p = I fs2 R n  (50 A) 2 (800 k)  2 mW


By current division, i 0 = 5/(5 + 5) (2 mA) = 1 mA V 0 = (4 k) i 0 = 4 x 103 x 10-3 = 4 V

(b)

4k 6k  2.4k. By current division,

5 (2mA)  1.19 mA 1  2.4  5 v '0  (2.4 k)(1.19 mA)  2.857 V

i '0 

v 0  v '0 1.143 x 100% = x100  28.57% v0 4

(c)

% error =

(d)

4k 36 k  3.6 k. By current division,

5 (2mA)  1.042mA 1  3.6  5 v '0 (3.6 k)(1.042 mA)  3.75V

i '0 

% error =

v  v '0 0.25x100 x100%   6.25% v0 4


40  24 60

4  100 mA 16  24 4 i'   97.56 mA 16  1  24 0.1  0.09756 % error = x100%  2.44% 0.1

i=

(b) (c)

Chapter 2, Solution 69 With the voltmeter in place, R2 Rm V0  VS R1  R S  R 2 R m where R m = 100 k without the voltmeter, R2 V0  VS R1  R 2  R S

100 k 101

(a)

When R 2 = 1 k, R m R 2 

(b)

100 V 0 = 101 (40)  1.278 V (with) 100  30 101 1 V0 = (40)  1.29 V (without) 1  30 1000 When R 2 = 10 k, R 2 R m   9.091k 110 9.091 V0 = (40)  9.30 V (with) 9.091  30 10 V0 = (40)  10 V (without) 10  30 When R 2 = 100 k, R 2 R m  50k

(c)

50 (40)  25 V (with) 50  30 100 V0 = (40)  30.77 V (without) 100  30 V0 


(a) Using voltage division, 12 (25)  15V 12  8 10 vb  (25)  10V 10  15  va  vb  15  10  5V va 

vab

(b)

c 8k 

+ 25 V –

15k  a

12k 

b 10k 

v a = 0; v ac = –(8/(8+12))25 = –10V; v cb = (15/(15+10))25 = 15V. v ab = v ac + v cb = –10 + 15 = 5V. v b = –v ab = –5V.

Chapter 2, Solution 71 R1 iL Vs + −

RL

Given that v s = 30 V, R 1 = 20 , I L = 1 A, find R L .

v s  i L ( R1  R L )

 

RL 

vs 30  R1   20  10 iL 1

Chapter 2, Solution 72 Converting the delta subnetwork into wye gives the circuit below.

1 ⅓ 1

⅓

Z in 10 V

+ _

1 1 1 1 1 4 Z in   (1  ) //(1  )   ( )  1  3 3 3 3 2 3

Vo 

Z in 1 (10)  (10)  5 V 1  Z in 11

⅓ 1

Chapter 2, Solution 73 By the current division principle, the current through the ammeter will be one-half its previous value when R = 20 + R x 65 = 20 + R x

R x = 45 

Chapter 2, Solution 74 With the switch in high position, 6 = (0.01 + R 3 + 0.02) x 5

R 3 = 1.17 

At the medium position, 6 = (0.01 + R 2 + R 3 + 0.02) x 3

R 2 + R 3 = 1.97

or R 2 = 1.97 - 1.17 = 0.8  At the low position, 6 = (0.01 + R 1 + R 2 + R 3 + 0.02) x 1 R 1 = 5.97 - 1.97 = 4 

R 1 + R 2 + R 3 = 5.97

Chapter 2, Solution 75 Converting delta-subnetworks to wye-subnetworks leads to the circuit below. 1 ⅓ 1 ⅓ ⅓ 1 1

1

1 ⅓

1

⅓

⅓

1 1 1 1 1 4  (1  ) //(1  )   ( )  1 3 3 3 3 2 3 With this combination, the circuit is further reduced to that shown below.

1

1

1 1

1

1

1 1 1 Z ab  1   (1  ) //(1  )  1  1 = 2  3 3 3

1

1 Chapter 2, Solution 76 Z ab = 1 + 1 = 2 


5  = 10 10  20 20 20 20

i.e., four 20  resistors in parallel. (b)

311.8 = 300 + 10 + 1.8 = 300 + 20 20  1.8 i.e., one 300 resistor in series with 1.8 resistor and a parallel combination of two 20 resistors.

(c)

40k = 12k + 28k = (24 24k )  (56k 56k ) i.e., Two 24k resistors in parallel connected in series with two 56k resistors in parallel.

(a)

42.32k = 42l + 320 = 24k + 28k = 320 = 24k = 56k 56k  300  20 i.e., A series combination of a 20 resistor, 300 resistor, 24k resistor, and a parallel combination of two 56k resistors.

Chapter 2, Solution 78 The equivalent circuit is shown below: R

VS

V0 =

+

+

(1-)R

V0

-

-

(1  )R 1  VS  VS R  (1  )R 2 V0 1    VS 2  

Chapter 2, Solution 79 Since p = v2/R, the resistance of the sharpener is R = v2/(p) = 62/(240 x 10-3) = 150 I = p/(v) = 240 mW/(6V) = 40 mA Since R and R x are in series, I flows through both. IR x = V x = 9 - 6 = 3 V R x = 3/(I) = 3/(40 mA) = 3000/(40) = 75 

Chapter 2, Solution 80 The amplifier can be modeled as a voltage source and the loudspeaker as a resistor:

V

+

V

R1

-

R2

-

Case 1

V 2 p2 R1  , Hence p  R p1 R 2

+

Case 2

p2 

R1 10 p1  (12)  30 W R2 4

Chapter 2, Solution 81 Let R 1 and R 2 be in k. R eq  R 1  R 2 5

(1)

5 R2 V0  VS 5 R 2  R 1

(2)

From (1) and (2), 0.05 

From (1), 40 = R 1 + 2

5 R1 40

2 = 5 R2 

5R 2 or R 2 = 3.333 k 5 R2

R 1 = 38 k

Thus, R 1 = 38 k, R 2 = 3.333 k


10  40  10  80  1

2 R 12

R 12 = 80 + 10 (10  40)  80 

50  88.33  6

(b) 3 10  10 

20  40  R 13

80  1

R 13 = 80 + 10 (10  40)  20  100  10 50  108.33  4

(c)

20  10  R 14 10  80  1

R 14 = 80  0 (10  40  10)  20  80  0  20  100 

40 

Chapter 2, Solution 83 The voltage across the fuse should be negligible when compared with 24 V (this can be checked later when we check to see if the fuse rating is exceeded in the final circuit). We can calculate the current through the devices. p1 45mW   5mA V1 9V p 480mW I2 = 2   20mA V2 24

I1 =

i 2 = 20 mA

I fuse i R1 24 V

+

-

R1

i 1 = 5 mA

R2 i R2

Let R 3 represent the resistance of the first device, we can solve for its value from knowing the voltage across it and the current through it. R 3 = 9/0.005 = 1,800 Ω This is an interesting problem in that it essentially has two unknowns, R 1 and R 2 but only one condition that need to be met and that the voltage across R 3 must equal 9 volts. Since the circuit is powered by a battery we could choose the value of R 2 which draws the least current, R 2 = ∞. Thus we can calculate the value of R 1 that give 9 volts across R3. 9 = (24/(R 1 + 1800))1800 or R 1 = (24/9)1800 – 1800 = 3 kΩ This value of R 1 means that we only have a total of 25 mA flowing out of the battery through the fuse which means it will not open and produces a voltage drop across it of 0.05V. This is indeed negligible when compared with the 24-volt source.

Chapter 3, Solution 1 Using Fig. 3.50, design a problem to help other students to better understand nodal analysis. R1 R2 Ix 12 V

+ 

R3

9V

+ 

Figure 3.50 For Prob. 3.1 and Prob. 3.39. Solution Given R 1 = 4 kΩ, R 2 = 2 kΩ, and R 3 = 2 kΩ, determine the value of I x using nodal analysis. Let the node voltage in the top middle of the circuit be designated as V x . [(V x –12)/4k] + [(V x –0)/2k] + [(V x –9)/2k] = 0 or (multiply this by 4 k) (1+2+2)V x = 12+18 = 30 or V x = 30/5 = 6 volts and I x = 6/(2k) = 3 mA.

Chapter 3, Solution 2 At node 1, v  v2  v1 v1   6 1 10 5 2

60 = - 8v 1 + 5v 2

(1)

At node 2, v2 v  v2  36 1 4 2

Solving (1) and (2), v 1 = 0 V, v 2 = 12 V

36 = - 2v 1 + 3v 2

(2)

Chapter 3, Solution 3 Applying KCL to the upper node,

8

v0 vo vo v    20  0 = 0 or v 0 = –60 V 10 20 30 60 v0 v  –6 A , i 2 = 0  –3 A, 10 20 v0 v i3 =  –2 A, i 4 = 0  1 A. 30 60

i1 =


v1 i1 20 

6A

i2 10 

v2 i3

40 

i4 40 

At node 1, –6 – 3 + v 1 /(20) + v 1 /(10) = 0 or v 1 = 9(200/30) = 60 V At node 2, 3 – 2 + v 2 /(10) + v 2 /(5) = 0 or v 2 = –1(1600/80) = –20 V i 1 = v 1 /(20) = 3 A, i 2 = v 1 /(10) = 6 A, i 3 = v 2 /(40) = –500 mA, i 4 = v 2 /(40) = –500 mA.

2A

Chapter 3, Solution 5 Apply KCL to the top node. 30  v 0 20  v 0 v 0   2k 5k 4k

v 0 = 20 V

Chapter 3, Solution 6. Solve for V 1 using nodal analysis. 10 

10 V

– +

5

10  + V1

10 

20 V

+ 

– Figure 3.55 For Prob. 3.6. Step 1. The first thing to do is to select a reference node and to identify all the unknown nodes. We select the bottom of the circuit as the reference node. The only unknown node is the one connecting all the resistors together and we will call that node V 1 . The other two nodes are at the top of each source. Relative to the reference, the one at the top of the 10-volt source is –10 V. At the top of the 20-volt source is +20 V. Step 2. unknown).

Setup the nodal equation (there is only one since there is only one

Step 3.

Simplify and solve.

or V 1 = –2 V. The answer can be checked by calculating all the currents and see if they add up to zero. The top two currents on the left flow right to left and are 0.8 A and 1.6 A respectively. The current flowing up through the 10-ohm resistor is 0.2 A. The current flowing right to left through the 10-ohm resistor is 2.2 A. Summing all the currents flowing out of the node, V 1 , we get, +0.8+1.6 –0.2–2.2 = 0. The answer checks.

Chapter 3, Solution 7 2

Vx  0 Vx  0   0.2Vx  0 10 20

0.35V x = 2 or V x = 5.714 V. Substituting into the original equation for a check we get, 0.5714 + 0.2857 + 1.1428 = 1.9999 checks!

Chapter 3, Solution 8 6

i1

v1

i3

20

i2 + V0

60V 4

+

–

+ 5V 0 –

– 20

i1 + i2 + i3 = 0

But

v1 ( v1  60)  0 v1  5v 0   0 10 20 20

2 v1 so that 2v 1 + v 1 – 60 + v 1 – 2v 1 = 0 5 or v 1 = 60/2 = 30 V, therefore v o = 2v 1 /5 = 12 V. v0 

Chapter 3, Solution 9 Let V 1 be the unknown node voltage to the right of the 250-Ω resistor. Let the ground reference be placed at the bottom of the 50-Ω resistor. This leads to the following nodal equation: V1  24 V1  0 V1  60I b  0   0 250 50 150 simplifying we get 3V1  72  15V1  5V1  300I b  0

But I b 

24  V1 . Substituting this into the nodal equation leads to 250 24.2V1  100.8  0 or V 1 = 4.165 V.

Thus,

I b = (24 – 4.165)/250 = 79.34 mA.


v1

v2

v3

At node 1.

[(v 1 –0)/8] + [(v 1 –v 3 )/1] + 4 = 0

At node 2.

–4 + [(v 2 –0)/2] + 2i o = 0

At node 3.

–2i o + [(v 3 –0)/4] + [(v 3 –v 1 )/1] = 0

Finally, we need a constraint equation,

i o = v 1 /8

This produces, 1.125v 1 – v 3 = 4

(1)

0.25v 1 + 0.5v 2 = 4

(2)

–1.25v 1 + 1.25v 3 = 0 or v 1 = v 3

(3)

Substituting (3) into (1) we get (1.125–1)v 1 = 4 or v 1 = 4/0.125 = 32 volts. This leads to, i o = 32/8 = 4 amps.

Chapter 3, Solution 11 Find V o and the power absorbed by all the resistors in the circuit of Fig. 3.60.

12 

60 V

Vo

+ _

12 

6

– +

24 V

Figure 3.60 For Prob. 3.11. Solution At the top node, KCL produces

Vo  60 Vo  0 Vo  (24)   0 12 12 6

(1/3)V o = 1 or V o = 3 V. P 12Ω = (3–60)2/1 = 293.9 W (this is for the 12 Ω resistor in series with the 60 V source) P 12Ω = (V o )2/12 = 9/12 = 750 mW (this is for the 12 Ω resistor connecting V o to ground) P 4Ω = (3–(–24))2/6 = 121.5 W.

Chapter 3, Solution 12 There are two unknown nodes, as shown in the circuit below.

20 

10 

V1

Vo

Ix

40 V

At node 1,

At node o,

+ _

20 

10 

4 Ix

V1  40 V1  0 V1  Vo    0 or 20 20 10 (0.05+0.05+.1)V 1 – 0.1V o = 0.2V 1 – 0.1V o = 2

(1)

Vo  V1 V 0  4I x  o  0 and I x = V 1 /20 10 10 –0.1V 1 – 0.2V 1 + 0.2V o = –0.3V 1 + 0.2V o = 0 or

(2)

V 1 = (2/3)V o

(3)

Substituting (3) into (1), 0.2(2/3)V o – 0.1V o = 0.03333V o = 2 or V o = 60 V.

Chapter 3, Solution 13 Calculate v 1 and v 2 in the circuit of Fig. 3.62 using nodal analysis. 10 V

15 A

Figure 3.62 For Prob. 3.13. Solution At node number 2, [((v 2 + 10) – 0)/10] + [(v 2 –0)/4] – 15 = 0 or (0.1+0.25)v 2 = 0.35v 2 = –1+15 = 14 or v 2 = 40 volts. Next, I = [(v 2 + 10) – 0]/10 = (40 + 10)/10 = 5 amps and v 1 = 8x5 = 40 volts.

Chapter 3, Solution 14 Using nodal analysis, find v o in the circuit of Fig. 3.63. 12.5 A

8 2

1

+ 4

vo 100 V

+ –

50 V

– +



Figure 3.63 For Prob. 3.14. 12.5 A

Solution

v0

v1 1

2

+ 4

vo 100 V

+

–

8

50 V

–

+



At node 1, [(v 1 –100)/1] + [(v 1 –v o )/2] + 12.5 = 0 or 3v 1 – v o = 200–25 = 175 At node o, [(v o –v 1 )/2] – 12.5 + [(v o –0)/4] + [(v o +50)/8] = 0 or –4v 1 + 7v o = 50 (2) Adding 4x(1) to 3x(2) yields,

(1)

4(1) + 3(2) = –4v o + 21v o = 700 + 150 or 17v o = 850 or v o = 50 V. Checking, we get v 1 = (175+v o )/3 = 75 V. At node 1, [(75–100)/1] + [(75–50)/2] + 12.5 = –25 + 12.5 + 12.5 = 0! At node o, [(50–75)/2] + [(50–0)/4] + [(50+50)/8] – 12.5 = –12.5 + 12.5 + 12.5 – 12.5 = 0!


5A

v0

v1 1

2 4

40 V

8

20 V

–

+

+

–

Nodes 1 and 2 form a supernode so that v 1 = v 2 + 10 At the supernode, 2 + 6v 1 + 5v 2 = 3 (v 3 - v 2 ) At node 3, 2 + 4 = 3 (v 3 - v 2 )

2 + 6v 1 + 8v 2 = 3v 3

v3 = v2 + 2

Substituting (1) and (3) into (2), 2 + 6v 2 + 60 + 8v 2 = 3v 2 + 6

v 1 = v 2 + 10 =

v2 =

54 11

i 0 = 6v i = 29.45 A 2

P 65 =

v12  54   v12 G    6  144.6 W R  11 

  56  = v G  5  129.6 W  11  2

P 55

2 2

P 35 = v L  v 3  G  (2) 2 3  12 W 2

(1)

 56 11

(2) (3)

Chapter 3, Solution 16 2S

v2

v1 i0 2A

+

1S

v0

4S

8S

v3 13 V

–

+

–

At the supernode, 2 = v 1 + 2 (v 1 - v 3 ) + 8(v 2 – v 3 ) + 4v 2 , which leads to 2 = 3v 1 + 12v 2 - 10v 3 (1) But v 1 = v 2 + 2v 0 and v 0 = v 2 . Hence v 1 = 3v 2 v 3 = 13V Substituting (2) and (3) with (1) gives, v 1 = 18.858 V, v 2 = 6.286 V, v 3 = 13 V

(2) (3)


v1 i0

4

2 10 

v2

60 V 60 V

+

8

3i 0

–

60  v1 v1 v1  v 2   4 8 2 60  v 2 v1  v 2 At node 2, 3i 0 +  0 10 2

At node 1,

But i 0 =

120 = 7v 1 - 4v 2

(1)

60  v1 . 4

Hence 360  v1  60  v 2 v1  v 2   0 4 10 2

1020 = 5v 1 + 12v 2

Solving (1) and (2) gives v 1 = 53.08 V. Hence i 0 =

60  v1  1.73 A 4

(2)


–+ v2

v1 2

15A

v3 2 8

4

30 V

+

+

v1

v3

–

–

(a)

(b)

v 2  v1 v 2  v 3  –15 = 0 or –0.5v 1 + v 2 – 0.5v 3 = 15 (1) 2 2 v  v1 v 2  v 3 v1 v 3 At the supernode, 2    = 0 and (v 1 /4) – 15 + (v 3 /8) = 0 or 2 2 4 8 2v 1 +v 3 = 120 (2) At node 2, in Fig. (a),

From Fig. (b), – v 1 – 30 + v 3 = 0 or v 3 = v 1 + 30 Solving (1) to (3), we obtain, v 1 = 30 V, v 2 = 60 V = v 3

(3)

Chapter 3, Solution 19 At node 1, V1  V3 V1  V2 V1   2 8 4 At node 2, 5  3

V1  V2 V2 V2  V3   8 2 4 At node 3,

12  V3

 

 

0  V1  7V2  2V3

V1  V3 V2  V3  0 8 2 4 From (1) to (3), 3



 7  1  4  V1   16         1 7  2 V2    0  4 2  7  V3    36   Using MATLAB,

 10  V  A 1 B   4.933  12.267 

16  7V1  V2  4V3

 

 

 

(1)

(2)

 36  4V1  2V2  7V3 (3)

AV  B

V1  10 V, V2  4.933 V, V3  12.267 V


Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence V1 V2 V3   0   V1  4V2  V3  0 (1) 4 1 4

. V1

V2

.

4

2

1

Between nodes 1 and 3,  V1  12  V3  0   V3  V1  12 Similarly, between nodes 1 and 2, V1  V2  2i But i  V3 / 4 . Combining this with (2) and (3) gives . V2

V3

 6  V1 / 2

4

(2) (3) (4)

Solving (1), (2), and (4) leads to V1  3V, V2  4.5V, V3  15V

Chapter 3, Solution 21 4 k

v1

2 k

v3

3v 0 +

3v 0 v2

+ v0

3 mA

–

1 k

+ +

+

v3

v2

–

–

(b)

(a)

Let v 3 be the voltage between the 2k resistor and the voltage-controlled voltage source. At node 1, v  v 2 v1  v 3 3x10 3  1 12 = 3v 1 - v 2 - 2v 3 (1)  4000 2000 At node 2, v1  v 2 v1  v 3 v 2   4 2 1

3v 1 - 5v 2 - 2v 3 = 0

(2)

Note that v 0 = v 2 . We now apply KVL in Fig. (b) - v 3 - 3v 2 + v 2 = 0 From (1) to (3), v 1 = 1 V, v 2 = 3 V

v 3 = - 2v 2

(3)

Chapter 3, Solution 22 At node 1,

12  v 0 v1 v  v0  3 1 2 4 8

At node 2, 3 +

24 = 7v 1 - v 2

(1)

v 1  v 2 v 2  5v 2  8 1

But, v 1 = 12 - v 1 Hence, 24 + v 1 - v 2 = 8 (v 2 + 60 + 5v 1 ) = 4 V 456 = 41v 1 - 9v 2 Solving (1) and (2), v 1 = - 10.91 V, v 2 = - 100.36 V

(2)

Chapter 3, Solution 23 We apply nodal analysis to the circuit shown below.

1

30 V

+ _

4

Vo

2

2 Vo +

+ Vo

V1

–

16 

_

3A

At node o, Vo  30 Vo  0 Vo  (2Vo  V1 )  0  1.25Vo  0.25V1  30   1 2 4

(1)

At node 1, (2Vo  V1 )  Vo V1  0   3  0  5V1  4Vo  48 4 16

From (1), V 1 = 5V o – 120. Substituting this into (2) yields 29V o = 648 or V o = 22.34 V.

(2)

Chapter 3, Solution 24 Consider the circuit below. 8 + Vo _ 4A V1

1

4

V2

2A V3

2

V4

2

V1  0 V  V4 4 1  0  1.125V1  0.125V4  4 8 1 V  0 V2  V3 4 2   0  0.75V2  0.25V3  4 2 4 V3  V2 V3  0   2  0  0.25V2  0.75V3  2 4 2 V  V1 V4  0 2 4   0  0.125V1  1.125V4  2 8 1

1

(1) (2) (3) (4)

0 0  0.125  1.125 4  0    4 0.75  0.25 0   V   0   2  0.25 0.75 0      0 0 1.125   0.125 2

Now we can use MATLAB to solve for the unknown node voltages.

>> Y=[1.125,0,0,-0.125;0,0.75,-0.25,0;0,-0.25,0.75,0;-0.125,0,0,1.125] Y=

1.1250 0 0 -0.1250 0 0.7500 -0.2500 0 0 -0.2500 0.7500 0 -0.1250 0 0 1.1250 >> I=[4,-4,-2,2]' I= 4 -4 -2 2 >> V=inv(Y)*I V= 3.8000 -7.0000 -5.0000 2.2000

V o = V 1 – V 4 = 3.8 – 2.2 = 1.6 V.

Chapter 3, Solution 25 Consider the circuit shown below. 20



4



 10 1  

1

2



10 





3   30

8

4

20

At node 1. V  V2 V1  V4 4 1    80  21V1  20V2  V4 1 20 At node 2, V1  V2 V2 V2  V3     0  80V1  98V2  8V3 1 8 10 At node 3, V2  V3 V3 V3  V4     0  2V2  5V3  2V4 10 20 10 At node 4, V1  V4 V3  V4 V4     0  3V1  6V3  11V4 20 10 30 Putting (1) to (4) in matrix form gives:

(1)

(2)

(3)

(4)

1   V1   80   21 20 0       0    80 98 8 0  V2  0  0 2 5 2  V3       0 6 11 V4  0  3

B =A V

V = A-1 B

Using MATLAB leads to V 1 = 25.52 V,

V 2 = 22.05 V, V 3 = 14.842 V, V 4 = 15.055 V


At node 1, V  V3 V1  V2 15  V1  3 1     45  7V1  4V2  2V3 20 10 5 At node 2, V1  V2 4 I o  V2 V2  V3   5 5 5 V1  V3 But I o  . Hence, (2) becomes 10 0  7V1  15V2  3V3 At node 3, V  V3  10  V3 V2  V3 3 1   0   70  3V1  6V2  11V3 10 15 5 Putting (1), (3), and (4) in matrix form produces

 7  4  2  V1    45        7  15 3  V2    0    3  6 11  V   70    3    Using MATLAB leads to   7.19    V  A 1B    2.78   2.89   

 

AV  B

Thus, V 1 = –7.19V; V 2 = –2.78V; V 3 = 2.89V.

(1)

(2)

(3)

(4)

Chapter 3, Solution 27 At node 1, 2 = 2v 1 + v 1 – v 2 + (v 1 – v 3 )4 + 3i 0 , i 0 = 4v 2 . Hence, 2 = 7v 1 + 11v 2 – 4v 3

(1)

At node 2, v 1 – v 2 = 4v 2 + v 2 – v 3

0 = – v 1 + 6v 2 – v 3

(2)

At node 3, 2v 3 = 4 + v 2 – v 3 + 12v 2 + 4(v 1 – v 3 ) – 4 = 4v 1 + 13v 2 – 7v 3

or

(3)

In matrix form, 7 11  4  v 1   2  1  6 1   v    0   2     4 13  7  v 3    4

7

11

  1 6 4

13

7

2

4

2

11

1  176,  1  0

6

7

4

4

7

2  1 0 1  66, 4 4 7

v1 =

13

11

4 1  110 7

2

 3  1  6 0  286 4 13  4

 1 110  66   0.625V, v 2 = 2   0.375V   176 176

v3 =

3 286   1.625V. 176 

v 1 = 625 mV, v 2 = 375 mV, v 3 = 1.625 V.


At node c, V d  V c V c  Vb V c     0  5Vb  11Vc  2Vd (1) 10 4 5 At node b, Va  90  Vb Vc  Vb Vb      90  Va  4Vb  2Vc (2) 8 4 8 At node a, Va  60  Vd Va Va  90  Vb   0   60  7Va  2Vb  4Vd (3) 4 16 8 At node d, Va  60  Vd Vd Vd  Vc     300  5Va  2Vc  8Vd (4) 4 20 10 In matrix form, (1) to (4) become  0  5 11  2  Va   0        1  4 2 0  Vb    90    AV  B  7  2 0  4  V    60  c       5 0 2  8  V   300    d    We use MATLAB to invert A and obtain   10.56     20.56  1 VA B 1.389      43.75    Thus, V a = –10.56 V; V b = 20.56 V; V c = 1.389 V; VC d = –43.75 V.

Chapter 3, Solution 29 At node 1, 5  V1  V4  2V1  V1  V2  0    5  4V1  V2  V4 At node 2, V1  V2  2V2  4(V2  V3 )  0   0  V1  7V2  4V3 At node 3, 6  4(V2  V3 )  V3  V4   6  4V2  5V3  V4 At node 4, 2  V3  V4  V1  V4  3V4   2  V1  V3  5V4 In matrix form, (1) to (4) become  4  1 0  1 V1    5         1 7  4 0 V2   0  AV  B    0  4 5  1 V    6  3        1 0  1 5 V   2   4     Using MATLAB,

(1)

(2)

(3) (4)

  0.7708     1.209  1 V  A B 2.309     0.7076    i.e. V1  0.7708 V, V 2  1.209 V, V 3  2.309 V, V 4  0.7076 V


i0 v1 10  +

80 V

–+ 96 V

20 

v0

1

2 4v 0

–

40 

+ –

2i 0

80 

At node 1, [(v 1 –80)/10]+[(v 1 –4v o )/20]+[(v 1 –(v o –96))/40] = 0 or (0.1+0.05+0.025)v 1 – (0.2+0.025)v o = 0.175v 1 – 0.225v o = 8–2.4 = 5.6

(1)

At node 2, –2i o + [((v o –96)–v 1 )/40] + [(v o –0)/80] = 0 and i o = [(v 1 –(v o –96))/40] –2[(v 1 –(v o –96))/40] + [((v o –96)–v 1 )/40] + [(v o –0)/80] = 0 –3[(v 1 –(v o –96))/40] + [(v o –0)/80] = 0 or –0.0.075v 1 + (0.075+0.0125)v o = 7.2 = –0.075v 1 + 0.0875v o = 7.2 (2) Using (1) and (2) we get,  0.175  0.225  v1  5.6  0.075 0.0875  v   7.2 or   o    0.0875 0.225 0.0875 0.225    v1  0.075 0.175 5.6  0.075 0.175 5.6  v   0.0153125  0.016875 7.2   0.0015625 7.2      o

v 1 = –313.6–1036.8 = –1350.4 v o = –268.8–806.4 = –1.0752 kV and i o = [(v 1 –(v o –96))/40] = [(–1350.4–(–1075.2–96))/40] = –4.48 amps.


+ v0 – v2

v1

2v 0

v3

2

i0 1A

4

1

10 V

4

–

At the supernode, 1 + 2v 0 =

v1 v 2 v1  v 3   4 1 1

(1)

But v o = v 1 – v 3 . Hence (1) becomes, 4 = -3v 1 + 4v 2 +4v 3

(2)

At node 3, 2v o + or

v3 10  v 3  v1  v 3  4 2

20 = 4v 1 + 0v 2 – v 3

At the supernode, v 2 = v 1 + 4i o . But i o =

(3)

v3 . Hence, 4

v2 = v1 + v3 Solving (2) to (4) leads to,

v 1 = 4.97V, v 2 = 4.85V, v 3 = –0.12V.

+

(4)


5 k

v1

v3

v2

+ 10 k

v1

10 V

20 V

–+

+–

12 V

–

4 mA

+

loop 1 +

loop 2

–

v3

–

(b)

(a)

We have a supernode as shown in figure (a). It is evident that v 2 = 12 V, Applying KVL to loops 1and 2 in figure (b), we obtain, -v 1 – 10 + 12 = 0 or v 1 = 2 and -12 + 20 + v 3 = 0 or v 3 = -8 V Thus,

v 1 = 2 V, v 2 = 12 V, v 3 = -8V.

Chapter 3, Solution 33 (a) This is a planar circuit. It can be redrawn as shown below.

5 3

1

2

4

6

2A

(b) This is a planar circuit. It can be redrawn as shown below.

4

3 5 12 V

+

– 1

2


This is a planar circuit because it can be redrawn as shown below, 7 2

1

3

6 10 V

5

+

– 4

(b)

This is a non-planar circuit.


30 V

20 V

+

–

+

– i1

2 k

+

i2

v0

4 k

–

5 k

Assume that i 1 and i 2 are in mA. We apply mesh analysis. For mesh 1, -30 + 20 + 7i 1 – 5i 2 = 0 or 7i 1 – 5i 2 = 10

(1)

For mesh 2, -20 + 9i 2 – 5i 1 = 0 or -5i 1 + 9i 2 = 20 Solving (1) and (2), we obtain, i 2 = 5. v 0 = 4i 2 = 20 volts.

(2)

Chapter 3, Solution 36 10 V

4

+–

i1

12 V

i2

+

I1

–

6

i3

I2

2

Applying mesh analysis gives, 10I 1 – 6I 2 = 12 and –6I 1 + 8I 2 = –10

or

4 3    6   5  3  I1   6   I1  3 5  6   5   3 4  I    5 or I   11  5    2     2     I 1 = (24–15)/11 = 0.8182 and I 2 = (18–25)/11 = –0.6364 i 1 = –I 1 = –818.2 mA; i 2 = I 1 – I 2 = 0.8182+0.6364 = 1.4546 A; and i 3 = I 2 = –636.4 mA.


60 V +

+ v0

20

4

–

i1

i2

–

5v 0

+ –

20

Applying mesh analysis to loops 1 and 2, we get, 30i 1 – 20i 2 + 60 = 0 which leads to i 2 = 1.5i 1 + 3

(1)

–20i 1 + 40i 2 – 60 + 5v 0 = 0

(2)

But, v 0 = –4i 1

(3)

Using (1), (2), and (3) we get –20i 1 + 60i 1 + 120 – 60 – 20i 1 = 0 or 20i 1 = –60 or i 1 = –3 amps and i 2 = 7.5 amps. Therefore, we get, v 0 = –4i 1 = 12 volts.

Chapter 3, Solution 38 Consider the circuit below with the mesh currents. 4

+ _

60 V

3

I3

I4

1

10 A 2

2 Io 1 1

I1

I2

+ _

22.5V

4 5A

I 1 = –5 A

(1)

1(I 2 –I 1 ) + 2(I 2 –I 4 ) + 22.5 + 4I 2 = 0 7I 2 – I 4 = –27.5

(2)

–60 + 4I 3 + 3I 4 + 1I 4 + 2(I 4 –I 2 ) + 2(I 3 – I 1 ) = 0 (super mesh) –2I 2 + 6 I 3 + 6I 4 = +60 – 10 = 50

(3)

But, we need one more equation, so we use the constraint equation –I 3 + I 4 = 10. This now gives us three equations with three unknowns. 0  1 I 2   27.5 7  2 6 6   I    50    3     0  1 1  I 4   10 

We can now use MATLAB to solve the problem. >> Z=[7,0,-1;-2,6,6;0,-1,0]

Z= 7 0 -1 -2 6 6 0 -1 0 >> V=[–27.5,50,10]' V= –27.5 50 10 >> I=inv(Z)*V I= –1.3750 –10.0000 17.8750 I o = I 1 – I 2 = –5 – 1.375 = –6.375 A. Check using the super mesh (equation (3)): –2I 2 + 6 I 3 + 6I 4 = 2.75 – 60 + 107.25 = 50!

Chapter 3, Solution 39 Using Fig. 3.50 from Prob. 3.1, design a problem to help other students to better understand mesh analysis. Solution Given R 1 = 4 kΩ, R 2 = 2 kΩ, and R 3 = 2 kΩ, determine the value of I x using mesh analysis. R1

R2 Ix

12 V

+ 

I2

I1 R3

9V

+ 

Figure 3.50 For Prob. 3.1 and 3.39. For loop 1 we get –12 +4kI 1 + 2k(I 1 –I 2 ) = 0 or 6I 1 – 2I 2 = 0.012 and at loop 2 we get 2k(I 2 –I 1 ) + 2kI 2 + 9 = 0 or –2I 1 + 4I 2 = –0.009. Now 6I 1 – 2I 2 = 0.012 + 3[–2I 1 + 4I 2 = –0.009] leads to, 10I 2 = 0.012 – 0.027 = –0.015 or I 2 = –1.5 mA and I 1 = (–0.003+0.012)/6 = 1.5 mA. Thus, I x = I 1 –I 2 = (1.5+1.5) mA = 3 mA.


6 k

6 k 56V

+

–

i2 2 k

i1 i3

4 k

4 k

Assume all currents are in mA and apply mesh analysis for mesh 1. –56 + 12i 1 – 6i 2 – 4i 3 = 0 or 6i 1 – 3i 2 – 2i 3 = 28

(1)

for mesh 2, –6i 1 + 14i 2 – 2i 3 = 0 or –3i 1 + 7i 2 – i 3

=0

(2)

=0

(3)

for mesh 3, –4i 1 – 2i 2 + 10i 3 = 0 or –2i 1 – i 2 + 5i 3

Solving (1), (2), and (3) using MATLAB, we obtain, i o = i 1 = 8 mA.

Chapter 3, Solution 41 10 

i1 6V

2

+– 1

i2 4

5

i3 8V

+

– i i2

i3 0

For loop 1, 6 = 12i 1 – 2i 2

3 = 6i 1 – i 2

(1)

For loop 2, -8 = – 2i 1 +7i 2 – i 3

(2)

For loop 3, -8 + 6 + 6i 3 – i 2 = 0

2 = – i 2 + 6i 3

We put (1), (2), and (3) in matrix form, 6  1 0  i1   3  2  7 1  i    8    2    0  1 6 i 3  2 6

1 0

6 3 0

  2  7 1  234,  2  2 8 1  240 0

1 6

0 2 6

(3)

6

1 3

 3  2  7 8  38 0 1 2

At node 0, i + i 2 = i 3 or i = i 3 – i 2 =

3  2  38  240  = 1.188 A   234

Chapter 3, Solution 42 Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine the mesh currents in the circuit of Fig. 3.88.

Figure 3.88 Solution For mesh 1,  12  50I 1  30I 2  0   12  50I1  30I 2 For mesh 2,  8  100 I 2  30 I 1  40 I 3  0   8  30 I 1  100 I 2  40 I 3 For mesh 3,  6  50 I 3  40 I 2  0   6  40 I 2  50 I 3 Putting eqs. (1) to (3) in matrix form, we get 0  I 1  12   50  30        30 100  40  I 2    8   0  40 50  I 3   6  

 

AI  B

Using Matlab,  0.48    I  A B   0.40   0.44    1

i.e. I 1 = 480 mA, I 2 = 400 mA, I 3 = 440 mA

(1) (2) (3)

Chapter 3, Solution 43 20  a 80 V

+

i1

–

30 

+ 30 

i3 20  80 V

+

i2

–

30 

20 

V ab

– b

For loop 1, 80 = 70i 1 – 20i 2 – 30i 3

8 = 7i 1 – 2i 2 – 3i 3

(1)

80 = 70i 2 – 20i 1 – 30i 3

8 = -2i 1 + 7i 2 – 3i 3

(2)

0 = -30i 1 – 30i 2 + 90i 3

0 = i 1 + i 2 – 3i 3

For loop 2,

For loop 3,

(3) Solving (1) to (3), we obtain i 3 = 16/9 I o = i 3 = 16/9 = 1.7778 A V ab = 30i 3 = 53.33 V.

Chapter 3, Solution 44 90 V +

2

4

i3

i2

1 180V +

– 5

i1 45 A i1

i2

Loop 1 and 2 form a supermesh. For the supermesh, 6i 1 + 4i 2 – 5i 3 + 180 = 0

(1)

For loop 3,

–i 1 – 4i 2 + 7i 3 + 90 = 0

(2)

Also,

i 2 = 45 + i 1

(3)

Solving (1) to (3), i 1 = –46, i 3 = –20;

i o = i 1 – i 3 = –26 A


30V

+

–

8

i3

i4

2

6

i1

3

i2

1

For loop 1,

30 = 5i 1 – 3i 2 – 2i 3

(1)

For loop 2,

10i 2 - 3i 1 – 6i 4 = 0

(2)

For the supermesh,

6i 3 + 14i 4 – 2i 1 – 6i 2 = 0

(3)

But

i 4 – i 3 = 4 which leads to i 4 = i 3 + 4

Solving (1) to (4) by elimination gives i = i 1 = 8.561 A.

(4)

Chapter 3, Solution 46 For loop 1,  12  3i1  8(i1  i 2 )  12  11i1  8i 2  0 For loop 2, 8(i 2  i1 )  6i 2  2v o  8i1  14i 2  2v o  0 But vo  3i1 ,

 

11i1  8i 2  12

 8i1  14i2  6i1  0   i1  7i2 Substituting (2) into (1), 77i2  8i2  12   i 2  0.1739 A and i1  7i2  1.217 A

(1)

(2)

Chapter 3, Solution 47 First, transform the current sources as shown below. - 6V +

2

8

V1

4

V2

I3 V3

4 2

I1

8 I2

+ 20V -

+ 12V -

For mesh 1,

 20  14 I 1  2 I 2  8I 3  0 For mesh 2,

10  7 I 1  I 2  4 I 3

(1)

 6   I1  7 I 2  2I 3

(2)

 6  14 I 3  4 I 2  8I 1  0   3  4 I 1  2 I 2  7 I 3 Putting (1) to (3) in matrix form, we obtain

(3)

12  14 I 2  2 I 1  4 I 3  0 For mesh 3,

 

 

 7  1  4  I 1   10         1 7  2  I 2     6    4  2 7  I   3   3     Using MATLAB,  2  I  A 1 B  0.0333 1.8667  But

 

AI  B

  I 1  2.5, I 2  0.0333, I 3  1.8667

I1 

20  V 4

 

V1  20  4 I 1  10 V

V 2  2( I 1  I 2 )  4.933 V

Also, I2 

V3  12 8

 

V3  12  8I 2  12.267 V.


We apply mesh analysis and let the mesh currents be in mA. 3k  I4 4k 

2k 

Io

1k 

I3 I2

I1 + 6V -

5k 

+ 4V -

10k 

For mesh 1,  6  8  5I1  I 2  4I 4  0   2  5I1  I 2  4I 4 For mesh 2,  4  13I 2  I1  10I 3  2I 4  0   4  I1  13I 2  10I 3  2I 4 For mesh 3,  3  15I 3  10I 2  5I 4  0   3  10I 2  15I 3  5I 4 For mesh 4,  4 I 1  2 I 2  5I 3  14 I 4  0 Putting (1) to (4) in matrix form gives 1  4  I1   2  0  5        1 13  10  2  I 2   4    AI B  0  10 15  5  I    3  3        4  2  5 14  I   0    4    Using MATLAB,

 3.608     4.044  1 IA B 0.148 3.896     3    The current through the 10k  resistor is I o = I 2 – I 3 = 148 mA.

3V +

(1)

(2)

(3) (4)


i3 2

1

2

i1

27 V

i2

+

–

2i 0 i1

i2

0 (a)

2

1

2

i1

+

+

v0 or

v0

–

i2

27V +

–

(b)

For the supermesh in figure (a), 3i 1 + 2i 2 – 3i 3 + 27 = 0

(1)

At node 0,

i 2 – i 1 = 2i 0 and i 0 = –i 1 which leads to i 2 = –i 1

(2)

For loop 3,

–i 1 –2i 2 + 6i 3 = 0 which leads to 6i 3 = –i 1

(3)

Solving (1) to (3), i 1 = (–54/3)A, i 2 = (54/3)A, i 3 = (27/9)A i 0 = –i 1 = 18 A, from fig. (b), v 0 = i 3 –3i 1 = (27/9) + 54 = 57 V.


i1

4

2

i3

10  8 35 V

+

–

i2 3i 0 i2

For loop 1,

i3

16i 1 – 10i 2 – 2i 3 = 0 which leads to 8i 1 – 5i 2 – i 3 = 0

(1)

For the supermesh, –35 + 10i 2 – 10i 1 + 10i 3 – 2i 1 = 0 or

–6i 1 + 5i 2 + 5i 3 = 17.5

Also, 3i 0 = i 3 – i 2 and i 0 = i 1 which leads to 3i 1 = i 3 – i 2 Solving (1), (2), and (3), we obtain i 1 = 1.0098 and i 0 = i 1 = 1.0098 A

(2) (3)


i1 8 2

i3

1

+

i2 40 V

4

+

v0

20V

–

+



–

For loop 1,

i 1 = 5A

(1)

For loop 2,

-40 + 7i 2 – 2i 1 – 4i 3 = 0 which leads to 50 = 7i 2 – 4i 3

(2)

For loop 3,

-20 + 12i 3 – 4i 2 = 0 which leads to 5 = - i 2 + 3 i 3

(3)

Solving with (2) and (3), And,

i 2 = 10 A, i 3 = 5 A

v 0 = 4(i 2 – i 3 ) = 4(10 – 5) = 20 V.


+ v0 2 

i2

– VS

+

–

8

3A

i2

i1

i3 4

i3

+ –

2V 0

For mesh 1, 2(i 1 – i 2 ) + 4(i 1 – i 3 ) – 12 = 0 which leads to 3i 1 – i 2 – 2i 3 = 6

(1)

For the supermesh, 2(i 2 – i 1 ) + 8i 2 + 2v 0 + 4(i 3 – i 1 ) = 0 But v 0 = 2(i 1 – i 2 ) which leads to -i 1 + 3i 2 + 2i 3 = 0 (2) For the independent current source, i 3 = 3 + i 2 Solving (1), (2), and (3), we obtain,

i 1 = 3.5 A, i 2 = -0.5 A, i 3 = 2.5 A.

(3)

Chapter 3, Solution 53 Applying mesh analysis leads to; –12 + 4kI 1 – 3kI 2 – 1kI 3 = 0 –3kI 1 + 7kI 2 – 4kI 4 = 0 –3kI 1 + 7kI 2 = –12 –1kI 1 + 15kI 3 – 8kI 4 – 6kI 5 = 0 –1kI 1 + 15kI 3 – 6k = –24 I 4 = –3mA –6kI 3 – 8kI 4 + 16kI 5 = 0 –6kI 3 + 16kI 5 = –24 Putting these in matrix form (having substituted I 4

(1) (2) (3) (4) (5) = 3mA in the above),

 4  3  1 0   I1   12   3 7 0 0  I 2    12    k   1 0 15  6  I 3   24       0  6 16   I 5   24 0

ZI = V Using MATLAB,

>> Z = [4,-3,-1,0;-3,7,0,0;-1,0,15,-6;0,0,-6,16] Z= 4 -3 -1 0 -3 7 0 0 -1 0 15 -6 0 0 -6 16 >> V = [12,-12,-24,-24]' V= 12 -12 -24 -24 We obtain, >> I = inv(Z)*V

I= 1.6196 mA –1.0202 mA –2.461 mA 3 mA –2.423 mA


Let the mesh currents be in mA. For mesh 1,  12  10  2 I 1  I 2  0   2  2I1  I 2 For mesh 2,  10  3I 2  I 1  I 3  0   10   I 1  3I 2  I 3 For mesh 3,  12  2 I 3  I 2  0   12   I 2  2 I 3 Putting (1) to (3) in matrix form leads to  2  1 0  I 1   2         1 3  1 I 2   10   0  1 2  I  12   3     Using MATLAB,

 5.25  I  A B   8.5  10.25 1

 

AI  B

  I 1  5.25 mA, I 2  8.5 mA, I 3  10.25 mA

I 1 = 5.25 mA, I 2 = 8.5 mA, and I 3 = 10.25 mA.

(1)

(2)

(3)

Chapter 3, Solution 55 10 V

I2

b

i1 4A

c

+

1A

I2

6

1A

2

i2 i3

I4 4A

I3

d

I1

12 

4 +–

a

I3

0

I4 8V

It is evident that I 1 = 4 For mesh 4,

(2)

12(I 4 – I 1 ) + 4(I 4 – I 3 ) – 8 = 0

For the supermesh

At node c,

(1)

6(I 2 – I 1 ) + 10 + 2I 3 + 4(I 3 – I 4 ) = 0 or -3I 1 + 3I 2 + 3I 3 – 2I 4 = -5

I2 = I3 + 1

(3) (4)

Solving (1), (2), (3), and (4) yields, I 1 = 4A, I 2 = 3A, I 3 = 2A, and I 4 = 4A At node b,

i 1 = I 2 – I 1 = -1A

At node a,

i 2 = 4 – I 4 = 0A

At node 0,

i 3 = I 4 – I 3 = 2A

Chapter 3, Solution 56 + v1 – 2 2

i2

2

2 12 V

+

i1

–

2

i3

+ v2

–

For loop 1, 12 = 4i 1 – 2i 2 – 2i 3 which leads to 6 = 2i 1 – i 2 – i 3

(1)

For loop 2, 0 = 6i 2 –2i 1 – 2 i 3 which leads to 0 = -i 1 + 3i 2 – i 3

(2)

For loop 3, 0 = 6i 3 – 2i 1 – 2i 2 which leads to 0 = -i 1 – i 2 + 3i 3

(3)

In matrix form (1), (2), and (3) become,  2  1  1  i1  6   1 3  1 i   0  2       1  1 3  i 3  0 2

1 1

2

6 1

 =  1 3  1  8,  2 =  1 3  1  24 1 1 3 1 0 3 2

1 6

 3 =  1 3 0  24 , therefore i 2 = i 3 = 24/8 = 3A, 1 1 0 v 1 = 2i 2 = 6 volts, v = 2i 3 = 6 volts

Chapter 3, Solution 57 Assume R is in kilo-ohms. V2  4kx15mA  60V, V1  90  V2  90  60  30V Current through R is 3 3   30  (15)R iR  i o, V1  i R R 3 R 3 R This leads to R = 90/15 = 6 kΩ.


i2 30 

10 

i1

10 

30 

i3

+

– 120 V

For loop 1, 120 + 40i 1 – 10i 2 = 0, which leads to -12 = 4i 1 – i 2

(1)

For loop 2, 50i 2 – 10i 1 – 10i 3 = 0, which leads to -i 1 + 5i 2 – i 3 = 0 For loop 3, -120 – 10i 2 + 40i 3 = 0, which leads to 12 = -i 2 + 4i 3 Solving (1), (2), and (3), we get, i 1 = -3A, i 2 = 0, and i 3 = 3A

(2) (3)


–+ I0

96 V

i2

10  20 

+ 80V

i1

+

4v 0

–

i3

+ –

80 

v0

– 2I 0 i2

i3

For loop 1, -80 + 30i 1 – 20i 2 + 4v 0 = 0, where v 0 = 80i 3 or 4 = 1.5i 1 – i 2 + 16i 3

(1)

For the supermesh, 60i 2 – 20i 1 – 96 + 80i 3 – 4 v 0 = 0, where v 0 = 80i 3 or 4.8 = -i 1 + 3i 2 – 12i 3

(2)

Also, 2I 0 = i 3 – i 2 and I 0 = i 2 , hence, 3i 2 = i 3 (3)

From (1), (2), and (3),

3

 = 1 0

2

32

 3  2 32    1 3  12   3  1   0 3

8

 i1   8  i   4.8  2   i3   0 

32

3

2

3

 12  5,  2 =  1 4.8  12  22.4,  3 =  1

3

3

1

3

0

0

1

I 0 = i 2 =  2 / = -28/5 = –4.48 A v 0 = 8i 3 = (-84/5)80 = –1.0752 kvolts

0

8 4.8  67.2 0

Chapter 3, Solution 60 0.5i 0

4

56 V

8

v1

v2 1

56 V

+

2

– i0

At node 1, [(v 1 –0)/1] + [(v 1 –56)/4] + 0.5[(v 1 –0)/1] = 0 or 1.75v 1 = 14 or v 1 = 8 V At node 2, [(v 2 –56)/8] – 0.5[8/1] + [(v 2 –0)/2] = 0 or 0.625v 2 = 11 or v 2 = 17.6 V P 1 = (v 1 )2/1 = 64 watts, P 2 = (v 2 )2/2 = 154.88 watts, P 4 = (56 – v 1 )2/4 = 576 watts, P 8 = (56 – v 2 )2/8 = 1.84.32 watts.

Chapter 3, Solution 61 v1

20 

v2

10  i0

is

+ v0

–

30 

– + 5v 0

At node 1, i s = (v 1 /30) + ((v 1 – v 2 )/20) which leads to 60i s = 5v 1 – 3v 2 But v 2 = -5v 0 and v 0 = v 1 which leads to v 2 = -5v 1 Hence, 60i s = 5v 1 + 15v 1 = 20v 1 which leads to v 1 = 3i s , v 2 = -15i s i 0 = v 2 /50 = -15i s /50 which leads to i 0 /i s = -15/50 = –0.3

40 

(1)


100V +

–

A

i1

8 k

i2

B

2 k

i3

+

–

40 V

We have a supermesh. Let all R be in k, i in mA, and v in volts. For the supermesh, -100 +4i 1 + 8i 2 + 2i 3 + 40 = 0 or 30 = 2i 1 + 4i 2 + i 3 (1) At node A,

i1 + 4 = i2

(2)

At node B,

i 2 = 2i 1 + i 3

(3)

Solving (1), (2), and (3), we get i 1 = 2 mA, i 2 = 6 mA, and i 3 = 2 mA.


A

5 50 V

+

–

i1

i2 + –

4i x

For the supermesh, -50 + 10i 1 + 5i 2 + 4i x = 0, but i x = i 1 . Hence, 50 = 14i 1 + 5i 2 At node A, i 1 + 3 + (v x /4) = i 2 , but v x = 2(i 1 – i 2 ), hence, i 1 + 2 = i 2 Solving (1) and (2) gives i 1 = 2.105 A and i 2 = 4.105 A v x = 2(i 1 – i 2 ) = –4 volts and i x = i 2 – 2 = 2.105 amp

(1) (2)

Chapter 3, Solution 64 i1

50 

A

i0 i1

i 2 10  +  v0

10 

i2

+ –

4i 0

i3

40 

250V +

– 5A

0.2V 0

i1 For mesh 2,

B

i3

20i 2 – 10i 1 + 4i 0 = 0

(1)

But at node A, i o = i 1 – i 2 so that (1) becomes i 1 = (16/6)i 2 (2) For the supermesh, –250 + 50i 1 + 10(i 1 – i 2 ) – 4i 0 + 40i 3 = 0 or

28i 1 – 3i 2 + 20i 3 = 125 (3)

At node B, But,

i 3 + 0.2v 0 = 2 + i 1 v 0 = 10i 2 so that (4) becomes i 3 = 5 + (2/3)i 2

Solving (1) to (5), i 2 = 0.2941 A, v 0 = 10i 2 = 2.941 volts, i 0 = i 1 – i 2 = (5/3)i 2 = 490.2mA.

(4) (5)


For mesh 1, –12 + 12I 1 – 6I 2 – I 4 = 0 or 12  12 I 1  6 I 2  I 4

(1)

–6I 1 + 16I 2 – 8I 3 – I 4 – I 5 = 0

(2)

–8I 2 + 15I 3 – I 5 – 9 = 0 or 9 = –8I 2 + 15I 3 – I 5

(3)

For mesh 2, For mesh 3,

For mesh 4, –I 1 – I 2 + 7I 4 – 2I 5 – 6 = 0 or 6 = –I 1 – I 2 + 7I 4 – 2I 5

(4)

For mesh 5, –I 2 – I 3 – 2I 4 + 8I 5 – 10 = 0 or 10   I 2  I 3  2 I 4  8 I 5 Casting (1) to (5) in matrix form gives 1 0  I1  12   12  6 0        6 16  8  1  1  I 2   0   0  8 15 0  1  I    9      3    7  2  I 4   6   1 1 0  0  1  1  2 8  I  10    5   

(5)

AI  B

Using MATLAB we input: Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8] and V=[12;0;9;6;10] This leads to >> Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8] Z= 12 -6 0 -1 0

-6 0 -1 0 16 -8 -1 -1 -8 15 0 -1 -1 0 7 -2 -1 -1 -2 8

>> V=[12;0;9;6;10] V= 12

0 9 6 10 >> I=inv(Z)*V I= 2.1701 1.9912 1.8119 2.0942 2.2489 Thus, I = [2.17, 1.9912, 1.8119, 2.094, 2.249] A.

Chapter 3, Solution 66 The mesh equations are obtained as follows. 12  24  30I1  4I2  6I3  2I4  0 or 30I 1 – 4I 2 – 6I 3 – 2I 4 = –12 24  40  4I1  30I2  2I4  6I5  0 or –4I 1 + 30I 2 – 2I 4 – 6I 5 = –16

(1)

(2)

–6I 1 + 18I 3 – 4I 4 = 30

(3)

–2I 1 – 2I 2 – 4I 3 + 12I 4 –4I 5 = 0

(4)

–6I 2 – 4I 4 + 18I 5 = –32

(5)

Putting (1) to (5) in matrix form  30  4  6  2 0    12   4 30 0  2  6   16        6 0 18  4 0  I   30       2  2  4 12  4  0   0  6 0  4 18   32

ZI = V

Using MATLAB, >> Z = [30,-4,-6,-2,0; -4,30,0,-2,-6; -6,0,18,-4,0; -2,-2,-4,12,-4; 0,-6,0,-4,18] Z= 30 -4 -6 -2 0 -4 30 0 -2 -6 -6 0 18 -4 0 -2 -2 -4 12 -4

0

-6

0

-4

18

>> V = [-12,-16,30,0,-32]' V= -12 -16 30 0 -32 >> I = inv(Z)*V I= -0.2779 A -1.0488 A 1.4682 A -0.4761 A -2.2332 A

Chapter 3, Solution 67 Consider the circuit below.

5A

V1

4

V2

2

V3

+ Vo 3 Vo

10 

5

10 A

0   0.35  0.25  5  3Vo   0.25 0.95  0.5 V    0      0  15   0.5 0.5 

Since we actually have four unknowns and only three equations, we need a constraint equation. Vo = V2 – V3 Substituting this back into the matrix equation, the first equation becomes, 0.35V 1 – 3.25V 2 + 3V 3 = –5 This now results in the following matrix equation,

3   5  0.35  3.25  0.25 0.95  0.5 V   0       15   0 0.5   0.5

Now we can use MATLAB to solve for V.

>> Y=[0.35,-3.25,3;-0.25,0.95,-0.5;0,-0.5,0.5] Y= 0.3500 -3.2500 3.0000 -0.2500 0.9500 -0.5000 0 -0.5000 0.5000 >> I=[–5,0,15]' I= –5 0 15 >> V=inv(Y)*I V= –410.5262 –194.7368 –164.7368

V o = V 2 – V 3 = –77.89 + 65.89 = –30 V. Let us now do a quick check at node 1. –3(–30) + 0.1(–410.5) + 0.25(–410.5+194.74) + 5 = 90–41.05–102.62+48.68+5 = 0.01; essentially zero considering the accuracy we are using. The answer checks.

Chapter 3, Solution 68 Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the voltage V o in the circuit of Fig. 3.112. 3A

10 

25  +

4A

40 

20 

Vo

+ _

24 V

+ _

24 V

_

Figure 3.112 For Prob. 3.68. Solution Consider the circuit below. There are two non-reference nodes. 3A

V1

10 

Vo

25 

+ 4A

40 

Vo _

20 

0.125  0.1  43   7  V    0.1 0.19   3  24 / 25   2.04       Using MATLAB, we get, >> Y=[0.125,-0.1;-0.1,0.19] Y= 0.1250 -0.1000 -0.1000 0.1900 >> I=[7,-2.04]' I= 7.0000 -2.0400 >> V=inv(Y)*I V= 81.8909 32.3636 Thus, V o = 32.36 V. We can perform a simple check at node V o , 3 + 0.1(32.36–81.89) + 0.05(32.36) + 0.04(32.36–24) = 3 – 4.953 + 1.618 + 0.3344 = – 0.0004; answer checks!

Chapter 3, Solution 69 Assume that all conductances are in mS, all currents are in mA, and all voltages are in volts. G 11 = (1/2) + (1/4) + (1/1) = 1.75, G 22 = (1/4) + (1/4) + (1/2) = 1, G 33 = (1/1) + (1/4) = 1.25, G 12 = -1/4 = -0.25, G 13 = -1/1 = -1, G 21 = -0.25, G 23 = -1/4 = -0.25, G 31 = -1, G 32 = -0.25 i 1 = 20, i 2 = 5, and i 3 = 10 – 5 = 5 The node-voltage equations are:  1   v 1   20  1.75  0.25   0.25 1  0.25 v 2    5     1  0.25 1.25  v 3   5 

Chapter 3, Solution 70  4I x  20  3 0 0 5  V    4 I  7    x  

With two equations and three unknowns, we need a constraint equation, I x = 2V 1 , thus the matrix equation becomes,   5 0  20   8 5 V    7      This results in V 1 = 20/(–5) = –4 V and V 2 = [–8(–4) – 7]/5 = [32 – 7]/5 = 5 V.


 9  4  5  30   4 7  1 I   15       5  1 9   0 

We can now use MATLAB solve for our currents.

>> R=[9,–4,–5;–4,7,–1;–5,–1,9] R= 9 –4 –5 –4 7 –1 –5 –1 9 >> V=[30,–15,0]' V= 30 –15 0 >> I=inv(R)*V I= 6.255 A 1.9599 A 3.694 A

Chapter 3, Solution 72 R 11 = 5 + 2 = 7, R 22 = 2 + 4 = 6, R 33 = 1 + 4 = 5, R 44 = 1 + 4 = 5, R 12 = -2, R 13 = 0 = R 14 , R 21 = -2, R 23 = -4, R 24 = 0, R 31 = 0, R 32 = -4, R 34 = -1, R 41 = 0 = R 42 , R 43 = -1, we note that R ij = R ji for all i not equal to j. v 1 = 8, v 2 = 4, v 3 = -10, and v 4 = -4 Hence the mesh-current equations are:

0   i1   8   7 2 0  2 6  4 0  i   4    2      0  4 5  1  i 3    10      0  1 5  i4    4   0

Chapter 3, Solution 73 R 11 = 2 + 3 +4 = 9, R 22 = 3 + 5 = 8, R 33 = 1+1 + 4 = 6, R 44 = 1 + 1 = 2, R 12 = -3, R 13 = -4, R 14 = 0, R 23 = 0, R 24 = 0, R 34 = -1 v 1 = 6, v 2 = 4, v 3 = 2, and v 4 = -3 Hence,

 9  3  4 0   i1   6   3 8 0 0   i 2   4     4 0 6  1  i 3   2       0  1 2   i 4    3  0

Chapter 3, Solution 74 R 11 = R 1 + R 4 + R 6 , R 22 = R 2 + R 4 + R 5 , R 33 = R 6 + R 7 + R 8 , R 44 = R 3 + R 5 + R 8 , R 12 = -R 4 , R 13 = -R 6 , R 14 = 0, R 23 = 0, R 24 = -R 5 , R 34 = -R 8 , again, we note that R ij = R ji for all i not equal to j.

 V1   V  2 The input voltage vector is =   V3      V4   R 1  R4  R6   R4    R6  0 

 R4 R2  R4  R5 0  R5

 R6 0 R6  R7  R8  R8

  i1   V1   i   V   2    2    i 3   V3      R3  R5  R8   i 4    V4  0  R5  R8

Chapter 3, Solution 75 * Schematics Netlist * R_R4 R_R2 R_R1 R_R3 R_R5 V_V4 v_V3 v_V2 v_V1

$N_0002 $N_0001 30 $N_0001 $N_0003 10 $N_0005 $N_0004 30 $N_0003 $N_0004 10 $N_0006 $N_0004 30 $N_0003 0 120V $N_0005 $N_0001 0 0 $N_0006 0 0 $N_0002 0

i3

i1

i2

Clearly, i 1 = –3 amps, i 2 = 0 amps, and i 3 = 3 amps, which agrees with the answers in Problem 3.44.

Chapter 3, Solution 76 * Schematics Netlist * I_I2 R_R1 R_R3 R_R2 F_F1 VF_F1 R_R4 R_R6 I_I1 R_R5

0 $N_0001 DC 4A $N_0002 $N_0001 0.25 $N_0003 $N_0001 1 $N_0002 $N_0003 1 $N_0002 $N_0001 VF_F1 3 $N_0003 $N_0004 0V 0 $N_0002 0.5 0 $N_0001 0.5 0 $N_0002 DC 2A 0 $N_0004 0.25

Clearly, v 1 = 625 mVolts, v 2 = 375 mVolts, and v 3 = 1.625 volts, which agrees with the solution obtained in Problem 3.27.

1 Chapter 3, Solution 77

As a check we can write the nodal equations,  1.7  0.2 5  1.2 1.2  V   2     Solving this leads to V 1 = 3.111 V and V 2 = 1.4444 V. The answer checks!

Chapter 3, Solution 78 The schematic is shown below. When the circuit is saved and simulated the node voltages are displayed on the pseudocomponents as shown. Thus, V1  3V, V2  4.5V, V3  15V,

.

Chapter 3, Solution 79 The schematic is shown below. When the circuit is saved and simulated, we obtain the node voltages as displayed. Thus, V a = –10.556 volts; V b = 20.56 volts; V c = 1.3889 volts; and V d = –43.75 volts. 1.3889 V R3 10

–43.75 V R6 4

R4 5

R5 4

R1

R2

20

8

20.56 V

R7 16

R8 8 V1

V2 60Vdc

90Vdc

–10.556 V

Chapter 3, Solution 80 * Schematics Netlist * H_H1 VH_H1 I_I1 V_V1 R_R4 R_R1 R_R2 R_R5 R_R3

$N_0002 $N_0003 VH_H1 6 0 $N_0001 0V $N_0004 $N_0005 DC 8A $N_0002 0 20V 0 $N_0003 4 $N_0005 $N_0003 10 $N_0003 $N_0002 12 0 $N_0004 1 $N_0004 $N_0001 2

Clearly, v 1 = 84 volts, v 2 = 4 volts, v 3 = 20 volts, and v 4 = -5.333 volts


Clearly, v 1 = 26.67 volts, v 2 = 6.667 volts, v 3 = 173.33 volts, and v 4 = –46.67 volts which agrees with the results of Example 3.4.

This is the netlist for this circuit. * Schematics Netlist * R_R1 R_R2 R_R3 R_R4 R_R5 I_I1 V_V1 E_E1

0 $N_0001 2 $N_0003 $N_0002 6 0 $N_0002 4 0 $N_0004 1 $N_0001 $N_0004 3 0 $N_0003 DC 10A $N_0001 $N_0003 20V $N_0002 $N_0004 $N_0001 $N_0004 3

Chapter 3, Solution 82 2i 0

+ v0 – 3 k

1

2 k

3v 0

2

3

6 k

4

+

4A 4 k

8 k

100V +

– 0

This network corresponds to the Netlist.

Chapter 3, Solution 83 The circuit is shown below. 1

20 V

+

20 

70 

2

50 

2A

3

30 

– 0

When the circuit is saved and simulated, we obtain v 2 = –12.5 volts

Chapter 3, Solution 84 From the output loop, v 0 = 50i 0 x20x103 = 106i 0

(1)

From the input loop, 15x10-3 + 4000i 0 – v 0 /100 = 0

(2)

From (1) and (2) we get, i 0 = 2.5 A and v 0 = 2.5 volt.


The amplifier acts as a source. Rs + Vs -

RL

For maximum power transfer, R L  Rs  9

Chapter 3, Solution 86 Let v 1 be the potential across the 2 k-ohm resistor with plus being on top. Then, Since i = [(0.047–v 1 )/1k] [(v 1 –0.047)/1k] – 400[(0.047–v 1 )/1k] + [(v 1 –0)/2k] = 0 or 401[(v 1 –0.047)] + 0.5v 1 = 0 or 401.5v 1 = 401x0.047 or v 1 = 0.04694 volts and i = (0.047–0.04694)/1k = 60 nA Thus, v 0 = –5000x400x60x10-9 = –120 mV.

Chapter 3, Solution 87 v 1 = 500(v s )/(500 + 2000) = v s /5 v 0 = -400(60v 1 )/(400 + 2000) = -40v 1 = -40(v s /5) = -8v s , Therefore, v 0 /v s = –8

Chapter 3, Solution 88 Let v 1 be the potential at the top end of the 100-ohm resistor. (v s – v 1 )/200 = v 1 /100 + (v 1 – 10-3v 0 )/2000

(1)

For the right loop, v 0 = -40i 0 (10,000) = -40(v 1 – 10-3)10,000/2000, or, v 0 = -200v 1 + 0.2v 0 = -4x10-3v 0

(2)

Substituting (2) into (1) gives, (v s + 0.004v 1 )/2 = -0.004v 0 + (-0.004v 1 – 0.001v 0 )/20 This leads to 0.125v 0 = 10v s or (v 0 /v s ) = 10/0.125 = –80

Chapter 3, Solution 89 Consider the circuit below.

_

0.7 V

C +

100 k

|

|

+ IC V CE _

2.25 V

15 V

+ _

1 k

E

For the left loop, applying KVL gives –2.25 – 0.7 + 105I B + V BE = 0 but V BE = 0.7 V means 105I B = 2.25 or I B = 22.5 µA. For the right loop, –V CE + 15 – I C x103 = 0. Addition ally, I C = βI B = 100x22.5x10–6 = 2.25 mA. Therefore, V CE = 15–2.25x10–3x103 = 12.75 V.


10 k

vs

+

-

i1

IB

i2

+ V CE + V BE – –

18V

500 

+

IE

–

+

-

V0

For loop 1, -v s + 10k(I B ) + V BE + I E (500) = 0 = -v s + 0.7 + 10,000I B + 500(1 + )I B which leads to v s + 0.7 = 10,000I B + 500(151)I B = 85,500I B But, v 0 = 500I E = 500x151I B = 4 which leads to I B = 5.298x10-5 Therefore, v s = 0.7 + 85,500I B = 5.23 volts

Chapter 3, Solution 91 We first determine the Thevenin equivalent for the input circuit. R Th = 6||2 = 6x2/8 = 1.5 k and V Th = 2(3)/(2+6) = 0.75 volts

5 k

IC 1.5 k

+ 0.75 V

-

i1

IB

i2

+ V CE + V BE – –

9V

400 

+

IE

–

+

-

V0

For loop 1, -0.75 + 1.5kI B + V BE + 400I E = 0 = -0.75 + 0.7 + 1500I B + 400(1 + )I B I B = 0.05/81,900 = 0.61 A v 0 = 400I E = 400(1 + )I B = 49 mV For loop 2, -400I E – V CE – 5kI C + 9 = 0, but, I C = I B and I E = (1 + )I B V CE = 9 – 5kI B – 400(1 + )I B = 9 – 0.659 = 8.641 volts

Chapter 3, Solution 92 Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find I B and V C for the circuit in Fig. 3.128. Let  = 100, V BE = 0.7V.

Figure 3.128 Solution I1

5 k

10 k

VC

IB

IC

+ V CE + V BE – – 4 k

IE

12V

+ V0

–

I 1 = I B + I C = (1 + )I B and I E = I B + I C = I 1

+

-

Applying KVL around the outer loop, 4kI E + V BE + 10kI B + 5kI 1 = 12 12 – 0.7 = 5k(1 + )I B + 10kI B + 4k(1 + )I B = 919kI B I B = 11.3/919k = 12.296 A Also, 12 = 5kI 1 + V C which leads to V C = 12 – 5k(101)I B = 5.791 volts

Chapter 3, Solution 93 1 4

v1 i1

24V

+

3v 0

i 2

2

2

v2 i3 + 8

i

i2

–

3v 0

4

+

+

+

v0

v1

v2

–

–

–

(a)

+

(b)

From (b), -v 1 + 2i – 3v 0 + v 2 = 0 which leads to i = (v 1 + 3v 0 – v 2 )/2 At node 1 in (a), ((24 – v 1 )/4) = (v 1 /2) + ((v 1 +3v 0 – v 2 )/2) + ((v 1 – v 2 )/1), where v 0 = v 2 or 24 = 9v 1 which leads to v 1 = 2.667 volts At node 2, ((v 1 – v 2 )/1) + ((v 1 + 3v 0 – v 2 )/2) = (v 2 /8) + v 2 /4, v 0 = v 2 v 2 = 4v 1 = 10.66 volts Now we can solve for the currents, i 1 = v 1 /2 = 1.333 A, i 2 = 1.333 A, and i 3 = 2.6667 A.

Chapter 4, Solution 1. 5

30V

+ 

25  i o

i

40 

15 

40 (25  15)  20 , i = [30/(5+20)] = 1.2 and i o = i20/40 = 600 mA.

Since the resistance remains the same we get can use linearity to find the new value of the voltage source = (30/0.6)5 = 250 V.

4.2 Using Fig. 4.70, design a problem to help other students better understand linearity.

Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find v o in the circuit of Fig. 4.70. If the source current is reduced to 1 A, what is v o ?

Figure 4.70

Solution 6 (4  2)  3, i1  i 2 

io 

1 A 2

1 1 i1  , v o  2i o  0.5V 2 4

5

4

i1

io

i2 1A

If i s = 1A, then v o = 0.5V

8

6

2

Chapter 4, Solution 3. R 3R io 3R Vs

3R

+ 

+ R

vo

1V

(a) We transform the Y sub-circuit to the equivalent  .

3 3 3 3R 2 3  R, R  R  R 4 4 2 4R 4

vs independent of R 2 i o = v o /(R) vo 

When v s = 1V, v o = 0.5V, i o = 0.5A (b) (c)

3R

 (a)

R 3R 

+ 

When v s = 10V, v o = 5V, i o = 5A When v s = 10V and R = 10, v o = 5V, i o = 10/(10) = 500mA

(b)

1.5R

Chapter 4, Solution 4. If I o = 1, the voltage across the 6 resistor is 6V so that the current through the 3 resistor is 2A. 2

2

2A 1A

3A

3A

i1 +

3

6

4

Is

2

4

v1 

(a) 3 6  2 , v o = 3(4) = 12V, i1 

(b) vo  3A. 4

Hence I s = 3 + 3 = 6A If

I s = 6A I s = 9A

Io = 1 I o = 9/6 = 1.5A

Is


Vs

If v o = 1V,

If v s =

10 3

Then v s = 15

3

v1

+ 

6

1 V1     1  2V 3 10 2 Vs  2   v1  3 3

vo = 1 vo =

3 x15  4.5V 10

vo

6

6

Chapter 4, Solution 6. Due to linearity, from the first experiment, 1 Vo  Vs 3 Applying this to other experiments, we obtain:

Experiment 2 3 4

Vs

Vo

48 1V -6 V

16 V 0.333 V -2V

Chapter 4, Solution 7. If V o = 1V, then the current through the 2- and 4- resistors is ½ = 0.5. The voltage across the 3- resistor is ½ (4 + 2) = 3 V. The total current through the 1- resistor is 0.5 +3/3 = 1.5 A. Hence the source voltage vs  1x1.5  3  4.5 V If vs  4.5   1V Then vs  4

 

1 x4  0.8889 V = 888.9 mV. 4.5

1 Chapter 4, Solution 8. Let V o = V 1 + V 2 , where V 1 and V 2 are due to 9-V and 3-V sources respectively. To find V 1 , consider the circuit below.

V1

3 9

1 + _

9  V1 V1 V1   3 9 1

9V

  V1  27 /13  2.0769

To find V 2 , consider the circuit below. V1

9

V2 V2 3  V2   9 3 1

3

  V2  27 /13  2.0769

V o = V 1 + V 2 = 4.1538 V

+ _

3V

Chapter 4. Solution 9. Given that I = 4 amps when V s = 40 volts and I s = 4 amps and I = 1 amp when V s = 20 volts and I s = 0, determine the value of I when V s = 60 volts and I s = –2 amps.

VS

+ 

I

IS

At first this appears to be a difficult problem. However, if you take it one step at a time then it is not as hard as it seems. The important thing to keep in mind is that it is linear! If I = 1 amp when V s = 20 and I s = 0 then I = 2 amps when V s = 40 volts and I s = 0 (linearity). This means that I is equal to 2 amps (4–2) when I s = 4 amps and V s = 0 (superposition). Thus, I = (60/20)1 + (–2/4)2 = 3–1 = 2 amps.

Chapter 4, Solution 10. Using Fig. 4.78, design a problem to help other students better understand superposition. Note, the letter k is a gain you can specify to make the problem easier to solve but must not be zero. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the circuit in Fig. 4.78, find the terminal voltage V ab using superposition.

Figure 4.78 For Prob. 4.10. Solution Let v ab = v ab1 + v ab2 where v ab1 and v ab2 are due to the 4-V and the 2-A sources respectively.

3v ab1

10 

10 

+

3v ab2 +

+ 4V

+ 

v ab1

+ 2A

 (a)

 (b)

For v ab1 , consider Fig. (a). Applying KVL gives, - v ab1 – 3 v ab1 + 10x0 + 4 = 0, which leads to v ab1 = 1 V For v ab2 , consider Fig. (b). Applying KVL gives,

v ab2

–v ab2 – 3v ab2 + 10x2 = 0, which leads to v ab2 = 5 v ab = 1 + 5 = 6 V

Chapter 4, Solution 11. Let v o = v 1 + v 2 , where v 1 and v 2 are due to the 6-A and 80-V sources respectively. To find v 1 , consider the circuit below.

I1

va

+ V1 _

40 

6A

20 

10  vb

4 i1

At node a, 6

va va  vb  40 10

  240  5va  4vb

(1)

At node b, –I 1 – 4I 1 + (v b – 0)/20 = 0 or v b = 100I 1 But

i1 

va  vb 10

which leads to 100(v a –v b )10 = v b or v b = 0.9091v a

Substituting (2) into (1), 5v a – 3.636v a = 240 or v a = 175.95 and v b = 159.96 However,

v 1 = v a – v b = 15.99 V.

To find v 2 , consider the circuit below.

(2)

io

10  + v2 _

40 

20 

vc  

4 io

– +

30 V

0  vc (30  vc )  4io  0 50 20 (0  vc ) But io  50 5vc (30  vc )  0   50 20 0  vc 0  10 1 i2    50 50 5 v2  10i2  2 V



vc  10 V

v o = v 1 + v 2 =15.99 + 2 = 17.99 V and i o = v o /10= 1.799 A.

Chapter 4, Solution 12. Let v o = v o1 + v o2 + v o3 , where v o1 , v o2 , and v o3 are due to the 2-A, 12-V, and 19-V sources respectively. For v o1 , consider the circuit below. 2A

2A

5

4

+ v o1  6

io 5  + v o1 

12 

3

5

6||3 = 2 ohms, 4||12 = 3 ohms. Hence, i o = 2/2 = 1, v o1 = 5io = 5 V For v o2 , consider the circuit below. 6

5

4

6

+ v o2  12V

+ 

3

12 

12V

+ 

5 +

+ v o2 

v1

3

3



3||8 = 24/11, v 1 = [(24/11)/(6 + 24/11)]12 = 16/5 v o2 = (5/8)v 1 = (5/8)(16/5) = 2 V For v o3 , consider the circuit shown below. 5

4

+ v o3  6

3

12 

5 + 

+ v o3  19V

2

12 

4 + v2

+  19V

 7||12 = (84/19) ohms, v 2 = [(84/19)/(4 + 84/19)]19 = 9.975 v = (-5/7)v2 = -7.125 v o = 5 + 2 – 7.125 = -125 mV

Chapter 4, Solution 13. Let vo  v1  v2  v 3 , where v 1 , v 2 , and v 3 are due to the independent sources. To find v 1 , consider the circuit below. 8

+

5

10 

2A

v1 _

10 x2  4.3478 10  8  5 To find v 2 , consider the circuit below. v1  5 x

4A

8

+ 10 

5

v2 _

8 x4  6.9565 8  10  5 To find v 3 , consider the circuit below. v2  5 x

8

12 V + –

10 

5

+ v3 _

5   v3  12    2.6087  5  10  8  vo  v1  v2  v 3  8.6956 V =8.696V.

Chapter 4, Solution 14. Let v o = v o1 + v o2 + v o3 , where v o1 , v o2 , and v o3 , are due to the 20-V, 1-A, and 2-A sources respectively. For v o1 , consider the circuit below. 6 4

2 +

+  20V

3

v o1 

6||(4 + 2) = 3 ohms, v o1 = (½)20 = 10 V For v o2 , consider the circuit below. 6 4

6 4V

2

2

+

+ 1A

4

+

3

v o2

v o2





3||6 = 2 ohms, v o2 = [2/(4 + 2 + 2)]4 = 1 V For v o3 , consider the circuit below. 6 2A 4

2A

2

3 +

v o3

3



3  v o3 +

6||(4 + 2) = 3, v o3 = (-1)3 = –3 v o = 10 + 1 – 3 = 8 V

3

Chapter 4, Solution 15. Let i = i 1 + i 2 + i 3 , where i 1 , i 2 , and i 3 are due to the 20-V, 2-A, and 16-V sources. For i 1 , consider the circuit below. io

20V

+ 

1 i1

2

4

3

4||(3 + 1) = 2 ohms, Then i o = [20/(2 + 2)] = 5 A, i 1 = i o /2 = 2.5 A For i 3 , consider the circuit below. + 2

vo’

1

4

i3

 +

3

16V

 2||(1 + 3) = 4/3, v o ’ = [(4/3)/((4/3) + 4)](-16) = -4 i 3 = v o ’/4 = -1 For i 2 , consider the circuit below.

2

1

1

2A

2A

(4/3) i2

4

3

2||4 = 4/3, 3 + 4/3 = 13/3

i2 3

Using the current division principle. i 2 = [1/(1 + 13/2)]2 = 3/8 = 0.375 i = 2.5 + 0.375 - 1 = 1.875 A p = i2R = (1.875)23 = 10.55 watts

Chapter 4, Solution 16. Let i o = i o1 + i o2 + i o3 , where i o1 , i o2 , and i o3 are due to the 12-V, 4-A, and 2-A sources. For i o1 , consider the circuit below.

12V

3

4

i o1 + 

10 

2

5

10||(3 + 2 + 5) = 5 ohms, i o1 = 12/(5 + 4) = (12/9) A 4A

For i o2 , consider the circuit below.

3

i o2 4

2

5

10

i1 2 + 5 + 4||10 = 7 + 40/14 = 69/7 i 1 = [3/(3 + 69/7)]4 = 84/90, i o2 =[-10/(4 + 10)]i 1 = -6/9 For i o3 , consider the circuit below.

3

i o3

2

i2 4

10 

5 2A

3 + 2 + 4||10 = 5 + 20/7 = 55/7 i 2 = [5/(5 + 55/7)]2 = 7/9, i o3 = [-10/(10 + 4)]i 2 = -5/9 i o = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA

Chapter 4, Solution 17. Let v x = v x1 + v x2 + v x3 , where v x1 ,v x2 , and v x3 are due to the 90-V, 6-A, and 40-V sources. For v x1 , consider the circuit below. 30  + 90V

+ 

20 

10 

60 

v x1

 i o 10  +  v x1

30  20 

3A

12 

20||30 = 12 ohms, 60||30 = 20 ohms By using current division, i o = [20/(22 + 20)]3 = 60/42, v x1 = 10i o = 600/42 = 14.286 V For v x2 , consider the circuit below. 10  i ’ o + 30 

10  i ’ o

v x2 

+ v x2 

60  6A

30 

20 

6A

20 

12 

i o ’ = [12/(12 + 30)]6 = 72/42, v x2 = –10i o ’ = –17.143 V For v x3 , consider the circuit below. 10  + 30 

60 

v x3

20 

10 



30 

+ 40V

+ 

20 

v x3

io  12

4A

i o ” = [12/(12 + 30)]2 = 24/42, v x3 = -10i o ” = -5.714= [12/(12 + 30)]2 = 24/42, v x3 = -10i o ” = -5.714 = [12/(12 + 30)]2 = 24/42, v x3 = -10i o ” = -5.714 v x = 14.286 – 17.143 – 5.714 = -8.571 V

Chapter 4, Solution 18. Let V o = V 1 + V 2, where V 1 and V 2 are due to 10-V and 2-A sources respectively. To find V 1 , we use the circuit below. 1 0.5 V 1

2

+ 10 V

+ _

V1 _

2

1

0.5 V 1 - + +

10 V

i

+ _

4

-10 + 7i – 0.5V 1 = 0 But V 1 = 4i `10  7i  2i  5i

  i  2,

V1  8 V

V1 _

To find V 2 , we use the circuit below. 1 0.5 V 2

2

+ 4

2A

2

V2 _

1

0.5 V 2 - + +

4V

+ _

- 4 + 7i – 0.5V 2 =0 But V 2 = 4i 4  7i  2 i  5 i   i  0.8,

i

4

V2  4i  3.2

V o = V 1 + V 2 = 8 +3.2 =11.2 V

V2 _

Chapter 4, Solution 19. Let v x = v 1 + v 2 , where v 1 and v 2 are due to the 4-A and 6-A sources respectively. v1

ix

ix

v2

+ 2

4A

8

v1

+ 2

6A 8



+

+

4i x

v2 

4i x (a)

(b)

To find v 1 , consider the circuit in Fig. (a). v 1 /8 – 4 + (v 1 – (–4i x ))/2 = 0 or (0.125+0.5)v 1 = 4 – 2i x or v 1 = 6.4 – 3.2i x But,

i x = (v 1 – (–4i x ))/2 or i x = –0.5v 1 . Thus, v 1 = 6.4 + 3.2(0.5v 1 ), which leads to v 1 = –6.4/0.6 = –10.667

To find v 2 , consider the circuit shown in Fig. (b). v 2 /8 – 6 + (v 2 – (–4i x ))/2 = 0 or v 2 + 3.2i x = 9.6 But i x = –0.5v 2 . Therefore, v 2 + 3.2(–0.5v 2 ) = 9.6 which leads to v 2 = –16 Hence,

v x = –10.667 – 16 = –26.67V.

Checking, i x = –0.5v x = 13.333A Now all we need to do now is sum the currents flowing out of the top node. 13.333 – 6 – 4 + (–26.67)/8 = 3.333 – 3.333 = 0

Chapter 4, Solution 20. Convert the voltage sources to current sources and obtain the circuit shown below.

3A

10 

0.6

1 1 1 1     0.1 0.05  0.025  0.175 Req 10 20 40

20 

0.4

40 

= 5.7143 5.714 Ω eq    RReq

I eq = 3 + 0.6 + 0.4 = 4 Thus, the circuit is reduced as shown below. Please note, we that this is merely an exercise in combining sources and resistors. The circuit we have is an equivalent circuit which has no real purpose other than to demonstrate source transformation. In a practical situation, this would need some kind of reference and a use to an external circuit to be of real value. 5.714 

18.285 V 4A

5.714 

+ _

4.21 Using Fig. 4.89, design a problem to help other students to better understand source transformation. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Apply source transformation to determine v o and i o in the circuit in Fig. 4.89.

Figure 4.89

Solution To get i o , transform the current sources as shown in Fig. (a). io

6

3 i

+  12V

+ 

6V 2 A

6

3

+ vo 2 A 

(a) From Fig. (a),

(b) -12 + 9i o + 6 = 0, therefore i o = 666.7 mA

To get v o , transform the voltage sources as shown in Fig. (b). i = [6/(3 + 6)](2 + 2) = 8/3 v o = 3i = 8 V

Chapter 4, Solution 22. We transform the two sources to get the circuit shown in Fig. (a). 5  + 10V

5

4

10

2A

(a)

i 1A

10

4

10

2A

(b) We now transform only the voltage source to obtain the circuit in Fig. (b). 10||10 = 5 ohms, i = [5/(5 + 4)](2 – 1) = 5/9 = 555.5 mA

Chapter 4, Solution 23 If we transform the voltage source, we obtain the circuit below.

8

10 

6

3

5A

3A

3//6 = 2-ohm. Convert the current sources to voltages sources as shown below. 10  8 2

+ 10V -

+ 30V -

Applying KVL to the loop gives  30  10  I (10  8  2)  0   I = 1 A p  VI  I 2 R  8 W

Chapter 4, Solution 24. Transform the two current sources in parallel with the resistors into their voltage source equivalents yield, a 30-V source in series with a 10-Ω resistor and a 20V x -V sources in series with a 10-Ω resistor. We now have the following circuit,

8 +

40 V

10 

Vx –

– + 30 V

+ _ I

10  20V x

+ –

We now write the following mesh equation and constraint equation which will lead to a solution for V x , 28I – 70 + 20V x = 0 or 28I + 20V x = 70, but V x = 8I which leads to 28I + 160I = 70 or I = 0.3723 A or V x = 2.978 V.

Chapter 4, Solution 25. Transforming only the current source gives the circuit below. 18 V

9

+ – +

12V

5 i

4 +

vo 2



 +

30 V

+ 30 V

Applying KVL to the loop gives, –(4 + 9 + 5 + 2)i + 12 – 18 – 30 – 30 = 0 20i = –66 which leads to i = –3.3 v o = 2i = –6.6 V

Chapter 4, Solution 26. Transforming the current sources gives the circuit below. 2

15 V

5

io

4

– +

12 V

+ _

–12 + 11i o –15 +20 = 0 or 11i o = 7 or i o = 636.4 mA.

+ _

20 V

Chapter 4, Solution 27. Transforming the voltage sources to current sources gives the circuit in Fig. (a). 10||40 = 8 ohms Transforming the current sources to voltage sources yields the circuit in Fig. (b). Applying KVL to the loop,

-40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4 v x 12i = -48 V

12  + vx  5A

10

40

8A

20

2A

(a)

8 + 

12  + vx 

40V

i

(b)

20  + 

200V

Chapter 4, Solution 28. Convert the dependent current source to a dependent voltage source as shown below. 1

4

io

3

+ Vo _

8V

+ _

– +

Applying KVL, 8  io(1 4  3)  Vo  0 But Vo  4io 8  8io  4io  0

  io  2 A

Vo

Chapter 4, Solution 29. Transform the dependent voltage source to a current source as shown in Fig. (a). 2||4 = (4/3) k ohms 4 k 2 k

2v o

(4/3) k

+

1.5v o + 3 mA

1 k

3 mA

i

1 k

+ vo 

vo 

(a)

(b)

It is clear that i = 3 mA which leads to v o = 1000i = 3 V If the use of source transformations was not required for this problem, the actual answer could have been determined by inspection right away since the only current that could have flowed through the 1 k ohm resistor is 3 mA.

Chapter 4, Solution 30 Transform the dependent current source as shown below.

ix

24 

60 

+ 12V -

10 

+

30 

7i x -

Combine the 60-ohm with the 10-ohm and transform the dependent source as shown below. ix

24 

+ 12V -

30 

70 

0.1i x

Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the dependent current source as shown below. 24  21  ix

+ 12V -

+ 2.1i x -

Applying KVL to the loop gives 45i x  12  2.1i x  0

 

ix 

12 = 254.8 mA. 47.1

Chapter 4, Solution 31. Transform the dependent source so that we have the circuit in Fig. (a). 6||8 = (24/7) ohms. Transform the dependent source again to get the circuit in Fig. (b). 3 + 12V

+ 

vx

 8

6

v x /3

(a) 3 + 12V

+ 

vx

(24/7)   i

+ 

(8/7)v x

(b) From Fig. (b), v x = 3i, or i = v x /3. Applying KVL, -12 + (3 + 24/7)i + (24/21)v x = 0 12 = [(21 + 24)/7]v x /3 + (8/7)v x , leads to v x = 84/23 = 3.652 V

Chapter 4, Solution 32. As shown in Fig. (a), we transform the dependent current source to a voltage source, 15 

10 

5i x +

60V

+ 

50 

40 

(a) 15 

60V

+ 

50 

50 

0.1i x

(b)

ix

60V

15 

+ 

25 

ix

 +

(c) In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c), -60 + 40i x – 2.5i x = 0, or i x = 1.6 A

2.5i x

Chapter 4, Solution 33. Determine the Thevenin equivalent circuit as seen by the 5-ohm resistor. Then calculate the current flowing through the 5-ohm resistor. 10 

10 

4A

5

Solution Step 1. We need to find V oc and I sc . To do this, we will need two circuits, label the appropriate unknowns and solve for V oc , I sc , and then R eq which is equal to V oc /I sc . V1

10 

V2

10 

+ 4A

10 

V oc

4A

10 

I sc

–

Note, in the first case V 1 = V oc and the nodal equation at 1 produces –4+(V 1 –0)/10 = 0. In the second case, I sc = (V 2 –0)/10 where the nodal equation at 2 produces, –4+[(V 2 –0)/10]+[(V 2 –0)/10] = 0. Step 2. 0.1V 1 = 4 or V 1 = 40 V = V oc = V Thev . Next, (0.1+0.1)V 2 = 4 or 0.2V 2 = 4 or V 2 = 20 V. Thus, I sc = 20/10 = 2 A. This leads to R eq = 40/2 = 20 Ω. We can check our results by using source transformation. The 4-amp current source in parallel with the 10-ohm resistor can be replaced by a 40-volt voltage source in series with a 10-ohm resistor which in turn is in series with the other 10-ohm resistor yielding the same Thevenin equivalent circuit. Once the 5-ohm resistor is connected to the Thevenin equivalent circuit, we now have 40 V across 25 Ω producing a current of 1.6 A.

4.34 Using Fig. 4.102, design a problem that will help other students better understand Thevenin equivalent circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.102.

Figure 4.102

Solution To find R Th , consider the circuit in Fig. (a). 3A

10 

10 

20 

20 

v1

v2

+ R Th

40 

+ 

(a)

40V

V Th

40 

(b) R Th = 20 + 10||40 = 20 + 400/50 = 28 ohms

To find V Th , consider the circuit in Fig. (b).

At node 1,

(40 – v 1 )/10 = 3 + [(v 1 – v 2 )/20] + v 1 /40, 40 = 7v 1 – 2v 2

(1)

At node 2,

3 + (v1- v2)/20 = 0, or v1 = v2 – 60

(2)

Solving (1) and (2),

v 1 = 32 V, v 2 = 92 V, and V Th = v 2 = 92 V

Chapter 4, Solution 35. To find R Th , consider the circuit in Fig. (a). R Th = R ab = 6||3 + 12||4 = 2 + 3 =5 ohms To find V Th , consider the circuit shown in Fig. (b). R Th a 6

b 12 

3

4

(a)

2A 6

v1

v2 4 

+ V Th

+ + 

12V v 1

+ 3

12



v2

+ 

19V



(b)

At node 1,

2 + (12 – v 1 )/6 = v 1 /3, or v 1 = 8

At node 2,

(19 – v 2 )/4 = 2 + v 2 /12, or v 2 = 33/4

But,

-v 1 + V Th + v 2 = 0, or V Th = v 1 – v 2 = 8 – 33/4 = -0.25

a +

vo



b

10  R Th = 5  + V Th = (-

v o = V Th /2 = -0.25/2 = –125 mV

Chapter 4, Solution 36. Remove the 30-V voltage source and the 20-ohm resistor.

a R Th

10 a

10

+ + 

40

V Th

40

50V

b

b

(a)

(b)

From Fig. (a),

R Th = 10||40 = 8 ohms

From Fig. (b),

V Th = (40/(10 + 40))50 = 40V 8

+ 

i a 12 

40V + 

30V b

(c) The equivalent circuit of the original circuit is shown in Fig. (c). Applying KVL, 30 – 40 + (8 + 12)i = 0, which leads to i = 500mA

Chapter 4, Solution 37 R N is found from the circuit below. 20  a 40 

12 

b

R N  12 //( 20  40)  10 Ω. I N is found from the circuit below. 2A 20 

a 40 

+ 120V -

12  IN b

Applying source transformation to the current source yields the circuit below. 20 

40 

+ 80 V -

+ 120V -

Applying KVL to the loop yields  120  80  60 I N  0  

IN

I N  40 / 60  666.7 mA.

Chapter 4, Solution 38 We find Thevenin equivalent at the terminals of the 10-ohm resistor. For R Th , consider the circuit below. 1 4 5 R Th

16 

RTh  1  5 //( 4  16)  1  4  5 For V Th , consider the circuit below. V1

4

1

V2 5

3A

+

16 

V Th + 12 V -

At node 1, V V  V2 3 1  1   48  5V1  4V2 16 4 At node 2, V1  V2 12  V2  0   48  5V1  9V2 4 5 Solving (1) and (2) leads to VTh  V2  19.2

-

(1)

(2)

Thus, the given circuit can be replaced as shown below. 5

+ 19.2V -

+ Vo -

10 

Using voltage division, Vo 

10 (19.2) = 12.8 V. 10  5

Chapter 4, Solution 39. We obtain R Th using the circuit below.

10 

16

5

10 

R Th

R Thev = 16 + (20||5) = 16 + (20x5)/(20+5) = 20 Ω To find V Th , we use the circuit below. 3A

10

16

V1

+

10  24

V2 +

+ _

V2

5

V Th

_

At node 1, 24  V1 V  V2 3 1   54  2V1  V2 10 10 At node 2, V1  V2 V 3 2   60  2V1  6V2 10 5

_

(1)

(2)

Substracting (1) from (2) gives 6 = -5V 2 or V 2 = -1.2 V But -V 2 + 16x3 + V Thev = 0 or V Thev = -(48 + 1.2) = –49.2 V

Chapter 4, Solution 40. To obtain V Th , we apply KVL to the loop. 70  (10  20)kI  4Vo  0 But Vo  10kI 70  70kI   I  1mA 70  10kI  VTh  0   VTh  60 V To find R Th , we remove the 70-V source and apply a 1-V source at terminals a-b, as shown in the circuit below.

a



I2 

– Vo

I1

10 k

1V

+

+ _ b

We notice that V o = -1 V. 1 20kI1  4Vo  0   I1  0.25 mA I2  I1 

1V  0.35 mA 10k

RTh 

1V 1  k  2.857 k I2 0.35

20  + –

4 Vo


To find R Th , consider the circuit below 14  a

6

5

b RTh  5 //(14  6)  4  R N Applying source transformation to the 1-A current source, we obtain the circuit below. 6

- 14V +

14 

V Th a

+ 6V

3A

5

b At node a, 14  6  VTh V  3  Th 6  14 5 IN 

 

VTh  8 V

VTh  (8) / 4  2 A RTh

Thus, RTh  RN  4 ,

VTh  8V,

I N  2 A

Chapter 4, Solution 42. To find R Th , consider the circuit in Fig. (a). 20  10 

30 

20 

a

a

b 30  30 

b 10

10 

10 

10 

(a)

10 

(b)

20||20 = 10 ohms. Transform the wye sub-network to a delta as shown in Fig. (b). 10||30 = 7.5 ohms. R Th = R ab = 30||(7.5 + 7.5) = 10 ohms. To find V Th , we transform the 20-V (to a current source in parallel with the 20 Ω resistor and then back into a voltage source in series with the parallel combination of the two 20 Ω resistors) and the 5-A sources. We obtain the circuit shown in Fig. (c).

a

10 

+

+

10 

b

10  i1

30V

10 V

10 

+ 

i2

50V

10 

+ 

(c) For loop 1,

-30 + 50 + 30i 1 – 10i 2 = 0, or -2 = 3i 1 – i 2

(1)

For loop 2,

-50 – 10 + 30i 2 – 10i 1 = 0, or 6 = -i 1 + 3i 2

(2)

Solving (1) and (2),

i 1 = 0, i 2 = 2 A

Applying KVL to the output loop, -v ab – 10i 1 + 30 – 10i 2 = 0, v ab = 10 V V Th = v ab = 10 volts

Chapter 4, Solution 43. To find R Th , consider the circuit in Fig. (a). R Th a 10

b 5

10

(a) 10 

a +

+ 

20V v a

b

+ V Th 10 

+ vb



5

 (b)

R Th = 10||10 + 5 = 10 ohms To find V Th , consider the circuit in Fig. (b). v b = 2x5 = 10 V, v a = 20/2 = 10 V But,

-v a + V Th + v b = 0, or V Th = v a – v b = 0 volts

2A

Chapter 4, Solution 44. (a)

For R Th , consider the circuit in Fig. (a). R Th = 1 + 4||(3 + 2 + 5) = 3.857 ohms

For V Th , consider the circuit in Fig. (b). Applying KVL gives, 10 – 24 + i(3 + 4 + 5 + 2), or i = 1 V Th = 4i = 4 V 1

3

a

+ 3

1

a

+ 

24V + 

R Th

4 2

2 i

b 5

b 10V

5

(b)

(a) (b)

V Th

4

For R Th , consider the circuit in Fig. (c). 3

1

4

3

b

24V

2 R Th

5

1

4 vo

+ 

b

+

2 5

2A

c (c)

c (d)

R Th = 5||(2 + 3 + 4) = 3.214 ohms

V Th

To get V Th , consider the circuit in Fig. (d). At the node, KCL gives, [(24 – vo)/9] + 2 = vo/5, or vo = 15 V Th = vo = 15 V

Chapter 4, Solution 45. For R N , consider the circuit in Fig. (a).

6

6

6

4

RN

4A

(a)

6

4

IN

(b) R N = (6 + 6)||4 = 3 Ω

For I N , consider the circuit in Fig. (b). The 4-ohm resistor is shorted so that 4-A current is equally divided between the two 6-ohm resistors. Hence, I N = 4/2 = 2 A

Chapter 4, Solution 46. Using Fig. 4.113, design a problem to help other students better understand Norton equivalent circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the Norton equivalent at terminals a-b of the circuit in Fig. 4.113. 10 

 a 20 

10 

4A

b Figure 4.113 For Prob. 4.46.

Solution R N is found using the circuit below.

10 

 a

10 

20  RN

b

R N = 20//(10+10) = 10  To find I N , consider the circuit below. 10 

4A

10 

The 20- resistor is short-circuited and can be ignored. IN = ½ x 4 = 2 A

20 



IN

Chapter 4, Solution 47 Since V Th = V ab = V x , we apply KCL at the node a and obtain 30  VTh VTh   2VTh   VTh  150 / 126  1.1905 V 12 60 To find R Th , consider the circuit below.

12 

Vx

a

60 

2V x 1A

At node a, KCL gives V V 1  2V x  x  x   V x  60 / 126  0.4762 60 12 V V RTh  x  0.4762, I N  Th  1.19 / 0.4762  2.5 1 RTh Thus, V Thev = 1.1905 V, R eq = 476.2 mΩ, and I N = 2.5 A.

Chapter 4, Solution 48. To get R Th , consider the circuit in Fig. (a). 10I o 10I o

+

2

+

+

Io +

Io 4

V Th

4

V 

1A



2A

(a) From Fig. (a),

2

(b)

I o = 1,

6 – 10 – V = 0, or V = -4 R eq = V/1 = -4 ohms

Note that the negative value of R eq indicates that we have an active device in the circuit since we cannot have a negative resistance in a purely passive circuit. To solve for I N we first solve for V Th , consider the circuit in Fig. (b), I o = 2, V Th = -10I o + 4I o = -12 V I N = V Th /R Th = 3A

Chapter 4, Solution 49. R N = R Th = 28 ohms To find I N , consider the circuit below, 3A

10 

vo

20  io

40V

At the node,

+ 

40 

I sc = I N

(40 – v o )/10 = 3 + (v o /40) + (v o /20), or v o = 40/7 i o = v o /20 = 2/7, but I N = I sc = i o + 3 = 3.286 A

Chapter 4, Solution 50. From Fig. (a), R N = 6 + 4 = 10 ohms

6

6

I sc = I N 4

4

2A

(a) From Fig. (b),

+ 12V 

(b) 2 + (12 – v)/6 = v/4, or v = 9.6 V

-I N = (12 – v)/6 = 0.4, which leads to I N = -0.4 A Combining the Norton equivalent with the right-hand side of the original circuit produces the circuit in Fig. (c).

i 0.4A

10 

5

4A

(c) i = [10/(10 + 5)] (4 – 0.4) = 2.4 A

Chapter 4, Solution 51.

(a)

From the circuit in Fig. (a),

R N = 4||(2 + 6||3) = 4||4 = 2 ohms R Th 6

V Th

+

4

6

3

2

120V

4

+ 

3

(a)

6A

2

(b)

For I N or V Th , consider the circuit in Fig. (b). After some source transformations, the circuit becomes that shown in Fig. (c).

+ V Th 2

40V

+ 

4

i

2

12V

+ 

(c) Applying KVL to the circuit in Fig. (c), -40 + 8i + 12 = 0 which gives i = 7/2 V Th = 4i = 14 therefore I N = V Th /R N = 14/2 = 7 A

(b)

To get R N , consider the circuit in Fig. (d). R N = 2||(4 + 6||3) = 2||6 = 1.5 ohms 6

2

4

i

+ 3

2

(d)

RN

V Th

12V

+ 

(e)

To get I N , the circuit in Fig. (c) applies except that it needs slight modification as in Fig. (e). i = 7/2, V Th = 12 + 2i = 19, I N = V Th /R N = 19/1.5 = 12.667 A

Chapter 4, Solution 52. For the transistor model in Fig. 4.118, obtain the Thevenin equivalent at terminals a-b.

12 V

Figure 4.118 For Prob. 4.52. Solution Step 1.

To find the Thevenin equivalent for this circuit we need to find v oc and i sc . Then V Thev = v oc and R eq = v oc /i sc . 3 k

12

+ 

+ v oc

2 k I

20I o

–

b For v oc , I o = (12–0)/3k = 4 mA and 20I o + (v oc –0)/2k = 0. For i sc , i sc = –20I o . Step 2.

v oc = –2k(20I o ) = –40x4 = –160 volts = V Thev i sc = –20x4x10–3 = –80 mA or R eq = –160/(80x10–3) = 2 kΩ.

i sc

Chapter 4, Solution 53. To get R Th , consider the circuit in Fig. (a). 0.25v o

0.25v o

2

2

a

+ 6

3

+

vo

2

1A



1/2

a 1/2

vo

v ab



 b

b (a)

(b)

From Fig. (b), v o = 2x1 = 2V, -v ab + 2x(1/2) +v o = 0 v ab = 3V R N = v ab /1 = 3 ohms To get I N , consider the circuit in Fig. (c). 0.25v o 6

2

a

+ 18V

+ 

3

vo

I sc =

 b (c)

[(18 – v o )/6] + 0.25v o = (v o /2) + (v o /3) or v o = 4V But,

(v o /2) = 0.25v o + I N , which leads to I N = 1 A

+ 1A

Chapter 4, Solution 54 To find V Th =V x , consider the left loop.

 3  1000io  2V x  0   For the right loop, V x  50 x 40i o  2000io Combining (1) and (2), 3  1000io  4000io  3000io

V x  2000io  2

 

3  1000io  2V x

(1) (2)

 

io  1mA

VTh  2

To find R Th , insert a 1-V source at terminals a-b and remove the 3-V independent source, as shown below. 1 k

ix

. io

+ 2V x -

2V x  2mA 1000 V 1 i x  40io  x  80mA  A  -60mA 50 50

V x  1,

RTh 

io  

1  1 / 0.060   16.67 ix

40i o

+ Vx -

50 

+ 1V -

Chapter 4, Solution 55. To get R N , apply a 1 mA source at the terminals a and b as shown in Fig. (a). 8 k

a

I

+ 2V

+ 

v ab /100

80I

+ 

50 k

IN

v ab  b

(b) We assume all resistances are in k ohms, all currents in mA, and all voltages in volts. At node a, (1) (v ab /50) + 80I = 1 Also, (2) -8I = (v ab /1000), or I = -v ab /8000 From (1) and (2),

(v ab /50) – (80v ab /8000) = 1, or v ab = 100 R N = v ab /1 = 100 k ohms

To get I N , consider the circuit in Fig. (b). a

I

+

v ab /100 8 k

80I

+ 

50 k

v ab  b

(a) Since the 50-k ohm resistor is shorted, I N = -80I, v ab = 0 Hence,

8i = 2 which leads to I = (1/4) mA I N = -20 mA

1mA

Chapter 4, Solution 56. We remove the 1-k resistor temporarily and find Norton equivalent across its terminals. R eq is obtained from the circuit below. 12 k

2 k

10 k

 RN

24 k



R eq = 10 + 2 + (12//24) = 12+8 = 20 k I N is obtained from the circuit below. 12 k

36 V

+ _

2k

10 k

3 mA

24 k



IN

We can use superposition theorem to find I N . Let I N = I 1 + I 2 , where I 1 and I 2 are due to 16-V and 3-mA sources respectively. We find I 1 using the circuit below. 12 k

36 V

+ _

24 k

2k

10 k



I1

Using source transformation, we obtain the circuit below. 12 k

3 mA

24 k

12 k



I1



I2

12//24 = 8 k 8 (3mA)  1.2 mA 8  12 To find I 2 , consider the circuit below. I1 

2k

24 k

12 k

3 mA

2k + 12k//24 k = 10 k I 2 =0.5(-3mA) = -1.5 mA I N = 1.2 –1.5 = -0.3 mA The Norton equivalent with the 1-k resistor is shown below a +

In

20 k

10 k

Vo

1 k

– b V o = 1k(20/(20+1))(-0.3 mA) = -285.7 mV.

Chapter 4, Solution 57. To find R Th , remove the 50V source and insert a 1-V source at a – b, as shown in Fig. (a). 2

B

a

A

i

+ 3

6

vx

10  0.5v x



+ 

1V

b

(a) We apply nodal analysis. At node A,

i + 0.5v x = (1/10) + (1 – v x )/2, or i + v x = 0.6

(1)

At node B, (1 – v o )/2 = (v x /3) + (v x /6), and v x = 0.5 From (1) and (2),

(2)

i = 0.1 and R Th = 1/i = 10 ohms

To get V Th , consider the circuit in Fig. (b). 3

2

v1

v2

a

+ 50V

+ 

vx

+ 6 0.5v x

 (b)

10  V Th  b

At node 1,

(50 – v 1 )/3 = (v 1 /6) + (v 1 – v 2 )/2, or 100 = 6v 1 – 3v 2

(3)

At node 2,

0.5v x + (v 1 – v 2 )/2 = v 2 /10, v x = v 1 , and v 1 = 0.6v 2

(4)

From (3) and (4), v 2 = V Th = 166.67 V I N = V Th /R Th = 16.667 A R N = R Th = 10 ohms

Chapter 4, Solution 58. This problem does not have a solution as it was originally stated. The reason for this is that the load resistor is in series with a current source which means that the only equivalent circuit that will work will be a Norton circuit where the value of R N = infinity. I N can be found by solving for I sc .

ib

VS

R1

 ib

vo

+ 

R2

I sc

Writing the node equation at node vo, i b + i b = v o /R 2 = (1 + )i b But

i b = (V s – v o )/R 1 vo = Vs – ibR1 V s – i b R 1 = (1 + )R 2 i b , or i b = V s /(R 1 + (1 + )R 2 ) I sc = I N = -i b = -V s /(R 1 + (1 + )R 2 )

Chapter 4, Solution 59. R Th = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms To find V Th , consider the circuit below.

i1

i2 10 

20 

+ V Th 8A

50 

40 

i 1 = i 2 = 8/2 = 4, 10i 1 + V Th – 20i 2 = 0, or V Th = 20i 2 –10i 1 = 10i 1 = 10x4 V Th = 40V, and I N = V Th /R Th = 40/22.5 = 1.7778 A

Chapter 4, Solution 60. The circuit can be reduced by source transformations. 2A

18 V

12 V

+ 

+ 

10 V

10 

5

+ 

2A

10 

a

b

3A

5 2A

3A

a

3.333

Norton Equivalent Circuit

b

a

3.333

10 V + 

Thevenin Equivalent Circuit

b

Chapter 4, Solution 61. To find R Th , consider the circuit in Fig. (a). Let

R Th = 2R||R = (2/3)R = 1.2 ohms.

R = 2||18 = 1.8 ohms,

To get V Th , we apply mesh analysis to the circuit in Fig. (d). 2

a 6

6

2

2 6 b (a)

2

a 18  1.8 

2

18 

a

2

18 

1.8 

1.8 

R Th

b

b

(b)

(c)

2

a 6 12V

6

i3

+ 

12V

+ + 

V Th

6 2

i1

i2

 +

2

12V b (d)

-12 – 12 + 14i 1 – 6i 2 – 6i 3 = 0, and 7 i 1 – 3 i 2 – 3i 3 = 12

(1)

12 + 12 + 14 i 2 – 6 i 1 – 6 i 3 = 0, and -3 i 1 + 7 i 2 – 3 i 3 = -12

(2)

14 i 3 – 6 i 1 – 6 i 2 = 0, and

(3)

-3 i 1 – 3 i 2 + 7 i 3 = 0

This leads to the following matrix form for (1), (2) and (3),  7  3  3  i1   12   3 7  3 i    12   2     3  3 7  i 3   0  7   3

3 3  3  100 ,

7

3 3

7

7

12

3

 2   3  12  3  120 3

0

7

i 2 = / 2 = -120/100 = -1.2 A V Th = 12 + 2i 2 = 9.6 V, and I N = V Th /R Th = 8 A

Chapter 4, Solution 62. Since there are no independent sources, V Th = 0 V To obtain R Th , consider the circuit below. 0.1i o

ix

2

+ vo 10   v1 1 io 40 

VS

+ 

20  2v o + 

At node 2, i x + 0.1i o = (1 – v 1 )/10, or 10i x + i o = 1 – v 1

(1)

(v 1 /20) + 0.1i o = [(2v o – v 1 )/40] + [(1 – v 1 )/10]

(2)

At node 1,

But i o = (v 1 /20) and v o = 1 – v 1 , then (2) becomes, 1.1v 1 /20 = [(2 – 3v 1 )/40] + [(1 – v 1 )/10] 2.2v 1 = 2 – 3v 1 + 4 – 4v 1 = 6 – 7v 1 v 1 = 6/9.2

or

(3)

From (1) and (3), 10i x + v 1 /20 = 1 – v 1 10i x = 1 – v 1 – v 1 /20 = 1 – (21/20)v 1 = 1 – (21/20)(6/9.2) i x = 31.52 mA, R Th = 1/i x = 31.73 ohms.

Chapter 4, Solution 63. Because there are no independent sources, I N = I sc = 0 A R N can be found using the circuit below.

10  + vo 

Applying KCL at node 1,

v1

io

20  0.5v o

+ 

1V

v 1 = 1, and v o = (20/30)v 1 = 2/3 i o = (v 1 /30) – 0.5v o = (1/30) – 0.5x2/3 = 0.03333 – 0.33333 = – 0.3 A.

Hence, R N = 1/(–0.3) = –3.333 ohms

Chapter 4, Solution 64. With no independent sources, V Th = 0 V. To obtain R Th , consider the circuit shown below. 4

1

vo

io

ix + –

2

+ 

1V

10i x i x = [(1 – v o )/1] + [(10i x – v o )/4], or 5v o = 4 + 6i x But i x = v o /2. Hence, 5v o = 4 + 3v o , or v o = 2, i o = (1 – v o )/1 = -1 Thus, R Th = 1/i o = –1 ohm

(1)

Chapter 4, Solution 65 At the terminals of the unknown resistance, we replace the circuit by its Thevenin equivalent. 12 Req  2  (4 || 12)  2  3  5, VTh  (32)  24 V 12  4 Thus, the circuit can be replaced by that shown below.

5

Io

+ 24 V -

+ Vo -

Applying KVL to the loop,

 24  5I o  Vo  0

 

V o = 24 – 5I o .

Chapter 4, Solution 66. We first find the Thevenin equivalent at terminals a and b. We find R Th using the circuit in Fig. (a). 2 10V  +

3

2

a

b

+ 3

a

V Th

5

b R Th

+ 

5

20V

 +

i 30V

(a)

(b) R Th = 2||(3 + 5) = 2||8 = 1.6 ohms

By performing source transformation on the given circuit, we obatin the circuit in (b). We now use this to find V Th . 10i + 30 + 20 + 10 = 0, or i = –6 V Th + 10 + 2i = 0, or V Th = 2 V p = V Th 2/(4R Th ) = (2)2/[4(1.6)] = 625 m watts

Chapter 4, Solution 67. We first find the Thevenin equivalent. We find R Th using the circuit below.

80 

20  R Th

10 

90 

RTh  20 // 80  90 //10  16  9  25 

We find V Th using the circuit below. We apply mesh analysis.

80 

I1

20 

40 V +–

10 

I2

(80  20)i1  40  0 (10  90)i2  40  0 90i2  20i1  VTh  0

+ V TH

90 

  i1  0.4   i2  0.4   VTh  28 V

(a) R = R Th = 25  V2 (28)2 (b) Pmax  Th   7.84 W 4RTh 100

_

Chapter 4, Solution 68. This is a challenging problem in that the load is already specified. This now becomes a "minimize losses" style problem. When a load is specified and internal losses can be adjusted, then the objective becomes, reduce R Thev as much as possible, which will result in maximum power transfer to the load. R

12 V

10 

+

20 

+

-

8V

Removing the 10 ohm resistor and solving for the Thevenin Circuit results in: R Th = (Rx20/(R+20)) and a V oc = V Th = 12x(20/(R +20)) + (-8) As R goes to zero, R Th goes to zero and V Th goes to 4 volts, which produces the maximum power delivered to the 10-ohm resistor. P = vi = v2/R = 4x4/10 = 1.6 watts Notice that if R = 20 ohms which gives an R Th = 10 ohms, then V Th becomes -2 volts and the power delivered to the load becomes 0.1 watts, much less that the 1.6 watts. It is also interesting to note that the internal losses for the first case are 122/20 = 7.2 watts and for the second case are = to 12 watts. This is a significant difference.

Chapter 4, Solution 69. We need the Thevenin equivalent across the resistor R. To find R Th , consider the circuit below. 22 k v 1 + 10 k

40 k

vo

30 k

0.003v o



1mA

Assume that all resistances are in k ohms and all currents are in mA. 10||40 = 8, and 8 + 22 = 30 1 + 3v o = (v 1 /30) + (v 1 /30) = (v 1 /15) 15 + 45v o = v 1 But v o = (8/30)v 1 , hence, 15 + 45x(8v 1 /30) v 1 , which leads to v 1 = 1.3636 R Th = v 1 /1 = –1.3636 k ohms R Th being negative indicates an active circuit and if you now make R equal to 1.3636 k ohms, then the active circuit will actually try to supply infinite power to the resistor. The correct answer is therefore: 2

2

VTh V    pR =   1363.6   Th  1363.6 =  0    1363.6  1363.6 

∞

It may still be instructive to find V Th . Consider the circuit below. 10 k v o 22 k

v1

+ 100V

+ 

vo 

+ 40 k

0.003v o

(100 – v o )/10 = (v o /40) + (v o – v 1 )/22

30 k

V Th 

(1)

[(v o – v 1 )/22] + 3v o = (v 1 /30) Solving (1) and (2), v 1 = V Th = -243.6 volts

(2)

Chapter 4, Solution 70 We find the Thevenin equivalent across the 10-ohm resistor. To find V Th , consider the circuit below. 3V x

5

5 +

+ 15 

4V -

V Th

6

– +

Vx

-

From the figure,

15 (4)  3V 15  5 consider the circuit below:

V x  0, To find R eq,

VTh 

3V x

5

5 V1

+ 4V –

15 

+

I sc

6

Vx

-

At node 1, [(V 1 –V x )/15] + [(V 1 –(4+V x ))/5] + [(V 1 –0)/5] + 3V x = 0 or

0.4667V 1 + 2.733V x = 0.8

(1)

At node x, [(V x –0)/6] + [((V x +4)–V 1 )/5] + [(V x –V 1 )/15] = 0 or –(0.2667)V 1 + 0.4333V x = –0.8

(2)

Adding (1) and (2) together lead to, (0.4667–0.2667)V 1 + (2.733+0.4333)V x = 0 or V 1 = –(3.166/0.2)V x = –15.83V x Now we can put this into (1) and we get, 0.4667(–15.83V x ) + 2.733V x = 0.8 = (–7.388+2.733)V x = –4.655V x or V x = –0.17186 V. I sc = –V x /6 = 0.02864 and R eq = 3/(0.02864) = 104.75 Ω An alternate way to find R eq is replace I sc with a 1 amp current source flowing up and setting the 4 volts source to zero. We then find the voltage across the 1 amp current source which is equal to R eq . First we note that V x = 6 volts ; V 1 = 6+3.75 = –9.75; V 2 = 19x5 + V 1 = 95+9.75 = 104.75 or R eq = 104.75 Ω. Clearly setting the load resistance to 104.75 Ω means that the circuit will deliver maximum power to it. Therefore, p max = [3/(2x104.75)]2x104.75 = 21.48 mW

Chapter 4, Solution 71. We need R Th and V Th at terminals a and b. To find R Th , we insert a 1-mA source at the terminals a and b as shown below. 10 k + 3 k

vo

 +

1 k



120v o

a 40 k 1mA b

Assume that all resistances are in k ohms, all currents are in mA, and all voltages are in volts. At node a, 1 = (v a /40) + [(v a + 120v o )/10], or 40 = 5v a + 480v o

(1)

The loop on the left side has no voltage source. Hence, v o = 0. From (1), v a = 8 V. R Th = v a /1 mA = 8 kohms To get V Th , consider the original circuit. For the left loop, v o = (1/4)8 = 2 V For the right loop,

v R = V Th = (40/50)(-120v o ) = -192

The resistance at the required resistor is R = R Th = 8 kΩ p = V Th 2/(4R Th ) = (-192)2/(4x8x103) = 1.152 watts

Chapter 4, Solution 72. (a) R Th and V Th are calculated using the circuits shown in Fig. (a) and (b) respectively. From Fig. (a),

R Th = 2 + 4 + 6 = 12 ohms

From Fig. (b),

-V Th + 12 + 8 + 20 = 0, or V Th = 40 V

4

2

6

4

12V

6

 +

+ 2

R Th

+ 

V Th

8V 20V

(a) (b)

+ 

(b) i = V Th /(R Th + R) = 40/(12 + 8) = 2A

(c)

For maximum power transfer,

R L = R Th = 12 ohms

(d)

p = V Th 2/(4R Th ) = (40)2/(4x12) = 33.33 watts.



Chapter 4, Solution 73 Find the Thevenin’s equivalent circuit across the terminals of R.

10 

25  R Th

20 

5

RTh  10 // 20  25 // 5  325 / 30  10.833

10  + 60 V -

25  +

V Th -

+

+

Va

Vb 20 

5

-

20 (60)  40, 30  Va  VTh  Vb  0

Va 

-

5 (60)  10 30   VTh  Va  Vb  40  10  30 V Vb 

2

p max

V 30 2 = 20.77 W.  Th  4 RTh 4 x10.833

Chapter 4, Solution 74. When R L is removed and V s is short-circuited, R Th = R 1 ||R 2 + R 3 ||R 4 = [R 1 R 2 /( R 1 + R 2 )] + [R 3 R 4 /( R 3 + R 4 )] R L = R Th = (R 1 R 2 R 3 + R 1 R 2 R 4 + R 1 R 3 R 4 + R 2 R 3 R 4 )/[( R 1 + R 2 )( R 3 + R 4 )] When R L is removed and we apply the voltage division principle, V oc = V Th = v R2 – v R4 = ([R 2 /(R 1 + R 2 )] – [R 4 /(R 3 + R 4 )])V s = {[(R 2 R 3 ) – (R 1 R 4 )]/[(R 1 + R 2 )(R 3 + R 4 )]}V s p max = V Th 2/(4R Th ) = {[(R 2 R 3 ) – (R 1 R 4 )]2/[(R 1 + R 2 )(R 3 + R 4 )]2}V s 2[( R 1 + R 2 )( R 3 + R 4 )]/[4(a)] where a = (R 1 R 2 R 3 + R 1 R 2 R 4 + R 1 R 3 R 4 + R 2 R 3 R 4 ) p max = [(R 2 R 3 ) – (R 1 R 4 )]2V s 2/[4(R 1 + R 2 )(R 3 + R 4 ) (R 1 R 2 R 3 + R 1 R 2 R 4 + R 1 R 3 R 4 + R 2 R 3 R 4 )]

Chapter 4, Solution 75. We need to first find R Th and V Th . R R R R

R

R

vo +

R Th 1V

+ 

2V

+ 

+ 

3V

V Th 

(a) (b) Consider the circuit in Fig. (a). (1/R eq ) = (1/R) + (1/R) + (1/R) = 3/R R eq = R/3 From the circuit in Fig. (b), ((1 – v o )/R) + ((2 – v o )/R) + ((3 – v o )/R) = 0 v o = 2 = V Th For maximum power transfer, R L = R Th = R/3 P max = [(V Th )2/(4R Th )] = 3 mW R Th = [(V Th )2/(4P max )] = 4/(4xP max ) = 1/P max = R/3 R = 3/(3x10-3) = 1 kΩ 1 kΩ, 3 mW

Chapter 4, Solution 76. Follow the steps in Example 4.14. The schematic and the output plots are shown below. From the plot, we obtain, V = 92 V [i = 0, voltage axis intercept] R = Slope = (120 – 92)/1 = 28 ohms

Chapter 4, Solution 77. (a) The schematic is shown below. We perform a dc sweep on a current source, I1, connected between terminals a and b. We label the top and bottom of source I1 as 2 and 1 respectively. We plot V(2) – V(1) as shown. V Th = 4 V [zero intercept] R Th = (7.8 – 4)/1 = 3.8 ohms

(b)

Everything remains the same as in part (a) except that the current source, I1, is connected between terminals b and c as shown below. We perform a dc sweep on I1 and obtain the plot shown below. From the plot, we obtain, V = 15 V [zero intercept] R = (18.2 – 15)/1 = 3.2 ohms

Chapter 4, Solution 78. The schematic is shown below. We perform a dc sweep on the current source, I1, connected between terminals a and b. The plot is shown. From the plot we obtain, V Th = -80 V [zero intercept] R Th = (1920 – (-80))/1 = 2 k ohms

Chapter 4, Solution 79. After drawing and saving the schematic as shown below, we perform a dc sweep on I1 connected across a and b. The plot is shown. From the plot, we get, V = 167 V [zero intercept] R = (177 – 167)/1 = 10 ohms

Chapter 4, Solution 80. The schematic in shown below. We label nodes a and b as 1 and 2 respectively. We perform dc sweep on I1. In the Trace/Add menu, type v(1) – v(2) which will result in the plot below. From the plot, V Th = 40 V [zero intercept] R Th = (40 – 17.5)/1 = 22.5 ohms [slope]

Chapter 4, Solution 81. The schematic is shown below. We perform a dc sweep on the current source, I2, connected between terminals a and b. The plot of the voltage across I2 is shown below. From the plot, V Th = 10 V [zero intercept] R Th = (10 – 6.7)/1 = 3.3 ohms. Note that this is in good agreement with the exact value of 3.333 ohms.

Chapter 4, Solution 82. V Th = V oc = 12 V, I sc = 20 A R Th = V oc /I sc = 12/20 = 0.6 ohm. 0.6  i 12V

i = 12/2.6 ,

+ 

2

p = i2R = (12/2.6)2(2) = 42.6 watts

Chapter 4, Solution 83. V Th = V oc = 12 V, I sc = I N = 1.5 A R Th = V Th /I N = 8 ohms, V Th = 12 V, R Th = 8 ohms

Chapter 4, Solution 84 Let the equivalent circuit of the battery terminated by a load be as shown below. R Th IL +

+

V Th VL

-

RL

-

For open circuit,

R L  ,   VTh  Voc  VL  10.8 V When R L = 4 ohm, V L =10.5, IL 

VL  10.8 / 4  2.7 RL

But VTh  VL  I L RTh

 

RTh 

VTh  V L 12  10.8   0.4444 2.7 IL

= 444.4 mΩ.

Chapter 4, Solution 85 (a) Consider the equivalent circuit terminated with R as shown below. R Th a

+ V Th -

+ V ab -

R

b

Vab

R  VTh R  RTh

 

10 6 VTh 10  RTh

or 60  6 RTh  10VTh where R Th is in k-ohm.

(1)

Similarly,

30   VTh 30  RTh Solving (1) and (2) leads to 12 

360  12 RTh  30VTh

VTh  24 V, RTh  30k

(b) V ab 

20 ( 24)  9.6 V 20  30

(2)

Chapter 4, Solution 86. We replace the box with the Thevenin equivalent. R Th + V Th

+ 

i R

v 

V Th = v + iR Th When i = 1.5, v = 3, which implies that V Th = 3 + 1.5R Th

(1)

When i = 1, v = 8, which implies that V Th = 8 + 1xR Th

(2)

From (1) and (2), R Th = 10 ohms and V Th = 18 V. (a)

When R = 4, i = V Th /(R + R Th ) = 18/(4 + 10) = 1.2857 A

(b)

For maximum power, R = R TH Pmax = (V Th )2/4R Th = 182/(4x10) = 8.1 watts


i m = 9.876 mA

i m = 9.975 mA +

Is

vm

Rs Rm

Is

Rs

Rs Rm

 (a)

(b)

From Fig. (a), v m = R m i m = 9.975 mA x 20 = 0.1995 V I s = 9.975 mA + (0.1995/R s )

(1)

From Fig. (b), v m = R m i m = 20x9.876 = 0.19752 V I s = 9.876 mA + (0.19752/2k) + (0.19752/R s ) = 9.975 mA + (0.19752/R s ) Solving (1) and (2) gives, R s = 8 k ohms,

I s = 10 mA

(b) i m ’ = 9.876

Is

Rs

Rs Rm

(b) 8k||4k = 2.667 k ohms i m ’ = [2667/(2667 + 20)](10 mA) = 9.926 mA

(2)

Chapter 4, Solution 88 To find R Th, consider the circuit below. 5k 

R Th A

B

30k 

20k 

10k  RTh  30  10  20 // 5  44k To find V Th , consider the circuit below. 5k  A

B io

30k 

20k 

+

4mA

60 V -

10k 

V A  30 x 4  120,

VB 

20 (60)  48, 25

VTh  V A  VB  72 V

The Thevenin equivalent circuit is shown below. 44k  I

Ri + 72 V -

2k 

72 mA 44  2  Ri assuming R i is in k-ohm. I

(a) When R i =500  , I

72  1.548 mA 44  2  0.5

(b) When R i = 0  , I

72  1.565 mA 44  2  0

Chapter 4, Solution 89 It is easy to solve this problem using Pspice. (a) The schematic is shown below. We insert IPROBE to measure the desired ammeter reading. We insert a very small resistance in series IPROBE to avoid problem. After the circuit is saved and simulated, the current is displaced on IPROBE as 99.99A .

(b) By interchanging the ammeter and the 12-V voltage source, the schematic is shown below. We obtain exactly the same result as in part (a).

Chapter 4, Solution 90. R x = (R 3 /R 1 )R 2 = (4/2)R 2 = 42.6, R 2 = 21.3 which is (21.3ohms/100ohms)% = 21.3%

Chapter 4, Solution 91. R x = (R 3 /R 1 )R 2 (a) Since 0 < R 2 < 50 ohms, to make 0 < R x < 10 ohms requires that when R 2 = 50 ohms, R x = 10 ohms. 10 = (R 3 /R 1 )50 or R 3 = R 1 /5 so we select R 1 = 100 ohms and R 3 = 20 ohms (b)

For 0 < R x < 100 ohms 100 = (R 3 /R 1 )50, or R 3 = 2R 1

So we can select R 1 = 100 ohms and R 3 = 200 ohms

Chapter 4, Solution 92. For a balanced bridge, v ab = 0. We can use mesh analysis to find v ab . Consider the circuit in Fig. (a), where i 1 and i 2 are assumed to be in mA. 2 k

3 k 220V

+ 

a

i1

+

6 k

i2 b v ab

5 k

10 k

0 (a) 220 = 2i 1 + 8(i 1 – i 2 ) or 220 = 10i 1 – 8i 2 (1) 0 = 24i 2 – 8i 1 or i 2 = (1/3)i 1

(2)

From (1) and (2), i 1 = 30 mA and i 2 = 10 mA Applying KVL to loop 0ab0 gives 5(i 2 – i 1 ) + v ab + 10i 2 = 0 V Since v ab = 0, the bridge is balanced. When the 10 k ohm resistor is replaced by the 18 k ohm resistor, the gridge becomes unbalanced. (1) remains the same but (2) becomes 0 = 32i 2 – 8i 1 , or i 2 = (1/4)i 1 Solving (1) and (3), i 1 = 27.5 mA, i 2 = 6.875 mA v ab = 5(i 1 – i 2 ) – 18i 2 = -20.625 V V Th = v ab = -20.625 V

(3)

To obtain R Th , we convert the delta connection in Fig. (b) to a wye connection shown in Fig. (c). 2 k

3 k

6 k

a

5 k

R Th

R2

6 k R1

b

18 k

a R Th

R3

(b)

b

18 k

(c)

R 1 = 3x5/(2 + 3 + 5) = 1.5 k ohms, R 2 = 2x3/10 = 600 ohms, R 3 = 2x5/10 = 1 k ohm. R Th = R 1 + (R 2 + 6)||(R 3 + 18) = 1.5 + 6.6||9 = 6.398 k ohms R L = R Th = 6.398 k ohms P max = (V Th )2/(4R Th ) = (20.625)2/(4x6.398) = 16.622 mWatts

Chapter 4, Solution 93. Rs

ix

VS

+ 

Ro R o i x ix

+ 

-V s + (R s + R o )i x + R o i x = 0 i x = V s /(R s + (1 + )R o )


V o /V g = R p /(R g + R s + R p )

(1)

R eq = R p ||(R g + R s ) = R g R g = R p (R g + R s )/(R p + R g + R s ) RgRp + Rg2 + RgRs = RpRg + RpRs R p R s = R g (R g + R s ) From (1),

(2)

R p / = R g + R s + R p R g + R s = R p ((1/) – 1) = R p (1 - )/

(1a)

Combining (2) and (1a) gives, R s = [(1 - )/]R eq = (1 – 0.125)(100)/0.125 = 700 ohms From (3) and (1a), R p (1 - )/ = R g + [(1 - )/]R g = R g / R p = R g /(1 - ) = 100/(1 – 0.125) = 114.29 ohms

(b) R Th I V Th

+ 

RL

V Th = V s = 0.125V g = 1.5 V R Th = R g = 100 ohms I = V Th /(R Th + R L ) = 1.5/150 = 10 mA

(3)

Chapter 4, Solution 95. Let 1/sensitivity = 1/(20 k ohms/volt) = 50 A For the 0 – 10 V scale, R m = V fs /I fs = 10/50 A = 200 k ohms For the 0 – 50 V scale, R m = 50(20 k ohms/V) = 1 M ohm R Th

V Th

+ 

Rm

V Th = I(R Th + R m ) (a)

A 4V reading corresponds to I = (4/10)I fs = 0.4x50 A = 20 A V Th = 20 A R Th + 20 A 250 k ohms = 4 + 20 A R Th

(b)

(1)

A 5V reading corresponds to I = (5/50)I fs = 0.1 x 50 A = 5 A V Th = 5 A x R Th + 5 A x 1 M ohm V Th = 5 + 5 A R Th

(2)

From (1) and (2) 0 = -1 + 15 A R Th which leads to R Th = 66.67 k ohms From (1), V Th = 4 + 20x10-6x(1/(15x10-6)) = 5.333 V


The resistance network can be redrawn as shown in Fig. (a), 10 

10 

8

R Th 9V

+ 

i1

40

+

i2

60 

8

10 

V Th 

R

+ V Th

+ 

(a)

Vo

R

 (b)

R Th = 10 + 10 + [60||(8 + 8 + [10||40])] = 20 + (60||24) = 37.14 ohms Using mesh analysis, -9 + 50i 1 - 40i 2 = 0 116i 2 – 40i 1 = 0 or i 1 = 2.9i 2 From (1) and (2),

(1) (2)

i 2 = 9/105 = 0.08571 V Th = 60i 2 = 5.143 V

From Fig. (b), V o = [R/(R + R Th )]V Th = 1.8 V R/(R + 37.14) = 1.8/5.143 = 0.35 or R = 0.35R + 13 or R = (13)/(1–0.35) which leads to R = 20 Ω (note, this is just for the V o = 1.8 V) (b)

Asking for the value of R for maximum power would lead to R = R Th = 37.14 Ω. However, the problem asks for the value of R for maximum current. This happens when the value of resistance seen by the source is a minimum thus R = 0 is the correct value. I max = V Th /(R Th ) = 5.143/(37.14) = 138.48 mA.


6 k 12V

+ 

B + 4 k

V Th  E

R Th = R 1 ||R 2 = 6||4 = 2.4 k ohms V Th = [R 2 /(R 1 + R 2 )]v s = [4/(6 + 4)](12) = 4.8 V

Chapter 4, Solution 98. The 20-ohm, 60-ohm, and 14-ohm resistors form a delta connection which needs to be connected to the wye connection as shown in Fig. (b),

20 

30 

30 

R2 R1

14  60 

a

b R Th

a

b

R3

(a)

R Th

(b)

R 1 = 20x60/(20 + 60 + 14) = 1200/94 = 12.766 ohms R 2 = 20x14/94 = 2.979 ohms R 3 = 60x14/94 = 8.936 ohms R Th = R 3 + R 1 ||(R 2 + 30) = 8.936 + 12.766||32.98 = 18.139 ohms To find V Th , consider the circuit in Fig. (c). IT 30 

20 

I1

14  b

60 

a

+ IT 16 V + 

(c)

V Th

I T = 16/(30 + 15.745) = 349.8 mA I 1 = [20/(20 + 60 + 14)]I T = 74.43 mA V Th = 14I 1 + 30I T = 11.536 V I 40 = V Th /(R Th + 40) = 11.536/(18.139 + 40) = 198.42 mA P 40 = I 40 2R = 1.5748 watts

Chapter 5, Solution 1. (a) (b) (c)

R in = 1.5 M R out = 60  A = 8x104 Therefore A dB = 20 log 8x104 = 98.06 dB

Chapter 5, Solution 2. v 0 = Av d = A(v 2 - v 1 ) = 105 (20-10) x 10-6 = 1V

Chapter 5, Solution 3. v 0 = Av d = A(v 2 - v 1 ) = 2 x 105 (30 + 20) x 10-6 = 10V

Chapter 5, Solution 4. v 0 = Av d = A(v 2 - v 1 ) v 4  2V v2 - v1 = 0  A 2 x10 6 v 2 - v 1 = -2 µV = –0.002 mV 1 mV - v 1 = -0.002 mV v 1 = 1.002 mV


I R0 R in

vd + vi

Av d

+ v0

-

+

-

-v i + Av d + (R i + R 0 ) I = 0 But

+ -

(1)

v d = R i I, -v i + (R i + R 0 + R i A) I = 0 I=

vi R 0  (1  A)R i

(2)

-Av d - R 0 I + v 0 = 0 v 0 = Av d + R 0 I = (R 0 + R i A)I =

(R 0  R i A) v i R 0  (1  A)R i

v0 R 0  RiA 100  10 4 x10 5    10 4 v i R 0  (1  A)R i 100  (1  10 5 )



10 9 100,000  10 4   0.9999990 5 100,001 1  10





Chapter 5, Solution 6. vi + -

R0 I

R in

vd

+ -

+

Av d

+ vo

-

(R 0 + R i )R + v i + Av d = 0 But

v d = R i I, v i + (R 0 + R i + R i A)I = 0 I=

 vi R 0  (1  A)R i

(1)

-Av d - R 0 I + v o = 0 v o = Av d + R 0 I = (R 0 + R i A)I Substituting for I in (1),  R 0  R iA   v i v 0 =    R 0  (1  A)R i  50  2x10 6 x 2 x10 5  10 3 =  50  1  2x10 5 x 2 x10 6











 200,000 x 2 x10 mV 200,001x 2 x10 6 6

v 0 = -0.999995 mV

Chapter 5, Solution 7. 100 k

10 k

VS

+

–

R out = 100 

1

2

+ Vd

R in

– +

AV d

–

At node 1,

+ V out

–

(V S – V 1 )/10 k = [V 1 /100 k] + [(V 1 – V 0 )/100 k] 10 V S – 10 V 1 = V 1 + V 1 – V 0 which leads to V 1 = (10V S + V 0 )/12

At node 2,

(V 1 – V 0 )/100 k = (V 0 – (–AV d ))/100

But V d = V 1 and A = 100,000, V 1 – V 0 = 1000 (V 0 + 100,000V 1 ) 0= 1001V 0 + 99,999,999[(10V S + V 0 )/12] 0 = 83,333,332.5 V S + 8,334,334.25 V 0 which gives us (V 0 / V S ) = –10 (for all practical purposes) If V S = 1 mV, then V 0 = –10 mV Since V 0 = A V d = 100,000 V d , then V d = (V 0 /105) V = –100 nV


If v a and v b are the voltages at the inverting and noninverting terminals of the op amp. va = vb = 0 1mA =

0  v0 2k

v 0 = –2 V

(b) 10 k

2V

+

ia

va

2V

+

vb 1V

+ vo

+

+2 k

-

-

10 k

+ va

-

+ ia

vo

-

(b)

(a)

Since v a = v b = 1V and i a = 0, no current flows through the 10 k resistor. From Fig. (b), -v a + 2 + v 0 = 0

v 0 = v a – 2 = 1 – 2 = –1V

Chapter 5, Solution 9. (a) Let v a and v b be respectively the voltages at the inverting and noninverting terminals of the op amp v a = v b = 4V At the inverting terminal, 1mA =

4  v0 2k

v 0 = 2V

1V

(b)

+-

+

+

vb

vo

-

-

Since v a = v b = 3V, -v b + 1 + v o = 0

v o = v b – 1 = 2V

Chapter 5, Solution 10. Since no current enters the op amp, the voltage at the input of the op amp is v s . Hence

 10  v o vs = vo    10  10  2

vo =2 vs

5.11 Using Fig. 5.50, design a problem to help other students to better understand how ideal op amps work. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find v o and i o in the circuit in Fig. 5.50.

Figure 5.50 for Prob. 5.11

Solution 8 k 2 k

3V

vb =

+ 

5 k

a b

 + 10 k

+ 4 k

vo 

10 (3)  2V 10  5

At node a, 3  va va  vo  2 8

io

12 = 5v a – v o

But v a = v b = 2V, 12 = 10 – v o –i o =

v o = –2V

va  vo 0  vo 2  2 2     1mA 8 4 8 4

i o = –1mA

Chapter 5, Solution 12. Step 1. Label the unknown nodes in the op amp circuit. Next we write the node equations and then apply the constraint, V a = V b . Finally, solve for V o in terms of V s . 25 k 5 k

VS

+ 

a b

 + 10 k

Step 2.

+ Vo 

[(V a -V s )/5k] + [(V a -V o )/25k] + 0 = 0 and [(V b -0)/10k] + 0 = 0 or V b = 0 = V a ! Thus, [(-V s )/5k] + [(-V o )/25k] = 0 or, V o = ( –25/5)V s or V o /V s = –5.


10 k a b 1V

+ 

+ 

io 100 k i 2

10 k

90 k 50 k

By voltage division, va =

90 (1)  0.9V 100

vb =

v 50 vo  o 150 3

But v a = v b

io = i1 + i2 =

v0  0.9 3

i1

v o = 2.7V

vo v  o  0.27mA + 0.018mA = 288 A 10k 150k

+ vo 

Chapter 5, Solution 14. Transform the current source as shown below. At node 1, 10  v1 v1  v 2 v1  v o   5 20 10

10 k

vo 10 k

5 k

20 k v1

10V

 +

v2

+ 

+ vo 

But v 2 = 0. Hence 40 – 4v 1 = v 1 + 2v 1 – 2v o At node 2,

v1  v 2 v 2  v o  , 20 10

v 2  0 or v 1 = –2v o

From (1) and (2), 40 = –14v o - 2v o

v o = –2.5V

40 = 7v 1 – 2v o

(1) (2)

Chapter 5, Solution 15 (a) Let v 1 be the voltage at the node where the three resistors meet. Applying KCL at this node gives  1 v1 v1  vo 1  vo     v1   R2 R3  R2 R3  R3 At the inverting terminal, is 

0  v1   v1  i s R1 R1 Combining (1) and (2) leads to  v R R  i s  1  1  1    o   R2 R3  R3 

(1)

is 

(2)

 vo RR   R1  R3  1 3 is R2 

(b) For this case, vo 20 x 40     20  40   k  - 92 k is 25  

= –92 kΩ

  

Chapter 5, Solution 16 Using Fig. 5.55, design a problem to help students better understand inverting op amps. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Obtain i x and i y in the op amp circuit in Fig. 5.55.

Figure 5.55 Solution

10k 

5k 

ix va vb

+ 0.5V -

iy + 2k 

8k 

vo

Let currents be in mA and resistances be in k  . At node a, 0.5  v a v a  v o    1  3v a  vo 5 10

(1)

But

8 10 vo   vo  v a (2) 82 8 Substituting (2) into (1) gives 10 8 1  3v a  v a   v a  8 14 Thus, 0.5  v a ix   1 / 70 mA   14.28 A 5 v  vb vo  va 10 0.6 8 iy  o   0.6(v o  v a )  0.6( v a  v a )  x mA 2 10 8 4 14 v a  vb 

= 85.71 µA

Chapter 5, Solution 17. (a) (b) (c)

G=

R vo 12   f    –2.4 vi Ri 5

vo 80  = –16 vi 5 vo 2000   –400 vi 5

(a) –2.4, (b) –16, (c) –400

Chapter 5, Solution 18. For the circuit, shown in Fig. 5.57, solve for the Thevenin equivalent circuit looking into terminals A and B. 10 k 10 k 

a b

7.5 V

+ 

 +

c

A 2.5  B


Write a node equation at a. Since node b is tied to ground, v b = 0. We cannot write a node equation at c, we need to use the constraint equation, v a = v b . Once, we know v c , we then proceed to solve for V open circuit and I short circuit . This will lead to V Thev (t) = V open circuit and R equivalent = V open circuit /I short circuit . [(v a – 7.5)/10k] + [(v a – v c )/10k] + 0 = 0 Our constraint equation leads to, v a = v b = 0 or v c = –7.5 volts This is also the open circuit voltage (note, the op-amp keeps the output voltage at –5 volts in spite of any connection between A and B. Since this means that even a short from A to B would theoretically then produce an infinite current, R equivalent = 0. In real life, the short circuit current will be limited to whatever the op-amp can put out into a short circuited output. V Thev = –7.5 volts; R equivalent = 0-ohms.

Chapter 5, Solution 19. We convert the current source and back to a voltage source.

24

(4/3) k

4 k

4 3 10 k

0V

(1.5/3)V

+ 

 +

vo 2 k

10k  1.5     –937.5 mV. 4  3    4  k 3  v v 0  –562.5 µA. io  o  o 2k 10k vo  


2 k 4 k

9V

a

+ 

4 k

vs

b

 +

+ 

+ vo 

At node a, 9  va va  vo va  vb   4 8 4

18 = 5v a – v o – 2v b

(1)

At node b, va  vb vb  vo  4 2

v a = 3v b – 2v o

But v b = v s = 2 V; (2) becomes v a = 6 –2v o and (1) becomes –18 = 30–10v o – v o – 4

v o = –44/(–11) = 4 V.

(2)

Chapter 5, Solution 21. Let the voltage at the input of the op amp be v a . va  1 V,

3-v a va  vo  4k 10k

 

3-1 1 vo  4 10

v o = –4 V.

Chapter 5, Solution 22. A v = -R f /R i = -15. If R i = 10k, then R f = 150 k.

Chapter 5, Solution 23 At the inverting terminal, v=0 so that KCL gives vs  0 0 0  vo   R1 R2 Rf



vo vs



Rf R1

Chapter 5, Solution 24 v1

Rf

R1

R2 - vs +

+ + R4

R3

vo -

v2

We notice that v 1 = v 2 . Applying KCL at node 1 gives

v1 (v1  v s ) v1  vo   0 R1 R2 Rf

 

 1    1  1 v1  v s  vo R R R2 R f R f  2  1

Applying KCL at node 2 gives R3 v1 v1  v s  0   v1  vs R3 R4 R3  R4 Substituting (2) into (1) yields

 R R R  R3  1     v s vo  R f  3  3  4   R1 R f R2  R3  R4  R2  i.e.  R R R  R3  1     k  R f  3  3  4   R1 R f R2  R3  R4  R2 

(2)

(1)

Chapter 5, Solution 25. This is a voltage follower. If v 1 is the output of the op amp, v 1 = 3.7 V v o = [20k/(20k+12k)]v 1 = [20/32]3.7 = 2.312 V.

Chapter 5, Solution 26 Using Fig. 5.64, design a problem to help other students better understand noninverting op amps.

Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine i o in the circuit of Fig. 5.64.

Figure 5.64 Solution + vb + 0.4V -

-

io + 2k 

8k 

vo -

vb  0.4 

8 vo  0.8vo 82

 

vo  0.4 / 0.8  0.5 V

Hence, io 

vo 0.5   0.1 mA 5k 5k

5k 

Chapter 5, Solution 27. This is a voltage follower. v 1 = [24/(24+16)]7.5 = 4.5 V; v 2 = v 1 = 4.5 V; and v o = [12/(12+8)]4.5 = 2.7 V.


v1

 +

va 10 k

At node 1,

+ 

10 V

0  v1 v1  v o  10k 50k

But v 1 = 10V, –5v 1 = v 1 – v o , leads to

v o = 6v 1 = 60V

Alternatively, viewed as a noninverting amplifier, v o = (1 + (50/10)) (10V) = 60V i o = v o /(20k) = 60/(20k) = 3 mA.

vo

20 k

Chapter 5, Solution 29 R1

va vb

+ vi -

+ -

R2

+

R2

vo

R1 -

va 

R2 vi , R1  R2

But v a  vb

vb 

 

R1 vo R1  R2 R2 R1 vi  vo R1  R2 R1  R2

Or

v o R2  vi R1

Chapter 5, Solution 30. The output of the voltage becomes v o = v i = 1.2 V (30k 20k )  12k By voltage division,

vx 

12 (1.2)  0.2V 12  60 ix 

vx 20 0.2    10A 20k 20k 2 x10 6 v 2x 0.04   2W. p R 20k

Chapter 5, Solution 31. After converting the current source to a voltage source, the circuit is as shown below: 12 k 3 k

1

6 k v o

v1

12 V

+ 

+ 

2

vo

6 k

At node 1, 12  v1 v1  v o v1  v o   3 6 12

48 = 7v 1 - 3v o

(1)

At node 2, v1  v o v o  0   ix 6 6

From (1) and (2),

48 11 v i x  o  727.2μA 6k vo 

v 1 = 2v o

(2)

Chapter 5, Solution 32. Let v x = the voltage at the output of the op amp. The given circuit is a non-inverting amplifier.

 50  v x  1   (4 mV) = 24 mV  10  60 30  20k By voltage division, v 20 v x  x  12mV 20  20 2 vx 24mV   600 A ix = 20  20k 40k

vo =

p=

v o2 144x10 6   204 W. R 60x10 3

Chapter 5, Solution 33. After transforming the current source, the current is as shown below: 1 k

4 k

vi

+ 

va 4V

+ 

2 k

vo 3 k

This is a noninverting amplifier.

3  1 v o  1   v i  v i 2  2 Since the current entering the op amp is 0, the source resistor has a 0 V potential drop. Hence v i = 4V.

vo 

3 (4)  6V 2

Power dissipated by the 3k resistor is v o2 36   12mW R 3k ix 

12mW, –2mA

va  vo 4  6   –2mA. R 1k


v1  vin v1  vin  0 R1 R2

(1)

R3 vo R3  R 4

(2)

but

va 

Combining (1) and (2), v1  va 

R1 R v 2  1 va  0 R2 R2

 R  R v a 1  1   v1  1 v 2 R2  R2  R 3v o  R  R 1  1   v1  1 v 2 R3  R 4  R 2  R2

vo 

 R3  R 4  R  v1  1 v 2  R2   R  R 3 1  1    R2 

vO =

R3  R4 (v1 R2  v2 ) R3 ( R1  R2 )


Av 

vo R  1  f  7.5 vi Ri

R f = 6.5R i

If R i = 60 k, R f = 390 k.

Chapter 5, Solution 36 VTh  Vab But

R1 Vab . Thus, R1  R2 R R  R2  1 v s  (1  2 )v s R1 R1

vs 

VTh  Vab

To get R Th , apply a current source I o at terminals a-b as shown below. v1 v2

+ -

a + R2 vo

io

R1 b Since the noninverting terminal is connected to ground, v 1 = v 2 =0, i.e. no current passes through R 1 and consequently R 2 . Thus, v o =0 and RTh 

vo 0 io

Chapter 5, Solution 37. R  R R v o    f v1  f v 2  f v 3  R2 R3   R1

30 30  30     (2)  (2)  (4.5) 20 30  10  v o = 1.5 V.

Chapter 5, Solution 38. Using Fig. 5.75, design a problem to help other students better understand summing amplifiers. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem

Calculate the output voltage due to the summing amplifier shown in Fig. 5.75.

Figure 5.75

Solution R  R R R v o    f v1  f v 2  f v 3  f v 4  R2 R3 R4   R1 50 50 50  50     (10)  (20)  (50)  (100) 20 10 50  25 

= -120mV


This is a summing amplifier. Rf Rf   Rf 50 50  50  vo   v1  v2  v3    (2)  v 2  (1)   9  2.5v 2 R2 R3  20 50  10   R1 Thus, v o  16.5  9  2.5v 2   v2  3 V

Chapter 5, Solution 40 Determine V o in terms of V 1 and V 2 . 200 k

100 k

100 k

V1

Va

+ 

V2

+ 

 +

Vc

10 

+ 40 

Vb

Vo 

Step 1. Label the reference and node voltages in the circuit, see above. Note we now can consider nodes a and b, we cannot write a node equation at c without introducing another unknown. The node equation at a is [(V a –V 1 )/105] + [(V a –V 2 )/105] + 0 + [(V a –V c )/2x105] = 0. At b it is clear that V b = 0. Since we have two equations and three unknowns, we need another equation. We do get that from the constraint equation, V a = V b . After we find V c in terms of V 1 and V 2 , we then can determine V o which is equal to [(V c –0)/50] times 40. Step 2.

Letting V a = V b = 0, the first equation can be simplified to, [–V 1 /105] + [–V 2 /105] + [–V c /2x105] = 0

Taking V c to the other side of the equation and multiplying everything by 2x105, we get, V c = –2V 1 – 2V 2 Now we can find V o which is equal to (40/50)V c = 0.8[–2V 1 –2V 2 ] V o = –1.6V 1 –1.6V 2 .

Chapter 5, Solution 41. R f /R i = 1/(4)

R i = 4R f = 40k

The averaging amplifier is as shown below: R 1 = 40 k

10 k

v1 R 2 = 40 k v2 R 3 = 40 k v3 R 4 = 40 k v4

 +

vo

Chapter 5, Solution 42 Since the average of three numbers is the sum of those numbers divided by three, the value of the feedback resistor needs to be equal to one-third of the input resistors or,

1 R f  R 1  25 kΩ. 3

Chapter 5, Solution 43. In order for  R R R R v o   f v1  f v 2  f v 3  f v 4  R2 R3 R4   R1

to become

1 v 1  v 2  v 3  v 4  4 Rf 1 R 80k  Rf  i   20 k. Ri 4 4 4

vo  

Chapter 5, Solution 44. R4 R3 a R1 v1

 +

b

R2

vo

v2

At node b,

v b  v1 v b  v 2  0 R1 R2

At node a,

0  va va  vo  R3 R4

v1 v 2  R1 R 2 vb  1 1  R1 R 2

(1)

vo 1 R4 / R3

(2)

va 

But v a = v b . We set (1) and (2) equal. vo R v  R 1v 2  2 1 1 R4 / R3 R1  R 2

or vo =

R3  R4   R2 v 1  R1 v 2  R3  R1  R2 

Chapter 5, Solution 45. This can be achieved as follows:

 R  v1   R v 2  v o   R/2  R / 3 R  R    f  v1   f v 2  R2   R1

i.e. R f = R, R 1 = R/3, and R 2 = R/2 Thus we need an inverter to invert v 1 , and a summer, as shown below (R<100k).

R R

R v1

 +

R/3 -v 1 R/2 v2

 +

vo

Chapter 5, Solution 46. v1 1 R R R 1  (  v 2 )  v 3  f v1  x (  v 2 )  f v 3 3 3 2 R1 R2 R3 i.e. R 3 = 2R f , R 1 = R 2 = 3R f . To get -v 2 , we need an inverter with R f = R i . If R f = 10k, a solution is given below.  vo 

30 k

10 k

v1

10 k v2

 +

10 k 30 k

-v 2 20 k v3

 +

vo

Chapter 5, Solution 47. Using eq. (5.18), R1  2k, R2  30k, R3  2k, R4  20k

vo 

30(1 2 / 30) 30 32 v2  V1  (2)  15(1)  14.09 V 2(1 2 / 20) 2 2.2 = 14.09 V.

Chapter 5, Solution 48. We can break this problem up into parts. The 5 mV source separates the lower circuit from the upper. In addition, there is no current flowing into the input of the op amp which means we now have the 40-kohm resistor in series with a parallel combination of the 60-kohm resistor and the equivalent 100-kohm resistor.

+10 mV

Thus,

40k + (60x100k)/(160) = 77.5k

which leads to the current flowing through this part of the circuit, i = 10 m/77.5k = 129.03x10–9 A The voltage across the 60k and equivalent 100k is equal to, v = ix37.5k = 4.839 mV We can now calculate the voltage across the 80-kohm resistor. v 80 = 0.8x4.839 m = 3.87 mV which is also the voltage at both inputs of the op amp and the voltage between the 20-kohm and 80-kohm resistors in the upper circuit. Let v 1 be the voltage to the left of the 20-kohm resistor of the upper circuit and we can write a node equation at that node.

(v 1 –10m)/(10k) + v 1 /30k + (v 1 –3.87m)/20k = 0 or

6v 1 – 60m + 2v 1 + 3v 1 – 11.61m = 0

or

v 1 = 71.61/11 = 6.51 mV.

The current through the 20k-ohm resistor, left to right, is, i 20 = (6.51m–3.87m)/20k = 132 x10–9 A thus,

v o = 3.87m – 132 x10–9x80k = –6.69 mV.

Chapter 5, Solution 49. R 1 = R 3 = 20k, R 2 /(R 1 ) = 4 i.e.

R 2 = 4R 1 = 80k = R 4 vo 

Verify:

4

R 2 1  R1 / R 2 R v 2  2 v1 R1 1  R 3 / R 4 R1

(1  0.25) v 2  4v1  4v 2  v1  1  0.25

Thus, R 1 = R 3 = 20 k, R 2 = R 4 = 80 k.

Chapter 5, Solution 50. (a) We use a difference amplifier, as shown below: R1

R2

v1

 + R1

vo

R2

v2

vo 

R2 v 2  v1   2.5v 2  v1 , i.e. R 2 /R 1 = 2.5 R1 If R 1 = 100 k then R 2 = 250k

(b) We may apply the idea in Prob. 5.35. v 0  2.5v1  2.5v 2

 R  v1   R v 2    R/2  R / 2 R  R    f  v1   f v 2  R2   R1 i.e. R f = R, R 1 = R/2.5 = R 2 We need an inverter to invert v 1 and a summer, as shown below. We may let R = 100 k. R R

R v1

 +

R/2.5 -v 1 R/2.5 v2

 +

vo

Chapter 5, Solution 51. We achieve this by cascading an inverting amplifier and two-input inverting summer as shown below:

R R

R v1

 +

R va R v2

Verify:

But

v o = -v a - v 2 v a = -v 1 . Hence vo = v1 - v2.

 +

vo

Chapter 5, Solution 52 Design an op amp circuit such that v o = 4v 1 + 6v 2 – 3v 3 – 5v 4 Let all the resistors be in the range of 20 to 200 k. Solution A summing amplifier shown below will achieve the objective. An inverter is inserted to invert v 2 . Since the smallest resistance must be at least 20 kΩ, then let R/6 = 20kΩ therefore let R = 120 kΩ. R/4

R

v1 R/6 v2

R

 +

R

R/3 v3 R/5 v4

 +

Chapter 5, Solution 53. (a) R1

R2

v1 va vb

 +

R1

vo

R2

v2

At node a,

At node b,

v1  v a v a  v o  R1 R2 R2 vb  v2 R1  R 2

va 

R 2 v1  R 1 v o R1  R 2

(1) (2)

But v a = v b . Setting (1) and (2) equal gives R v  R 1v o R2 v2  2 1 R1  R 2 R1  R 2 R v 2  v1  1 v o  v i R2 vo R 2  vi R1 (b)



v1

vi + v2

R 1 /2 v A

R 1 /2

R2 va

Rg R 1 /2

R 1 /2 vB

vb

 +

R2

+ vo 

At node A,

v1  v A v B  v A v A  v a   R1 / 2 Rg R1 / 2

or

v1  v A 

At node B,

v2  vB vB  vA vB  vb   R1 / 2 R1 / 2 Rg

or

v2  vB 

R1 v B  v A   v A  v a 2R g

R1 (v B  v A )  v B  v b 2R g

(1)

(2)

Subtracting (1) from (2), v 2  v1  v B  v A 

2R 1 v B  v A   v B  v A  v b  v a 2R g

R v 2  v1   1 1  2R 2 g 

 v B  v A   v i  2 

Since, v a = v b ,

or

vB  vA 

vi  2

1 R 1 1 2R g

(3)

But for the difference amplifier, R2 v B  v A  R1 / 2 R vB  vA  1 vo 2R 2 vo 

or

Equating (3) and (4),

R1 v vo  i  2R 2 2 vo R 2   vi R1

(4)

1 R 1 1 2R g

1 R 1 1 2R g

(c)

At node a,

At node b,

v1  v a v a  v A  R1 R2 /2 2R 1 2R 1 v1  v a  va  vA R2 R2 2R 1 2R 1 v2  vb  vb  vB R2 R2

(1) (2)

Since v a = v b , we subtract (1) from (2), v  2R 1 (v B  v A )  i R2 2  R2 vB  vA  vi 2R 1 v 2  v1 

or

(3)

At node A, va  vA vB  vA vA  vo   R2 /2 Rg R/2 va  vA 

At node B,

R2 v B  v A   v A  v o 2R g

(4)

vb  vB vB  vA vB  0   R/2 Rg R/2 vb  vB 

R2 v B  v A   v B 2R g

(5)

Subtracting (5) from (4), v B v A 

R2 v B  v A   v A  v B  v o Rg

 R  2v B  v A 1  2    v o  2R  g   Combining (3) and (6),  R 2  R  v i 1  2   v o  2R  R1 g   v o R2  R  1 2   vi R1  2 R g 

(6)

Chapter 5, Solution 54. The first stage is a summer (please note that we let the output of the first stage be v 1 ).

R  R v1   v s  v o  = –v s – v o R  R The second stage is a noninverting amplifier v o = (1 + R/R)v 1 = 2v 1 = 2(–v s – v o ) or 3v o = –2v s v o /v s = –0.6667.

Chapter 5, Solution 55. Let A 1 = k, A 2 = k, and A 3 = k/(4) A = A 1 A 2 A 3 = k3/(4) 20Log10 A  42 Log10 A  2.1 A = 102 1 = 125.89 k3 = 4A = 503.57 k = 3 503.57  7.956 Thus A 1 = A 2 = 7.956, A 3 = 1.989

Chapter 5, Solution 56. Using Fig. 5.83, design a problem to help other students better understand cascaded op amps. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem

Calculate the gain of the op amp circuit shown in Fig. 5.83. 10 k

40 k

1 k 20 k

– +

+ vi –

Figure 5.83

– +

For Prob. 5.56.

Solution

Each stage is an inverting amplifier. Hence, vo 10 40  ( )( )  20 vs 1 20

Chapter 5, Solution 57. Let v 1 be the output of the first op amp and v 2 be the output of the second op amp. The first stage is an inverting amplifier. 50 v1   vs1  2vs1 25 The second state is a summer. v 2 = –(100/50)v s2 – (100/100)v 1 = –2v s2 + 2v s1 The third state is a noninverting amplifier vo  (1

100 )v2  3v2  6vs1  6vs2 50

Chapter 5, Solution 58. Looking at the circuit, the voltage at the right side of the 5-kΩ resistor must be at 0V if the op amps are working correctly. Thus the 1-kΩ is in series with the parallel combination of the 3-kΩ and the 5-kΩ. By voltage division, the input to the voltage follower is: v1 

35 1 3 5

(0.6)  0.3913V = to the output of the first op amp.

Thus, v o = –10((0.3913/5)+(0.3913/2)) = –2.739 V.

io 

0  vo  684.8 µA. 4k

Chapter 5, Solution 59. The first stage is a noninverting amplifier. If v 1 is the output of the first op amp, v 1 = (1 + 2R/R)v s = 3v s The second stage is an inverting amplifier v o = –(4R/R)v 1 = –4v 1 = –4(3v s ) = –12v s v o /v s = –12.

Chapter 5, Solution 60. The first stage is a summer. Let V 1 be the output of the first stage. 10 10 vi  vo   v1  2vi  2.5vo 5 4 By voltage division, 10 5 v1  vo  vo 10  2 6 Combining (1) and (2), 5 10 vo  2v1  2.5v0   v0  2vi 6 3 v1  

vo  6 /10  0.6 vi

(1)

(2)

Chapter 5, Solution 61. The first op amp is an inverter. If v 1 is the output of the first op amp, V 1 = –(200/100)(0.4) = –0.8 V The second op amp is a summer V o = –(40/10)(–0.2) – (40/20)(–0.8) = 0.8 + 1.6 = 2.4 V.

Chapter 5, Solution 62. Let v 1 = output of the first op amp v 2 = output of the second op amp The first stage is a summer v1  

R2 R vi – 2 vo R1 Rf

(1)

The second stage is a follower. By voltage division vo  v2 

R4 v1 R3  R4

v1 

R3  R4 vo R4

(2)

From (1) and (2),  R3  R R  v o   2 v i  2 v o 1  Rf R1  R4   R3 R2  R  v o   2 v i 1   R1  R4 Rf   R2 R4 R f vo R 1  2   R R R1 R2 R4  R3 R f  R4 R f vi R1 1 3  2 R4 Rf





Chapter 5, Solution 63. The two op amps are summers. Let v 1 be the output of the first op amp. For the first stage,

v1  

R2 R vi  2 vo R1 R3

(1)

For the second stage, vo  

R4 R v1  4 v i R5 R6

(2)

Combining (1) and (2),  R2  R R  R  v i  4  2  v o  4 v i  R5  R3  R6  R1   R R  R R R  v o 1  2 4    2 4  4  v i  R 3 R 5   R 1R 5 R 6  vo 

R4 R5

R2 R4 R4  v o R1 R5 R6  R R vi 1 2 4 R3 R5


G4 G

G3

G1 1 +

0V

G +

v

2 0V +

G2

vs -

-

At node 1, v 1 =0 so that KCL gives G1v s  G4 vo  Gv

(1)

At node 2, G2 v s  G3 v o  Gv From (1) and (2), G1v s  G4 v o  G2 v s  G3 vo or

+ vo

(2)

 

(G1  G2 )v s  (G3  G4 )vo

vo G1  G2  v s G3  G4


The output of the first op amp (to the left) is 6 mV. The second op amp is an inverter so that its output is

30 (6mV)  -18 mV 10 The third op amp is a noninverter so that vo '  

vo ' 

40 vo 40  8

 

vo 

48 v o '   21.6 mV 40

Chapter 5, Solution 66. We can start by looking at the contributions to v o from each of the sources and the fact that each of them go through inverting amplifiers. The 6 V source contributes –[100k/25k]6; the 4 V source contributes –[40k/20k][–(100k/20k)]4; and the 2 V source contributes –[100k/10k]2 or

vo 

40  100  100  100 (6)    (2) (4)  25 20  20  10  24  40  20  –4V


80  80  80   (0.3)  (0.7) 20 40  20   4.8  2.8  2 V.

vo = 

Chapter 5, Solution 68. If R q = , the first stage is an inverter.

Va  

15 (15)  45 mV 5

when V a is the output of the first op amp. The second stage is a noninverting amplifier.

 6 v o  1   v a  (1  3)(45)  –180mV.  2

Chapter 5, Solution 69. In this case, the first stage is a summer

va  

15 15 (15)  v o  45  1.5v o 5 10

For the second stage,

 6 v o  1   v a  4v a  4 45  1.5v o   2 180 7 v o  180 v o    –25.71 mV. 7

Chapter 5, Solution 70. The output of amplifier A is

vA  

30 30 (1)  (2)  9 10 10

The output of amplifier B is

vB  

20 20 (3)  (4)  14 10 10

40 k 20 k vA a

60 k

 +

vB b

vo

10 k

vb 

10 (14)  2V 60  10

At node a,

vA  va va  vo  20 40

But v a = v b = -2V, 2(-9+2) = -2-v o Therefore,

v o = 12V

Chapter 5, Solution 71 20k 

5k 

100k  40k 

+ + 1.5 V –

v2 10k 

80k 

+ 20k 

+ vo -

+

10k  v1

+ 2.25V –

v1  2.25,

+ -

30k 

v2  

v3

50k 

20 50 (1.5)  6, v 3  (1  ) v1  6 5 30 100   100 v o   v2  v 3   (15  7.5)  7.5 V. 80   40

Chapter 5, Solution 72. Since no current flows into the input terminals of ideal op amp, there is no voltage drop across the 20 k resistor. As a voltage summer, the output of the first op amp is v 01 = 1.8 V The second stage is an inverter

v2  

250 v 01 100

 2.5(1.8)  –4.5 V.

Chapter 5, Solution 73. The first stage is a noninverting amplifier. The output is 50 vo1  (1.8)  1.8  10.8V 10

The second stage is another noninverting amplifier whose output is vL  v01  10.8V

Chapter 5, Solution 74. Let v 1 = output of the first op amp v 2 = input of the second op amp. The two sub-circuits are inverting amplifiers

100 (0.9)  9V 10 32 v2   (0.6)  12V 1.6  9  12 v  v2  io  1  150 A. 20k 20k v1  

Chapter 5, Solution 75. The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE are involved as shown to measure v o and i respectively. Once the circuit is saved, we click Analysis | Simulate. The values of v and i are displayed on the pseudo-components as: i = 200 A (v o /v s ) = -4/2 = –2 The results are slightly different than those obtained in Example 5.11.

Chapter 5, Solution 76. The schematic is shown below. IPROBE is inserted to measure i o . Upon simulation, the value of i o is displayed on IPROBE as i o = –562.5 A 11.25V –19.358uV

0.750V 375mV

750 mV

–936.8mV

2 kΩ

–11.25V

Chapter 5, Solution 77. The schematic for the PSpice solution is shown below. Note that the output voltage, –6.686 mV, agrees with the answer to problem, 5.48.

6.510mV

3.872mV

–6.686mV 3.872mV 0.0100V

4.838mV

Chapter 5, Solution 78. The circuit is constructed as shown below. We insert a VIEWPOINT to display v o . Upon simulating the circuit, we obtain, v o = 667.75 mV

Chapter 5, Solution 79. The schematic is shown below.

v o = –4.992 V R3

R4

R5

20k

10k

40k

V5

V3 20Vdc

1Vdc

0

1.000V 1

OS1 6 2.000V OUT 5 OS2 7

V-

4

1.000V

+

3

7

U1 +

OS2

OUT 2 uA741 R1

R2

20k

10k

-

4

0 0

20.00V

0V

V+

0V

OS1 V-

-20.00V

V4 20Vdc

1.666V

V6 5Vdc

3

V+

0

V2 20Vdc

-

U2 uA741 0V 2

R6 100k

5 6 1 20.00V

V1 20Vdc

-4.992V

1.666V 5.000V

0

Checking using nodal analysis we get, For the first op-amp we get v a1 = [5/(20+10)]10 = 1.6667 V = v b1 . For the second op-amp, [(v b1 – 1)/20] + [(v b1 – v c2 )/10] = 0 or v c2 = 10[1.6667–1)/20] + 1.6667 = 2 V; [(v a2 – v c2 )/40] + [(v a2 – v c1 )/100] = 0; and v b2 = 0 = v a2 . This leads to v c1 = –2.5v c2 . Thus, = –5 V.


The schematic is as shown below. After it is saved and simulated, we obtain v o = 2.4 V.

Chapter 5, Solution 81. The schematic is shown below. We insert one VIEWPOINT and one IPROBE to measure v o and i o respectively. Upon saving and simulating the circuit, we obtain, v o = 343.4 mV i o = 24.51 A

Chapter 5, Solution 82. The maximum voltage level corresponds to 11111 = 25 – 1 = 31 Hence, each bit is worth

(7.75/31) = 250 mV

Chapter 5, Solution 83. The result depends on your design. Hence, let R G = 10 k ohms, R 1 = 10 k ohms, R 2 = 20 k ohms, R 3 = 40 k ohms, R 4 = 80 k ohms, R 5 = 160 k ohms, R 6 = 320 k ohms, then, -v o = (R f /R 1 )v 1 + --------- + (R f /R 6 )v 6 = v 1 + 0.5v 2 + 0.25v 3 + 0.125v 4 + 0.0625v 5 + 0.03125v 6 (a)

|v o | = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1/8) + (1/16) which implies, [v 1 v 2 v 3 v 4 v 5 v 6 ] = [100110]

(b)

|v o | = 0 + (1/2) + (1/4) + 0 + (1/16) + (1/32) = (27/32) = 843.75 mV

(c)

This corresponds to [1 1 1 1 1 1]. |v o | = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32) = 63/32 = 1.96875 V


(a)

The easiest way to solve this problem is to use superposition and to solve for each term letting all of the corresponding voltages be equal to zero. Also, starting with each current contribution (i k ) equal to one amp and working backwards is easiest.

2R

v1

+ 

R

R

R

2R

2R

2R

ik v2

+ 

v3

+ 

v4

2R

+ 

For the first case, let v 2 = v 3 = v 4 = 0, and i 1 = 1A. Therefore,

v 1 = 2R volts or i 1 = v 1 /(2R).

Second case, let v 1 = v 3 = v 4 = 0, and i 2 = 1A.

2R

R

R

2R

R

2R

2R

i2 v2

Simplifying, we get,

+ 

2R

2R R 1A 2R

v2

+ 

Therefore, v 2 = 1xR + (3/2)(2R) = 4R volts or i 2 = v 2 /(4R) or i 2 = 0.25v 2 /R. Clearly this is equal to the desired 1/4th. Now for the third case, let v 1 = v 2 = v 4 = 0, and i 3 = 1A.

2R 5R/3 1.5 2R

v3

+ 

The voltage across the 5R/3-ohm resistor is 5R/2 volts. The current through the 2R resistor at the top is equal to (5/4) A and the current through the 2R-ohm resistor in series with the source is (3/2) + (5/4) = (11/4) A. Thus,

v 3 = (11/2)R + (5/2)R = (16/2)R = 8R volts or i 3 = v 3 /(8R) or 0.125v 3 /R. Again, we have the desired result. For the last case, v 1 = v 2 = v 3 and i 4 = 1A. Simplifying the circuit we get,

R 1A

R

2R

R

2R

v4

5R/3 1.5A

2R 2R

+ 

R

2R

2R

v4

+ 

2R

21R/11 11/4A

2R

v4

+ 

2R

Since the current through the equivalent 21R/11-ohm resistor is (11/4) amps, the voltage across the 2R-ohm resistor on the right is (21/4)R volts. This means the current going through the 2R-ohm resistor is (21/8) A. Finally, the current going through the 2R resistor in series with the source is ((11/4)+(21/8)) = (43/8) A. Now, v 4 = (21/4)R + (86/8)R = (128/8)R = 16R volts or i 4 = v 4 /(16R) or 0.0625v 4 /R. This is just what we wanted.

(b)

If R f = 12 k ohms and R = 10 k ohms, -v o = (12/20)[v 1 + (v 2 /2) + (v 3 /4) + (v 4 /8)] = 0.6[v 1 + 0.5v 2 + 0.25v 3 + 0.125v 4 ] For

[v 1 v 2 v 3 v 4 ] = [1 0 11], |v o | = 0.6[1 + 0.25 + 0.125] = 825 mV

For

[v 1 v 2 v 3 v 4 ] = [0 1 0 1], |v o | = 0.6[0.5 + 0.125] = 375 mV

Chapter 5, Solution 85. This is a noninverting amplifier. v o = (1 + R/40k)v s = (1 + R/40k)2 The power being delivered to the 10-kΩ give us P = 10 mW = (v o )2/10k or v o = 10  2 x10 4 = 10V

Returning to our first equation we get 10 = (1 + R/40k)2 or R/40k = 5 – 1 = 4 Thus,

R = 160 kΩ.

Chapter 5, Solution 86. Design a voltage controlled ideal current source (within the operating limits of the op amp) where the output current is equal to 200v s (t) µA. The easiest way to solve this problem is to understand that the op amp creates an output voltage so that the current through the feedback resistor remains equal to the input current. In the following circuit, the op amp wants to keep the voltage at a equal to zero. So, the input current is v s /R = 200v s (t) µA = v s (t)/5k. Thus, this circuit acts like an ideal voltage controlled current source no matter what (within the operational parameters of the op amp) is connected between a and b. Note, you can change the direction of the current between a and b by sending v s (t) through an inverting op amp circuit. b

a 5kΩ v s (t)

+ –

 +

Chapter 5, Solution 87. The output, v a , of the first op amp is,

Also,

v a = (1 + (R 2 /R 1 ))v 1

(1)

v o = (-R 4 /R 3 )v a + (1 + (R 4 /R 3 ))v 2

(2)

Substituting (1) into (2), v o = (-R 4 /R 3 ) (1 + (R 2 /R 1 ))v 1 + (1 + (R 4 /R 3 ))v 2 Or, If

v o = (1 + (R 4 /R 3 ))v 2 – (R 4 /R 3 + (R 2 R 4 /R 1 R 3 ))v 1 R 4 = R 1 and R 3 = R 2 , then, v o = (1 + (R 4 /R 3 ))(v 2 – v 1 )

which is a subtractor with a gain of (1 + (R 4 /R 3 )).

Chapter 5, Solution 88. We need to find V Th at terminals a – b, from this, v o = (R 2 /R 1 )(1 + 2(R 3 /R 4 ))V Th = (500/25)(1 + 2(10/2))V Th = 220V Th Now we use Fig. (b) to find V Th in terms of v i . a a 30 k

20 k

20 k vi

vi

30 k

+ 40 k

80 k

40 k

b

80 k

b

(a)

(b) v a = (3/5)v i , v b = (2/3)v i V Th = v b – v a (1/15)v i (v o /v i ) = A v = -220/15 = -14.667

Chapter 5, Solution 89. A summer with v o = –v 1 – (5/3)v 2 where v 2 = 6-V battery and an inverting amplifier with v 1 = –12v s .

Chapter 5, Solution 90. The op amp circuit in Fig. 5.107 is a current amplifier. Find the current gain i o /i s of the amplifier.

Figure 5.107 For Prob. 5.90. Solution Transforming the current source to a voltage source produces the circuit below, v b = (2/(2 + 4))v o = v o /3

At node b,

20 k 5 k a

5i s

+ 

b

 + 4 k io 2 k

+ vo 

At node a,

(5i s – v a )/5 = (v a – v o )/20

But v a = v b = v o /3.

20i s – (4/3)v o = (1/3)v o – v o , or i s = v o /30 i o = [(2/(2 + 4))/2]v o = v o /6 i o /i s = (v o /6)/(v o /30) = 5


 +

vo R2

R1

is

i2 i1

But

io

io = i1 + i2

(1)

i1 = is

(2)

R 1 and R 2 have the same voltage, v o , across them. R 1 i 1 = R 2 i 2 , which leads to i 2 = (R 1 /R 2 )i 1 Substituting (2) and (3) into (1) gives, i o = i s (1 + R 1 /R 2 ) i o /i s = 1 + (R 1 /R 2 ) = 1 + 8/1 = 9

(3)

Chapter 5, Solution 92 The top op amp circuit is a non-inverter, while the lower one is an inverter. The output at the top op amp is v 1 = (1 + 60/30)v i = 3v i while the output of the lower op amp is v 2 = -(50/20)v i = -2.5v i Hence,

v o = v 1 – v 2 = 3v i + 2.5v i = 5.5v i v o /v i = 5.5

Chapter 5, Solution 93. R3

R1 v a vb

 +

io

+

R4

vi

+



R2

iL

vL

RL

+ vo 



At node a,

(v i – v a )/R 1 = (v a – v o )/R 3 v i – v a = (R 1 /R 2 )(v a – v o ) v i + (R 1 /R 3 )v o = (1 + R 1 /R 3 )v a

(1)

But v a = v b = v L . Hence, (1) becomes v i = (1 + R 1 /R 3 )v L – (R 1 /R 3 )v o

(2)

i o = v o /(R 4 + R 2 ||R L ), i L = (R 2 /(R 2 + R L ))i o = (R 2 /(R 2 + R L ))(v o /( R 4 + R 2 ||R L )) Or,

v o = i L [(R 2 + R L )( R 4 + R 2 ||R L )/R 2

(3)

But,

vL = iLRL

(4)

Substituting (3) and (4) into (2), v i = (1 + R 1 /R 3 ) i L R L – R 1 [(R 2 + R L )/(R 2 R 3 )]( R 4 + R 2 ||R L )i L

= [((R 3 + R 1 )/R 3 )R L – R 1 ((R 2 + R L )/(R 2 R 3 )(R 4 + (R 2 R L /(R 2 + R L ))]i L = (1/A)i L

Thus, A =

1   R  RL R   1  1  R L  R 1  2 R3    R 2R 3

 R 2RL  R 4  R2  RL 

  

Please note that A has the units of mhos. An easy check is to let every resistor equal 1ohm and v i equal to one amp. Going through the circuit produces i L = 1A. Plugging into the above equation produces the same answer so the answer does check.


iC





dv  7.5 2e 3t  6te 3t  15(1 – 3t)e-3t A dt

p = vi = 15(1–3t)e–3t  2t e–3t = 30t(1 – 3t)e–6t W. 15(1 – 3t)e-3t A, 30t(1 – 3t)e–6t W

Chapter 6, Solution 2. w(t) = (1/2)C(v(t))2 or (v(t))2 = 2w(t)/C = (20cos2(377t))/(50x10–6) = 0.4x106cos2(377t) or v(t) = ±632.5cos(377t) V. Let us assume that v(t) = 632.5cos(377t) V, which leads to i(t) = C(dv/dt) = 50x10–6(632.5)(–377sin(377t)) = –11.923sin(377t) A.

Please note that if we had chosen the negative value for v, then i(t) would have been positive.

Chapter 6, Solution 3. Design a problem to help other students to better understand how capacitors work. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem In 5 s, the voltage across a 40-mF capacitor changes from 160 V to 220 V. Calculate the average current through the capacitor.

Solution

i=C

dv 220  160  40x10 3  480 mA dt 5


v

1 t idt  v(0) C o t

1 t  0.8    4 sin( 4t )dt  1    cos(4t )   1  0.2 cos(4t )  0.2  1 0 5  4 0

= [1.2 – 0.2 cos(4t)] V.

Chapter 6, Solution 5.  5000 t , 0  t  2ms  v =  20  5000 t , 2  t  6ms  40  5000 t , 6  t  8ms 

dv 4 x10 6  iC dt 10 3

 5,  5,   5,

0  t  2ms  20 mA,  2  t  6ms  20 mA, 6  t  8ms  20 mA,

0  t  2ms 2  t  6ms 6  t  8ms


dv  55 x10 6 times the slope of the waveform. dt For example, for 0 < t < 2, iC

dv 10  dt 2x10 3 dv 10  (55 x106 )  275mA dt 2 x103 Thus the current i(t) is sketched below.

i= C

i(t) (mA) 275

4 2

–275

8 6

t (msec) 10

12


v

=

1 1 idt  v(t o )   25 x10 3 C

t

 5tx10 o

2.5t 2 10  [0.1t2 + 10] V. 25

3

dt  10

Chapter 6, Solution 8. (a) i  C

dv  100 ACe 100t  600 BCe 600t dt

i (0)  2  100 AC  600 BC

 

(1) 5   A  6B

v (0  )  v (0  )   50  A  B Solving (2) and (3) leads to A=61, B=-11

(b) Energy 

(2) (3)

1 2 1 Cv (0)  x 4 x10 3 x 2500  5 J 2 2

(c ) From (1), i  100 x 61 x 4 x10 3 e 100 t  600 x11 x 4 x10 3 e 600 t   24.4e 100 t  26.4e 600 t A

Chapter 6, Solution 9. v(t) =









t 1 t 6 1  e  t dt  0  12 t  e  t V = 12(t + e-t) – 12  12 o 0

v(2) = 12(2 + e-2) – 12 = 13.624 V p = iv = [12 (t + e-t) – 12]6(1-e-t) p(2) = [12 (2 + e-2) – 12]6(1-e-2) = 70.66 W


iC

dv dv  5 x10 3 dt dt

 16t , 0  t  1s  v   16, 1  t  3 s 64 - 16t, 3  t  4s 

 16 x10 6 , 0  t  1s dv    0, 1  t  3 s dt  6 - 16x10 , 3  t  4 s

0  t  1s  80 kA,  i ( t )   0, 1  t  3 s - 80 kA, 3  t  4s 

Chapter 6, Solution 11. t

t

1 1 v   idt  v(0)  10  i(t)dt C0 4 x10 3 0 t

For 0
v  10 

103 15dt  10  3.76t 4 x10 3 0

v(2) = 10+7.5 =17.5 For 2 < t <4, i(t) = –10 mA t t 1 10 x10 3 v(t)  i(t)dt  v(2)   dt  17.5  22.5  2.5t 4 x10 3 2 4 x10 3 2 v(4)=22.5-2.5x4 =12.5 t

v(t) 

For 4
1 0dt  v(4) 12.5 4 x103 2

For 6
v(t) 

10 x103 dt  v(6) 2.5(t  6)  12.5  2.5t  2.5 4 x103 4

Hence,  10  3.75t V, 22.5  2.5t V,  v(t) =   12.5 V,  2.5t  2.5 V, which is sketched below. v(t)

0  t  2s 2  t  4s 4  t  6s 6  t  8s

20 15 10 5 t (s) 0

2

4

6

8

Chapter 6, Solution 12. i R = V/R = (30/12)e–2000t = 2.5 e–2000t and i C = C(dv/dt) = 0.1x30(–2000) e–2000t = –6000 e–2000t A. Thus, i = i R + i C = –5,997.5 e–2000t. The power is equal to: vi = –179.925 e–4000t W.

Chapter 6, Solution 13. Under dc conditions, the circuit becomes that shown below: i1

10 

i2

20 

+ 70 

50 

+ v2

v1  60V

+ 

i 2 = 0, i 1 = 60/(70+10+20) = 0.6 A v 1 = 70i 1 = 42 V, v 2 = 60–20i 1 = 48 V Thus, v 1 = 42 V, v 2 = 48 V.



Chapter 6, Solution 14. 20 pF is in series with 60pF = 20*60/80=15 pF 30-pF is in series with 70pF = 30x70/100=21pF 15pF is in parallel with 21pF = 15+21 = 36 pF

Chapter 6, Solution 15. Arranging the capacitors in parallel results in circuit shown in Fig. (1) (It should be noted that the resistors are in the circuits only to limit the current surge as the capacitors charge. Once the capacitors are charged the current through the resistors are obviously equal to zero.): v 1 = v 2 = 100 R

R + 100V

+ 

C1

v1

+ v2



C2 100V

+ 

+

C1

+ v2

 (1)

v1 

 (2)

1 2 1 Cv  x 25 x10 6 x100 2  125 mJ 2 2 1 w 30 = x75 x10 6 x100 2  375 mJ 2

w 20 =

(b)

Arranging the capacitors in series results in the circuit shown in Fig. (2): v1 

C2 75 V x100  75 V, v 2 = 25 V C1  C2 100

w 25 =

1 x 25 x10 6 x75 2  70.31 mJ 2

w 75 =

1 x75 x10 6 x 25 2  23.44 mJ. 2

(a) 125 mJ, 375 mJ (b) 70.31 mJ, 23.44 mJ

C2


C eq  14 

Cx 80  30 C  80

 

C  20 F


(b) (c)

4F in series with 12F = 4 x 12/(16) = 3F 3F in parallel with 6F and 3F = 3+6+3 = 12F 4F in series with 12F = 3F i.e. C eq = 3F C eq = 5 + [6x(4 + 2)/(6+4+2)] = 5 + (36/12) = 5 + 3 = 8F 3F in series with 6F = (3 x 6)/9 = 2F 1 1 1 1    1 C eq 2 6 3 C eq = 1F

Chapter 6, Solution 18. 4 F in parallel with 4 F = 8F 4 F in series with 4 F = 2 F 2 F in parallel with 4 F = 6 F Hence, the circuit is reduced to that shown below. 8F

6 F

6 F

C eq 1 1 1 1     0.4583 Ceq 6 6 8

  Ceq  2.1818 F

Chapter 6, Solution 19. We combine 10-, 20-, and 30-  F capacitors in parallel to get 60  F. The 60 -  F capacitor in series with another 60-  F capacitor gives 30  F. 30 + 50 = 80  F, 80 + 40 = 120  F The circuit is reduced to that shown below. 12

120

12

80

120-  F capacitor in series with 80  F gives (80x120)/200 = 48 48 + 12 = 60 60-  F capacitor in series with 12  F gives (60x12)/72 = 10 µF

Chapter 6, Solution 20. Consider the circuit shown below.

C1 C2

C3

C1  1 1  2  F C2  2  2  2  6 F

C3  4 x3  12  F 1/C eq = (1/C 1 ) + (1/C 2 ) + (1/C 3 ) = 0.5 + 0.16667 + 0.08333 = 0.75x106 C eq = 1.3333 µF.

Chapter 6, Solution 21. 4F in series with 12F = (4x12)/16 = 3F 3F in parallel with 3F = 6F 6F in series with 6F = 3F 3F in parallel with 2F = 5F 5F in series with 5F = 2.5F Hence C eq = 2.5F

Chapter 6, Solution 22. Combining the capacitors in parallel, we obtain the equivalent circuit shown below: a

b 40 F

60 F

30 F

20 F

Combining the capacitors in series gives C1eq , where 1 1 1 1 1     1 C eq 60 20 30 10 Thus C eq = 10 + 40 = 50 F

C1eq = 10F

Chapter 6, Solution 23. Using Fig. 6.57, design a problem to help other students better understand how capacitors work together when connected in series and parallel. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the circuit in Fig. 6.57, determine: (a) the voltage across each capacitor, (b) the energy stored in each capacitor.

Figure 6.57

Solution (a)

(b)

3F is in series with 6F v 4F = 1/2 x 120 = 60V v 2F = 60V 3 v 6F = (60)  20V 63 v 3F = 60 - 20 = 40V

3x6/(9) = 2F

Hence w = 1/2 Cv2 w 4F = 1/2 x 4 x 10-6 x 3600 = 7.2mJ w 2F = 1/2 x 2 x 10-6 x 3600 = 3.6mJ w 6F = 1/2 x 6 x 10-6 x 400 = 1.2mJ w 3F = 1/2 x 3 x 10-6 x 1600 = 2.4mJ

Chapter 6, Solution 24. 20F is series with 80F = 20x80/(100) = 16F 14F is parallel with 16F = 30F (a) v 30F = 90V v 60F = 30V v 14F = 60V 80 v 20F = x 60  48V 20  80 v 80F = 60 - 48 = 12V

1 2 Cv 2 w 30F = 1/2 x 30 x 10-6 x 8100 = 121.5mJ w 60F = 1/2 x 60 x 10-6 x 900 = 27mJ w 14F = 1/2 x 14 x 10-6 x 3600 = 25.2mJ w 20F = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ w 80F = 1/2 x 80 x 10-6 x 144 = 5.76mJ

(b) Since w =

Chapter 6, Solution 25. (a) For the capacitors in series, Q1 = Q2 vs = v1 + v2 =

Similarly, v 1 

v1 C 2  v 2 C1

C1v1 = C2v2 C2 C  C2 v2  v2  1 v2 C1 C1

C2 vs C1  C 2

(b) For capacitors in parallel Q1 Q 2  C1 C 2 C  C2 C Qs = Q1 + Q2 = 1 Q 2  Q 2  1 Q2 C2 C2

v1 = v2 =

or C2 C1  C 2 C1 Q1  Qs C1  C 2

Q2 =

i=

dQ dt

i1 

C1 is , C1  C 2

i2 

C2 is C1  C 2

v2 

C1 vs C1  C 2


(a)

C eq = C 1 + C 2 + C 3 = 35F

(b)

Q 1 = C 1 v = 5 x 150C = 0.75mC Q 2 = C 2 v = 10 x 150C = 1.5mC Q 3 = C 3 v = 20 x 150 = 3mC

(c)

w=

1 1 C eq v 2  x35x150 2 J = 393.8mJ 2 2

Chapter 6, Solution 27. If they are all connected in parallel, we get CT  4 x4  F  16 F If they are all connected in series, we get 1 4    CT  1 F CT 4  F All other combinations fall within these two extreme cases. Hence, C min = 1 µF, C max = 16 µF

Chapter 6, Solution 28. We may treat this like a resistive circuit and apply delta-wye transformation, except that R is replaced by 1/C.

Cb

50 F

Cc

20 F

Ca

 1  1   1  1   1  1             1 10 40 10 30 30 40           1 Ca 30 3 1 1 2 =    40 10 40 10

C a = 5F 1 1 1   1 2  400 300 1200  1 Cb 30 10 C b = 15F 1 1 1   1 4  400 300 1200  1 Cc 15 40 C c = 3.75F

C b in parallel with 50F = 50 + 15 = 65F C c in series with 20F = 23.75F 65x 23.75 65F in series with 23.75F =  17.39F 88.75 17.39F in parallel with C a = 17.39 + 5 = 22.39F

Hence C eq = 22.39F


C in series with C = C/(2) C/2 in parallel with C = 3C/2

3C in series with C = 2

3

3C 2  3C C 5 5 2

Cx

C C in parallel with C = C + 3  1.6 C 5 5

(b) 2C

C eq 2C

1 1 1 1    C eq 2C 2C C

C eq = 1 C


1 t idt  i(0) C o For 0 < t < 1, i = 90t mA, 10 3 t vo  90tdt  0  15t 2 kV 3x10  6 o v o (1) = 15 kV

vo =

For 1< t < 2, i = (180 – 90t) mA, 10 3 t vo = (180  90t )dt  vo (1) 3x10t 6 1 = [60t – 15t2 ] 1  15kV

= [60t – 15t2 – (60–15) + 15] kV = [60t – 15t2 – 30] kV 15t 2 kV , 0t 1 vo ( t )   2 [60t  15t  30]kV , 1  t  2

Chapter 6, Solution 31. 0  t 1 30tmA,  is (t )  30mA, 1 t  3  75  15t , 3  t  5

C eq = 4 + 6 = 10F 1 t v idt  v(0) C eq o For 0 < t < 1, v

10 3 10 x10 6

t

 30t dt + 0 = 1.5t o

2

kV

For 1 < t < 3, 103 t v 20dt  v(1)  [3(t  1)  1.5]kV 10 1  [3t  1.5]kV For 3 < t < 5, 103 t v 15(t  5)dt  v(3) 10 3  t2   1.5  7.5t  t3 7.5kV  [0.75t 2  7.5t  23.25]kV  2 

1.5t 2 kV , 0  t  1s  v ( t )  [3t  1.5]kV , 1  t  3s [0.75t 2  7.5t  23.25]kV , 3  t  5 s  dv dv  6x10 6 dt dt 0  t  1s 18tmA,  i1  18mA, 1  t  3s [9t  45]mA, 3  t  5 s

i 1  C1

i2  C2

dv dv  4x10 6 dt dt

0  t  1s 12tmA,  i2  12mA, 1  t  3s [6t  30]mA, 3  t  5 s

Chapter 6, Solution 32. (a) C eq = (12x60)/72 = 10  F

103 v1  12 x10 6

103 v2  60 x10 6

t

 50e

 2t

dt  v1 (0)   2083e  2 t

t 0

 50   2083e  2 t  2133V

0

t

 50e

 2t

dt  v2 (0)   416.7e  2 t

t 0

 20   416.7e  2 t  436.7V

0

(b) At t=0.5s, v1  2083e 1  2133  1366.7, w12 F 

1 x12 x10  6 x(1366.7 )2  11.207 J 2

1 x 20 x10 6 x ( 283.4)2  803.2 mJ 2 1  x 40 x10 6 x ( 283.4)2  1.6063 J 2

w20 F  w40 F

v2  416.7e 1  436.7  283.4

Chapter 6, Solution 33 Because this is a totally capacitive circuit, we can combine all the capacitors using the property that capacitors in parallel can be combined by just adding their values and we combine capacitors in series by adding their reciprocals. However, for this circuit we only have the three capacitors in parallel. 3 F + 2 F = 5 F (we need this to be able to calculate the voltage) C Th = C eq = 5+3+2 = 10 F The voltage will divide equally across the two 5 F capacitors. Therefore, we get: V Th = 15 V, C Th = 10 F. 15 V, 10 F

Chapter 6, Solution 34. i = 10e–t/2 di 1 v  L  10 x10 3 (10) e t / 2 dt 2 -t/2 = –50e mV

v(3) = –50e–3/2 mV = –11.157 mV p = vi = –500e–t mW p(3) = –500e–3 mW = –24.89 mW.

Chapter 6, Solution 35. di vL dt

v 160 x10 3   L   6.4 mH di / dt (100  50)x10 3 2 x10 3

Chapter 6, Solution 36. Design a problem to help other students to better understand how inductors work. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem The current through a 12-mH inductor is i(t)  30te2t A, t  0. Determine: (a) the voltage across the inductor, (b) the power being delivered to the inductor at t = 1 s, (c) the energy stored in the inductor at t = 1 s. Solution

di  12 x10 3(30e2t  60te2 t )  (0.36  0.72t)e2t V dt (b) p  vi  (0.36  0.72 x1)e2 x30 x1e2  0.36 x30e4  0.1978 W 1 (c) w  Li 2 = 0.5x12x10–3(30x1xe–2)2 = 98.9 mJ. 2

(a) v  L


di  12 x10 3 x 4(100) cos100t dt = 4.8 cos (100t) V

vL

p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t w=

t

11 / 200

o

o

 pdt  

9.6 sin 200 t

9.6 / 200 cos 200t 11 J o 200  48(cos   1)mJ  96 mJ



Please note that this problem could have also been done by using (½)Li2.


vL





di  40x10 3 e  2 t  2te  2 t dt dt = 40(1  2t )e 2 t mV , t  0


vL

i

di 1   i   0t idt  i(0) dt L 1

200x10

t (3t 2 3  0

 5( t 3  t 2  4t )

t 0

 2 t  4)dt  1

1

i(t) = [5t3 + 5t2 + 20t + 1] A

Chapter 6, Solution 40.  5t, 0  t  2ms  i   10, 2  t  4ms 30  5t, 4  t  6ms   5, di 5 x10 3  vL   0, dt 10 3  5,

0  t  2ms

 25, 0  t  2ms  2  t  4ms   0, 2  t  4ms 4  t  6ms  25, 4  t  6ms

At t = 1ms, v = 25 V At t = 3ms, v = 0 V At t = 5ms, v= –25 V


i





1 t 1 t vdt  C     20 1  e  2 t dt  C  L 0  2 o

 1  = 10 t  e 2 t  ot  C  10t  5e 2 t  4.7A 2   Note, we get C = –4.7 from the initial condition for i needing to be 0.3 A. We can check our results be solving for v = Ldi/dt. v = 2(10 – 10e–2t)V which is what we started with. At t = l s, i = 10 + 5e-2 – 4.7 = 10 + 0.6767 – 4.7 = 5.977 A w

1 2 L i = 35.72J 2


1 t 1 t vdt  i(0)   v( t )dt  1  5 o L o 10 t For 0 < t < 1, i   dt  1  2t  1 A 5 0 i

For 1 < t < 2, i = 0 + i(1) = 1A 1 10dt  i(2)  2t 2t 1 5 = 2t - 3 A

For 2 < t < 3, i =

For 3 < t < 4, i = 0 + i(3) = 3 A 1 t 10dt  i(4)  2 t 4t 3  4 5 = 2t - 5 A

For 4 < t < 5, i =

Thus,  2t  1 A, 1 A,  i ( t )   2t  3 A,   3 A,  2t  5,

0t 1 1 t  2 2t 3 3t 4 4t 5

Chapter 6, Solution 43. w = L

t



idt 

1 2 1 2 Li ( t )  Li () 2 2





2 1 x80x10 3 x 60x10 3  0 2 = 144 J.



Chapter 6, Solution 44. di  100 x10 3(400)x50 x10 3 e400t  2e400t V dt (b) Since R and L are in parallel, vR  vL  2e400t V (c) No 1 (d) w  Li 2 = 0.5x100x10–3(0.05)2 = 125 µJ. 2

(a) vL  L

Chapter 6, Solution 45. i(t) =

1 t v( t )  i(0) L o

For 0 < t < 1, v = 5t

i

1 10x10 3

t

 5t dt + 0 o

= 250t2 A For 1 < t < 2, v = -10 + 5t i

1 10x10 3

t

 (10  5t )dt  i(1) 1

t

  (0.5t  1)dt  0.25kA 1

= [1 – t + 0.25t2 ] kA  250t 2 A, 0  t  1s i(t )   2 1  t  2s [1  t  0.25t ] kA,

Chapter 6, Solution 46. Under dc conditions, the circuit is as shown below: 2 iL + vC

4 3A



By current division,

iL 

4 (3)  2A, v c = 0V 42

wL 

1 2 11 2 L i L   (2)  1J 2 22

wc 

1 1 C v c2  (2)( v)  0J 2 2

Chapter 6, Solution 47. Under dc conditions, the circuit is equivalent to that shown below: R

+

vC



2 5A

iL 

2 10 10R , v c  Ri L  (5)  R2 R2 R2

1 2 100R 2 6 w c  Cv c  80x10 x 2 (R  2) 2 1 100 w L  Li12  2x10 3 x 2 (R  2) 2 If w c = w L , 80x10

6

x

100R 2 (R  2)

2



2x10 3 x100 (R  2)

2

R = 5

80 x 10-3R2 = 2

iL

Chapter 6, Solution 48. Under steady-state, the inductor acts like a short-circuit, while the capacitor acts like an open circuit as shown below. i + 5 mA

30k

v

20 k

–

Using current division, i = (30k/(30k+20k))(5mA) = 3 mA v = 20ki = 60 V

Chapter 6, Solution 49. Converting the wye-subnetwork to its equivalent delta gives the circuit below.

30 mH

30mH 5mH 30 mH

30//0 = 0, 30//5 = 30x5/35=4.286 Leq  30 // 4.286 

30 x4.286  3.75 mH 34.286

Chapter 6, Solution 50. 16mH in series with 14 mH = 16+14=30 mH 24 mH in series with 36 mH = 24+36=60 mH 30mH in parallel with 60 mH = 30x60/90 = 20 mH


1 1 1 1 1     L 60 20 30 10

L eq  10 25  10 

L = 10 mH

10x35 45 = 7.778 mH

Chapter 6, Solution 52. Using Fig. 6.74, design a problem to help other students better understand how inductors behave when connected in series and when connected in parallel. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find L eq in the circuit of Fig. 6.74. 10 H

4H

6H

5H

3H

L eq 7H

Figure 6.74

For Prob. 6.52.

Solution Leq  5 //(7  3  10 //(4  6))  5 //(7  3  5)) 

5 x15  3.75 H 20

Chapter 6, Solution 53. L eq  6  10  8 5 (8  12)  6 (8  4)  16  8 (4  4)  16  4

L eq = 20 mH


L eq  4  (9  3) 10 0  6 12  4  12 (0  4)  4  3

L eq = 7H


(a) L//L = 0.5L, L + L = 2L

Leq  L  2 L // 0.5L  L 

2 Lx0.5 L  1.4 L = 1.4 L. 2 L  0.5L

(b) L//L = 0.5L, L//L + L//L = L L eq = L//L = 500 mL


1 L  3 3 L Hence the given circuit is equivalent to that shown below: LLL

L

L/3

L/3 L

L eq

5 Lx L 2   3  5L  L L  L  5 8 3   L L 3


Let v  L eq

di dt

v  v1  v 2  4 i = i1 + i2 v2  3

(1)

di  v2 dt i2 = i – i1

di1 di1 v 2 or  dt dt 3

(2) (3) (4)

and di di 5 2  0 dt dt di di v2  2  5 2 dt dt

 v2  2

Incorporating (3) and (4) into (5), v2  2

di v di di di 5 5 1  7 5 2 dt dt dt dt 3

di  5 v 2 1    7 dt  3 21 di v2  8 dt Substituting this into (2) gives v4



di 21 di  dt 8 dt

53 di 8 dt

Comparing this with (1), L eq 

53  6.625 H 8

(5)


vL

di di  3  3 x slope of i(t). dt dt

Thus v is sketched below: v(t) (V) 6

t (s) 1

-6

2

3

4

5

6

7


(a) v s  L1  L 2 

di dt

vs di  dt L1  L 2 di di v 1  L1 , v 2  L 2 dt dt v1 

(b)

v i  v 2  L1

L1 L2 vs , vL  vs L1  L2 L1  L2

di1 di  L2 2 dt dt

i s  i1  i 2 di s di1 di 2 L  L 2  v v     v 1 dt dt dt L1 L 2 L1 L 2 L1 L 2 1 1 vdt    L1 L1  L 2 L1 L1 L 2 1 1 i2  vdt    L2 L 2 L1  L 2 i1 

L2 di s dt  is L1  L2 dt L1 di s dt  is L1  L2 dt


15 8 di 15 d vo  Leq  4e  2t   15e 2t dt 8 dt Leq  3 // 5 



t



t

I 1 io   vo (t )dt  io (0)  2   (15)e  2t dt  2  1.5e  2t L0 50

i o = (0.5 + 1.5e–2t) A

t 0

Chapter 6, Solution 61. (a) Leq  20 //(4  6)  20 x10 / 30  6.667 mH Using current division, i1(t) 

10 i s  et mA 10  20

i2(t)  2e t mA (b) vo  Leq (c ) w 

dis 20  x10 3(3e t x10 3 )  20e t  V dt 3

1 2 1 Li1  x20 x10 3 xe2 x10 6  1.3534 nJ 2 2


(a)

Leq  25  20 // 60  25 

v  Leq

di dt

20 x60  40 mH 80

1 10 3 ( ) ( 0 ) v t dt  i  12e 3t dt  i (0)  0.1(e 3t  1)  i (0) Leq  40 x10 3 0 t

 

i

Using current division and the fact that all the currents were zero when the circuit was put together, we get, 60 3 1 i1  i  i, i 2  i 80 4 4 3 i1 (0)  i (0)   0.75i (0)  0.01   i (0)  0.01333 4

i2 

1 (0.1e 3t  0.08667) A  - 25e -3t  21.67 mA 4 i 2 (0)  25  21.67   3.33 mA

(b) i1 

3 ( 0.1e  3 t  0.08667 ) A  - 75e - 3t  65 mA 4 i 2  - 25e -3t  21.67 mA


We apply superposition principle and let vo  v1  v 2 where v 1 and v 2 are due to i 1 and i 2 respectively. di1 di  2, 0  t  3 2 1  dt dt  2, 3  t  6 0t 2  4, di2 di2  2   0, 2t 4 v2  L dt dt  4t6  4, v1 v1  L

v2 2

4

0 3

6

t

0

-2

2

4

-4

Adding v 1 and v 2 gives v o , which is shown below. v o (t) V 6

2 0 2 -2

-6

3

4

6

t (s)

6

t


(a) When the switch is in position A, i= –6 = i(0) When the switch is in position B, i ()  12 / 4  3,   L / R  1/ 8 i (t )  i ()  [i (0)  i ()]e  t /  i(t) = (3 – 9e–8t) A (b) -12 + 4i(0) + v=0, i.e. v = 12 – 4i(0) = 36 V (c) At steady state, the inductor becomes a short circuit so that v=0V


1 1 L1i12  x5x (4) 2  40 J 2 2 1 w 20  (20)(2) 2  40 J 2 (b) w = w 5 + w 20 = 80 J 1 t (c) i1   50e  200 t dt  i1 (0)   0 L1 = [5x10-5(e-200t – 1) + 4] A (a)

w5 





t 1 1   200 t x10  3  4   50e 0 5  200 





1 t 1  1   200 t  200 t 3 t 50 e dt i ( 0 )    50 e x 10 2   2 L2  0 20  200  0 = [1.25x10-5 (e-200t – 1) – 2] A

i2 

(d)

i = i 1 + i 2 = [6.25x10-5 (e-200t– 1) + 2] A

Chapter 6, Solution 66. If v=i, then di dt di iL    dt L i Integrating this gives  i  t   i = C o et/L  ln(i)  ln(C o )  ln L  Co  i(0) = 2 = C o i(t) = 2et/0.02 = 2e50t A.


1 vi dt, RC = 50 x 103 x 0.04 x 10-6 = 2 x 10-3 RC   10 3 10 sin 50t dt vo  2  v o = 100cos(50t) mV

vo  


1 vi dt + v(0), RC = 50 x 103 x 100 x 10-6 = 5 RC  1 t v o =   10dt  0  2t 5 o The op amp will saturate at v o =  12 vo  

-12 = -2t

t = 6s

Chapter 6, Solution 69. RC = 4 x 106 x 1 x 10-6 = 4

vo  

1 1 v i dt    v i dt  4 RC

For 0 < t < 1, v i = 20, v o  

1 t 20dt  -5t mV 4 o

1 t 10dt  v(1)  2.5( t  1)  5 4 1 = -2.5t - 2.5mV

For 1 < t < 2, v i = 10, v o  

1 t 20dt  v(2)  5( t  2)  7.5 4 2 = 5t - 17.5 mV

For 2 < t < 4, v i = - 20, v o  

1 t 10dt  v(4)  2.5( t  4)  2.5 4 4 = 2.5t - 7.5 mV

For 4 < t < 5m, v i = -10, v o 

1 t 20dt  v(5)  5( t  5)  5 4 5 = - 5t + 30 mV

For 5 < t < 6, v i = 20, v o  

Thus v o (t) is as shown below: v(t) (V) 5 2.5 t (s)

0 1 -2.5 -5 -7.5

2

3

4

5

6

7

Chapter 6, Solution 70. One possibility is as follows:

1  50 RC Let R = 100 k, C 

1  0.2 F 50 x100 x10 3

Chapter 6, Solution 71. By combining a summer with an integrator, we have the circuit below:

R1 v1

C

R2 v2 R3

 +

vo

v3

vo  

1 1 1 v1dt  v 2 dt  v 2 dt   R 1C R 2C R 2C 

For the given problem, C = 2F, R1C = 1 R 2 C = 1/(4) R 3 C = 1/(10)

R 1 = 1/(C) = 106/(2) = 500 k R 2 = 1/(4C) = 500k/(4) = 125 k R 3 = 1/(10C) = 50 k

Chapter 6, Solution 72. The output of the first op amp is

v1  

t 1 1 100t v i dt =  v i dt     RC 2 10x10 3 x 2 x10 6 o

= - 50t

vo  

1 1 v i dt =  3  RC 20x10 x 0.5x10 6

= 2500t2 At t = 1.5ms, v o  2500(1.5) 2 x10 6  5.625 mV

t

 (50t )dt o

Chapter 6, Solution 73. Consider the op amp as shown below: Let v a = v b = v At node a,

0  v v  vo  R R

2v - v o = 0

(1) R

R

v a

R

 + R

v

vo

b vi

At node b,

+ 

C

vi  v v  vo dv  C R R dt dv v i  2v  v o  RC dt

Combining (1) and (2), v i v o v o 

RC dv o 2 dt

or vo 

+

2 v i dt RC 

showing that the circuit is a noninverting integrator.



(2)

Chapter 6, Solution 74. RC = 0.01 x 20 x 10-3 sec v o   RC

dv i dv  0.2 m sec dt dt

  2V, v o  2V,  2V,

0  t 1 1 t  3 3 t 4

Thus v o (t) is as sketched below: v o (t) (V) 2

t (ms) 1

-2

2

3

Chapter 6, Solution 75. v 0   RC

dv i , RC  250 x10 3 x10x10 6  2.5 dt

v o  2.5

d (12t )  –30 mV dt

Chapter 6, Solution 76. dv i , RC = 50 x 103 x 10 x 10-6 = 0.5 dt  10, 0  t  5 dv v o  0.5 i   5  t  15 dt 5,

v o   RC

The input is sketched in Fig. (a), while the output is sketched in Fig. (b). v o (t) (V)

v i (t) (mV) 100

5

t (ms) 0

5

10

t (ms)

15

0

5

10

(a)

-10 (b)

15

Chapter 6, Solution 77. i = iR + iC vi  0 0  v0 d   C 0  v o  R RF dt

R F C  10 6 x10 6  1 dv   Hence v i   v o  o  dt   Thus v i is obtained from v o as shown below:

–dv o (t)/dt

– v o (t) (V)

4

4

t (s)

t (s) 0

1

2

3

-4

-4

v i (t) (V) 8

t (s) -4 -8

0

4

1

2

3

4

1

2

3

4

Chapter 6, Solution 78. d 2 vo 2dv o  10 sin 2 t   vo dt dt Thus, by combining integrators with a summer, we obtain the appropriate analog computer as shown below:

2v o  +

C

C

R 2

d v o /dt 2

t=0 R

R

 +

R

 +

-dv o /dt

 +

vo R

2

R  +

R/2 dv o /dt

R R

sin2t

+ 

 +

d2v o /dt

R/10

-sin2t


We can write the equation as dy  f (t )  4 y (t ) dt

which is implemented by the circuit below. 1V

t=0

C

R

R

R R/4 dy/dt

+

+

-y R f(t)

R + dy/dt

Chapter 6, Solution 80. From the given circuit, d 2vo 1000k 1000k dv o  f (t)  vo  2 5000k 200k dt dt or d 2vo dv  5 o  2v o  f ( t ) 2 dt dt

Chapter 6, Solution 81 We can write the equation as d 2v  5v  2 f (t ) dt 2 which is implemented by the circuit below.

C

C R

R 2

2

d v/dt

+

R

R/5 -

-dv/dt

+

v

+ R/2

f(t)

d2v/dt2


The circuit consists of a summer, an inverter, and an integrator. Such circuit is shown below. 10R

R

R

R +

+ R

C=1/(2R)

R + vs -

+

vo

Chapter 6, Solution 83. Since two 10F capacitors in series gives 5F, rated at 600V, it requires 8 groups in parallel with each group consisting of two capacitors in series, as shown below:

+ 600 

Answer: 8 groups in parallel with each group made up of 2 capacitors in series.

Chapter 6, Solution 84. v = L(di/dt) = 8x10–3x5x2πsin(πt)cos(πt)10–3 = 40πsin(2πt) µV p = vi = 40πsin(2πt)5sin2(πt)10–9 W, at t=0 p = 0W w

1 2 1 Li  x8 x10 3 x[5 sin2( / 2)x10 3 ]2  4 x25 x10 9  100 nJ 2 2

= 100 ηJ

Chapter 6, Solution 85. It is evident that differentiating i will give a waveform similar to v. Hence, di vL dt 4t ,0  t  1ms i 8  4 t ,1  t  2ms  di 4000L,0  t  1ms  v  L     dt  4000L,1  t  2ms

But,

5V,0  t  1ms v  5V,1  t  2ms

Thus, 4000L = 5

L = 1.25 mH in a 1.25 mH inductor

Chapter 6, Solution 86. v  vR  vL  Ri  L

di  12 x2te10t  200 x10 3 x(20te10t  2e10t )  (0.4  20t)e10t V dt


=RC = 1/200

For the resistor, V=iR= 56e200 t  8Re 200t x10 3 C

 

R

56  7 k 8

1 1   0.7143  F 200R 200 X7 X103

(b)  =1/200= 5 ms (c) If value of the voltage at = 0 is 56 . 1 x56  56e200t 2

 

200to  ln2

e200 t  2

 

to 

1 ln2  3.466 ms 200

Chapter 7, Solution 2.   R th C where R th is the Thevenin equivalent at the capacitor terminals. R th  120 || 80  12  60 

  60  200  10 -3  12 s.

Chapter 7, Solution 3. R = 10 +20//(20+30) =10 + 40x50/(40 + 50)=32.22 k

  RC  32.22 X103 X100 X10 12  3.222  S

Chapter 7, Solution 4. For t<0, v(0-)=40 V. For t >0. we have a source-free RC circuit.   RC  2 x103 x10 x10 6  0.02 v(t)  v(0)et /   40e50t V

Chapter 7, Solution 5. Using Fig. 7.85, design a problem to help other students to better understand source-free RC circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the circuit shown in Fig. 7.85, find i(t), t>0. 2

t=0 i

+

5 4Ω

24 V 4  _

1/3 F

Figure 7.85

For Prob. 7.5.

Solution Let v be the voltage across the capacitor. For t <0, 4 v(0  )  (24)  16 V 24 For t >0, we have a source-free RC circuit as shown below.

i 5 + v – 1 3

  RC  (4  5)  3 s

1/3 F

4

v(t)  v(0)et /   16e t / 3 V dv 1 1 i(t)  C   ( )16e t / 3  1.778et / 3 A dt 3 3


v o  v ( 0) 

2 (40)  6.667 V 10  2

v( t )  v o e  t /  ,   RC  40 x10 6 x 2 x10 3  v( t )  6.667e 12.5 t V

2 25

Chapter 7, Solution 7. Assuming that the switch in Fig. 7.87 has been in position A for a long time and is moved to position B at t=0. Then at t = 1second, the switch moves from B to C. Find v C (t) for t  0. 10 k

12V

+ 

A B

500 

C 1 k

2 mF

Figure 7.87 For Prob. 7.7 Solution Step 1. Determine the initial voltage on the capacitor. Clearly it charges to 12 volts when the switch is at position A because the circuit has reached steady state. This then leaves us with two simple circuits, the first a 500 Ω resistor in series with a 2 mF capacitor and an initial charge on the capacitor of 12 volts. The second circuit which exists from t = 1 sec to infinity. The initial condition for the second circuit will be v C (1) from the first circuit. The time constant for the first circuit is (500)(0.002) = 1 sec and the time constant for the second circuit is (1,000)(0.002) = 2 sec. v C (∞) = 0 for both circuits. Step 1. v C (t) = 12e-t volts for 0 < t < 1 sec and = 12e-1e-2(t-1) at t = 1 sec, and = 4.415e-2(t-1) volts for 1 sec < t < ∞. 12e-t volts for 0 < t < 1 sec, 4.415e-2(t-1) volts for 1 sec < t < ∞.


  RC 

dv dt -4t - 0.2 e  C (10)(-4) e-4t   C  5 mF 1 R  50  4C 1   RC   0.25 s 4 1 1 w C (0)  CV02  (5  10 -3 )(100)  250 mJ 2 2 1 1 1 w R   CV02  CV02 1  e -2t 0   2 2 2 1  e -8t 0  0.5  1  e -8t 0  2 8t 0 e 2 or 1 t 0  ln (2)  86.6 ms 8 -i  C

(b) (c) (d)

1 4

Chapter 7, Solution 9. For t < 0, the switch is closed so that 4 vo(0)  (6)  4 V 24 For t >0, we have a source-free RC circuit.   RC  3 x103 x4 x103  12 s v o (t) = v o (0)e–t/τ = 4e–t/12 V.

Chapter 7, Solution 10. 3 (36V )  9 V 39 For t>0, we have a source-free RC circuit   RC  3 x103 x20 x106  0.06 s

For t<0,

v(0  ) 

v o (t) = 9e–16.667t V Let the time be t o . 3 = 9e–16.667to or e16.667to = 9/3 = 3 t o = ln(3)/16.667 = 65.92 ms.

Chapter 7, Solution 11. For t<0, we have the circuit shown below. 4

4H

4 24 V

8

+ _

4H io 4 6A

4

4||4= 4x4/8=2 i o (0–) = [2/(2+8)]6 = 1.2 A For t >0, we have a source-free RL circuit. L 4    1/ 3 thus, R 48 i o (t) = 1.2e–3t A.

8

Chapter 7, Solution 12. Using Fig. 7.92, design a problem to help other students better understand source-free RL circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem The switch in the circuit in Fig. 7.90 has been closed for a long time. At t = 0, the switch is opened. Calculate i(t) for t > 0.

Figure 7.90

Solution When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4  resistor is short-circuited so that the resulting circuit is as shown in Fig. (a). 3

12 V

i(0-)

+ 

4

(a)

12 4A 3 Since the current through an inductor cannot change abruptly, i(0)  i(0  )  i(0  )  4 A i (0  ) 

2H

(b)

When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b). L 2     0.5 R 4 Hence, i( t )  i(0) e - t   4 e -2t A

Chapter 7, Solution 13. (a)  

1  1ms 103 = 1 ms.

v(t) = i(t)R = 80e–1000t V = R5e–1000tx10–3 or R = 80,000/5 = 16 kΩ. But τ = L/R = 1/103 or L = 16x103/103 = 16 H. (b) The energy dissipated in the resistor is

= 200(1–e–1)x10–6 = 126.42 µJ. (a) 16 kΩ, 16 H, 1 ms

(b) 126.42 µJ

Chapter 7, Solution 14. 60 x40  24 k 100 5 x10 3   L/R   0.2083  s 24 x103

RTh  (40  20)//(10  30) 


(a) R Th  2  10 // 40  10,

(b) RTh  40 // 160  48  40,

L  5 / 10  0.5s R Th L   (20 x10 3 ) / 80  0.25 ms RTh



(a) 10 Ω, 500 ms

(b) 40 Ω, 250 µs


 (a)

L eq R eq

L eq  L and R eq  R 2 



(b)

R 1R 3 R 2 (R 1  R 3 )  R 1 R 3  R1  R 3 R1  R 3

L( R 1  R 3 ) R 2 (R 1  R 3 )  R 1 R 3 R 3 (R 1  R 2 )  R 1 R 2 L1 L 2 R 1R 2 and R eq  R 3   L1  L 2 R1  R 2 R1  R 2 L1L 2 (R 1  R 2 )  (L 1  L 2 ) ( R 3 ( R 1  R 2 )  R 1 R 2 )

where L eq 

Chapter 7, Solution 17. i( t )  i(0) e - t  ,



14 1 L   R eq 4 16

i (t )  6 e -16t

vo (t )  3i  L

di  18 e -16t  (1 4)(-16) 6 e -16t dt v o ( t )  –6e–16tu(t) V

Chapter 7, Solution 18. If v( t )  0 , the circuit can be redrawn as shown below. + 0.4 H

R eq i(t)

6 L 2 5 1 ,     5 R 5 6 3 -t  -3t i (t )  i (0) e  5e di - 2 vo (t )  -L  (-3)5 e -3t  6 e -3t V dt 5 R eq  2 || 3 

v o (t)



Chapter 7, Solution 19. i

1V  +

10 

i1

i1

i2

i/2

i2

40 

To find R th we replace the inductor by a 1-V voltage source as shown above. 10 i1  1  40 i 2  0 But i  i2  i 2 and i  i1 i.e. i1  2 i 2  i 1 10 i  1  20 i  0   i  30

1 R th   30  i L 6    0.2 s R th 30

i( t )  6 e -5t u (t ) A


(b) (c)

L 1    R  50L R 50 di v  L dt -50t 90 e   L(30)(-50) e -50t   L  60 mH R  50L  3 Ω L 1    20 ms R 50 1 1 w  L i 2 (0)  (0.06)(30) 2  27 J 2 2 The value of the energy remaining at 10 ms is given by: 

w 10 = 0.03(30e–0.5)2 = 0.03(18.196)2 = 9.933 J. So, the fraction of the energy dissipated in the first 10 ms is given by: (27–9.933)/27 = 0.6321 or 63.21%.

Chapter 7, Solution 21. The circuit can be replaced by its Thevenin equivalent shown below. R th

V th

+ 

2H

80 (60)  40 V 80  40 80 R R th  40 || 80  R  3 Vth 40 I  i(0)  i()   R th 80 3  R

Vth 

  1 3 40 40 1   R  3 R  80 3

1  40 1 w  L I 2  (2) 2  R  80 2

2

R  13.333 Ω

Chapter 7, Solution 22. i( t )  i(0) e - t  ,



L R eq

R eq  5 || 20  1  5  ,



2 5

i( t )  10e–2.5t A Using current division, the current through the 20 ohm resistor is 5 -i io  (-i)   -2 e -2.5t 5  20 5 v( t )  20 i o  –40e–2.5t V

Chapter 7, Solution 23. Since the 2  resistor, 1/3 H inductor, and the (3+1)  resistor are in parallel, they always have the same voltage.

10 10   7.5   i (0)  -7.5 2 3 1 The Thevenin resistance R th at the inductor’s terminals is 13 1 L 4    R th  2 || (3  1)  , 3 R th 4 3 4 -i 

i (t )  i (0) e - t   -7.5 e -4t , t  0 di v L  vo  L  -7.5(-4)(1/3) e - 4t dt -4t v o  10e V , t  0 1 vx  v L  2.5 e-4t V , t  0 3 1

Chapter 7, Solution 24. (a) v( t )  - 5 u(t)

(b) i( t )  -10  u ( t )  u ( t  3)  10 u ( t  3)  u ( t  5) = - 10 u(t )  20 u(t  3)  10 u(t  5) (c) x ( t )  ( t  1)  u ( t  1)  u ( t  2)   u ( t  2)  u ( t  3)  (4  t )  u ( t  3)  u ( t  4) = ( t  1) u ( t  1)  ( t  2) u ( t  2)  ( t  3) u ( t  3)  ( t  4) u ( t  4) = r(t  1)  r(t  2)  r(t  3)  r(t  4) (d) y( t )  2 u (-t )  5  u ( t )  u ( t  1) = 2 u(-t )  5 u(t )  5 u(t  1)

Chapter 7, Solution 25. Design a problem to help other students to better understand singularity functions. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Sketch each of the following waveforms. (a) i(t) = [u(t–2)+u(t+2)] A (b) v(t) = [r(t) – r(t–3) + 4u(t–5) – 8u(t–8)] V Solution The waveforms are sketched below. (a)

i(t) (A) 2 1

-2 -1 0

1

2

3

4

t

(b) v(t) (V)

7

3

0 –1

1

2

3

4

5

6

7

8

t


(b)

(c)

(d)

v1 ( t )  u ( t  1)  u ( t )   u ( t  1)  u ( t ) v1 ( t )  u(t  1)  2 u(t )  u(t  1) v 2 ( t )  ( 4  t )  u ( t  2)  u ( t  4)  v 2 ( t )  -( t  4) u ( t  2)  ( t  4) u ( t  4) v 2 ( t )  2 u(t  2)  r(t  2)  r(t  4)

v 3 ( t )  2  u(t  2)  u(t  4)  4  u(t  4)  u(t  6) v 3 ( t )  2 u(t  2)  2 u(t  4)  4 u(t  6) v 4 ( t )  -t  u ( t  1)  u ( t  2)  -t u(t  1)  t u ( t  2) v 4 ( t )  (-t  1  1) u ( t  1)  ( t  2  2) u ( t  2) v 4 ( t )  - r(t  1)  u(t  1)  r(t  2)  2 u(t  2)

Chapter 7, Solution 27. v(t)= [5u(t+1)+10u(t)–25u(t–1)+15u(t-2)] V

Chapter 7, Solution 28. i(t) is sketched below. i(t) 1

0 -1

1

2

3

4

t

Chapter 7, Solution 29 x(t) (a)

3.679

0 (b)

1

t

y(t) 27.18

0 (c)

t

z (t )  cos 4t (t  1)  cos 4 (t  1)  0.6536 (t  1) , which is sketched below. z(t) 1 0

t

–0.653δ(t–1)

Chapter 7, Solution 30. 

(a)

  4t

(b)

- 4t



2

2

( t  1) dt  4t 2 t 1  4

cos(2t ) ( t  0.5) dt  4t 2 cos(2t ) t 0.5  cos   - 1


(a) (b)

 e  112  10   e (t  2) dt  e   5 (t )  e (t )  cos 2t (t ) dt   5  e  cos(2t ) 

- 

-

- 4t 2

- 4t 2

-t

t2

-16

-9

-t

t 0

 5 11  7


t

(a)

t

t

1

1

 u ( )d   1d   1 4

1

 t 1

4

t2 (b)  r (t  1)dt   0dt   (t  1)dt   t 14  4.5 2 0 0 1 5

(c )

 (t  6) 1

2

 (t  2)dt  (t  6) 2

t 2

 16


i( t ) 

i (t ) 

1 t  v(t ) dt  i(0) L 0 t 10 -3 15 (t  2) dt  0 -3 0 10  10

i( t )  1.5 u( t  2) A


(a)

d u(t  1) u(t  1)   (t  1)u(t  1)  dt u( t  1) ( t  1)   ( t  1)1  0 ( t  1)   ( t  1)

(b)

d r (t  6) u(t  2)  u(t  6)u(t  2)  dt r ( t  6) ( t  2)  u( t  6)1  0 ( t  2)  u( t  6)

(c)

d sin 4t u(t  3)  4 cos 4t u(t  3)  sin 4t (t  3) dt  4 cos 4t u( t  3)  sin 4 x 3 ( t  3)  4 cos 4t u( t  3)  0.5366 ( t  3)

Chapter 7, Solution 35. (a) v  Ae2t ,

v(0)  A  1 v(t) = –e–2tu(t) V

i  Ae3 t / 2 ,

i(0)  A  2

(b) i(t) = 2e–1.5tu(t) A


(a)

(b)

v( t )  A  B e-t , t  0 A  1, v(0)  0  1  B v( t )  1  e -t V , t  0

v( t )  A  B e t 2 , t  0 A  -3 , v(0)  -6  -3  B v( t )  - 3  1  e

t 2

 V,

t0

or

B  -1

or

B  -3


Let v = v h + v p , v p =10. 

1

vh  4 v

h

0

 

v h  Ae t / 4

v  10  Ae 0.25t v(0)  2  10  A v  10  8e 0.25t (a)   4 s (b) v ()  10 V





(c ) v  10  8e 0.25 t u( t ) V

 

A  8


Let i = i p +i h 

i h  3ih  0

Let i p  ku (t ),



ip  0,

 

3ku (t )  2u (t )

ip 

ih  Ae 3t u (t )

 

2 u (t ) 3

2 i  ( Ae 3t  )u (t ) 3 If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus,

i

2 (1  e  3 t )u( t ) 3

k

2 3


(a)

Before t = 0,

v( t ) 

1 (20)  4 V 4 1

After t = 0, v( t )  v()   v(0)  v() e - t    RC  (4)(2)  8 , v(0)  4 ,

v()  20

v( t )  20  (4  20) e -t 8 v( t )  20  16 e - t 8 V Before t = 0, v  v1  v 2 , where v1 is due to the 12-V source and v 2 is due to the 2-A source. v1  12 V To get v 2 , transform the current source as shown in Fig. (a). v 2  -8 V Thus, v  12  8  4 V After t = 0, the circuit becomes that shown in Fig. (b).

(b)

4

2F +

v2

2F

 + 

8V

12 V

+ 

3

3

(a)

(b)

v( t )  v()   v(0)  v() e - t  v()  12 , v(0)  4 ,   RC  (2)(3)  6 -t 6 v( t )  12  (4  12) e v( t )  12  8 e -t 6 V


(a)

Before t = 0, v  12 V . After t = 0, v( t )  v()   v(0)  v() e - t  v()  4 , v(0)  12 ,   RC  (2)(3)  6 v( t )  4  (12  4) e - t 6 v( t )  4  8 e - t 6 V

(b)

Before t = 0, v  12 V . After t = 0, v( t )  v()   v(0)  v() e - t  After transforming the current source, the circuit is shown below. t=0 2

12 V

v(0)  12 , v  12 V

+ 

v()  12 ,

4 5F

  RC  (2)(5)  10

Chapter 7, Solution 41. Using Fig. 7.108, design a problem to help other students to better understand the step response of an RC circuit. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the circuit in Fig. 7.108, find v(t) for t > 0.


v(0)  0 ,

v ( ) 

R eq C  (6 || 30)(1) 

30 (12)  10 36

(6)(30) 5 36

v( t )  v()   v(0)  v() e - t  v( t )  10  (0  10) e - t 5

v( t )  10 (1  e -0.2t ) u ( t )V


(a)

v o ( t )  v o ()   v o (0)  v o () e - t  4 v o ()  (12)  8 v o (0)  0 , 42 4   R eq C eq , R eq  2 || 4  3 4   (3)  4 3 v o ( t )  8  8 e -t 4 v o ( t )  8 ( 1  e -0.25t ) V

(b)

For this case, v o ()  0 so that v o ( t )  v o (0) e - t  4 (12)  8 , v o (0)  42

  RC  (4)(3)  12 v o ( t )  8 e -t 12 V


Before t = 0, the circuit has reached steady state so that the capacitor acts like an open circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage source.

0.5i

vo i

40 

2A

80 

0.5i (a)

vo vo i , 40 80 vo 1 vo 320  2   v o   64 Hence, 2 80 40 5 vo i  0.8 A 80 0.5i  2 

After t = 0, the circuit is as shown in Fig. (b).

0.5i

vC i

3 mF

80 

0.5i (b)

v C ( t )  v C (0) e - t  ,

  R th C

To find R th , we replace the capacitor with a 1-V voltage source as shown in Fig. (c). 0.5i

vC i

1V

+ 

0.5i (c)

80 

vC 1 0.5  , i o  0.5 i  80 80 80 1 80 R th    160  ,   R th C  480 i o 0.5 v C (0)  64 V

i

v C ( t )  64 e - t 480 dv C  1   64 e - t 480 0.5 i  -i C  -C  -3   480  dt i( t )  800 e - t 480 u( t ) mA


R eq  6 || 3  2  ,

  RC  4

v( t )  v()   v(0)  v() e - t 

Using voltage division, 3 v(0)  (60)  20 V , 36

v ( ) 

3 (24)  8 V 3 6

Thus, v(t )  8  (20  8) e -t 4  8  12 e -t 4 dv  -1 i (t )  C  (2)(12)   e -t 4  - 6 e -0.25t A dt 4

Chapter 7, Solution 45. To find R Th , consider the circuit shown below. 20 k

10 k

40 k

RTh  10  20 // 40  10 

  RThC 

R Th

20 x40 70  k 60 3

70 x103 x3 x10 6  0.07 3

To find vo(), consider the circuit below.

20 k

10 k +

30V

+ _

40 k

vo –

v o (∞) = [40/(40+20)]30 = 20 V v o (t) = v o (∞) + [v o (0)– v o (∞)]e–t/0.07 = [20 –15e–14.286t]u(t) V.


  RTh C  (2  6) x0.25  2s,

v(0)  0,

v()  6i s  6 x5  30

v ( t )  v ( )  [v (0)  v ( )]e  t /   30(1  e  t / 2 ) u( t ) V

Chapter 7, Solution 47. For t < 0, u ( t )  0 ,

u ( t  1)  0 ,

v(0)  0

For 0 < t < 1,   RC  (2  8)(0.1)  1 v(0)  0 , v()  (8)(3)  24 v( t )  v()   v(0)  v() e - t  v( t )  24 1  e - t  For t > 1,

v(1)  24 1  e -1   15.17

- 6  v() - 24  0   v()  30 v( t )  30  (15.17  30) e -(t-1) v( t )  30  14.83 e -(t-1) Thus,





 24 1  e - t V , 0t1 v( t )   -(t -1) V, t 1  30  14.83 e


For t < 0,

u (-t)  1 ,

For t > 0, u (-t)  0 , R th  20  10  30 ,

v()  0   R th C  (30)(0.1)  3

v( t )  v()   v(0)  v() e - t 

v( t )  10 e -t 3 V i( t )  C

 - 1 dv  (0.1)  10 e - t 3 3 dt

i( t ) 

- 1 -t 3 e A 3

Chapter 7, Solution 49. For 0 < t < 1, v(0)  0 , R eq  4  6  10 ,

v()  (2)(4)  8   R eq C  (10)(0.5)  5

v( t )  v()   v(0)  v() e - t  v( t )  8  1  e - t 5  V

v(1)  8  1  e -0.2   1.45 , For t > 1, v( t )  v()   v(1)  v() e -( t 1)  v( t )  1.45 e -( t 1) 5 V Thus,





 8 1  e -t 5 V , 0  t  1 v( t )   - ( t 1 ) 5 V, t 1  1.45 e

v()  0


For the capacitor voltage, v( t )  v()   v(0)  v() e- t  v(0)  0

For t > 0, we transform the current source to a voltage source as shown in Fig. (a). 1 k

1 k

+ 30 V

+ 

2 k

v

 (a)

2 (30)  15 V 2 11 R th  (1  1) || 2  1 k 1 1   R th C  10 3   10 -3  4 4 -4t v( t )  15  1  e  , t  0 v() 

We now obtain i x from v(t). Consider Fig. (b). i T 1 k

v

ix 30 mA

1 k

1/4 mF (b)

But

i x  30 mA  i T dv v C iT  dt R3

i T ( t )  7.5  1  e -4t  mA  i T ( t )  7.5  1  e -4t  mA

1  10 -3 (-15)(-4) e -4t A 4

Thus, i x ( t )  30  7.5  7.5 e -4t mA i x ( t )  7.5  3  e -4t  mA , t  0

2 k


Consider the circuit below. t=0

R

+ VS

+ 

i

L

v

 After the switch is closed, applying KVL gives di VS  Ri  L dt  VS  di  or L  -R  i   dt R di -R  dt i  VS R L

Integrating both sides,  V  i( t ) - R ln  i  S  I 0  t  R L  i  VS R  - t  ln   I0  VS R   or

i  VS R  e- t  I0  VS R i( t ) 

VS  VS  -t  e   I0  R  R which is the same as Eq. (7.60).

Chapter 7, Solution 52. Using Fig. 7.118, design a problem to help other students to better understand the step response of an RL circuit. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the circuit in Fig. 7.118, find i(t) for t > 0.


20 i()  2 A  2 A, 10 i( t )  i()   i(0)  i() e- t 

i(0) 

i( t )  2 A


25 5A 3 2 i( t )  i(0) e - t  After t = 0, L 4     2, i(0)  5 R 2 i( t )  5 e - t 2 u( t ) A i

(a)

Before t = 0,

(b)

Before t = 0, the inductor acts as a short circuit so that the 2  and 4  resistors are short-circuited. i( t )  6 A After t = 0, we have an RL circuit. L 3 i( t )  i(0) e - t  ,   R 2 -2t 3 u( t ) A i( t )  6 e


Before t = 0, i is obtained by current division or 4 (2)  1 A i( t )  44 After t = 0, i( t )  i()   i(0)  i() e- t  L  , Req  4  (4 || 12)  7  R eq



3.5 1  7 2

i(0)  1 ,

i ( ) 

(4 || 12) 3 6 (2)  (2)  4  (4 || 12) 43 7

6  6   1   e -2 t 7  7 1 i( t )   6  e - 2t  A 7 10 2A Before t = 0, i( t )  23 Req  3  (6 || 2)  4.5 After t = 0, i( t ) 

(b)

L 2 4   R eq 4.5 9 i(0)  2 To find i() , consider the circuit below, at t = when the inductor becomes a short circuit, 

v i 10 V 2

+ 

24 V 6

+ 

2H 3

10  v 24  v v v     v  9 i()   3 A and 2 6 3 3 i( t )  3  (2  3) e -9 t 4 i( t )  3  e - 9 t 4 A

Chapter 7, Solution 55. For t < 0, consider the circuit shown in Fig. (a).

0.5 H

io 3

i

io + 

24 V

0.5 H

+ 4i o

+ 

v

2



(a) 3i o  24  4i o  0   i o  24 v v( t )  4i o  96 V i   48 A 2

For t > 0, consider the circuit in Fig. (b). i( t )  i()   i(0)  i() e- t  i(0)  48 , i ( )  0 L 0.5 1 R th  2  ,     R th 2 4 i( t )  (48) e -4t v( t )  2 i( t )  96 e -4t u( t )V

8

20 V

+ v

+ 



(b)

2


R eq  6  20 || 5  10  ,



L  0.05 R

i( t )  i()   i(0)  i() e- t 

i(0) is found by applying nodal analysis to the following circuit. 5

vx

12 

2A

20 

i

6 + 0.5 H

v

 20  v x v x v x v x    5 12 20 6 v i(0)  x  2 A 6 2

  v x  12

Since 20 || 5  4 , 4 (4)  1.6 i()  46 i( t )  1.6  (2  1.6) e- t 0.05  1.6  0.4 e-20t di 1 v( t )  L  (0.4) (-20) e -20t dt 2 v( t )  - 4 e -20t V

+ 

20 V

Chapter 7, Solution 57. At t  0  , the circuit has reached steady state so that the inductors act like short circuits. 6

30 V

i

+ 

i1

i2

5

20 

20 30 30   3, i1  (3)  2.4 , 25 6  (5 || 20) 10 i 1 ( 0 )  2 .4 A , i 2 ( 0 )  0 .6 A i

i 2  0 .6

For t > 0, the switch is closed so that the energies in L1 and L 2 flow through the closed switch and become dissipated in the 5  and 20  resistors. L 2.5 1 i1 ( t )  i1 (0) e - t 1 , 1  1   R1 5 2 i1 ( t )  2.4e–2tu(t) A

i 2 ( t )  i 2 (0) e - t  2 ,

2 

L2 4 1   R 2 20 5 i 2 ( t )  600e–5tu(t) mA

Chapter 7, Solution 58. For t < 0,

v o (t)  0

For t > 0,

i(0)  10 ,

R th  1  3  4  ,

20 5 1 3 L 14 1    R th 4 16

i() 

i( t )  i()   i(0)  i() e- t  i( t )  5  1  e-16t  A

di 1  15  1  e-16t   (-16)(5) e-16t dt 4 -16t v o ( t )  15  5 e V vo (t )  3i  L

Chapter 7, Solution 59. Let i(t) be the current through the inductor. i(0)  0 For t < 0, vs  0 , For t > 0,

R eq  4  (6 || 3)  6 Ω and  

L 1.5   0.25 sec. R eq 6

At t = ∞, the inductor becomes a short and the current delivered by the 18 volts source is I s = 18/[6+(3||4)] = 18/7.714 = 2.333 amps. The voltage across the 4ohm resistor is equal to 18–6(2.333) = 18–14 = 4 volts. Therefore the current through the inductor is equal to i(∞) = 4/4 = 1 amp. i( t )  i()   i(0)  i() e- t  i( t )  1(1  e -4t ) amps. v o (t)  L

di  (1.5)(1)(-4)(-e- 4t ) dt v o ( t )  [6e–4t]u(t) volts.

Chapter 7, Solution 60. Let I be the inductor current. u(t)  0   i(0)  0 For t < 0, For t > 0,

R eq  5 || 20  4  ,

i()  4 i( t )  i()   i(0)  i() e- t  i( t )  4  1  e - t 2   - 1 di  (8)(-4)  e - t 2 2 dt v( t )  16 e -0.5t V v( t )  L



L 8  2 R eq 4

Chapter 7, Solution 61. The current source is transformed as shown below. 4

20u(-t) + 40u(t)

+ 

0.5 H

L 12 1 i(0)  5 ,   , R 4 8 i( t )  i()   i(0)  i() e - t  

i()  10

i( t )  (10 – 5e–8t)u(t) A

v( t )  L

di  1    (-5)(-8) e -8t dt  2 

v( t )  20e–8tu(t) V

Chapter 7, Solution 62. 

L 2  1 R eq 3 || 6

For 0 < t < 1, u ( t  1)  0 so that 1 i(0)  0 , i()  6 1 i( t )   1  e - t  6

1  1  e -1   0.1054 6 1 1 1 i()    3 6 2 i( t )  0.5  (0.1054  0.5) e-(t -1) i( t )  0.5  0.3946 e-(t -1)

For t > 1,

i(1) 

Thus,  1  1  e -t  A 0t1 i( t )   6  0.5  0.3946 e -(t -1) A t1


10 2 5

For t < 0,

u (- t )  1 ,

i(0) 

For t > 0,

u (-t)  0 ,

i()  0 L 0.5 1    R th 4 8

R th  5 || 20  4  ,

i( t )  i()   i(0)  i() e - t 

i( t )  2e–8tu(t) A

v( t )  L

di  1    (-8)(2) e -8t dt  2  v( t )  –8e–8tu(t) V

2e–8tu(t) A, –8e–8tu(t) V

Chapter 7, Solution 64 Determine the value of i L (t) and the total energy dissipated by the circuit from t = 0 sec to t = ∞ sec. The value of v in (t) is equal to [40–40u(t)] volts. 40 

v in (t)

+ 

40 

R eq

v 1 i L (t) 10 H

v Thev (t

+ 

i L (t)

10 H

Solution Step 1. Determine the Thevenin equivalent circuit to the left of the inductor. This means we need to find v oc (t) and i sc (t) which gives us v Thev (t) = v oc (t) and R eq = v oc (t)/i sc (t) (note, this only works for resistor networks in the time domain). This leads to the second circuit shown above. Now, with this circuit, we can use the generalized solution to a first order differential equation, i L (t) = Ae–(t–0)/τ+B where, t 0 = 0, τ = L/R, A+B = i L (0) and 0+B = i L (∞). Finally, we can use w = (1/2)Li L (t)2 to calculate the energy dissipated by the circuit (w = [(1/2)Li L (∞)2–(1/2)Li L (0)2]. Step 2. We now determine the Thevenin equivalent circuit. First we need to pick a reference node and mark the unknown voltages, as seen above. With the inductor out of the circuit, the node equation is simply [(v 1 –v in (t))/40] + [(v 1 –0)/40] + 0 (since the inductor is out of the circuit, there is an open circuit where it was) = 0. This leads to [(1/40)+(1/40)]v 1 = (1/40)v in (t) or 2v 1 = v in (t) or v 1 = 0.5v in (t) = [20–20u(t)] v oc (t) = v Thev (t). Now to short the open circuit which produces v 1 = 0 and i sc = –[(0–v in (t))/40] = v in (t)/40 = 0.025v in (t) A. Step 3. Now, everything comes together, R eq = v oc (t)/i sc (t) = 0.5v in (t)/[0.025v in (t)] = 0.5/0.025 = 20 Ω. Next we find τ = L/R eq = 10/20 = (1/2) sec. At t = 0–, v in (0–) = [10– 0] V (note u(t) = 0 until t = 0). Since it has been at this value for a very long time, the inductor can be considered a short and the value of the current is equal to 20/20 or i L (0–) = 1 amp. Since you cannot change the current instantaneously, i L (0) = 1 amp = A+B. Since v Thev (t) = 20–20 = 0 for all t > 0, all the energy in the inductor will be dissipated by the circuit and i L (∞) = 0 = B which means that A = 1 and i L (t) = [e–2t] u(t) amps. The total energy dissipated from t = 0 to ∞ sec is equal to [(1/2)Li L (0)2–(1/2)Li L (∞)2] = (0.5)10(1)2–0 = 5 J.

Chapter 7, Solution 65. Since v s  10  u ( t )  u ( t  1) , this is the same as saying that a 10 V source is turned on at t = 0 and a -10 V source is turned on later at t = 1. This is shown in the figure below.

vs 10 1 t -10 For 0 < t < 1, i(0)  0 ,

R th  5 || 20  4 ,

10 2 5 L 2 1    R th 4 2

i() 

i( t )  i()   i(0)  i() e- t  i( t )  2  1  e -2t  A i(1)  2  1  e-2   1.729 For t > 1,

i()  0

since vs  0

i( t )  i(1) e- ( t 1)  i( t )  1.729 e-2( t 1) A Thus,

 2  1  e - 2t  A 0  t  1 i( t )   t1  1.729 e - 2( t 1) A

Chapter 7, Solution 66. Using Fig. 7.131, design a problem to help other students to better understand first-order op amp circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the op-amp circuit of Fig. 7.131, find v o . Assume that v s changes abruptly from 0 to 1 V at t=0. Find v o . 50 k 0.5 F 20 k – + vs

+ vo

+ _

–

Figure 7.131 For Prob. 7.66. Solution For t<0-, v s =0 so that v o (0)=0 :Let v be the capacitor voltage For t>0, v s =1. At steady state, the capacitor acts like an open circuit so that we have an inverting amplifier v o (∞) = –(50k/20k)(1V) = –2.5 V τ = RC = 50x103x0.5x10–6 = 25 ms v o (t) = v o (∞) + (v o (0) – v o (∞))e–t/0.025 = 2.5(e–40t – 1) V.

Chapter 7, Solution 67. The op amp is a voltage follower so that v o  v as shown below. R

R

 +

vo

v1

vo

+ R

vo

C



At node 1, v o  v1 v1  0 v1  v o   R R R

  v1 

2 v 3 o

At the noninverting terminal, dv v  v1 C o  o 0 dt R dv 2 1  RC o  v o  v1  v o  v o  v o dt 3 3 dv o v  o dt 3RC v o ( t )  VT e - t 3RC VT  v o (0)  5 V ,

  3RC  (3)(10  103 )(1  10- 6 )  v o ( t )  5e–100t/3u(t) V

3 100

Chapter 7, Solution 68. This is a very interesting problem which has both an ideal solution as well as a realistic solution. Let us look at the ideal solution first. Just before the switch closes, the value of the voltage across the capacitor is zero which means that the voltage at both terminals input of the op amp are each zero. As soon as the switch closes, the output tries to go to a voltage such that both inputs to the op amp go to 4 volts. The ideal op amp puts out whatever current is necessary to reach this condition. An infinite (impulse) current is necessary if the voltage across the capacitor is to go to 8 volts in zero time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal of the op amp). So v o will be equal to 8 volts for all t > 0. What happens in a real circuit? Essentially, the output of the amplifier portion of the op amp goes to whatever its maximum value can be. Then this maximum voltage appears across the output resistance of the op amp and the capacitor that is in series with it. This results in an exponential rise in the capacitor voltage to the steady-state value of 8 volts. vC(t) = V op amp max (1 – e-t/(RoutC)) volts, for all values of vC less than 8 V, = 8 V when t is large enough so that the 8 V is reached.

Chapter 7, Solution 69. Let v x be the capacitor voltage. For t < 0, v x ( 0)  0 For t > 0, the 20 k and 100 k resistors are in series and together, they are in parallel with the capacitor since no current enters the op amp terminals. As t   , the capacitor acts like an open circuit so that 4 v o ( )  (20  100)  48 10 R th  20  100  120 k ,   R th C  (120  103 )(25  10-3 )  3000

v o ( t )  v o ()   v o (0)  v o () e - t 





v o ( t )  48 1  e - t 3000 V = 48(e–t/3000–1)u(t)V

Chapter 7, Solution 70. Let v = capacitor voltage. For t < 0, the switch is open and v(0)  0 .

For t > 0, the switch is closed and the circuit becomes as shown below. 1

+ 

2 vS

+

+ 

vo v



C R

v1  v 2  v s 0  vs dv C R dt  v o  v s  v where v  v s  v o  From (1), dv v s  0 dt RC - t vs -1 v v s dt  v(0)   RC RC Since v is constant, RC  (20  10 3 )(5  10 -6 )  0.1 - 20 t v mV  -200 t mV 0.1

From (3), v o  v s  v  20  200 t v o  20 ( 1  10t ) mV

(1) (2) (3)

Chapter 7, Solution 71. We temporarily remove the capacitor and find the Thevenin equivalent at its terminals. To find R Th , we consider the circuit below. Ro 20 k

R Th

Since we are assuming an ideal op amp, R o = 0 and R Th =20k . The op amp circuit is a noninverting amplifier. Hence, 10 )vs  2vs  6V 10 The Thevenin equivalent is shown below. 20 k VTh  (1

+ 6V

+ _

v

10 F

–

Thus, v(t)  6(1 e t /  ) , t  0 where   RTH C  20 x10 3 x10 x10 6  0.2 v(t)  6(1 e5t ), t  0 V

Chapter 7, Solution 72. The op amp acts as an emitter follower so that the Thevenin equivalent circuit is shown below. C +

3u(t)

Hence,

+ 

v



io R

v( t )  v()   v(0)  v() e - t  v(0)  -2 V , v()  3 V ,   RC  (10  10 3 )(10  10 -6 )  0.1 v( t )  3  (-2 - 3) e -10t  3  5 e -10t

dv  (10  10 -6 )(-5)(-10) e -10t dt i o  0.5 e -10t mA , t  0

io  C

Chapter 7, Solution 73. Consider the circuit below. Rf v1

R1

v2 +

v1

+ 

C v

v3 

 +

+ vo



At node 2, v1  v 2 dv C R1 dt At node 3, dv v 3  v o C  dt Rf But v 3  0 and v  v 2  v 3  v 2 . Hence, (1) becomes v1  v dv C dt R1 dv v 1  v  R 1C dt v1 dv v or   dt R 1C R 1C which is similar to Eq. (7.42). Hence,  vT t0 v( t )   -t  t0  v1   v T  v1  e where v T  v(0)  1 and v1  4

  R 1C  (10  10 3 )(20  10 -6 )  0.2  1 t0 v( t )   -5t t0 4  3 e From (2), dv  (20  10 3 )(20  10 -6 )(15 e -5t ) dt v o  -6 e -5t , t  0 v o  -R f C

(1)

(2)

v o  - 6 e -5t u(t ) V

Chapter 7, Solution 74. Let v = capacitor voltage. For t < 0, v(0)  0 10 kΩ

2 µF is

 +

50 kΩ is

+ vo



For t > 0,

i s  10 A .

Since the current through the feedback resistor is i s , then v o = –i s x104 volts = –10–5x104 = –100 mV. It is interesting to look at the capacitor voltage.

dv v  dt R v( t )  v()   v(0)  v() e - t 

is  C

It is evident that   RC  (2  10 6 )(50  10 3 )  0.1 At steady state, the capacitor acts like an open circuit so that i s passes through R. Hence, v()  i s R  (10  10 6 )(50  10 3 )  0.5 V Then the voltage across the capacitor is, v(t) = 500(1–e–10t) mV.

Chapter 7, Solution 75. Let v1 = voltage at the noninverting terminal. Let v 2 = voltage at the inverting terminal. For t > 0, v1  v 2  v s  4 0  vs  i o , R 1  20 k R1 vo  -ioR Also, i o 

i.e.

dv v C , dt R2

(1)

R 2  10 k , C  2 F

- vs v dv  C R1 R 2 dt

(2)

This is a step response. v( t )  v()   v(0)  v() e - t  , where   R 2 C  (10  10 3 )(2  10 -6 ) 

v(0)  1

1 50

At steady state, the capacitor acts like an open circuit so that i o passes through R 2 . Hence, as t   - vs v()  io  R1 R2 -R2 - 10 (4)  -2 i.e. vs  v()  20 R1

v( t )  -2  (1  2) e -50t v( t )  -2  3 e -50t But

v  vs  vo

or

v o  v s  v  4  2  3 e -50 t v o  6  3 e -50 t u( t )V - vs -4   -0.2 mA R 1 20k v dv io  C  - 0.2 mA R2 dt io 

or

Chapter 7, Solution 76. The schematic is shown below. For the pulse, we use IPWL and enter the corresponding values as attributes as shown. By selecting Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s since the width of the input pulse is 1 s. After saving and simulating the circuit, we select Trace/Add and display –V(C1:2). The plot of V(t) is shown below.

Chapter 7, Solution 77. The schematic is shown below. We click Marker and insert Mark Voltage Differential at the terminals of the capacitor to display V after simulation. The plot of V is shown below. Note from the plot that V(0) = 12 V and V() = -24 V which are correct.


(a)

When the switch is in position (a), the schematic is shown below. We insert IPROBE to display i. After simulation, we obtain, i(0) = 7.714 A

from the display of IPROBE.

(b) When the switch is in position (b), the schematic is as shown below. For inductor I1, we let IC = 7.714. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s. After Simulation, we click Trace/Add in the probe menu and display I(L1) as shown below. Note that i() = 12A, which is correct.


When the switch is in position 1, i o (0) = 12/3 = 4A. When the switch is in position 2, 4 L i o ( )    0.5 A, R Th  (3  5) // 4  8 / 3,   3 / 80 53 R Th

i o ( t )  i o ()  [i o (0)  i o ()]e  t /    0.5  4.5e 80 t / 3 u(t)A


(a) When the switch is in position A, the 5-ohm and 6-ohm resistors are short-circuited so that i 1 ( 0 )  i 2 ( 0)  v o ( 0)  0

but the current through the 4-H inductor is i L (0) =30/10 = 3A. (b) When the switch is in position B,

R Th  3 // 6  2,



L  4 / 2  2 sec R Th

i L ( t )  i L ( )  [i L (0)  i L ( )]e  t /   0  3e  t / 2  3e  t / 2 A

(c) i1 () 

30  2 A, 10  5 v o (t )  L

3 i 2 ( )   i L ( )  0 A 9 di L dt

 

v o ( )  0 V

Chapter 7, Solution 81. The schematic is shown below. We use VPWL for the pulse and specify the attributes as shown. In the Analysis/Setup/Transient menu, we select Print Step = 25 ms and final Step = 3 S. By inserting a current marker at one terminal of LI, we automatically obtain the plot of i after simulation as shown below.

2.0A

1.5A

1.0A

0.5A

0A 0s

0.5s

1.0s

1.5s

-I(L1) Time

2.0s

2.5s

3.0s

Chapter 7, Solution 82. 3  10 -3    RC   R    30  C 100  10 -6

Chapter 7, Solution 83. v()  120,

v(0)  0,

  RC  34 x10 6 x15 x10 6  510s

v(t )  v()  [v(0)  v()]e t /  Solving for t gives

 

85.6  120(1  e t / 510 )

t  510 ln 3.488  637.16 s

speed = 4000m/637.16s = 6.278m/s


Let I o be the final value of the current. Then

i(t )  I o (1  e  t /  ), 0.6 I o  I o (1  e  50 t )

  R / L  0.16 / 8  1 / 50  

t

1 1 ln  18.33 ms. 50 0.4


The light is on from 75 volts until 30 volts. During that time we essentially have a 120-ohm resistor in parallel with a 6-µF capacitor. v(0) = 75, v(∞) = 0, τ = 120x6x10-6 = 0.72 ms v(t 1 ) = 75 e  t1 /   30 which leads to t 1 = –0.72ln(0.4) ms = 659.7 µs of lamp on time.

(b)

  RC  (4  106 )(6  10-6 )  24 s

Since v( t )  v()   v(0)  v() e - t  v( t 1 )  v()   v(0)  v() e - t1  v( t 2 )  v()   v(0)  v() e- t 2  Dividing (1) by (2), v( t1 )  v()  e( t 2  t1 )  v( t 2 )  v()  v( t )  v()   t 0  t 2  t1   ln  1  v( t 2 )  v() 

 75  120    24 ln (2)  16.636 s t 0  24 ln   30  120 

(1) (2)

Chapter 7, Solution 86. v( t )  v()   v(0)  v() e- t  v()  12 , v(0)  0 -t  v( t )  12  1  e  v( t 0 )  8  12  1  e- t 0   8 1  1  e- t 0    e- t 0   12 3 t 0   ln (3)

For R  100 k ,   RC  (100  103 )(2  10-6 )  0.2 s t 0  0.2 ln (3)  0.2197 s For R  1 M ,   RC  (1  106 )(2  10-6 )  2 s t 0  2 ln (3)  2.197 s Thus, 0.2197 s  t 0  2.197 s

Chapter 7, Solution 87. Let i be the inductor current. For t < 0,

i (0  ) 

120  1.2 A 100

For t > 0, we have an RL circuit L 50    0.1 , i()  0 R 100  400 i( t )  i()   i(0)  i() e - t  i( t )  1.2 e -10t

At t = 100 ms = 0.1 s, i(0.1)  1.2 e -1  441mA which is the same as the current through the resistor.


(b)

  RC  (300  10 3 )(200  10 -12 )  60 s As a differentiator, T  10   600 s  0.6 ms i.e. Tmin  0.6 ms   RC  60 s As an integrator, T  0.1  6 s i.e. Tmax  6 s

Chapter 7, Solution 89. Since   0.1 T  1 s L  1 s R

L  R  10 -6  (200  10 3 )(1  10 -6 )

L  200 mH

Chapter 7, Solution 90. We determine the Thevenin equivalent circuit for the capacitor C s . Rs v th  v, R th  R s || R p Rs  Rp i

R th

V th

+ 

Cs

The Thevenin equivalent is an RC circuit. Since Rs 1 1 v th  v i    10 10 R s  R p Rs 

6 2 1 R p   M 9 3 9

Also,   R th C s  15 s 6 (2 3)  0.6 M where R th  R p || R s  62 3  15  10 -6   25 pF Cs  R th 0.6  10 6


12  240 mA , i()  0 50 i( t )  i()   i(0)  i() e - t  i( t )  240 e - t 

i o (0) 

L 2  R R i( t 0 )  10  240 e - t 0 



e t 0   24   t 0   ln (24) t0 2 5    1.573  R ln (24) ln (24) 2 R  1.271  1.573

Chapter 7, Solution 92.  10  -3 dv  4  10 -9   2  10 iC - 10 dt   5  10 -6

0  t  tR tR  t  tD

 20 A 0  t  2 ms i( t )   - 8 mA 2 ms  t  2 ms  5 s which is sketched below. i(t)

5 s

20 A t

2 ms

-8 mA (not to scale)


(a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a).

6 VS

+ 

6

6 +

+ vL

10 H



(a)

10 F

v 

(b) i(0-) = 12/6 = 2A, v(0-) = 12V At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V (b)

For t > 0, we have the equivalent circuit shown in Figure (b). v L = Ldi/dt or di/dt = v L /L

Applying KVL at t = 0+, we obtain, v L (0+) – v(0+) + 10i(0+) = 0 v L (0+) – 12 + 20 = 0, or v L (0+) = -8 Hence,

di(0+)/dt = -8/2 = -4 A/s

Similarly,

i C = Cdv/dt, or dv/dt = i C /C i C (0+) = -i(0+) = -2 dv(0+)/dt = -2/0.4 = -5 V/s

(c)

As t approaches infinity, the circuit reaches steady state. i() = 0 A, v() = 0 V

Chapter 8, Solution 2. Using Fig. 8.63, design a problem to help other students better understand finding initial and final values. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem In the circuit of Fig. 8.63, determine: (a) i R (0+), i L (0+), and i C (0+), (b) di R (0+)/dt, di L (0+)/dt, and di C (0+)/dt, (c) i R (), i L (), and i C ().

Figure 8.63

Solution (a)

At t = 0-, the equivalent circuit is shown in Figure (a). 25 k

20 k iR

80V

+ 

+ 60 k v  (a)

iL

25 k

20 k iL

iR + 

80V

(b) 60||20 = 15 kohms, i R (0-) = 80/(25 + 15) = 2mA. By the current division principle, i L (0-) = 60(2mA)/(60 + 20) = 1.5 mA v C (0-) = 0 At t = 0+, v C (0+) = v C (0-) = 0 i L (0+) = i L (0-) = 1.5 mA 80 = i R (0+)(25 + 20) + v C (0-) i R (0+) = 80/45k = 1.778 mA iR = iC + iL

But,

1.778 = i C (0+) + 1.5 or i C (0+) = 0.278 mA (b)

v L (0+) = v C (0+) = 0 But,

v L = Ldi L /dt and di L (0+)/dt = v L (0+)/L = 0 di L (0+)/dt = 0 Again, 80 = 45i R + v C 0

But,

= 45di R /dt + dv C /dt

dv C (0+)/dt = i C (0+)/C = 0.278 mamps/1 F = 278 V/s Hence,

di R (0+)/dt = (-1/45)dv C (0+)/dt = -278/45

di R (0+)/dt = -6.1778 A/s Also, i R = i C + i L di R (0+)/dt = di C (0+)/dt + di L (0+)/dt -6.1788 = di C (0+)/dt + 0, or di C (0+)/dt = -6.1788 A/s (c)

As t approaches infinity, we have the equivalent circuit in Figure (b). i R () = i L () = 80/45k = 1.778 mA i C () = Cdv()/dt = 0.

Chapter 8, Solution 3. At t = 0-, u(t) = 0. Consider the circuit shown in Figure (a). i L (0-) = 0, and v R (0-) = 0. But, -v R (0-) + v C (0-) + 10 = 0, or v C (0-) = -10V. (a)

At t = 0+, since the inductor current and capacitor voltage cannot change abruptly, the inductor current must still be equal to 0A, the capacitor has a voltage equal to –10V. Since it is in series with the +10V source, together they represent a direct short at t = 0+. This means that the entire 2A from the current source flows through the capacitor and not the resistor. Therefore, v R (0+) = 0 V.

(b)

At t = 0+, v L (0+) = 0, therefore Ldi L (0+)/dt = v L (0+) = 0, thus, di L /dt = 0A/s, i C (0+) = 2 A, this means that dv C (0+)/dt = 2/C = 8 V/s. Now for the value of dv R (0+)/dt. Since v R = v C + 10, then dv R (0+)/dt = dv C (0+)/dt + 0 = 8 V/s.

40 

40  +

+ vC

+ vR

10 

+





2A

vR + 

 10V

iL

vC 

10 

+ 

10V

(b)

(a)

As t approaches infinity, we end up with the equivalent circuit shown in (c) Figure (b). i L () = 10(2)/(40 + 10) = 400 mA v C () = 2[10||40] –10 = 16 – 10 = 6V v R () = 2[10||40] = 16 V


At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in Figure (a). i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V. i(0+) = i(0-) = 5A

Hence,

v(0+) = v(0-) = 25V 3 i 40V

+

+ 

v

5 

(a) 3 i 40V

0.25 H + vL  iC

+ 

0.1F

iR

4A 5

(b) (b)

i C = Cdv/dt or dv(0+)/dt = i C (0+)/C

For t = 0+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b). Since i and v cannot change abruptly, i R = v/5 = 25/5 = 5A, i(0+) + 4 =i C (0+) + i R (0+) 5 + 4 = i C (0+) + 5 which leads to i C (0+) = 4 dv(0+)/dt = 4/0.1 = 40 V/s

Similarly,

v L = Ldi/dt which leads to di(0+)/dt = v L (0+)/L 3i(0+) + v L (0+) + v(0+) = 0

15 + v L (0+) + 25 = 0 or v L (0+) = -40 di(0+)/dt = -40/0.25 = –160 A/s (c)

As t approaches infinity, we have the equivalent circuit in Figure (c). 3 i

+

4A

v

5

 (c) i() = -5(4)/(3 + 5) = –2.5 A v() = 5(4 – 2.5) = 7.5 V


For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0). i L (0-) = 0 and v C (0-) = 0. For t = 0+, 4u(t) = 4. Consider the circuit below. A

iL

i 4A

+ 4  vC

1H

iC +

vL 

0.25F

+

6



v 

Since the 4-ohm resistor is in parallel with the capacitor, i(0+) = v C (0+)/4 = 0/4 = 0 A Also, since the 6-ohm resistor is in series with the inductor, v(0+) = 6i L (0+) = 0V.

(b)

di(0+)/dt = d(v R (0+)/R)/dt = (1/R)dv R (0+)/dt = (1/R)dv C (0+)/dt = (1/4)4/0.25 A/s = 4 A/s

v = 6i L or dv/dt = 6di L /dt and dv(0+)/dt = 6di L (0+)/dt = 6v L (0+)/L = 0 Therefore dv(0+)/dt = 0 V/s (c)

As t approaches infinity, the circuit is in steady-state. i() = 6(4)/10 = 2.4 A v() = 6(4 – 2.4) = 9.6 V


Let i = the inductor current. For t < 0, u(t) = 0 so that i(0) = 0 and v(0) = 0. For t > 0, u(t) = 1. Since, v(0+) = v(0-) = 0, and i(0+) = i(0-) = 0. v R (0+) = Ri(0+) = 0 V Also, since v(0+) = v R (0+) + v L (0+) = 0 = 0 + v L (0+) or v L (0+) = 0 V. (1)

(b)

Since i(0+) = 0, But,

i C (0+) = V S /R S

i C = Cdv/dt which leads to dv(0+)/dt = V S /(CR S ) (2)

From (1), (3)

dv(0+)/dt = dv R (0+)/dt + dv L (0+)/dt v R = iR or dv R /dt = Rdi/dt

(4) But,

v L = Ldi/dt, v L (0+) = 0 = Ldi(0+)/dt and di(0+)/dt = 0

(5)

From (4) and (5),

dv R (0+)/dt = 0 V/s

From (2) and (3),

dv L (0+)/dt = dv(0+)/dt = V s /(CR s )

(c) As t approaches infinity, the capacitor acts like an open circuit, while the inductor acts like a short circuit. v R () = [R/(R + R s )]V s v L () = 0 V

Chapter 8, Solution 7. α = [R/(2L)] = 20x103/(2x0.2x10–3) = 50x106 ω o = [1/(LC)0.5] = 1/(0.2x10–3x5x10–6)0.5 = 3.162 x104

  o

overdamped

 

overdamped

Chapter 8, Solution 8. Design a problem to help other students better understand source-free RLC circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem The branch current in an RLC circuit is described by the differential equation d 2i di  6  9i  0 2 dt dt and the initial conditions are i(0) = 0, di(0)/dt = 4. Obtain the characteristic equation and determine i(t) for t > 0. Solution s2 + 6s + 9 = 0, thus s 1,2 =

 6  6 2  36 = -3, repeated roots. 2

i(t) = [(A + Bt)e-3t], i(0) = 0 = A di/dt = [Be-3t] + [-3(Bt)e-3t] di(0)/dt = 4 = B. Therefore, i(t) = [4te-3t] A

Chapter 8, Solution 9. s2 + 10s + 25 = 0, thus s 1,2 =

 10  10  10 = –5, repeated roots. 2

i(t) = [(A + Bt)e-5t], i(0) = 10 = A di/dt = [Be-5t] + [-5(A + Bt)e-5t] di(0)/dt = 0 = B – 5A = B – 50 or B = 50. Therefore, i(t) = [(10 + 50t)e-5t] A


 5  25  16 = –4, –1. 2

v(t) = (Ae-4t + Be-t), v(0) = 0 = A + B, or B = -A dv/dt = (-4Ae-4t - Be-t) dv(0)/dt = 10 = – 4A – B = –3A or A = –10/3 and B = 10/3. Therefore, v(t) = (–(10/3)e-4t + (10/3)e-t) V


2 44 = –1, repeated roots. 2

v(t) = [(A + Bt)e-t], v(0) = 10 = A dv/dt = [Be-t] + [-(A + Bt)e-t] dv(0)/dt = 0 = B – A = B – 10 or B = 10. Therefore, v(t) = [(10 + 10t)e-t] V


(a)

Overdamped when C > 4L/(R2) = 4x1.5/2500 = 2.4x10-3, or C > 2.4 mF

(b)

Critically damped when C = 2.4 mF

(c)

Underdamped when C < 2.4 mF

Chapter 8, Solution 13. Let R||60 = R o . For a series RLC circuit, o =

1 LC

=

1 0.01x 4

= 5

For critical damping,  o =  = R o /(2L) = 5 or R o = 10L = 40 = 60R/(60 + R) which leads to R = 120 ohms

Chapter 8, Solution 14. When the switch is in position A, v(0-)= 0 and i L (0) = 80/40 = 2 A. When the switch is in position B, we have a source-free series RCL circuit. R 10    1.25 2L 2 x4 1 1 o   1 LC 1 x4 4 When the switch is in position A, v(0–)= 0. When the switch is in position B, we have a source-free series RCL circuit. R 10    1.25 2L 2 x4 1 1 o   1 LC 1 x4 4 Since   o , we have overdamped case.

–2 64, 0.9336 s1,2     2  o2  1.25  1.5625–0.5  1 and 1.56 v(t) = Ae–2t + Be–0.5t

(1)

v(0) = 0 = A + B

(2)

i C (0) = C(dv(0)/dt) = –2 or dv(0)/dt = –2/C = –8. But

dv( t )  2Ae  2 t  0.5Be  0.5t dt dv(0)  2 A  0.5B  8 dt

Solving (2) and (3) gives A= 1.3333 and B = –1.3333 v(t) = 5.333e–2t–5.333e–0.5t V.

(3)

Chapter 8, Solution 15. Given that s 1 = -10 and s 2 = -20, we recall that s 1,2 =     2  o2 = -10, -20 Clearly,

s 1 + s 2 = -2 = -30 or  = 15 = R/(2L) or R = 60L

(1)

s 1 =  15  15 2   o2 = -10 which leads to 152 –  o 2 = 25 or  o =

225  25 =

200  1

LC , thus LC = 1/200

Since we have a series RLC circuit, i L = i C = Cdv C /dt which gives, i L /C = dv C /dt = [200e-20t – 300e-30t] or i L = 100C[2e-20t – 3e-30t] But,

i is also = 20{[2e-20t – 3e-30t]x10-3} = 100C[2e-20t – 3e-30t] Therefore,

C = (0.02/102) = 200 F

L = 1/(200C) = 25 H R = 30L = 750 ohms

(2)

Chapter 8, Solution 16. At t = 0, i(0) = 0, v C (0) = 40x30/50 = 24V For t > 0, we have a source-free RLC circuit.  = R/(2L) = (40 + 60)/5 = 20 and  o =

1 LC

=

1 10 3 x 2.5

 o =  leads to critical damping i(t) = [(A + Bt)e-20t], i(0) = 0 = A di/dt = {[Be-20t] + [-20(Bt)e-20t]}, but di(0)/dt = -(1/L)[Ri(0) + v C (0)] = -(1/2.5)[0 + 24] Hence,

B = -9.6 or i(t) = [–9.6te–20t] A

= 20


i (0)  I 0  0, v(0)  V0  4 x5  20 di (0) 1   ( RI 0  V0 )  4(0  20)  80 dt L 1 1 o    10 LC 1 1 4 25 R 10    20, which is  o . 2L 2 1 4

s     2  o2  20  300  20  10 3  2.679,  37.32 i (t )  A1e  2.679t  A2 e 37.32t di (0)  2.679 A1  37.32 A2  80 dt This leads to A1  2.309   A2 i (0)  0  A1  A2 ,



i (t )  2.309 e 37.32t  e  2.679t Since, v(t ) 



1 t i (t )dt  20, we get C 0 v(t) = [21.55e-2.679t – 1.55e-37.32t] V

Chapter 8, Solution 18. When the switch is off, we have a source-free parallel RLC circuit.

o    o

1 LC



1 0.25 x1

 

 2,



1  0.5 2 RC

underdamped case  d   o   2  4  0.25  1.936 2

I o (0) = i(0) = initial inductor current = 100/5 = 20 A V o (0) = v(0) = initial capacitor voltage = 0 V

v(t )  e t ( A1 cos(d t )  A2 sin(d t ))  e 0.5t ( A1 cos(1.936t )  A2 sin(1.936t )) v(0) =0 = A 1

dv  e 0.5t (0.5)( A1 cos(1.936t )  A2 sin(1.936t ))  e 0.5t (1.936 A1 sin(1.936t )  1.936 A2 cos(1.936t )) dt (V  RI o ) (0  20) dv(0)  o   20  0.5 A1  1.936 A2 1 dt RC Thus, v ( t )  [10.333e 0.5 t sin(1.936t )]volts

 

A2  10.333

Chapter 8, Solution 19. For t < 0, the equivalent circuit is shown in Figure (a). 10 

i +

120V

+ 

i

+ v

L

v

C



 (b)

(a) i(0) = 120/10 = 12, v(0) = 0

For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = . o =

1 LC

=

1 4

= 0.5 =  d

i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t], which leads to -v(0)/L = 0 = B Hence,

i(t) = 12cos0.5t A and v = 0.5

However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin(0.5t) V

Chapter 8, Solution 20. For t < 0, the equivalent circuit is as shown below. 2

i

30 + 



vC

+

v(0) = –30 V and i(0) = 30/2 = 15 A For t > 0, we have a series RLC circuit.  = R/(2L) = 2/(2x0.5) = 2  o = 1/ LC  1 / 0.5x 1 4  2 2 Since  is less than  o , we have an under-damped response. d  o2   2  8  4  2

i(t) = (Acos(2t) + Bsin(2t))e-2t i(0) = 15 = A di/dt = –2(15cos(2t) + Bsin(2t))e-2t + (–2x15sin(2t) + 2Bcos(2t))e-t di(0)/dt = –30 + 2B = –(1/L)[Ri(0) + v C (0)] = –2[30 – 30] = 0 Thus, B = 15 and i(t) = (15cos(2t) + 15sin(2t))e-2t A

Chapter 8, Solution 21. By combining some resistors, the circuit is equivalent to that shown below. 60||(15 + 25) = 24 ohms. 12 

6

t=0

i

3H 24V

+ 

24 

+ (1/27)F

v 

At t = 0-,

i(0) = 0, v(0) = 24x24/36 = 16V

For t > 0, we have a series RLC circuit.

R = 30 ohms, L = 3 H, C = (1/27) F

 = R/(2L) = 30/6 = 5  o  1 / LC  1 / 3x1 / 27 = 3, clearly  >  o (overdamped response)

s 1,2 =     2  o2  5  5 2  3 2 = -9, -1 v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B

(1)

i = Cdv/dt = C[-Ae-t - 9Be-9t] i(0) = 0 = C[-A – 9B] or A = -9B From (1) and (2),

B = -2 and A = 18. Hence,

v(t) = (18e-t – 2e-9t) V

(2)

Chapter 8, Solution 22. Compare the characteristic equation with eq. (8.8), i.e. R 1 s2  s  0 L LC we obtain R R 2000  100   L   20H L 100 100 1  106 LC

 C

1 10 6   50 nF 20 106 L

Chapter 8, Solution 23. Let C o = C + 0.01. For a parallel RLC circuit,  = 1/(2RC o ),  o = 1/ LC o  = 1 = 1/(2RC o ), we then have C o = 1/(2R) = 1/20 = 50 mF  o = 1/ 0.02 x0.05 = 141.42 >  (underdamped) C o = C + 10 mF = 50 mF or C = 40 mF

Chapter 8, Solution 24. When the switch is in position A, the inductor acts like a short circuit so i(0  )  4 When the switch is in position B, we have a source-free parallel RCL circuit 1 1   5 2RC 2 x10 x10 x10 3 1 1 o    20 LC 1 x10 x10 3 4 Since   o , we have an underdamped case. s1,2  5  25  400  5  j19.365

i(t)  e5t  A1 cos19.365t  A2 sin19.365t  i(0)  4  A1 di vL   dt

di(0) v(0)  0 dt L

di  e5t  5 A1 cos19.365t  5 A2 sin19.365t  19.365 A1 sin19.365t  19.365 A2 cos19.365t  dt

0 = [di(0)/dt] = –5A 1 + 19.365A 2 or A 2 = 20/19.365 = 1.0328 i(t) = e–5t[4cos(19.365t) + 1.0328sin(19.365t)] A

Chapter 8, Solution 25. Using Fig. 8.78, design a problem to help other students to better understand source-free RLC circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem In the circuit in Fig. 8.78, calculate i o (t) and v o (t) for t > 0.

Figure 8.78 Solution In the circuit in Fig. 8.76, calculate i o (t) and v o (t) for t>0. 2

30V

+ 

1H

i o (t +

t=0, note this is a make before break switch so the inductor current is not interrupted.

Figure 8.78

8

For Problem 8.25.

At t = 0-, v o (0) = (8/(2 + 8)(30) = 24 For t > 0, we have a source-free parallel RLC circuit.  = 1/(2RC) = ¼  o = 1/ LC  1 / 1x 1 4  2

(1/4)F

v o (t) 

Since  is less than  o , we have an under-damped response. d  o2   2  4  (1 / 16)  1.9843

v o (t) = (A 1 cos d t + A 2 sin d t)e-t v o (0) = 30(8/(2+8)) = 24 = A 1 and i o (t) = C(dv o /dt) = 0 when t = 0. dv o /dt = -(A 1 cos d t + A 2 sin d t)e-t + (- d A 1 sin d t +  d A 2 cos d t)e-t at t = 0, we get dv o (0)/dt = 0 = -A 1 +  d A 2 Thus, A 2 = (/ d )A 1 = (1/4)(24)/1.9843 = 3.024 v o (t) = (24cos1.9843t + 3.024sin1.9843t)e-t/4 volts. i 0 (t) = Cdv/dt = 0.25[–24(1.9843)sin1.9843t + 3.024(1.9843)cos1.9843t – 0.25(24cos1.9843t) – 0.25(3.024sin1.9843t)]e–t/4 = [– 12.095sin1.9843t]e–t/4 A.

Chapter 8, Solution 26. s2 + 2s + 5 = 0, which leads to s 1,2 =

 2  4  20 = –1j4 2

These roots indicate an underdamped circuit which has the generalized solution given as: i(t) = I s + [(A 1 cos(4t) + A 2 sin(4t))e-t], At t = ∞, (di(t)/dt) = 0 and (d2i(t)/dt2) = 0 so that I s = 10/5 = 2 (from (d2i(t)/dt2)+2(di(t)/dt)+5 = 10) i(0) = 2 = 2 + A 1 , or A 1 = 0 di/dt = [(4A 2 cos(4t))e-t] + [(-A 2 sin(4t))e-t] = 4 = 4A 2 , or A 2 = 1 i(t) = [2 + sin(4te-t)] amps

Chapter 8, Solution 27. s2 + 4s + 8 = 0 leads to s =

 4  16  32  2  j2 2

v(t) = V s + (A 1 cos2t + A 2 sin2t)e-2t 8V s = 24 means that V s = 3 v(0) = 0 = 3 + A 1 leads to A 1 = -3 dv/dt = -2(A 1 cos2t + A 2 sin2t)e-2t + (-2A 1 sin2t + 2A 2 cos2t)e-2t 0 = dv(0)/dt = -2A 1 +2A 2 or A 2 = A 1 = -3 v(t) = [3 – 3(cos(2t) + sin(2t))e–2t] volts.

Chapter 8, Solution 28. The characteristic equation is 1 Ls2  Rs   0   C

s1,2 

1 2 1 s  4s  0 2 0.2

  s2  8 s  10  0

8  64  40 –6.45 and –1.5505  0.838, 7.162 2

i( t )  i s  Ae 6.45t  Be 1.5505t But

[i s /C] = 10 or i s = 0.2x10 = 2 i (t )  2  Ae 6.45t  Be 1.5505t

i(0) = 1 = 2 + A + B or A + B = –1 or A = –1–B

(1)

di (t )  6.45 Ae 6.45t  1.5505Be 1.5505t dt di (0) but  0  6.45 A  1.5505B dt

(2)

Solving (1) and (2) gives –6.45(–1–B) – 1.5505B = 0 or (6.45–1.5505)B = –6.45 B = –6.45/(4.9) = –1.3163 and A = –1–1.3163 = –2.3163

A= –2.3163, B= –1.3163 Hence, i(t) = [2–2.3163e–6.45t –1.3163e–1.5505t] A.


s2 + 4 = 0 which leads to s 1,2 = j2 (an undamped circuit) v(t) = V s + Acos2t + Bsin2t 4V s = 12 or V s = 3 v(0) = 0 = 3 + A or A = -3 dv/dt = -2Asin2t + 2Bcos2t dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 – 3cos2t + sin2t) V

(b)

s2 + 5s + 4 = 0 which leads to s 1,2 = -1, -4 i(t) = (I s + Ae-t + Be-4t) 4I s = 8 or I s = 2 i(0) = -1 = 2 + A + B, or A + B = -3

(1)

di/dt = -Ae-t - 4Be-4t di(0)/dt = 0 = -A – 4B, or B = -A/4 From (1) and (2) we get A = -4 and B = 1 i(t) = (2 – 4e-t + e-4t) A (c)

s2 + 2s + 1 = 0, s 1,2 = -1, -1 v(t) = [V s + (A + Bt)e-t], V s = 3. v(0) = 5 = 3 + A or A = 2 dv/dt = [-(A + Bt)e-t] + [Be-t] dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3 v(t) = [3 + (2 + 3t)e-t] V

(2)

(d)

s2 + 2s +5 = 0, s 1,2 = -1 + j2, -1 – j2 i(t) = [I s + (Acos2t + Bsin2t)e-t], where 5I s = 10 or I s = 2 i(0) = 4 = 2 + A or A = 2 di/dt = [-(Acos2t + Bsin2t)e-t] + [(-2Asin2t + 2Bcos2t)e-t] di(0)/dt = -2 = -A + 2B or B = 0 i(t) = [2 + (2cos2t)e-t] A

Chapter 8, Solution 30. The step responses of a series RLC circuit are v C (t) = [40–10e–2000t–10e–4000t] volts, t > 0 and i L (t) = [3e–2000t+6e–4000t ] m A, t > 0. (a) Find C. (b) Determine what type of damping exhibited by the circuit. Solution Step 1.

For a series RLC circuit, i R (t) = i L (t) = i C (t). We can determine C from i C (t) = i L (t) = C(dv C /dt) and we can determine that the circuit is overdamped since the exponent value are real and negative.

Step 2.

C(dv C /dt) = C[20,000e–2000t+40,000e–4000t] = 0.003e–2000t+0.006e–4000t or C = 0.003/20,000 = 150 F.

Chapter 8, Solution 31. For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent circuit is shown in Figure (b). By KVL, v(0+) = v(0-) = 40, i(0+) = i(0-) = 1 By KCL, 2 = i(0+) + i 1 = 1 + i 1 which leads to i 1 = 1. By KVL, -v L + 40i 1 + v(0+) = 0 which leads to v L (0+) = 40x1 + 40 = 80 v L (0+) = 80 V, 40 

i 1 40 

10  +

i

v C (0+) = 40 V

+

+

v

50V



+ 

(a)

v

vL 

10 

50V



0.5H

(b)

+ 

Chapter 8, Solution 32. For t = 0-, the equivalent circuit is shown below. 2A

i +

v



6 i(0-) = 0, v(0-) = -2x6 = -12V For t > 0, we have a series RLC circuit with a step input.  = R/(2L) = 6/2 = 3,  o = 1/ LC  1 / 0.04 s =  3  9  25  3  j4

Thus, v(t) = V f + [(Acos4t + Bsin4t)e-3t] where V f = final capacitor voltage = 50 V v(t) = 50 + [(Acos4t + Bsin4t)e-3t] v(0) = -12 = 50 + A which gives A = -62 i(0) = 0 = Cdv(0)/dt dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(–Asin4t + Bcos4t)e–3t] 0 = dv(0)/dt = -3A + 4B or B = (3/4)A = –46.5 v(t) = {50 + [(–62cos4t – 46.5sin4t)e–3t]} V

Chapter 8, Solution 33. We may transform the current sources to voltage sources. For t = 0-, the equivalent circuit is shown in Figure (a). 10 

i

i 1H

+ 30V

+ 

v

+ 5



5

v

20V

4F

+ 



(a)

(b)

i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V For t > 0, we have a series RLC circuit, shown in (b).  = R/(2L) = 5/2 = 2.5  o  1 / LC  1 / 4 = 0.5, clearly  >  o (overdamped response)

s 1,2 =     2   2o  2.5  6.25  0.25 = –4.949, –0.0505 v(t) = V s + [A 1 e–4.949t + A 2 e–0.0505t], V s = 20. v(0) = 10 = 20 + A 1 + A 2 or A 2 = –10 – A 1 (1) i(0) = Cdv(0)/dt or dv(0)/dt = –2/4 = –1/2 Hence,

–0.5 = – 4.949A 1 – 0.0505A 2

From (1) and (2),

–0.5 = –4.949A 1 + 0.0505(10 + A 1 ) or –4.898A 1 = –0.5–0.505 = –1.005 A 1 = 0.2052, A 2 = –10.205 v(t) = [20 + 0.2052e–4.949t – 10.205e-0.05t] V.

(2)

Chapter 8, Solution 34. Before t = 0, the capacitor acts like an open circuit while the inductor behaves like a short circuit. i(0) = 0, v(0) = 50 V For t > 0, the LC circuit is disconnected from the voltage source as shown below. +

Vx

 i

(1/16)F

(¼) H

This is a lossless, source-free, series RLC circuit.  = R/(2L) = 0,  o = 1/ LC = 1/

1 1  = 8, s = j8 16 4

Since  is equal to zero, we have an undamped response. Therefore, i(t) = A 1 cos(8t) + A 2 sin(8t) where i(0) = 0 = A 1 di(0)/dt = (1/L)v L (0) = –(1/L)v(0) = –4x50 = –200 However, di/dt = 8A 2 cos(8t), thus, di(0)/dt = –200 = 8A 2 which leads to A 2 = –25 Now we have

i(t) = –25sin(8t) A

Chapter 8, Solution 35. Using Fig. 8.83, design a problem to help other students to better understand the step response of series RLC circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine v(t) for t > 0 in the circuit in Fig. 8.83.

Figure 8.83 Solution At t = 0-, i L (0) = 0, v(0) = v C (0) = 8 V For t > 0, we have a series RLC circuit with a step input.  = R/(2L) = 2/2 = 1,  o = 1/ LC = 1/ 1 / 5 =

5

s 1,2 =     2   2o  1  j2 v(t) = V s + [(Acos2t + Bsin2t)e-t], V s = 12. v(0) = 8 = 12 + A or A = -4, i(0) = Cdv(0)/dt = 0. But dv/dt = [-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t] 0

= dv(0)/dt = -A + 2B or 2B = A = -4 and B = -2 v(t) = {12 – (4cos2t + 2sin2t)e-t V.

Chapter 8, Solution 36. For t = 0–, 3u(t) = 0. Thus, i(0) = 0, and v(0) = 20 V. For t > 0, we have the series RLC circuit shown below. 10 

i

10 

5H

+ 30V

+ 

2

40 V

0.2 F

+ 

v 

 = R/(2L) = (2 + 5 + 1)/(2x5) = 0.8  o = 1/ LC = 1/ 5x 0.2 = 1 s 1,2 =     2  2o  0.8  j0.6 v(t) = V s + [(Acos(0.6t) + Bsin(0.6t))e-0.8t] V s = 30 + 40 = 70 V and v(0) = 40 = 70 + A or A = –30 i(0) = Cdv(0)/dt = 0 But dv/dt = [–0.8(Acos(0.6t) + Bsin(0.6t))e–0.8t] + [0.6(–Asin(0.6t) + Bcos(0.6t))e–0.8t] 0

= dv(0)/dt = –0.8A + 0.6B which leads to B = 0.8x(–30)/0.6 = –40 v(t) = {70 – [(30cos(0.6t) + 40sin(0.6t))e–0.8t]} V i = Cdv/dt

= 0.2{[0.8(30cos(0.6t) + 40sin(0.6)t)e-0.8t] + [0.6(30sin(0.6t) – 40cos(0.6t))e-0.8t]} i(t) = 10sin(0.6t)e-0.8t A

Chapter 8, Solution 37. For t = 0–, the equivalent circuit is shown below.

+

i2

6

6

6 v(0) 45V

+ 

i1

15V

+ 



18i 2 – 6i 1 = 0 or i 1 = 3i 2 –45 + 6(i 1 – i 2 ) + 15 = 0 or i 1 – i 2 = 30/6 (2) From (1) and (2), (2/3)i 1 = 5 or i 1 = 7.5 and i 2 = i 1 – 5 = 2.5 i(0) = i 1 = 7.5A –15 – 6i 2 + v(0) = 0 v(0) = 15 + 6x2.5 = 30 For t > 0, we have a series RLC circuit. R = 6||12 = 4  o = 1/ LC = 1/ (1 / 2)(1 / 8) = 4  = R/(2L) = (4)/(2x(1/2)) = 4  =  o , therefore the circuit is critically damped v(t) = V s +[(A + Bt)e-4t], and V s = 15

(1) =5

v(0) = 30 = 15 + A, or A = 15 i C = Cdv/dt = C[–4(15 + Bt)e-4t] + C[(B)e-4t] To find i C (0) we need to look at the circuit right after the switch is opened. At this time, the current through the inductor forces that part of the circuit to act like a current source and the capacitor acts like a voltage source. This produces the circuit shown below. Clearly, i C (0+) must equal –i L (0) = –7.5A.

6

6

iC

6 30V

+ 

7.5A

i C (0) = –7.5 = C(–60 + B) which leads to –60 = –60 + B or B = 0 i C = Cdv/dt = (1/8)[–4(15 + 0t)e-4t] + (1/8)[(0)e-4t] i C (t) = [–(1/2)(15)e-4t] i(t) = –i C (t) = 7.5e-4t A

Chapter 8, Solution 38. At t = 0—, the equivalent circuit is as shown. 2A + i 10  v

i1 5 10 



i(0) = 2A, i 1 (0) = 10(2)/(10 + 15) = 0.8 A v(0) = 5i 1 (0) = 4V For t > 0, we have a source-free series RLC circuit. R = 5||(10 + 10) = 4 ohms  o = 1/ LC = 1/ (1 / 3)(3 / 4) = 2  = R/(2L) = (4)/(2x(3/4)) = 8/3 s 1,2 =     2   2o  -4.431, -0.903 i(t) = [Ae-4.431t + Be-0.903t] i(0) = A + B = 2

(1)

di(0)/dt = (1/L)[-Ri(0) + v(0)] = (4/3)(-4x2 + 4) = -16/3 = -5.333 Hence, -5.333 = -4.431A – 0.903B From (1) and (2), A = 1 and B = 1. i(t) = [e-4.431t + e-0.903t] A

(2)

Chapter 8, Solution 39. For t = 0–, the source voltages are equal to zero thus, the initial conditions are v(0) = 0 and i L (0) = 0. 30  0.5F 0.25H

60u(t)V

+ v  20 

+ 

+ 

30u(t)V

For t > 0, the circuit is shown above. R = 20||30 = 12 ohms  o = 1/ LC = 1/ (1 / 2)(1 / 4) =

8

 = R/(2L) = (12)/(0.5) = 24 Since  >  o , we have an overdamped response. s 1,2 =     2   2o  –47.83, –0.167 Thus,

v(t) = V s + [Ae–47.83t + Be–0.167t], where V s = [60/(30+20)]20–30 = –6 volts. v(0) = 0 = –6 + A + B or 6 = A + B

(1)

i(0) = Cdv(0)/dt = 0 But,

dv(0)/dt = –47.83A – 0.167B = 0 or

B = –286.4A From (1) and (2),

A + (–286.4)A = 6 or A = 6/(–285.4) = –0.02102 and B = –286.4x(–0.02102) = 6.02 v(t) = [–6 + (–0.021e-47.83t+6.02e-0.167t)] volts.

(2)

Chapter 8, Solution 40. At t = 0-, v C (0) = 0 and i L (0) = i(0) = (6/(6 + 2))4 = 3A For t > 0, we have a series RLC circuit with a step input as shown below. i

0.02 F

2H +

6

v

14  

24V

12V

+ 

+ 

 o = 1/ LC = 1/ 2x 0.02 = 5  = R/(2L) = (6 + 14)/(2x2) = 5 Since  =  o , we have a critically damped response. v(t) = V s + [(A + Bt)e-5t], V s = 24 – 12 = 12V v(0) = 0 = 12 + A or A = -12 i = Cdv/dt = C{[Be-5t] + [-5(A + Bt)e-5t]} i(0) = 3 = C[-5A + B] = 0.02[60 + B] or B = 90 Thus, i(t) = 0.02{[90e-5t] + [-5(-12 + 90t)e-5t]} i(t) = {(3 – 9t)e-5t} A

Chapter 8, Solution 41. At t = 0–, the switch is open. i(0) = 0, and v(0) = 5x100/(20 + 5 + 5) = 50/3 For t > 0, we have a series RLC circuit shown in Figure (a). After source transformation, it becomes that shown in Figure (b). 10 H 4

1H i

20 

5A

5

10 F

+ 20V

+ 

0.04F

v 

(a)

(b)

 o = 1/ LC = 1/ 1x1 / 25 = 5  = R/(2L) = (4)/(2x1) = 2 s 1,2 =     2   2o  -2  j4.583 Thus,

v(t) = V s + [(Acos( d t) + Bsin( d t))e-2t], where  d = 4.583 and V s = 20 v(0) = 50/3 = 20 + A or A = -10/3 i(t) = Cdv/dt = C(-2) [(Acos( d t) + Bsin( d t))e-2t] + C d [(-Asin( d t) + Bcos( d t))e-2t] i(0) = 0 = -2A +  d B B = 2A/ d = -20/(3x4.583) = -1.455 i(t) = C{[(0cos( d t) + (-2B -  d A)sin( d t))]e-2t} = (1/25){[(2.91 + 15.2767) sin( d t))]e-2t} i(t) = 727.5sin(4.583t)e-2t mA

Chapter 8, Solution 42. For t = 0-, we have the equivalent circuit as shown in Figure (a). i(0) = i(0) = 0, and v(0) = 4 – 12 = -8V 4V

5

 +

+ 

1

6

12V

i

1H

+ v(0)

+  +

v

12V



 (a)

0.04F

(b)

For t > 0, the circuit becomes that shown in Figure (b) after source transformation.  o = 1/ LC = 1/ 1x1 / 25 = 5  = R/(2L) = (6)/(2) = 3 s 1,2 =     2   2o  -3  j4 Thus,

v(t) = V s + [(Acos4t + Bsin4t)e-3t], V s = -12 v(0) = -8 = -12 + A or A = 4

i = Cdv/dt, or i/C = dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t] i(0) = -3A + 4B or B = 3 v(t) = {-12 + [(4cos4t + 3sin4t)e-3t]} A


For t>0, we have a source-free series RLC circuit.



R 2L



R  2L  2x8x 0.5  8Ω

d  o 2   2  30

o 

1 LC

 

 

C

1

 L 2 o

o  900  64  964



1  2.075 mF 964 x 0.5




R 1000   500, 2L 2 x1

o  

 

o 

1 LC



underdamped.

1 100 x10

9

 10 4

Chapter 8, Solution 45.  o = 1/ LC = 1/ 1x 0.5 =

2

 = 1/(2RC) = (1)/(2x2x0.5) = 0.5 Since  <  o , we have an underdamped response. s 1,2 =    o2   2  –0.5  j1.3229 i(t) = I s + [(Acos1.3229t + Bsin1.3229t)e-0.5t], I s = 4

Thus,

i(0) = 1 = 4 + A or A = -3 v = v C = v L = Ldi(0)/dt = 0 di/dt = [1.3229(-Asin1.3229t + Bcos1.3229t)e-0.5t] + [-0.5(Acos1.3229t + Bsin1.3229t)e-0.5t] di(0)/dt = 0 = 1.3229B – 0.5A or B = 0.5(–3)/1.3229 = –1.1339

Thus,

i(t) = {4 – [(3cos1.3229t + 1.1339sin1.3229t)e-t/2]} A

To find v(t) we use v(t) = v L (t) = Ldi(t)/dt. From above, di/dt = [1.3229(-Asin1.3229t + Bcos1.3229t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t] Thus, v(t) = Ldi/dt = [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t] = [1.3229(3sin1.3229t – 1.1339cos1.3229t)e-0.5t] + [(1.5cos1.3229t + 0.5670sin1.3229t)e-0.5t] v(t) = [(–0cos1.323t + 4.536sin1.323t)e-0.5t] V = [(4.536sin1.3229t)e-t/2] V Please note that the term in front of the cos calculates out to –3.631x10-5 which is zero for all practical purposes when considering the rounding errors of the terms used to calculate it.

Chapter 8, Solution 46. Using Fig. 8.93, design a problem to help other students to better understand the step response of a parallel RLC circuit. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find i(t) for t > 0 in the circuit in Fig. 8.93.

Figure 8.93 Solution For t = 0-, u(t) = 0, so that v(0) = 0 and i(0) = 0. For t > 0, we have a parallel RLC circuit with a step input, as shown below. +

i 8mH

5F

v 

2 k 6mA

 = 1/(2RC) = (1)/(2x2x103 x5x10-6) = 50  o = 1/ LC = 1/ 8x10  3 x 5x10  6 = 5,000

Since  <  o , we have an underdamped response. s 1,2 =     2  o2  -50  j5,000 Thus,

i(t) = I s + [(Acos5,000t + Bsin5,000t)e-50t], I s = 6mA i(0) = 0 = 6 + A or A = -6mA

v(0) = 0 = Ldi(0)/dt di/dt = [5,000(-Asin5,000t + Bcos5,000t)e-50t] + [-50(Acos5,000t + Bsin5,000t)e-50t] di(0)/dt = 0 = 5,000B – 50A or B = 0.01(-6) = -0.06mA Thus,

i(t) = {6 – [(6cos5,000t + 0.06sin5,000t)e-50t]} mA

Chapter 8, Solution 47. At t = 0-, we obtain,

i L (0) = 3x5/(10 + 5) = 1A

and v o (0) = 0. For t > 0, the 10-ohm resistor is short-circuited and we have a parallel RLC circuit with a step input.  = 1/(2RC) = (1)/(2x5x0.01) = 10  o = 1/ LC = 1/ 1x 0.01 = 10 Since  =  o , we have a critically damped response. s 1,2 = -10 Thus,

i(t) = I s + [(A + Bt)e-10t], I s = 3 i(0) = 1 = 3 + A or A = -2 v o = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t] v o (0) = 0 = B – 10A or B = -20 Thus, v o (t) = (200te-10t) V

Chapter 8, Solution 48. For t = 0–, we obtain i(0) = –6/(1 + 2) = –2 and v(0) = 2x1 = 2. For t > 0, the voltage is short-circuited and we have a source-free parallel RLC circuit.  = 1/(2RC) = (1)/(2x1x0.25) = 2  o = 1/ LC = 1/ 1x 0.25 = 2 Since  =  o , we have a critically damped response. s 1,2 = Thus,

–2

i(t) = [(A + Bt)e-2t], i(0) = –2 = A v = Ldi/dt = [Be-2t] + [–2(–2 + Bt)e–2t] v o (0) = 2 = B + 4 or B = –2

Thus, i(t) = [(–2 – 2t)e–2t] A and v(t) = [(2 + 4t)e–2t] V

Chapter 8, Solution 49. For t = 0-, i(0) = 3 + 12/4 = 6 and v(0) = 0. For t > 0, we have a parallel RLC circuit with a step input.  = 1/(2RC) = (1)/(2x5x0.05) = 2  o = 1/ LC = 1/ 5x 0.05 = 2 Since  =  o , we have a critically damped response. s 1,2 = Thus,

-2

i(t) = I s + [(A + Bt)e-2t], I s = 3 i(0) = 6 = 3 + A or A = 3 v = Ldi/dt or v/L = di/dt = [Be-2t] + [-2(A + Bt)e-2t] v(0)/L = 0 = di(0)/dt = B – 2x3 or B = 6 Thus, i(t) = {3 + [(3 + 6t)e-2t]} A

Chapter 8, Solution 50. For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A. For t > 0, we have a parallel RLC circuit. i

+ 10  3A

10 mF 6A

40 

v

10 H



I s = 3 + 6 = 9A and R = 10||40 = 8 ohms  = 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25  o = 1/ LC = 1/ 4x 0.01 = 5 Since  >  o , we have a overdamped response. s 1,2 =     2   o2  -10, -2.5 Thus,

i(t) = I s + [Ae-10t] + [Be-2.5t], I s = 9 i(0) = 3 = 9 + A + B or A + B = -6 di/dt = [-10Ae-10t] + [-2.5Be-2.5t], v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A – 2.5B or B = -4A Thus, A = 2 and B = -8 Clearly,

i(t) = { 9 + [2e-10t] + [-8e-2.5t]} A

Chapter 8, Solution 51. Let i = inductor current and v = capacitor voltage. At t = 0, v(0) = 0 and i(0) = i o . For t > 0, we have a parallel, source-free LC circuit (R = ).  = 1/(2RC) = 0 and  o = 1/ LC which leads to s 1,2 =  j o v = Acos o t + Bsin o t, v(0) = 0 A i C = Cdv/dt = -i dv/dt =  o Bsin o t = -i/C dv(0)/dt =  o B = -i o /C therefore B = i o /( o C) v(t) = -(i o /( o C))sin o t V where  o = 1 / LC


  300 

1 2 RC

d  o   2  400 2

(1)

 

o2  d2   2  160,000  90,000 

From (2), C

1  80 µF 250,000x 50x10 3

From (1), R

1 1   20.83Ω. 2C 2x 300x80x10 6

1 LC

(2)

Chapter 8, Solution 53. At t<0, i(0  )  0, vc(0  )  120V For t >0, we have the circuit as shown below.

80  i

120 V

+ _

120  V dv C i R dt

10 mF

 

120  V  RC

di dt Substituting (2) into (1) yields di d2 i 120  L  RCL 2  iR   dt dt or

But

dv  iR dt

vL  v  L

0.25 H

(1)

(2)

120 

1 di 1 d2 i  80 x x10 x10 3 2  80i 4 dt 4 dt

(d2i/dt2) + 0.125(di/dt) + 400i = 600

Chapter 8, Solution 54. Using Fig. 8.100, design a problem to help other students better understand general second-order circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the circuit in Fig. 8.100, let I = 9A, R 1 = 40 Ω, R 2 = 20 Ω, C = 10 mF, R 3 = 50 Ω,and L = 20 mH. Determine: (a) i(0+) and v(0+), (b) di(0+)/dt and dv(0+)/dt, (c) i(∞) and v(∞).

A

t=0

R3 i

B I

+

R1

R2

C

v

L



Figure 8.100 For Prob. 8.54. Solution (a) When the switch is at A, the circuit has reached steady state. Under this condition, the circuit is as shown below. t=0 50Ω A i

B 9A

40Ω

+ 20Ω

v 

(a) When the switch is at A, i(0–) = 9[(40x50)/(40+50)]/50 = 4 A and v(0–) = 50i(0–) = 200 V. Since the current flowing through the inductor cannot change in zero time, i(0+) = i(0–) = 4 A. Since the voltage across the capacitor cannot change in zero time, v(0–) = v(0–) = 200 V.

(b)

For the inductor, v L = L(di/dt) or di(0+)/dt = v L (0+)/0.02.

At t = 0+, the right hand loop becomes, –200 + 50x4 + v L (0+) = 0 or v L (0+) = 0 and (di(0+)/dt) = 0. For the capacitor, i C = C(dv/dt) or dv(0+)/dt = i C (0+)/0.01. At t = 0+, and looking at the current flowing out of the node at the top of the circuit, ((200–0)/20) + i C + 4 = 0 or i C = –14 A. Therefore,

dv(0+)/dt = –14/0.01 = –1.4 kV/s.

(c) When the switch is in position B, the circuit reaches steady state. Since it is source-free, i and v decay to zero with time. Thus, i(∞) = 0 A and v(∞) = 0 V.

Chapter 8, Solution 55. At the top node, writing a KCL equation produces, i/4 +i = C 1 dv/dt, C 1 = 0.1 5i/4 = C 1 dv/dt = 0.1dv/dt i = 0.08dv/dt But,

(1)

v =  (2i  (1 / C 2 )  idt ) , C 2 = 0.5

or

-dv/dt = 2di/dt + 2i

(2)

Substituting (1) into (2) gives, -dv/dt = 0.16d2v/dt2 + 0.16dv/dt 0.16d2v/dt2 + 0.16dv/dt + dv/dt = 0, or d2v/dt2 + 7.25dv/dt = 0 Which leads to s2 + 7.25s = 0 = s(s + 7.25) or s 1,2 = 0, -7.25

From (1),

v(t) = A + Be-7.25t

(3)

v(0) = 4 = A + B

(4)

i(0) = 2 = 0.08dv(0+)/dt or dv(0+)/dt = 25

But,

dv/dt = -7.25Be-7.25t, which leads to,

dv(0)/dt = -7.25B = 25 or B = -3.448 and A = 4 – B = 4 + 3.448 = 7.448 Thus, v(t) = {7.448 – 3.448e-7.25t} V

Chapter 8, Solution 56. For t < 0, i(0) = 0 and v(0) = 0. For t > 0, the circuit is as shown below. 4

i

6

i

0.04F 20

+ 

io

0.25H

Applying KVL to the larger loop, -20 +6i o +0.25di o /dt + 25  (i o  i)dt = 0

Taking the derivative, 6di o /dt + 0.25d2i o /dt2 + 25(i o + i) = 0

For the smaller loop,

4 + 25  (i  i o )dt = 0

Taking the derivative,

25(i + i o ) = 0 or i = -i o

From (1) and (2)

6di o /dt + 0.25d2i o /dt2 = 0

This leads to, 0.25s2 + 6s = 0 or s 1,2 = 0, -24 i o (t) = (A + Be-24t) and i o (0) = 0 = A + B or B = -A As t approaches infinity, i o () = 20/10 = 2 = A, therefore B = -2 Thus, i o (t) = (2 - 2e-24t) = -i(t) or i(t) = (-2 + 2e-24t) A

(1)

(2)

Chapter 8, Solution 57. (a) Let v = capacitor voltage and i = inductor current. At t = 0-, the switch is closed and the circuit has reached steady-state. v(0-) = 16V and i(0-) = 16/8 = 2A At t = 0+, the switch is open but, v(0+) = 16 and i(0+) = 2. We now have a source-free RLC circuit. R 8 + 12 = 20 ohms, L = 1H, C = 4mF.  = R/(2L) = (20)/(2x1) = 10  o = 1/ LC = 1/ 1x (1 / 36) = 6 Since  >  o , we have a overdamped response. s 1,2 =     2  o2  -18, -2 Thus, the characteristic equation is (s + 2)(s + 18) = 0 or s2 + 20s +36 = 0. i(t) = [Ae-2t + Be-18t] and i(0) = 2 = A + B

(b)

(1)

To get di(0)/dt, consider the circuit below at t = 0+. i 12  +

+ (1/36)F

v 

8 vL 

1H

-v(0) + 20i(0) + v L (0) = 0, which leads to, -16 + 20x2 + v L (0) = 0 or v L (0) = -24 Ldi(0)/dt = v L (0) which gives di(0)/dt = v L (0)/L = -24/1 = -24 A/s Hence -24 = -2A – 18B or 12 = A + 9B From (1) and (2),

B = 1.25 and A = 0.75

(2)

i(t) = [0.75e-2t + 1.25e-18t] = -i x (t) or i x (t) = [-0.75e-2t - 1.25e-18t] A v(t) = 8i(t) = [6e-2t + 10e-18t] A


(a) Let i =inductor current, v = capacitor voltage i(0) =0, v(0) = 4

dv(0) [v(0)  Ri (0)] (4  0)     8 V/s dt RC 0.5 (b) For t  0 , the circuit is a source-free RLC parallel circuit.



1 1   1, 2 RC 2 x0.5 x1

o 

1 LC



1 0.25 x1

2

 d   2 o   2  4  1  1.732 Thus, v(t )  e t ( A1 cos1.732t  A2 sin 1.732t ) v(0) = 4 = A 1 dv  e t A1 cos1.732t  1.732e t A1 sin 1.732t  e t A2 sin 1.732t  1.732e t A2 cos1.732t dt dv(0)  8   A1  1.732 A2   A2  2.309 dt v ( t )  e  t (4 cos 1.732t  2.309 sin 1.732t ) V


Let i = inductor current and v = capacitor voltage v(0) = 0, i(0) = 40/(4+16) = 2A For t>0, the circuit becomes a source-free series RLC with R 16 1 1   2,  o    2,      o  2 2L 2 x4 LC 4 x1 / 16 i (t )  Ae 2t  Bte 2t i(0) = 2 = A di  2 Ae  2t  Be  2t  2 Bte  2t dt di(0) 1 1  2A  B   [Ri(0)  v(0)]    2A  B   (32  0), dt L 4



B  4

i (t )  2e 2t  4te 2t t t t t 64 t 1  2t v    idt  v(0)  32  e dt  64  te  2 t dt  16e  2 t  e  2 t (2 t  1) C 0 4 0 0

0

0

v = –32te–2t V. Checking, v = Ldi/dt + Ri = 4(–4e–2t – 4e–2t + 8e–2t) + 16(2e–2t – 4te–2t) = –32te–2t V.

Chapter 8, Solution 60. At t = 0-, 4u(t) = 0 so that i 1 (0) = 0 = i 2 (0)

(1)

Applying nodal analysis, 4 = 0.5di 1 /dt + i 1 + i 2 (2) Also,

i 2 = [1di 1 /dt – 1di 2 /dt]/3 or 3i 2 = di 1 /dt – di 2 /dt

Taking the derivative of (2), 0 = d2i 1 /dt2 + 2di 1 /dt + 2di 2 /dt From (2) and (3),

(3) (4)

di 2 /dt = di 1 /dt – 3i 2 = di 1 /dt – 3(4 – i 1 – 0.5di 1 /dt) = di 1 /dt – 12 + 3i 1 + 1.5di 1 /dt

Substituting this into (4), d2i 1 /dt2 + 7di 1 /dt + 6i 1 = 24 which gives s2 + 7s + 6 = 0 = (s + 1)(s + 6) Thus, i 1 (t) = I s + [Ae-t + Be-6t], 6I s = 24 or I s = 4 i 1 (t) = 4 + [Ae-t + Be-6t] and i 1 (0) = 4 + [A + B]

(5)

i 2 = 4 – i 1 – 0.5di 1 /dt = i 1 (t) = 4 + -4 - [Ae-t + Be-6t] – [-Ae-t - 6Be-6t] = [-0.5Ae-t + 2Be-6t] and i 2 (0) = 0 = -0.5A + 2B From (5) and (6),

A = -3.2 and B = -0.8 i 1 (t) = {4 + [-3.2e-t – 0.8e-6t]} A i 2 (t) = [1.6e-t – 1.6e-6t] A

(6)

Chapter 8, Solution 61. For t > 0, we obtain the natural response by considering the circuit below. a

1H

iL

+ 4

vC

0.25F

6

 At node a,

v C /4 + 0.25dv C /dt + i L = 0

(1)

But,

v C = 1di L /dt + 6i L

(2)

Combining (1) and (2), (1/4)di L /dt + (6/4)i L + 0.25d2i L /dt2 + (6/4)di L /dt + i L = 0 d2i L /dt2 + 7di L /dt + 10i L = 0 s2 + 7s + 10 = 0 = (s + 2)(s + 5) or s 1,2 = -2, -5 Thus, i L (t) = i L () + [Ae-2t + Be-5t], where i L () represents the final inductor current = 4(4)/(4 + 6) = 1.6 i L (t) = 1.6 + [Ae-2t + Be-5t] and i L (0) = 1.6 + [A+B] or -1.6 = A+B (3) di L /dt = [-2Ae-2t - 5Be-5t] and di L (0)/dt = 0 = -2A – 5B or A = -2.5B

(4)

From (3) and (4), A = -8/3 and B = 16/15 i L (t) = 1.6 + [-(8/3)e-2t + (16/15)e-5t] v(t) = 6i L (t) = {9.6 + [-16e-2t + 6.4e-5t]} V v C = 1di L /dt + 6i L = [ (16/3)e-2t - (16/3)e-5t] + {9.6 + [-16e-2t + 6.4e-5t]} v C = {9.6 + [-(32/3)e-2t + 1.0667e-5t]} i(t) = v C /4 = {2.4 + [-2.667e-2t + 0.2667e-5t]} A

Chapter 8, Solution 62. This is a parallel RLC circuit as evident when the voltage source is turned off.  = 1/(2RC) = (1)/(2x3x(1/18)) = 3  o = 1/ LC = 1/ 2 x1 / 18 = 3 Since  =  o , we have a critically damped response. s 1,2 =

-3

Let v(t) = capacitor voltage Thus, v(t) = V s + [(A + Bt)e-3t] where V s = 0 But -10 + v R + v = 0 or v R = 10 – v Therefore v R = 10 – [(A + Bt)e-3t] where A and B are determined from initial conditions.

Chapter 8, Solution 63. vs  0 d(0  vo ) vs dv C    C o dt R dt R 2 dvo v di di vo  L   L 2  s dt dt RC dt Thus, v d 2 i (t )  s 2 RCL dt

Chapter 8, Solution 64. Using Fig. 8.109, design a problem to help other students to better understand second-order op amp circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Obtain the differential equation for v o (t) in the network of Fig. 8.109.

Figure 8.109 Solution C2 R2 R1 vs

1 v1

C1

2

 +

vo

At node 1,

(v s – v 1 )/R 1 = C 1 d(v 1 – 0)/dt or v s = v 1 + R 1 C 1 dv 1 /dt

At node 2,

C 1 dv 1 /dt = (0 – v o )/R 2 + C 2 d(0 – v o )/dt or

From (1) and (2), or

–R 2 C 1 dv 1 /dt = v o + R 2 C 2 dv o /dt

(1)

(2)

(v s – v 1 )/R 1 = C 1 dv 1 /dt = –(1/R 2 )(v o + R 2 C 2 dv o /dt) v 1 = v s + (R 1 /R 2 )(v o + R 2 C 2 dv o /dt)

(3)

Substituting (3) into (1) produces, v s = v s + (R 1 /R 2 )(v o + R 2 C 2 dv o /dt) + R 1 C 1 d{v s + (R 1 /R 2 )(v o + R 2 C 2 dv o /dt)}/dt = v s + (R 1 /R 2 )(v o )+ (R 1 C 2 )dv o /dt + R 1 C 1 dv s /dt + (R 1 R 1 C 1 /R 2 )dv o /dt + ((R 1 )2 C 1 C 2 )[d2v o /dt2] ((R 1 )2 C 1 C 2 )[d2v o /dt2] + [(R 1 C 2 ) + (R 1 R 1 C 1 /R 2 )]dv o /dt + (R 1 /R 2 )(v o ) = –R 1 C 1 dv s /dt Simplifying we get, [d2v o /dt2] + {[(R 1 C 2 ) + (R 1 R 1 C 1 /R 2 )]/((R 1 )2 C 1 C 2 )}dv o /dt + {(R 1 /R 2 )(v o )/ ((R 1 )2 C 1 C 2 )} = –{R 1 C 1 /((R 1 )2 C 1 C 2 )}dv s /dt d2v o /dt2 + [(1/ R 1 C 1 ) + (1/(R 2 C 2 ))]dv o /dt + [1/(R 1 R 2 C 1 C 2 )](v o ) = –[1/(R 1 C 2 )]dv s /dt Another way to successfully work this problem is to give actual values of the resistors and capacitors and determine the actual differential equation. Alternatively, one could give a differential equations and ask the other students to choose actual value of the differential equation.

Chapter 8, Solution 65. At the input of the first op amp, (v o – 0)/R = Cd(v 1 – 0)

(1)

At the input of the second op amp, (-v 1 – 0)/R = Cdv 2 /dt

(2)

Let us now examine our constraints. Since the input terminals are essentially at ground, then we have the following, v o = -v 2 or v 2 = -v o

(3)

Combining (1), (2), and (3), eliminating v 1 and v 2 we get, d 2 vo  1  d 2vo  v   100 v o  0   o dt 2  R 2 C 2  dt 2

Which leads to s2 – 100 = 0 Clearly this produces roots of –10 and +10. And, we obtain, v o (t) = (Ae+10t + Be-10t)V At t = 0, v o (0+) = – v 2 (0+) = 0 = A + B, thus B = –A This leads to v o (t) = (Ae+10t – Ae-10t)V. Now we can use v 1 (0+) = 2V. From (2), v 1 = –RCdv 2 /dt = 0.1dv o /dt = 0.1(10Ae+10t + 10Ae-10t) v 1 (0+) = 2 = 0.1(20A) = 2A or A = 1 Thus, v o (t) = (e+10t – e-10t) V It should be noted that this circuit is unstable (clearly one of the poles lies in the righthalf-plane).

Chapter 8, Solution 66. We apply nodal analysis to the circuit as shown below. v2 10pF

60 k

60 k v1

– v2 +

vs

20 pF

+ _

+ vo –

At node 1, vs  v1 v1  v2 d   10 pF (v1  vo ) dt 60k 60k But v2  vo d( v 1  v o ) v s  2 v1  v o  6x10  7 dt At node 2, v1  v2 d  20 pF (v2  0), v2  vo 60k dt dv v1  v o  1.2 x10  6 o dt Substituting (2) into (1) gives

(1)

(2)

 dv  d 2 v o   v s  2 v o  1.2 x10  6 o   v o  6 x10  7 1.2 x10  6  dt   dt 2  

v s = v o + 2.4x10–6(dv o /dt) + 7.2x10–13(d2v o /dt2).

Chapter 8, Solution 67. At node 1, d ( v1  v o ) v in  v1 d( v1  0)  C2  C1 R1 dt dt

At node 2,

C2

(1)

 vo d ( v 1  0) 0  v o dv1 , or   dt R2 dt C2R 2

(2)

From (1) and (2), v in  v1  

v1  v in 

dv v R 1C1 dv o  R 1 C1 o  R 1 o C 2 R 2 dt dt R2

dv v R 1C1 dv o  R 1C1 o  R 1 o C 2 R 2 dt dt R2

(3)

C1 R2 R1 v in

C2 1

v1

2 0V

 +

vo

From (2) and (3), 

vo d 2 v o R 1 dv o dv dv R C dv o  1  in  1 1  R 1 C1  C2R 2 dt dt C 2 R 2 dt R 2 dt dt 2 d 2 vo vo 1  1 1  dv o 1 dv in       2 R 2  C1 C 2  dt C1C 2 R 2 R 1 R 1C1 dt dt

But C 1 C 2 R 1 R 2 = 10-4 x10-4 x104 x104 = 1 1 R2

 1 1  2 2     4 2  4  C1 C 2  R 2 C1 10 x10

d 2 vo dv dv  2 o  v o   in 2 dt dt dt Which leads to s2 + 2s + 1 = 0 or (s + 1)2 = 0 and s = –1, –1 Therefore,

v o (t) = [(A + Bt)e-t] + V f

As t approaches infinity, the capacitor acts like an open circuit so that V f = v o () = 0 v in = 10u(t) mV and the fact that the initial voltages across each capacitor is 0 means that v o (0) = 0 which leads to A = 0. v o (t) = [Bte-t] dv o = [(B – Bt)e-t] dt dv o (0 ) v (0 )  o 0 dt C2R 2

From (2), From (1) at t = 0+,

dv (0) dv o (0 ) 1 0 1  C 1 o which leads to   1 R1 dt dt C1 R 1

Substituting this into (4) gives B = –1 Thus,

v(t) = –te-tu(t) V

(4)

Chapter 8, Solution 68. The schematic is as shown below. The unit step is modeled by VPWL as shown. We insert a voltage marker to display V after simulation. We set Print Step = 25 ms and final step = 6s in the transient box. The output plot is shown below.

Chapter 8, Solution 69. The schematic is shown below. The initial values are set as attributes of L1 and C1. We set Print Step to 25 ms and the Final Time to 20s in the transient box. A current marker is inserted at the terminal of L1 to automatically display i(t) after simulation. The result is shown below.


The schematic is shown below.

After the circuit is saved and simulated, we obtain the capacitor voltage v(t) as shown below.

Chapter 8, Solution 71. The schematic is shown below. We use VPWL and IPWL to model the 39 u(t) V and 13 u(t) A respectively. We set Print Step to 25 ms and Final Step to 4s in the Transient box. A voltage marker is inserted at the terminal of R2 to automatically produce the plot of v(t) after simulation. The result is shown below.


When the switch is in position 1, we obtain IC=10 for the capacitor and IC=0 for the inductor. When the switch is in position 2, the schematic of the circuit is shown below.

When the circuit is simulated, we obtain i(t) as shown below.

Chapter 8, Solution 73. Design a problem, using PSpice, to help other students to better understand source-free RLC circuits. Although there are many ways to work this problem, this is an example based on a somewhat similar problem worked in the third edition. Problem

The step response of an RLC circuit is given by

Given that i L (0) = 3 A and v C (0) = 24 V, solve for v C (t) and I C (t). Solution (a)

For t < 0, we have the schematic below. When this is saved and simulated, we obtain the initial inductor current and capacitor voltage as i L (0) = 3 A and v c (0) = 24 V.

(b) For t > 0, we have the schematic shown below. To display i(t) and v(t), we insert current and voltage markers as shown. The initial inductor current and capacitor voltage are also

incorporated. In the Transient box, we set Print Step = 25 ms and the Final Time to 4s. After simulation, we automatically have i o (t) and v o (t) displayed as shown below.

Chapter 8, Solution 74. The dual is constructed as shown below. 0.5 

0.25  2

9V

+ _

4

1/6  6

1

3A

1

9A

– +

The dual is redrawn as shown below. 1/6 

9A

1/2 

1

1/4 

– +

3V

3V

Chapter 8, Solution 75. The dual circuit is connected as shown in Figure (a). It is redrawn in Figure (b).

0.1  12V

+ 

10 

12A

24A

0.5 F 24V 0.25 

4 10 H

10 H 10 F

(a) 0.1  2F 0.5 H

24A

12A 0.25 

(b)

+ 

Chapter 8, Solution 77. The dual is obtained from the original circuit as shown in Figure (a). It is redrawn in Figure (b). 0.1 

0.05 

1/3 

10 

20 

60 A

30  120 A

+ 

– +

60 V

2V

120 V + 

4H

1F

1H

2A

4F

(a) 0.05 

60 A

120 A

1H 0.1 

1/30  1/4 F 2V

(b)

+ 

Chapter 8, Solution 77. The dual is constructed in Figure (a) and redrawn in Figure (b). – +

5A

5V

2

1/3 

1/2  1F

1

1/4 H

1H

3

1

1/4 F

12V

+ 

12 A

(a)

1 1/2  1/3 

1/4 F

12 A 1H

5V

+ 

(b)

Chapter 8, Solution 78. The voltage across the igniter is v R = v C since the circuit is a parallel RLC type. v C (0) = 12, and i L (0) = 0.  = 1/(2RC) = 1/(2x3x1/30) = 5  o  1 / LC  1 / 60 x10 3 x1 / 30 = 22.36  <  o produces an underdamped response. s1, 2     2   o2 = –5  j21.794

v C (t) = e-5t(Acos21.794t + Bsin21.794t)

(1)

v C (0) = 12 = A dv C /dt = –5[(Acos21.794t + Bsin21.794t)e-5t] + 21.794[(–Asin21.794t + Bcos21.794t)e-5t]

(2)

dv C (0)/dt = –5A + 21.794B But,

dv C (0)/dt = –[v C (0) + Ri L (0)]/(RC) = –(12 + 0)/(1/10) = –120

Hence,

–120 = –5A + 21.794B, leads to B (5x12 – 120)/21.794 = –2.753

At the peak value, dv C (t o )/dt = 0, i.e., 0

= A + Btan21.794t o + (A21.794/5)tan21.794t o – 21.794B/5 (B + A21.794/5)tan21.794t o = (21.794B/5) – A

tan21.794t o = [(21.794B/5) – A]/(B + A21.794/5) = –24/49.55 = –0.484 Therefore,

21.7945t o = |–0.451|

t o = |–0.451|/21.794 = 20.68 ms


For critical damping of a parallel RLC circuit,

  o

 

1  2 RC

Hence, C

L 0.25   434 F 2 4 x144 4R

1 LC

Chapter 8, Solution 80. t 1 = 1/|s 1 | = 0.1x10-3 leads to s 1 = –1000/0.1 = –10,000 t 2 = 1/|s 2 | = 0.5x10-3 leads to s 1 = –2,000 s1     2  o2 s 2     2   o2

s 1 + s 2 = –2 = –12,000, therefore  = 6,000 = R/(2L) L = R/12,000 = 50,000/12,000 = 4.167H s 2     2  o2 = –2,000    2   o2 = 2,000 6,000   2   o2 = 2,000  2   o2 = 4,000

2 – o2 = 16x106

o2 = 2 – 16x106 = 36x106 – 16x106  o = 103 20  1 / LC C = 1/(20x106x4.167) = 12 nF

Chapter 8, Solution 81. t = 1/ = 0.25 leads to  = 4 But,

 1/(2RC) or,

C = 1/(2R) = 1/(2x4x200) = 625 F d  o2   2

o2  d2   2  (24 x10 3 ) 2  16  (24 x10 3 0 2 = 1/(LC) This results in L = 1/(642x106x625x10-6) = 2.533 H

Chapter 8, Solution 82. For t = 0-, v(0) = 0. For t > 0, the circuit is as shown below. R1

a

+

+ C1

vo

R2

C2

v





At node a, (v o – v/R 1 = (v/R 2 ) + C 2 dv/dt v o = v(1 + R 1 /R 2 ) + R 1 C 2 dv/dt 60 = (1 + 5/2.5) + (5x106 x5x10-6)dv/dt 60 = 3v + 25dv/dt v(t) = V s + [Ae-3t/25] where

3V s = 60 yields V s = 20

v(0) = 0 = 20 + A or A = –20 v(t) = 20(1 – e-3t/25)V

Chapter 8, Solution 83. i = i D + Cdv/dt

(1)

–v s + iR + Ldi/dt + v = 0

(2)

Substituting (1) into (2), v s = Ri D + RCdv/dt + Ldi D /dt + LCd2v/dt2 + v = 0 LCd2v/dt2 + RCdv/dt + Ri D + Ldi D /dt = v s d2v/dt2 + (R/L)dv/dt + (R/LC)i D + (1/C)di D /dt = v s /LC

Chapter 9, Solution 1. (a) V m = 50 V.

2  0.2094s = 209.4ms  30 (c ) Frequency f = ω/(2π) = 30/(2π) = 4.775 H z . (d) At t=1ms, v(0.01) = 50cos(30x0.01rad + 10˚) = 50cos(1.72˚ + 10˚) = 44.48 V and ωt = 0.3 rad. (b) Period T 

2




(a)

amplitude = 15 A

(b)

 = 25 = 78.54 rad/s

(c)

f =

(d)

I s = 1525 A I s (2 ms) = 15 cos((500 )(2  10 -3 )  25) = 15 cos( + 25) = 15 cos(205) = –13.595 A

 = 12.5Hz 2


(a)

10 sin(t + 30) = 10 cos(t + 30 – 90) = 10cos(t – 60)

(b)

–9 sin(8t) = 9cos(8t + 90)

(c)

–20 sin(t + 45) = 20 cos(t + 45 + 90) = 20cos(t + 135)

(a) 10cos(t – 60), (b) 9cos(8t + 90), (c) 20cos(t + 135)

Chapter 9, Solution 4. Design a problem to help other students to better understand sinusoids. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem (a) Express v = 8 cos(7t + 15) in sine form. (b) Convert i = –10sin(3t - 85) to cosine form.

Solution

(a)

v = 8 cos(7t + 15) = 8 sin(7t + 15 + 90) = 8 sin(7t + 105)

(b)

i = –10 sin(3t – 85) = 10 cos(3t – 85 + 90) = 10 cos(3t + 5)

Chapter 9, Solution 5. v 1 = 45 sin(t + 30) V = 45 cos(t + 30  90) = 45 cos(t  60) V v 2 = 50 cos(t – 30) V

This indicates that the phase angle between the two signals is 30 and that v 1 lags v2.


v(t) = 10 cos(4t – 60) i(t) = 4 sin(4t + 50) = 4 cos(4t + 50 – 90) = 4 cos(4t – 40) Thus, i(t) leads v(t) by 20.

(b)

v 1 (t) = 4 cos(377t + 10) v 2 (t) = -20 cos(377t) = 20 cos(377t + 180) Thus, v 2 (t) leads v 1 (t) by 170.

(c)

x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90) X = 130 + 5-90 = 13 – j5 = 13.928-21.04 x(t) = 13.928 cos(2t – 21.04) y(t) = 15 cos(2t – 11.8) phase difference = -11.8 + 21.04 = 9.24 Thus, y(t) leads x(t) by 9.24.

Chapter 9, Solution 7. If f() = cos + j sin, df  -sin  j cos   j (cos   j sin )  j f ( ) d

df  j d f Integrating both sides ln f = j + ln A f = Aej = cos + j sin f(0) = A = 1 i.e. f() = ej = cos + j sin


(a)

(b)

6045 6045 + j2 = + j2 7.5  j10 12.5 - 53.13 = 4.898.13 + j2 = –0.6788+j4.752+j2 = –0.6788 + j6.752 (6–j8)(4+j2) = 24–j32+j12+16 = 40–j20 = 44.72–26.57˚

20 32  20 32  20 20 + + = (6  j8)(4  j2) - 10  j24 44.72  26.57 26112.62

= 0.71566.57˚+0.7692–112.62˚ = 0.7109+j0.08188–0.2958–j0.71 = 0.4151–j0.6281

(c)

20 + (16–50)(1367.38) = 20+20817.38 = 20 + 198.5+j62.13 = 218.5+j62.13


(530)(6  j8  1.1197  j0.7392)  (530)(7.13  j7.261)  (530)(10.176  45.52)  50.88–15.52˚.

(b)

(1060)(35  50)  60.02–110.96˚. (3  j5)  (5.83120.96)

Chapter 9, Solution 10. Design a problem to help other students to better understand phasors. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Given that z 1 = 6 – j8, z 2 = 10–30, and z 3 = 8e–j120, find: (a) z 1 + z 2 + z 3 (b) z 1 z 2 / z 3 Solution (a) z1  6  j8, z 2  8.66  j 5, and z 3  4  j 6.9282 z1  z 2  z 3 = (10.66 – j19.928)Ω (b)

z1 z 2 = [(10–53.13˚)(10–30˚)/(8–120˚)] = 12.536.87˚Ω = (10 + j7.5) Ω z3


V  21  15o V

(b) i(t )  8sin(10t  70o  180o )  8cos(10t  70o  180o  90o )  8cos(10t  160o )

I = 8 160° mA (c ) v(t )  120sin(103 t  50o )  120 cos(103 t  50o  90o )

V = 120 –140° V

(d)

i(t )  60 cos(30t  10o )  60 cos(30t  10o  180o )

I = 60 –170° mA

Chapter 9, Solution 12. Let X = 440 and Y = 20–30. Evaluate the following quantities and express your results in polar form. (X + Y)/X* (X - Y)* (X + Y)/X X = 3.064+j2.571; Y = 17.321–j10

(20.38  j 7.429)(4  40)  (21.69  20.03)(4  40)  86.76  60.03 = 86.76–60.03˚

(a)

(X + Y)X* =

(b)

(X – Y)* = (–14.257+j12.571)* = 19.41–139.63˚

(c)

(X + Y)/X = (21.69–20.03˚)/(440˚) = 5.422–60.03˚

Chapter 9, Solution 13. (a) ( 0.4324  j 0.4054) ( 0.8425  j 0.2534)   1.2749  j 0.1520

(b)

50  30 o   2.0833 = –2.083 24150 o

(c) (2+j3)(8-j5) –(-4) = 35 +j14


3  j14 14.318  77.91   0.7788169.71   0.7663  j0.13912  7  j17 18.385112.38

(b)

(62.116  j 231.82  138.56  j80)(60  j80) 24186  6944.9    1.922  j11.55 (67  j84)(16.96  j10.5983) 246.06  j 2134.7

(c) = [(22.3663.43˚)/(553.13˚)]2[(11.1826.57˚)(25.61–51.34˚)]0.5 = [4.47210.3˚]2[286.3–24.77˚]0.5 = (19.99920.6˚)(16.921–12.38˚) = 338.48.22˚ or 334.9+j48.38


(b)

(c)

10  j6 2  j3 = -10 – j6 + j10 – 6 + 10 – j15 -5 -1  j = –6 – j11 20  30 - 4 - 10 = 6015 + 64-10 160 345 = 57.96 + j15.529 + 63.03 – j11.114 = 120.99 + j4.415

1 j  j 0 j 1 j 1 j 1 j 1 j  j j 1

= 1  1  0  1  0  j2 (1  j)  j2 (1  j)

0 j = 1  1 (1  j  1  j) = 1 – 2 = –1


–20 cos(4t + 135) = 20 cos(4t + 135  180) = 20 cos(4t  45) The phasor form is 20–45

(b)

8 sin(20t + 30) = 8 cos(20t + 30 – 90) = 8 cos(20t – 60) The phasor form is 8–60

(c)

20 cos(2t) + 15 sin(2t) = 20 cos(2t) + 15 cos(2t – 90) The phasor form is 200 + 15–90 = 20 – j15 = 25-36.87

Chapter 9, Solution 17. V  V1  V2  10  60o  12  30o  5  j8.66  10.392  j 6  15.62  9.805o v(t) = 15.62cos(50t–9.8˚) V


v1 ( t ) = 60 cos(t + 15)

(b)

V2 = 6 + j8 = 1053.13 v 2 ( t ) = 10 cos(40t + 53.13)

(c)

i1 ( t ) = 2.8 cos(377t – /3)

(d)

I 2 = -0.5 – j1.2 = 1.3247.4 i 2 ( t ) = 1.3 cos(103t + 247.4)


310  5-30 = 2.954 + j0.5209 – 4.33 + j2.5 = -1.376 + j3.021 = 3.32114.49 Therefore, 3 cos(20t + 10) – 5 cos(20t – 30) = 3.32 cos(20t + 114.49)

(b)

40-90 + 30-45 = -j40 + 21.21 – j21.21 = 21.21 – j61.21 = 64.78-70.89 Therefore, 40 sin(50t) + 30 cos(50t – 45) = 64.78 cos(50t – 70.89)

(c)

Using sin = cos(  90), 20-90 + 1060  5-110 = -j20 + 5 + j8.66 + 1.7101 + j4.699 = 6.7101 – j6.641 = 9.44-44.7 Therefore, 20 sin(400t) + 10 cos(400t + 60) – 5 sin(400t – 20) = 9.44 cos(400t – 44.7)

Chapter 9, Solution 20. 7.5cos(10t+30˚) A can be represented by 7.530˚ and 120cos(10t+75˚) V can be represented by 12075˚. Thus, Z = V/I = (12075˚)/(7.530˚) = 1645˚ or (11.314+j11.314) Ω.

Chapter 9, Solution 21. (a) F  515 o  4 30 o  90 o  6.8296  j 4.758  8.323634.86 o f ( t )  8.324 cos( 30t  34.86 o )

(b) G  8  90 o  450 o  2.571  j 4.9358  5.565  62.49 o g ( t )  5.565 cos( t  62.49 o )

(c) H 





1 100 o  50  90 o , j

  40

i.e. H  0.25  90 o  1.25  180 o   j0.25  1.25  1.2748  168.69 o h(t) = 1.2748cos(40t – 168.69°)

Chapter 9, Solution 22. t

dv Let f(t) = 10v(t )  4  2  v(t )dt dt  2V F  10V  j 4V  ,   5, V  5545o j F  10V  j 20V  j 0.4V  (10  j 20.4)V  22.7263.89(5545)  1249.6108.89 o

f(t) = 1249.6cos(5t+108.89˚)

Chapter 9, Solution 23. (a) v = [110sin(20t+30˚) + 220cos(20t–90˚)] V leads to V = 110(30˚–90˚) + 220–90˚ = 55–j95.26 – j220 = 55–j315.3 = 320.1–80.11˚ or v = 320.1cos(20t–80.11˚) A.

(b) i = [30cos(5t+60˚)–20sin(5t+60˚)] A leads to I = 3060˚ – 20(60˚–90˚) = 15+j25.98 – (17.321–j10) = –2.321+j35.98 = 36.0593.69˚ or i = 36.05cos(5t+93.69˚) A.

(a) 320.1cos(20t–80.11˚) A, (b) 36.05cos(5t+93.69˚) A

Chapter 9, Solution 24. (a) V  10 0,   1 j V (1  j)  10 10 V  5  j5  7.07145 1 j V

Therefore, v(t) = 7.071cos(t + 45) V (b) 4V  20(10  90),   4 j  4 V  j4  5    20  - 80 j4   20 - 80  3.43 - 110.96 V 5  j3 jV  5V 

Therefore, v(t) = 3.43cos(4t – 110.96) V


2jI  3I  445,   2 I (3  j 4)  445 445 445 I   0.8 - 8.13 3  j4 553.13 Therefore, i(t) = 800cos(2t – 8.13) mA

(b) I  jI  6I  522,   5 j (- j2  j5  6) I  522 522 522   0.745 - 4.56 I 6  j3 6.70826.56 Therefore, i(t) = 745 cos(5t – 4.56) mA 10

Chapter 9, Solution 26. I  10,   2 j  1 I  j2  2    1 j2   1  0.4  - 36.87 I 2  j1.5 Therefore, i(t) = 0.4 cos(2t – 36.87) jI  2I 

Chapter 9, Solution 27. V  110  - 10,   377 j  j100    110 - 10 V  j377  50   377  V (380.682.45)  110 - 10 V  0.289  - 92.45 jV  50V  100

Therefore, v(t) = 289 cos(377t – 92.45) mV.


i (t ) 

vs (t ) 156 cos(377t  45)   10.4cos(377t+45˚) A. 15 R


Z

1 1   - j 0.5 6 jC j (10 )(2  10 -6 )

V  IZ  (4 25)(0.5 - 90)  2  - 65 Therefore

v(t) = 2 sin(106t – 65) V.

Chapter 9, Solution 30. Since R and C are in parallel, they have the same voltage across them. For the resistor, 100  20o V  IR R   IR  V / R   2.5  20o mA 40k iR  2.5cos(60t  20o ) mA For the capacitor,

iC  C

dv  50 x106 (60) x100sin(60t  20o )  300sin(60t  20o ) mA dt


j L  j 2 x 240 x103  j 0.48 1 1 C  5mF      j100 jC j 2 x5 x103 Z  80  j 0.48  j100  80  j 99.52 = L  240mH

 

V 10  00  0.0783  51.206o I  Z 80  j 99.52

i(t) = 78.3cos(2t+51.21˚) mA

Chapter 9, Solution 32. Using Fig. 9.40, design a problem to help other students to better understand phasor relationships for circuit elements. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Two elements are connected in series as shown in Fig. 9.40. If i = 12 cos (2t - 30) A, find the element values.


V = 18010,

Z

I = 12-30,

 = 2

V 18010  1540  11.49  j 9.642   I 12 - 30

One element is a resistor with R = 11.49 . The other element is an inductor with L = 9.642 or

L = 4.821 H.

Chapter 9, Solution 33. 110  v 2R  v 2L v L  110 2  v 2R v L  110 2  85 2  69.82 V

Chapter 9, Solution 34. v o  0 when jX L –jX C = 0 so X L = X C or L 



1 C

1 (5  10 3 )(20  10 3 )

   

 100 rad/s

1 LC

.


vs (t )  50 cos 200t

 

Vs  50  0o ,   200

1 1  j jC j 200 x5 x103   j L  j 20 x103 x 200  j 4

 

5mF

20mH

Z in  10  j  j 4  10  j 3 I

Vs 50  0o   4.789  16.7o Z in 10  j 3 i(t) = 4.789cos(200t–16.7°) A

Chapter 9, Solution 36. Using Fig. 9.43, design a problem to help other students to better understand impedance. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem In the circuit in Fig. 9.43, determine i. Let v s = 60 cos(200t - 10) V.

Figure 9.43 Solution Let Z be the input impedance at the source. 100 mH

10 F

 

 

jL  j 200 x100 x10 3  j 20

1 1    j 500 jC j10 x10 6 x 200

1000//-j500 = 200 –j400 1000//(j20 + 200 –j400) = 242.62 –j239.84 Z  2242.62  j 239.84  2255  6.104 o I

60  10 o  26.61  3.896 o mA o 2255  6.104 i  266.1 cos( 200t  3.896 o ) mA

Chapter 9, Solution 37. Y = (1/4) + (1/(j8)) + (1/(–j10)) = 0.25 – j0.025 = (250–j25) mS

Chapter 9, Solution 38. Using Fig. 9.45, design a problem to help other students to better understand admittance. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find i(t) and v(t) in each of the circuits of Fig. 9.45.

Figure 9.45 Solution (a)

1 F   6

1 1   - j2 jC j (3)(1 / 6)

- j2 (10 45)  4.472  - 18.43 4  j2 Hence, i(t) = 4.472 cos(3t – 18.43) A I

V  4I  (4)( 4.472  - 18.43)  17.89  - 18.43 Hence, v(t) = 17.89 cos(3t – 18.43) V

(b)

1 F   12

1 1   - j3 jC j ( 4)(1 / 12)

3H

 

jL  j ( 4)(3)  j12

V 50 0   10 36.87 Z 4  j3 Hence, i(t) = 10 cos(4t + 36.87) A I

j12 (50 0)  41.6 33.69 8  j12 Hence, v(t) = 41.6 cos(4t + 33.69) V V


Z eq  4  j 20  10 //(  j14  j 25)  9.135  j 27.47  = (9.135+j27.47) Ω I

V 12   0.4145  71.605o Z eq 9.135  j 27.47 i(t) = 414.5cos(10t–71.6˚) mA


For   1 , 1H  

jL  j (1)(1)  j 1 1 0.05 F     - j20 jC j (1)(0.05) - j40 Z  j  2 || (- j20)  j   1.98  j0.802 2  j20

4 0 40 V    1.872  - 22.05 Z 1.98  j0.802 2.13622.05 Hence, i o ( t )  1.872 cos(t – 22.05) A Io 

(b)

For   5 , 1H  

jL  j (5)(1)  j5 1 1 0.05 F     - j4 jC j (5)(0.05) - j4 Z  j5  2 || (- j4)  j5   1.6  j4.2 1  j2

40 40 V    0.89 - 69.14 Z 1.6  j4 4.49469.14 Hence, i o ( t )  890cos(5t – 69.14) mA Io 

(c)

For   10 , 1H   jL  j (10)(1)  j10 1 1 0.05 F     - j2 jC j (10)(0.05) - j4 Z  j10  2 || (- j2)  j10   1  j9 2  j2 V 4 0 4 0   0.4417  - 83.66  Z 1  j9 9.05583.66 Hence, i o ( t )  441.7cos(10t – 83.66) mA Io 


  1, 1H  

jL  j (1)(1)  j 1 1 1F     -j jC j (1)(1) - j1 Z  1  (1  j) || (- j)  1   2 j 1 I

Vs 10  , Z 2 j

I c  (1  j) I

V  (- j)(1  j) I  (1  j) I 

(1  j)(10)  6.325 - 18.43 2 j

Thus, v(t) = 6.325cos(t – 18.43) V

Chapter 9, Solution 42.   200

50 F  

1 1   - j100 jC j (200)(50  10 -6 )

0.1 H   jL  j (200)(0.1)  j20 (50)(-j100) - j100 50 || -j100    40  j20 50  j100 1 - j2 Vo 

j20 j20 (600)  (600)  17.14 90 j20  30  40  j20 70

Thus, v o ( t )  17.14 sin(200t + 90) V or v o ( t )  17.14 cos(200t) V

Chapter 9, Solution 43. Z in  50  j80 //(100  j 40)  50  Io 

j80(100  j 40)  105.71  j 57.93 100  j 40

60  0o  0.4377  0.2411  0.4997  28.85o A = 499.7–28.85˚ mA Z in

Chapter 9, Solution 44.   200 10 mH  

jL  j (200)(10  10 -3 )  j2 1 1    -j 5 mF  jC j (200)(5  10 -3 ) 3 j 1 1 1 Y    0.25  j0.5   0.55  j0.4 10 4 j2 3  j 1 1 Z   1.1892  j0.865 Y 0.55  j0.4 60 60 I   0.96  - 7.956 5  Z 6.1892  j0.865

Thus, i(t) = 960cos(200t – 7.956) mA

Chapter 9, Solution 45. We obtain I o by applying the principle of current division twice. I

I2

Z1

I2

-j2 

Z2

(a) Z 1  - j2 ,

Io

2

(b) Z 2  j4  (-j2) || 2  j4 

I2 

Z1 - j2 - j10 I (50)  Z1  Z 2 - j 2  1  j3 1 j

Io 

 - j   - j10  - 10 - j2     I 2    –5 A 2 - j2 1- j 1 j  11

- j4  1  j3 2 - j2

Chapter 9, Solution 46. i s  5 cos(10 t  40)   I s  540 1 1 0.1 F     -j jC j (10)(0.1)

0.2 H   Z1  4 || j2 

Let

jL  j (10)(0.2)  j2

j8  0.8  j1.6 , 4  j2

Z2  3  j

Z1 0.8  j1.6 Is  (540) Z1  Z 2 3.8  j0.6 (1.78963.43)(540) Io   2.32594.46 3.847 8.97

Io 

Thus, i o ( t )  2.325cos(10t + 94.46) A


First, we convert the circuit into the frequency domain.

Ix

50˚

Ix 

+ 

2

j4

-j10

20 

5 5 5    0.460752.63  j10(20  j4) 2  4.588  j8.626 10.854  52.63 2  j10  20  j4 i s (t) = 460.7cos(2000t +52.63˚) mA


Converting the circuit to the frequency domain, we get: 10 

V 1 30  Ix

20-40˚

+ 

j20

We can solve this using nodal analysis. V1  20  40 V1  0 V 0   1 0 10 j 20 30  j 20 V1 (0.1  j 0.05  0.02307  j 0.01538)  2  40 240  15.643  24.29 0.12307  j 0.03462 15.643  24.29 Ix   0.43389.4 30  j 20 i x  0.4338 sin(100t  9.4 ) A

V1 

-j20


Z T  2  j2 || (1  j)  2 

I

( j2)(1  j) 4 1 j

Ix

1

j2 

j2 j2 I I, j2  1  j 1 j 1 j 1 j I Ix  j2 j4

Ix 

Vs  I Z T 

-j 

where I x  0.50 

1 j 1 j  1  j  1.414 - 45 (4)  j j4 v s ( t )  1.4142 sin(200t – 45) V

1 2

Chapter 9, Solution 50. Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3) = -j10Ω. j10

Ix +

540˚

-j10

20 

vx 

Using the current dividing rule:

 j10 540   j 2.540  2.5  50  j10  20  j10 V x  20 I x  50  50 Ix 

v x (t) = 50cos(100t–50°) V


0.1 F  

1 1   - j5 jC j (2)(0.1)

0.5 H  

jL  j (2)(0.5)  j

The current I through the 2- resistor is Is 1 , I Is  1  j5  j  2 3  j4 I s  (5)(3  j4)  25 - 53.13

where I 

Therefore, i s ( t )  25cos(2t – 53.13) A

10 0  5 2

Chapter 9, Solution 52. We begin by simplifying the circuit. First we replace the parallel inductor and resistor with their series equivalent. 5 || j5 

j25 j5   2.5  j2.5 5  j5 1  j

Next let Z1  10 , and Z 2  - j 5  2.5  j 2.5  2.5  j 2.5 .

I2 Z1

IS

By current division I 2 

Z2

Z1 10 4 Is  Is  I . 12.5  j2.5 5 j s Z1  Z 2

Since Vo  I 2 (2.5  j2.5) we can now find I s .

 4  10 (1  j)  I s (2.5)(1  j)  830   I 5 j s  5  j Is 

(830)(5  j)  2.884–26.31 A. 10 (1  j)


Convert the delta to wye subnetwork as shown below. Z1 Io

Z2

2 Z3

+

10 

8

60  30 V o

-

Z  j2x 4 8  90 j6x 4   1  j1, Z2   3  j3, 4  j4 5.656945 4  j4 12 Z3   1.5  j1.5 4  j4 ( Z 3  8) //( Z 2  10)  (9.5  j1.5) //(13  j3)  5.6910.21  5.691  j0.02086

Z1 

Z  2  Z1  5.691  j0.02086  6.691  j0.9791

Io 

60  30 o 60  30 o   8.873  21.67 o A Z 6.7623  8.33 o

Chapter 9, Solution 54. Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below. + 

– 2Z

V2 +

Vs + V1

Z

–

V1  I o (1  j)  2 (1  j) V2  2V1  4 (1  j) V 2 + V s + V 1 = 0 or Vs  V1  V2  6 (1  j) = (6180)(1.4142–45) Vs  8.485135 V


12 

I

I1 I2

-j20 V

+ 

-j4 

Z + Vo

j8 

 Vo 4   -j0.5 j 8 j8 I (Z  j8) (-j0.5)(Z  j8) Z I2  1   j - j4 - j4 8 Z Z I  I 1  I 2  -j0.5   j   j0.5 8 8 - j20  12 I  I 1 (Z  j8) Z j - j - j20  12     (Z  j8)  8 2 2 3 1 - 4 - j26  Z   j  2 2 - 4 - j26 26.31261.25  16.64279.68  Z 1 1.5811 - 18.43 3 j 2 2 I1 

Z = (2.798 – j16.403) 

Chapter 9, Solution 56. 1 1    j 53.05 jC j 377 x50 x106 60mH   j L  j 377 x60 x103  j 22.62 Z in  12  j 53.05  j 22.62 // 40  21.692  j 35.91  50  F

 


2H

 

jL  j 2

1 j j C j2(2  j) Z  1  j2 //( 2  j)  1   2.6  j1.2 j2  2  j

1F

 

Y 1

Z

 0.3171  j 0.1463 S

Chapter 9, Solution 58. Using Fig. 9.65, design a problem to help other students to better understand impedance combinations. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem At  = 50 rad/s, determine Z in for each of the circuits in Fig. 9.65.


 10 mF 

1 1   - j2 jC j (50)(10  10 -3 )

10 mH  

jL  j (50)(10  10 -3 )  j0.5

Z in  j0.5  1 || (1  j2) 1  j2 Z in  j0.5  2  j2 Z in  j0.5  0.25 (3  j) Z in  0.75 + j0.25 

(b)

0.4 H  

jL  j (50)(0.4)  j20

0.2 H  

jL  j (50)(0.2)  j10 1 1   - j20 jC j (50)(1  10 -3 )

1 mF  

For the parallel elements, 1 1 1 1    Z p 20 j10 - j20 Z p  10  j10 Then, Z in  10 + j20 + Z p = 20 + j30 

1 Chapter 9, Solution 59. 0.25 F 0.5 H

   

1 1    j 0.4 jC j10 x0.25 j L  j10 x0.5  j 5

Z in  j5 (5  j0.4) 

(590)(5.016  4.57)  3.69142.82 6.79442.61 = (2.707+j2.509) Ω.


Z  (25  j15)  (20  j 50) //(30  j10)  25  j15  26.097  j 5.122

Z = (51.1+j9.878) Ω

Chapter 9, Solution 61. All of the impedances are in parallel. 1 1 1 1 1     Z eq 1  j 1  j2 j5 1  j3 1  (0.5  j0.5)  (0.2  j0.4)  (- j0.2)  (0.1  j0.3)  0.8  j0.4 Z eq

Z eq 

1  (1 + j0.5)  0.8  j0.4


2 mH  

jL  j (10  10 3 )(2  10 -3 )  j20 1 1 1 F     - j100 3 jC j (10  10 )(1  10 -6 )

50  + 10 A

+

V

j20   + 

V in 

-j100  V  (10)(50)  50 Vin  (10)(50  j20  j100)  (2)(50) Vin  50  j80  100  150  j80 Z in 

Vin  150 – j80  10

2V


First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with the corresponding delta. 200  j150  j300  20  j45 10 200  j450 200  j450 z2   30  j13.333, z3   10  j22.5 j15 20 z1 

8

–j12 

–j16 

z2

10 

z1

ZT

z3 –j16 

10 

Now all we need to do is to combine impedances.

z 2 (10  j16) 

(30  j13.333)(10  j16)  8.721  j8.938 40  j29.33

z3 (10  j16)  21.70  j3.821 ZT  8  j12  z1 (8.721  j8.938  21.7  j3.821)  34.69  j6.93

Chapter 9, Solution 64.  j10(6  j8)  19  j5 6  j2 3090 I  0.3866  j1.4767  1.527104.7 A ZT ZT  4 

Z T = (19–j5) Ω I = 1.527104.7° A

Chapter 9, Solution 65. Z T  2  (4  j6) || (3  j4) (4  j6)(3  j4) ZT  2  7  j2 Z T  6.83 + j1.094  = 6.9179.1 

I

120 10 V   17.350.9 A Z T 6.917 9.1

Chapter 9, Solution 66. (20  j5)(40  j10) 170  (12  j) 60  j5 145 Z T  14.069 – j1.172  = 14.118-4.76 Z T  (20  j5) || (40  j10) 

I

V 6090  4.2594.76  Z T 14.118 - 4.76 I I1

I2

20 

j10 

+

V ab

40  j10 8  j2 I I 60  j5 12  j 20  j5 4 j I2  I I 60  j5 12  j

I1 

Vab  -20 I 1  j10 I 2 - (160  j40) 10  j40 Vab  I I 12  j 12  j - 150 (-12  j)(150) Vab  I I 12  j 145 Vab  (12.457 175.24)(4.2597.76) Vab  52.94273 V




20 mH  

jL  j (10 3 )(20  10 -3 )  j20 1 1 12.5 F     - j80 3 jC j (10 )(12.5  10 -6 ) Z in  60  j20 || (60  j80) ( j20)(60  j80) Z in  60  60  j60 Z in  63.33  j23.33  67.494 20.22

Yin  (b)

1  14.8-20.22 mS Z in

10 mH   20 F  

jL  j (10 3 )(10  10 -3 )  j10 1 1   - j50 3 jC j (10 )(20  10 -6 )

30 || 60  20 Z in  - j50  20 || (40  j10)

Z in  - j50 

(20)(40  j10) = –j50 + 20(41.231 14.036°)/(60.828 9.462°) 60  j10

= –j50 + (13.5566 4.574° = –j50 + 13.51342 + j1.08109

= 13.51342 – j48.9189 = 50.751 –74.56° Z in  13.5  j48.92  50.75 - 74.56

Yin 

1  19.70474.56 mS = 5.246 + j18.993 mS Z in

Chapter 9, Solution 68. Yeq 

1 1 1   5  j2 3  j - j4

Yeq  (0.1724  j0.069)  (0.3  j0.1)  ( j0.25)

Yeq  (472.4 + j219) mS


1 1 1 1    (1  j2) Yo 4 - j2 4 Yo 

4 (4)(1  j2)   0.8  j1.6 1  j2 5

Yo  j  0.8  j0.6

1

1 1 1     (1)  ( j0.333)  (0.8  j0.6)  1 - j3 0.8  j0.6 Yo 1

Yo 

 1.8  j0.933  2.02827.41

Yo   0.4932 - 27.41  0.4378  j0.2271

Yo   j5  0.4378  j4.773 0.4378  j4.773 1 1 1    0.5  22.97 Yeq 2 0.4378  j4.773

1  0.5191  j0.2078 Yeq

Yeq 

0.5191  j0.2078  (1.661 + j0.6647) S 0.3126

Chapter 9, Solution 70. Make a delta-to-wye transformation as shown in the figure below. a

Z an

Z bn

Z eq

n

Z cn

b

c 8

2 -j5 

(- j10)(10  j15) (10)(15  j10)   7  j9 5  j10  10  j15 15  j5 (5)(10  j15)   4.5  j3.5 15  j5 (5)(- j10)   -1  j3 15  j5

Z an  Z bn Z cn

Z eq  Z an  (Z bn  2) || (Z cn  8  j5)

Z eq  7  j9  (6.5  j3.5) || (7  j8) (6.5  j3.5)(7  j8) 13.5  j4.5  7  j9  5.511  j0.2

Z eq  7  j9  Z eq

Z eq  12.51  j9.2  15.53-36.33 

Chapter 9, Solution 71. We apply a wye-to-delta transformation.

j4 

Z ab

b

a Z bc

Z ac

Z eq -j2 

1 c 2  j2  j4 2  j2   1 j j2 j2 2  j2 Z ac   1 j 2 2  j2 Z bc   -2  j2 -j Z ab 

( j4)(1  j)  1.6  j0.8 1  j3 (1)(1  j)  0.6  j0.2 1 || Z ac  1 || (1  j)  2 j j4 || Z ab  1 || Z ac  2.2  j0.6 j4 || Z ab  j4 || (1  j) 

1 1 1 1    Z eq - j2 - 2  j2 2.2  j0.6

 j0.5  0.25  j0.25  0.4231  j0.1154

 0.173  j0.3654  0.404364.66 Z eq  2.473-64.66  = (1.058 – j2.235) 

Chapter 9, Solution 72. Transform the delta connections to wye connections as shown below. a j2 

j2  -j18 

-j9 

j2 

R1

R2

R3 b

- j9 || - j18  - j6 , R1 

(20)(20)  8 , 20  20  10 (20)(10) R3   4 50

Z ab  j2  ( j2  8) || (j2  j6  4)  4 Z ab  4  j2  (8  j2) || (4  j4) (8  j2)(4  j4) Z ab  4  j2  12 - j2 Z ab  4  j2  3.567  j1.4054 Z ab  (7.567 + j0.5946) 

R2 

(20)(10)  4, 50

Chapter 9, Solution 73. Transform the delta connection to a wye connection as in Fig. (a) and then transform the wye connection to a delta connection as in Fig. (b). a j2 

j2  -j18 

-j9 

j2 

R1

R2

R3 b ( j8)(- j6) 48   - j4.8 j8  j8  j6 j10 Z 2  Z1  -j4.8 ( j8)( j8) - 64 Z3    j6.4 j10 j10 Z1 

(2  Z1 )(4  Z 2 )  (4  Z 2 )(Z 3 )  (2  Z1 )(Z 3 )  (2  j4.8)(4  j4.8)  (4  j4.8)( j6.4)  (2  j4.8)( j6.4)  46.4  j9.6

46.4  j9.6  1.5  j7.25 j6.4 46.4  j9.6 Zb   3.574  j6.688 4  j4.8 46.4  j9.6 Zc   1.727  j8.945 2  j4.8 Za 

(690)(7.58361.88)  07407  j3.3716 3.574  j12.688 (-j4)(1.5  j7.25) - j4 || Z a   0.186  j2.602 1.5  j11.25 j6 || Z b 

j12 || Z c 

(1290)(9.1179.07)  0.5634  j5.1693 1.727  j20.945

Z eq  ( j6 || Z b ) || (- j4 || Z a  j12 || Z c )

Z eq  (0.7407  j3.3716) || (0.7494  j2.5673) Z eq  1.50875.42  = (0.3796 + j1.46) 

Chapter 9, Solution 74. One such RL circuit is shown below.

20 

V

20 

+

+ j20 

Vi =

j20 

Vo 

10 Z

We now want to show that this circuit will produce a 90 phase shift. Z  j20 || (20  j20) 

V

( j20)(20  j20) - 20  j20   4 (1  j3) 20  j40 1  j2

Z 4  j12 1  j3 1 Vi  (10)   (1  j) Z  20 6  j3 3 24  j12

Vo 

 j  1  j j20  (1  j)    0.333390 V   3 20  j20 1  j  3

This shows that the output leads the input by 90.

Chapter 9, Solution 75. Since cos(t )  sin(t  90) , we need a phase shift circuit that will cause the output to lead the input by 90. This is achieved by the RL circuit shown below, as explained in the previous problem. 10 

10 

+ Vi 

+ j10 

j10 

Vo 

This can also be obtained by an RC circuit.

Chapter 9, Solution 76. (a) v2  8sin 5t  8cos(5t  90o ) v 1 leads v 2 by 70o. (b) v2  6sin 2t  6 cos(2t  90o ) v 1 leads v 2 by 180o. (c ) v1  4 cos10t  4 cos(10t  180o ) v2  15sin10t  15cos(10t  90o ) v 1 leads v 2 by 270o.


- jX c V R  jX c i 1 1 where X c    3.979 C (2)(2  10 6 )(20  10 -9 ) Vo 

Vo - j3.979   Vi 5 - j3.979

Vo  Vi

3.979 25  15.83

3.979 5  3.979 2

2

(-90  tan -1 (3.979 5))

(-90  38.51)

Vo  0.6227  - 51.49 Vi

Therefore, the phase shift is 51.49 lagging (b)

  -45  -90  tan -1 (X c R )  R  X c  45  tan -1 (X c R )    2f  f

1 C

1 RC

1 1   1.5915 MHz 2RC (2)(5)(20  10 -9 )


8+j6 R Z -jX

Z  R //8  j (6  X ) 

R[8  j (6  X )] 5 R  8  j (6  X )

i.e 8R + j6R – jXR = 5R + 40 + j30 –j5X Equating real and imaginary parts: 8R = 5R + 40 which leads to R=13.333Ω 6R-XR =30-5X which leads to X= 6 Ω.


Consider the circuit as shown. 20 

V2

40 

V1

30 

+ Vi

+ j10 

j30 

j60 



Vo 

Z2

Z1

( j30)(30  j60)  3  j21 30  j90 ( j10)(43  j21) Z 2  j10 || (40  Z1 )   1.535  j8.896  9.02880.21 43  j31

Z1  j30 || (30  j60) 

Let Vi  10 . Z2 (9.02880.21)(10) Vi  21.535  j8.896 Z 2  20 V2  0.387557.77

V2 

Z1 3  j21 (21.21381.87)(0.387557.77) V2  V2  43  j21 47.8526.03 Z1  40 V1  0.1718113.61 V1 

j60 j2 2 V1  V1  (2  j)V1 30  j60 1  j2 5 Vo  (0.8944 26.56)(0.1718113.6) Vo  0.1536 140.2 Vo 

Therefore, the phase shift is 140.2 (b)

The phase shift is leading.

(c)

If Vi  120 V , then Vo  (120)(0.1536 140.2)  18.43140.2 V and the magnitude is 18.43 V.


200 mH   Vo 

jL  j (2)(60)(200  10 -3 )  j75.4 

j75.4 j75.4 Vi  (1200) R  50  j75.4 R  50  j75.4

(a)

When R  100  , j75.4 (75.490)(1200) Vo  (120 0)  150  j75.4 167.8826.69 Vo  53.8963.31 V

(b)

When R  0  , j75.4 (75.490)(120 0) Vo  (1200)  50  j75.4 90.47 56.45 Vo  10033.55 V

(c)

To produce a phase shift of 45, the phase of Vo = 90 + 0   = 45. Hence,  = phase of (R + 50 + j75.4) = 45. For  to be 45, R + 50 = 75.4 Therefore, R = 25.4 

Chapter 9, Solution 81. Let

Z1  R 1 ,

Z2  R 2 

1 , jC 2

Zx 

Z3 Z Z1 2

Rx 

R3  1 1  R 2    jC x R 1  jC 2 

Rx 

R3 1200 (600)  1.8 k R2  400 R1

Z 3  R 3 , and Z x  R x 

R1  400  1 R3   1  (0.3  10 -6 )  0.1 F      C x  C2    1200  R3 Cx  R1   C2 

1 . jC x

Chapter 9, Solution 82. Cx 

R1  100  (40  10 -6 )  2 F Cs    2000  R2

Chapter 9, Solution 83. Lx 

R2  500  (250  10 -3 )  104.17 mH Ls   1200  R1


1 Z2  R 2 , , jC s R1 jC s R1 Z1   1 jR 1C s  1 R1  jC s

Z1  R 1 ||

Since Z x 

Z 3  R 3 , and Z x  R x  jL x .

Z3 Z , Z1 2

R x  jL x  R 2 R 3

jR 1C s  1 R 2 R 3  (1  jR 1C s ) R1 R1

Equating the real and imaginary components, R 2R 3 Rx  R1 L x 

R 2R 3 (R 1C s ) implies that R1 L x  R 2 R 3Cs

Given that R 1  40 k , R 2  1.6 k , R 3  4 k , and C s  0.45 F R 2 R 3 (1.6)(4)  k  0.16 k  160  R1 40 L x  R 2 R 3 C s  (1.6)(4)(0.45)  2.88 H

Rx 


1 , jC 2 R4 - jR 4 Z4   jR 4 C 4  1 R 4 C 4  j Z1  R 1 ,

Since Z 4 

Z2  R 2 

Z3 Z Z1 2

Z 3  R 3 , and Z 4  R 4 ||

  Z1 Z 4  Z 2 Z 3 ,

 - jR 4 R 1 j    R 3 R 2  R 4 C 4  j C 2   jR 3 - jR 4 R 1 (R 4 C 4  j)  R 3R 2  2 2 2  R 4C4  1 C 2

Equating the real and imaginary components, R 1R 4  R 2R 3 2  R 24 C 24  1 (1) R3 R 1 R 24 C 4  2 R 24 C 24  1 C 2 (2) Dividing (1) by (2), 1  R 2 C 2 R 4 C 4 1 2  R 2C2R 4C4 1   2f  R 2C2 R 4C4

f

1 2 R 2 R 4 C 2 C 4

1 . jC 4

Chapter 9, Solution 86. 1 1 1   240 j95 - j84 Y  4.1667  10 -3  j0.01053  j0.0119 Y

Z

1 1000 1000   Y 4.1667  j1.37 4.386118.2 Z = 228-18.2 

Chapter 9, Solution 87. 1 -j  50  jC (2)(2  10 3 )(2  10 -6 ) Z1  50  j39.79 Z1  50 

Z 2  80  jL  80  j (2)(2  10 3 )(10  10 -3 ) Z 2  80  j125.66 Z 3  100

1 1 1 1    Z Z1 Z 2 Z 3 1 1 1 1    Z 100 50  j39.79 80  j125.66 1  10 -3 (10  12.24  j9.745  3.605  j5.663) Z  (25.85  j4.082)  10 -3  26.17  10 -3 8.97 Z = 38.21–8.97 


Z  - j20  j30  120  j20 Z = (120 – j10) 

(b)

If the frequency were halved,

1 1  would cause the capacitive C 2f C impedance to double, while L  2f L would cause the inductive impedance to halve. Thus, Z  - j40  j15  120  j40 Z = (120 – j65) 

Chapter 9, Solution 89. An industrial load is modeled as a series combination of an inductor and a resistance as shown in Fig. 9.89. Calculate the value of a capacitor C across the series combination so that the net impedance is resistive at a frequency of 2 kHz.

10  C 5 mH

Figure 9.89 For Prob. 9.89. Solution Step 1. There are different ways to solve this problem but perhaps the easiest way is to convert the series R L elements into their parallel equivalents. Then all you need to do is to make the inductance and capacitance cancel each other out to result in a purely resistive circuit. X L = 2x103x5x10–3 = 10 which leads to Y = 1/(10+j10) = 0.05–j0.05 or a 20Ω resistor in parallel with a j20Ω inductor. X c = 1/(2x103C) and the parallel combination of the capacitor and inductor is equal to, [(–jX C )(j20)/(–jX C +j20)]. Step 2.

Now we just need to set X C = 20 = 1/(2x103C) which will create an open circuit. C = 1/(20x2x103) = 25 µF.


Vs  1450 , X  L  (2)(60) L  377 L Vs 1450 I  80  R  jX 80  R  jX

V1  80 I 

50 

(80)(145) 80  R  jX

(80)(145) 80  R  jX (1)

Vo  (R  jX) I  110 

(R  jX)(1450) 80  R  jX

(R  jX)(145) 80  R  jX (2)

From (1) and (2), 50 80  110 R  jX 11  R  jX  (80)   5

R 2  X 2  30976 (3) From (1),

(80)(145)  232 50 6400  160R  R 2  X 2  53824 160R  R 2  X 2  47424 80  R  jX 

(4) Subtracting (3) from (4), 160R  16448   R  102.8  From (3), X 2  30976  10568  20408 X  142.86  377L   L  378.9 mH

Chapter 9, Solution 91. 1  R || jL jC -j jLR Z in   C R  jL Z in 



- j  2 L2 R  jLR 2  C R 2   2 L2

To have a resistive impedance, Im(Z in )  0 . Hence, -1 LR 2  0 C R 2  2 L2 1 LR 2  C R 2  2 L2 R 2   2 L2 C 2 LR 2 where   2 f  2  10 7

9  10 4  (4 2  1014 )(400  10 12 ) (4 2  1014 )(20  10  6 )(9  10 4 ) 9  16 2 C nF 72 2 C

C = 235 pF


(a) Z o  (b)  

Z  Y

10075 o  471.413.5 o  45048 o x10  6

ZY  10075 o x 450 48 o x10 6  212.1 61.5 o mS

Chapter 9, Solution 93. Z  Zs  2 Z  ZL Z  (1  0.8  23.2)  j(0.5  0.6  18.9) Z  25  j20 IL 

VS 1150  Z 32.02 38.66

I L  3.592–38.66 A

Chapter 10, Solution 1. We first determine the input impedance. 1H

 

j L  j1x10  j10

1F

 

1

j C



1   j 0.1 j10 x1

1

 1 1 1 Zin  1      1.0101 j 0.1  1.015  5.653o  j10  j 0.1 1 I

2  0o  1.9704  5.653o o 1.015  5.653 i(t) = 1.9704cos(10t+5.65˚) A

Chapter 10, Solution 2. Using Fig. 10.51, design a problem to help other students better understand nodal analysis. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Solve for V o in Fig. 10.51, using nodal analysis. 2 + o

40 V

j4 

+ _

–j5 

Vo

–

Figure 10.51 For Prob. 10.2. Solution Consider the circuit shown below. 2 Vo –j5

+ 40 V- _

j4

o

At the main node, 4  Vo V V  o  o 2  j5 j 4

  40  Vo (10  j )

V o = 40/(10–j) = (40/10.05)5.71˚ = 3.985.71˚ V


 4 2 cos(4 t )   20 16 sin(4 t )   16 - 90  -j16 2H  

jL  j8 1 1 1 12 F     - j3 jC j (4)(1 12) The circuit is shown below. 4

-j16 V

-j3 

+ 

Vo

1

j8 

20 A

Applying nodal analysis,

- j16  Vo Vo Vo 2  4  j3 1 6  j8  1 1  - j16 V  2  1   4  j3  4  j3 6  j8  o Vo 

3.92  j2.56 4.682  - 33.15   3.835 - 35.02 1.22  j0.04 1.2207 1.88

Therefore, v o ( t )  3.835cos(4t – 35.02) V

6

Chapter 10, Solution 4. Step 1.

Convert the circuit into the frequency domain and solve for the node voltage, V 1 , using analysis. The find the current I C = V 1 /[1+(1/(j4x0.25)] which then produces V o = 1xI C . Finally, convert the capacitor voltage back into the time domain. Ix j1 –j1 Ω V1 160º V

0.5I x

Vo

Note that we represented 16sin(4t–10º) volts by 160º V. That will make our calculations easier and all we have to do is to offset our answer by a –10º. Our node equation is [(V 1 –16)/j] – (0.5I x ) + [(V 1 –0)/(1–j)] = 0. We have two unknowns, therefore we need a constraint equation. I x = [(16–V 1 )/j] = j(V 1 –16). Once we have V 1 , we can find I o = V 1 /(1–j) and V o = 1xI o . Step 2.

Now all we need to do is to solve our equations. [(V 1 –16)/j] – [0.5j(V 1 –16] + [(V 1 –0)/(1–j)] = [–j–j0.5+0.5+j0.5]V 1 +j16+j8 = 0 or [0.5–j]V 1 = –j24 or V 1 = j24/(–0.5+j) = (2490º)/(1.118116.57º) = 21.47–26.57º V. Finally, I x = V 1 /(1–j) = (21.47–26.57º) (0.707145º) = 15.18118.43º A and V o = 1xI o = 15.18118.43º V or v o (t) = 15.181sin(4t–10º+18.43º) = 15.181sin(4t–8.43º) volts.


0.25H

2 F

j L  j 0.25 x4 x103  j1000

 

 

1

j C



1   j125 j 4 x10 x2 x106 3

Consider the circuit as shown below.

2000

Io

Vo

250o V + _

-j125

j1000 + –

At node V o , Vo  25 Vo  0 Vo  10I o   0 2000 j1000  j125 Vo  25  j2Vo  j16Vo  j160I o  0 (1  j14)Vo  j160I o  25

But I o = (25–V o )/2000

(1  j14)Vo  j2  j0.08Vo  25 Vo 

25  j2 25.084.57  1.7768  81.37 1  j14.08 14.11558.94

Now to solve for i o , 25  Vo 25  0.2666  j1.7567   12.367  j0.8784 mA 2000 2000  12.3984.06

Io 

i o = 12.398cos(4x103t + 4.06˚) mA.

10I o


Let V o be the voltage across the current source. Using nodal analysis we get:

Vo  4Vx Vo 20 3  0 where Vx  Vo 20  j10 20  j10 20 Combining these we get:

Vo 4Vo Vo  3  0  (1  j0.5  3)Vo  60  j30 20 20  j10 20  j10 Vo 

60  j30 20(3) or Vx    2  j0.5  2  j0.5 29.11–166˚ V.

Chapter 10, Solution 7. At the main node,

120  15 o  V V V  630 o   40  j20  j30 50

 

115.91  j31.058  5.196  j3  40  j20

 1 j 1 V     40  j20 30 50  V

 3.1885  j4.7805  124.08  154 o V 0.04  j0.0233


  200,

100mH 50F

   

jL  j200x 0.1  j20 1 1    j100 jC j200x 50x10  6

The frequency-domain version of the circuit is shown below.

0.1 V o 40  V1

20 

615 o

+ Vo -

Io

V2

-j100  j20 

At node 1,

or

V V1 V  V2  1 615 o  0.1V1  1  20  j100 40 5.7955  j1.5529  (0.025  j 0.01)V1  0.025V2

(1)

At node 2, V1  V2 V  0.1V1  2 40 j20 From (1) and (2),



0  3V1  (1  j2)V2

(0.025  j0.01)  0.025 V1   (5.7955  j1.5529)      3 (1  j2)  V2   0  

Using MATLAB,

or

AV  B

(2)

V = inv(A)*B leads to V1  70.63  j127.23,

V2  110.3  j161.09

V  V2 Io  1  7.276  82.17 o 40

Thus, i o ( t )  7.276 cos( 200t  82.17 o ) A


10 cos(10 3 t )   10 0,   10 3

10 mH    50 F 

jL  j10 1 1   - j20 3 jC j (10 )(50  10 -6 )

Consider the circuit shown below. 20 

V1

-j20 

V2

j10 

Io

100 V

+ 

20 

+ 4 Io

30 

Vo 

At node 1, 10  V1 V1 V1  V2   20 20 - j20 10  (2  j) V1  jV2

(1)

At node 2, V1  V2 V V2 V1  (4) 1  , where I o  has been substituted. 20 - j20 20 30  j10 (-4  j) V1  (0.6  j0.8) V2 0.6  j0.8 (2) V1  V2 -4 j Substituting (2) into (1) (2  j)(0.6  j0.8) 10  V2  jV2 -4 j 170 or V2  0.6  j26.2 Vo  Therefore,

30 3 170 V2    6.154 70.26 30  j10 3  j 0.6  j26.2 v o ( t )  6.154 cos(103 t + 70.26) V


 

50 mH 2F

 

jL  j2000x50 x10  3  j100, 1 1    j250 jC j2000 x 2x10  6

  2000

Consider the frequency-domain equivalent circuit below.

V1

360o

2k 

-j250

j100

V2

0.1V 1

4k 

At node 1,

36 

V1 V V  V2  1  1 2000 j100  j250

 

36  (0.0005  j0.006)V1  j0.004V2

(1)

0  (0.1  j0.004)V1  (0.00025  j0.004)V2

(2)

At node 2, V1  V2 V  0.1V1  2  j250 4000



Solving (1) and (2) gives

Vo  V2  535.6  j893.5  8951.193.43o v o (t) = 8.951 sin(2000t +93.43o) kV

Chapter 10, Solution 11. Consider the circuit as shown below. –j5 

Io

2

2 V1

40o V

+ _

V2 j8  2I o

At node 1, V1  4 V  V2  2I o  1 0 2 2 V1  0.5V2  2I o  2

But, I o = (4–V 2 )/(–j5) = –j0.2V 2 + j0.8 Now the first node equation becomes, V 1 – 0.5V 2 + j0.4V 2 – j1.6 = 2 or V 1 + (–0.5+j0.4)V 2 = 2 + j1.6 At node 2,

V2  V1 V2  4 V2  0   0 2  j5 j8 –0.5V 1 + (0.5 + j0.075)V 2 = j0.8 Using MATLAB to solve this, we get, >> Y=[1,(-0.5+0.4i);-0.5,(0.5+0.075i)] Y=

1.0000 -0.5000

-0.5000 + 0.4000i 0.5000 + 0.0750i

>> I=[(2+1.6i);0.8i] I= 2.0000 + 1.6000i 0 + 0.8000i >> V=inv(Y)*I V= 4.8597 + 0.0543i 4.9955 + 0.9050i I o = –j0.2V 2 + j0.8 = –j0.9992 + 0.01086 + j0.8 = 0.01086 – j0.1992 = 199.586.89˚ mA.

Chapter 10, Solution 12. Using Fig. 10.61, design a problem to help other students to better understand Nodal analysis. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem By nodal analysis, find i o in the circuit in Fig. 10.61.


20 sin(1000t )   20 0,   1000 10 mH    50 F 

jL  j10 1 1   - j20 3 jC j (10 )(50  10 -6 )

The frequency-domain equivalent circuit is shown below. 2 Io

V1

10 

V2 Io

200 A

At node 1,

20 

-j20 

j10 

20  2 I o 

V1 V1  V2  , 20 10

V2 j10 2V2 V1 V1  V2 20    j10 20 10 400  3V1  (2  j4) V2 (1)

Io 

At node 2,

or

2V2 V1  V2 V V   2  2 j10 10 - j20 j10 j2 V1  (-3  j2) V2 V1  (1  j1.5) V2 (2)

Substituting (2) into (1), 400  (3  j4.5) V2  (2  j4) V2  (1  j0.5) V2

Therefore,

V2 

400 1  j0.5

Io 

V2 40   35.74 - 116.6 j10 j (1  j0.5)

i o ( t )  35.74 sin(1000t – 116.6) A

where


Nodal analysis is the best approach to use on this problem. We can make our work easier by doing a source transformation on the right hand side of the circuit.

–j2 

18 

j6 

+ 4030º V

+ 

Vx

3



Vx  4030 Vx Vx  50   0  j2 3 18  j6 which leads to V x = 29.3662.88˚ A.

500º V

+ 

Chapter 10, Solution 14. At node 1,

0  V1 0  V1 V2  V1    2030 - j2 10 j4 - (1  j2.5) V1  j2.5 V2  173.2  j100

(1)

V2 V2 V2  V1    20 30 j2 - j5 j4 - j5.5 V2  j2.5 V1  173.2  j100

(2)

At node 2,

Equations (1) and (2) can be cast into matrix form as 1  j2.5 j2.5  V1   - 200 30   j2.5 - j5.5 V2   200 30  



1  j2.5

j2.5

j2.5

- j5.5

 20  j5.5  20.74 - 15.38

1 

- 200 30 j2.5  j3 (20030)  600120 200 30 - j5.5

2 

1  j2.5 - 20030  (200 30)(1  j5)  1020108.7 j2.5 20030

1  28.93135.38 V  2 V2   49.18124.08 V  V1 

Chapter 10, Solution 15. We apply nodal analysis to the circuit shown below.

5A

2

j

V1

V2

I -j20 V

+ 

-j2 

4

2I

At node 1,

- j20  V1 V V  V2  5 1  1 2 - j2 j - 5  j10  (0.5  j0.5) V1  j V2

(1)

At node 2, V1  V2 V2 ,  j 4 V where I  1 - j2 5 V2  V1 0.25  j 5  2I 

(2)

Substituting (2) into (1), j5  0.5 (1  j) V1 0.25  j j40 (1  j) V1  -10  j20  1  j4 160 j40  ( 2  - 45) V1  -10  j20  17 17 V1  15.81313.5 - 5  j10 

I

V1  (0.590)(15.81313.5) - j2 I  7.90643.49 A

Chapter 10, Solution 16. Consider the circuit as shown in the figure below. j4 

V1

V2

+ Vx – –j3 

20o A

345o A

5

At node 1,

V  0 V1  V2 2 1  0 5 j4 (0.2  j0.25)V1  j0.25V2  2

(1)

At node 2,

V2  V1 V2  0   345  0 j4  j3 j0.25V1  j0.08333V2  2.121  j2.121 In matrix form, (1) and (2) become 0.2  j0.25  j0.25 

j0.25   V1   2      j0.08333 V2  2.121  j2.121

Solving this using MATLAB, we get, >> Y=[(0.2-0.25i),0.25i;0.25i,0.08333i] Y= 0.2000 - 0.2500i 0 + 0.2500i

0 + 0.2500i 0 + 0.0833i

>> I=[2;(2.121+2.121i)] I=

(2)

2.0000 2.1210 + 2.1210i >> V=inv(Y)*I V= 5.2793 - 5.4190i 9.6145 - 9.1955i V s = V 1 – V 2 = –4.335 + j3.776 = 5.749138.94˚ V.

Chapter 10, Solution 17. Consider the circuit below.

j4  10020 V

+ 

1 Io

2

V1

V2

3

-j2 

At node 1,

10020  V1 V1 V1  V2   j4 3 2 V1 100 20  (3  j10)  j2 V2 3 (1) At node 2, 10020  V2 V1  V2 V2   1 2 - j2 100 20  -0.5 V1  (1.5  j0.5) V2 (2) From (1) and (2), 10020  - 0.5 0.5 (3  j)  V1  10020  1  j10 3 - j2  V2    



- 0.5

1.5  j0.5

1  j10 3

- j2

 0.1667  j4.5

1 

10020 1.5  j0.5  -55.45  j286.2 10020 - j2

2 

- 0.5 10020  -26.95  j364.5 1  j10 3 10020

1  64.74  - 13.08  2 V2   81.17  - 6.35  V  V2  1   2 - 28.5  j78.31  Io  1  2 2 0.3333  j 9 V1 

I o  9.25-162.12 A


8

V1

j6  V 2

4

j5 

+ 2

445 A

Vx

+ 2 Vx

-j 

-j2 



Vo 

At node 1,

V1 V1  V2  2 8  j6 200 45  (29  j3) V1  (4  j3) V2 (1) 445 

At node 2, V1  V2 V V2  2Vx  2  , 8  j6 - j 4  j5  j2 (104  j3) V1  (12  j41) V2 12  j41 V1  V 104  j3 2 (2) Substituting (2) into (1),

(12  j41) V  (4  j3) V2 104  j3 2 200 45  (14.2189.17) V2 20045 V2  14.2189.17

200 45  (29  j3)

- j2 - j2 - 6  j8 V2  V2  V2 4  j5  j2 4  j3 25 10233.13 200 45 Vo   25 14.2189.17

Vo 

Vo  5.63189 V

where Vx  V1

Chapter 10, Solution 19. We have a supernode as shown in the circuit below.

j2 

V1

V2

4 V3

+ 2

-j4 

Vo

0.2 V o



Notice that

Vo  V1 .

At the supernode, V3  V2 V2 V1 V1  V3    4 - j4 2 j2 0  (2  j2) V1  (1  j) V2  (-1  j2) V3 At node 3, V1  V3 V3  V2 0.2V1   j2 4 (0.8  j2) V1  V2  (-1  j2) V3  0

(2)

Subtracting (2) from (1), 0  1.2V1  j V2

(3)

But at the supernode, V1  12 0  V2 or V2  V1  12

(4)

Substituting (4) into (3), 0  1.2V1  j (V1  12) j12 V1   Vo 1.2  j

Vo 

1290 1.56239.81 Vo  7.68250.19 V

(1)

Chapter 10, Solution 20. The circuit is converted to its frequency-domain equivalent circuit as shown below.

R + V m 0

+ 

jL

Vo

1 jC



Let

Z  jL ||

1  jC

L C jL 

1 jC

If

Vo  A , then A

and

jL 1  2 LC

jL jL 1  2 LC Vm  V jL R (1  2 LC)  jL m R 1  2 LC   L Vm L 90  tan -1   R (1  2 LC)  R 2 (1  2 LC) 2  2 L2 

Z Vo  V  RZ m

Vo 



L Vm R 2 (1   2 LC) 2   2 L2

  90  tan -1

L R (1   2 LC)


(a)

Vo  Vi

1 jC R  jL 

1 jC

As    ,

(b)

Vo  Vi

1 LC

Vo  Vi

,

jL R  jL 

As    , 1 LC

1 jC



1 jRC 

1



-j L R C

LC

 2 LC 1  2 LC  jRC

Vo  0 Vi Vo 1   1 Vi 1

At   0 ,

At  

1 1  2 LC  jRC

Vo 1   1 Vi 1 Vo  0 Vi

At   0 ,

At  



,

Vo  Vi

1 jRC 

1 LC



j L R C

Chapter 10, Solution 22. Consider the circuit in the frequency domain as shown below. R1

R2 Vs

1 jC

+ 

jL

Let

Z  (R 2  jL) ||

+ Vo 

1 jC

1 (R  jL) R 2  jL jC 2 Z  1 1  jR 2  2 LC R 2  jL  jC R 2  jL Vo 1  2 LC  jR 2 C Z   R 2  jL Vs Z  R 1 R1  1  2 LC  jR 2 C Vo R 2  jL  2 Vs R 1  R 2   LCR 1  j (L  R 1 R 2 C)


V  Vs V   jCV  0 1 R jL  j C V

jRCV  2LC  1

 jRCV  Vs

 1  2LC  jRC  jRC  j3RLC2    V  Vs 2   1   LC   V 

(1   2 LC )V s 1   2 LC  jRC ( 2   2 LC )

Chapter 10, Solution 24. Design a problem to help other students to better understand mesh analysis. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Use mesh analysis to find V o in the circuit in Prob. 10.2. Solution Consider the circuit as shown below. 2 + o

40 V

I1

+ _

j4  –j5 

Vo

I2 –

For mesh 1, 4  (2  j 5) I1  j 5 I1 For mesh 2,

(1)

0  j 5I 1  ( j 4  j 5) I 2

  I1 

1 I2 5

(2)

Substituting (2) into (1), 1 4  (2  j 5) I 2  j 5I 2 5

  I2 

1 0.1  j

V o = j4I 2 = j4/(0.1+j) = j4/(1.00499 84.29°) = 3.98 5.71° V


 2 10 cos(2 t )   100

6 sin(2t )   6  - 90  -j6 2H  

jL  j4 1 1 0.25 F     - j2 jC j (2)(1 4)

The circuit is shown below.

4

j4  Io

100 V

+ 

-j2 

I1

I2

+ 

6-90 V

For loop 1, - 10  (4  j2) I 1  j2 I 2  0 5  (2  j) I 1  j I 2 (1)

For loop 2, j2 I 1  ( j4  j2) I 2  (- j6)  0 I1  I 2  3

(2) In matrix form (1) and (2) become  2  j j   I 1   5  1 1   I    3   2   

  2 (1  j) ,

I o  I1  I 2 

 1  5  j3 ,

 2  1  j3

1   2 4   1  j  1.414245  2 (1  j )

Therefore, i o ( t )  1.4142cos(2t + 45) A

Chapter 10, Solution 26. 0.4 H

1 F

 

 

j L  j10 3 x 0.4  j 400

1

j C



1   j1000 j10 x10 6 3

The circuit becomes that shown below.

2 k –j1000 Io 100o

+ _

+ _

I1 j400

–j20

I2

For loop 1, 10  (12000  j 400) I1  j 400 I 2  0   1  (200  j 40) I1  j 40 I 2 (1) For loop 2,  j 20  ( j 400  j1000) I 2  j 400 I1  0   12  40 I1  60 I 2 (2) In matrix form, (1) and (2) become  1   200  j 40  j 40   I1   12    40 60   I 2     Solving this leads to I 1 =0.0025-j0.0075, I 2 = -0.035+j0.005 I o = I 1 – I 2 = 0.0375 – j0.0125 = 39.5 –18.43° mA

i o (t) = 39.5cos(103t–18.43°) mA

Chapter 10, Solution 27. For mesh 1, - 40 30  ( j10  j20) I 1  j20 I 2  0 4 30  - j I 1  j2 I 2

(1)

50 0  (40  j20) I 2  j20 I 1  0 5  - j2 I 1  (4  j2) I 2

(2)

For mesh 2,

From (1) and (2),  430  - j j2  I 1   5    - j2 - (4  j2)  I      2 

  -2  4 j  4.472116.56  1  -(4 30)(4  j2)  j10  21.01211.8  2  - j5  8120  4.44 154.27

I1 

1  4.69895.24 A 

I2 

2  992.837.71 mA 

Chapter 10, Solution 28. 1 1    j0.25 jC j1x 4 The frequency-domain version of the circuit is shown below, where 1H

 

V1  100 o ,

jL  j4,

 

1F

V2  20  30 o .

1

j4

j4

1

-j0.25 +

+

V1

I1

1

I2

V2 -

-

V1  100 o ,

V2  20  30 o

Applying mesh analysis, 10  (2  j3.75)I1  (1  j0.25)I 2

(1)

 20  30 o  (1  j0.25)I1  (2  j3.75)I 2

(2)

From (1) and (2), we obtain 10    2  j3.75  1  j0.25  I1          17.32  j10    1  j0.25 2  j3.75  I 2 

Solving this leads to I1  2.741  41.07 o ,

I 2  4.11492 o

Hence, i 1 (t) = 2.741cos(4t–41.07˚)A, i 2 (t) = 4.114cos(4t+92˚)A.

Chapter 10, Solution 29. Using Fig. 10.77, design a problem to help other students better understand mesh analysis. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem By using mesh analysis, find I 1 and I 2 in the circuit depicted in Fig. 10.77.


For mesh 1, (5  j5) I 1  (2  j) I 2  30 20  0 30 20  (5  j5) I 1  (2  j) I 2 (1)

For mesh 2, (5  j3  j6) I 2  (2  j) I 1  0 0  - (2  j) I 1  (5  j3) I 2 (2)

From (1) and (2), 3020  5  j5 - (2  j)  I 1   0    - (2  j) 5 - j3  I   2    

  37  j6  37.489.21  1  (30 20)(5.831 - 30.96)  175 - 10.96  2  (30 20)(2.356 26.56)  67.0846.56

I1 

1  4.67–20.17 A 

I2 

2  1.7937.35 A 


300mH

 

j L  j100 x300 x10 3  j30

200mH

 

j L  j100 x200 x10 3  j20

400mH

 

j L  j100 x400 x10 3  j40

1   j200 j100 x50 x10 6 The circuit becomes that shown below. 50  F

1

 

j C



j40

j20

20 

+ 12090

10 

o

+ _

I1

j30

–j200 I2

vo

I 3`

-

+ _

800o

For mesh 1, 120  90o  (20  j30) I1  j 30 I 2  0   j120  (20  j 30) I1  j 30 I 2 For mesh 2,  j 30 I1  ( j 30  j 40  j 200) I 2  j 200 I 3  0   0  3I1  13I 2  20 I 3 For mesh 3, 80  j200I 2  (10  j180)I 3  0  8  j20I 2  (1  j18)I 3 (3)

We put (1) to (3) in matrix form. 0   I1   j12 2  j3  j3   3  13 20  I 2    0    0 j20 1  j18  I 3    8

This is an excellent candidate for MATLAB. >> Z=[(2+3i),-3i,0;-3,-13,20;0,20i,(1-18i)] Z= 2.0000 + 3.0000i

0 - 3.0000i

0

(1) (2)

-3.0000 0

-13.0000 20.0000 0 +20.0000i 1.0000 -18.0000i

>> V=[12i;0;-8] V= 0 +12.0000i 0 -8.0000 >> I=inv(Z)*V I= 2.0557 + 3.5651i 0.4324 + 2.1946i 0.5894 + 1.9612i V o = –j200(I 2 – I 3 ) = –j200(–0.157+j0.2334) = 46.68 + j31.4 = 56.2633.93˚ v o = 56.26cos(100t + 33.93˚) V.

Chapter 10, Solution 31. Consider the network shown below.

80 

100120 V

+ 

I1

Io

-j40 

j60 

I2

-j40 

20 

I3

+ 

60-30 V

For loop 1, - 100120  (80  j40) I1  j40 I 2  0 10 20  4 (2  j) I 1  j4 I 2

(1)

j40 I 1  ( j60  j80) I 2  j40 I 3  0 0  2 I1  I 2  2 I 3

(2)

60  - 30  (20  j40) I 3  j40 I 2  0 - 6 - 30  j4 I 2  2 (1  j2) I 3

(3)

For loop 2,

For loop 3,

From (2), 2 I 3  I 2  2 I1 Substituting this equation into (3), - 6  - 30  -2 (1  j2) I 1  (1  j2) I 2

(4)

From (1) and (4),  10120   4 (2  j) j4  I 1   - 6  - 30   - 2 (1  j2) 1  j2 I      2 



2 

8  j4 - j4  32  j20  37.7432 - 2  j4 1  j2 8  j4 10120  -4.928  j82.11  82.2593.44 - 2  j4 - 6 - 30

Io  I2 

2  2.17961.44 A 


j4  Io

+ 2

4-30 V

Vo

I1

+ 

3 Vo

I2

 For mesh 1, where

(2  j4) I 1  2 (4  - 30)  3 Vo  0 Vo  2 (4  - 30  I 1 )

Hence, (2  j4) I 1  8 - 30  6 (4  - 30  I 1 )  0 4  - 30  (1  j) I 1

I 1  2 2 15

or

Io 

3 Vo 3  (2)(4 - 30  I 1 ) - j2 - j2

I o  j3 (4 - 30  2 2 15)

I o  8.48515 A Vo 

- j2 I o  5.657-75 V 3

-j2 

5A


I4 2

j I

-j20 V

+ 

I1

-j2 

I2

2I

I3

4

For mesh 1, j20  (2  j2) I 1  j2 I 2  0 (1  j) I 1  j I 2  - j10

(1)

For the supermesh, ( j  j2) I 2  j2 I 1  4 I 3  j I 4  0

(2)

Also, I 3  I 2  2 I  2 (I 1  I 2 ) I 3  2 I1  I 2

(3)

I4  5

(4)

For mesh 4,

Substituting (3) and (4) into (2), (8  j2) I 1  (- 4  j) I 2  j5

(5)

Putting (1) and (5) in matrix form,  1 j j  I 1   - j10  8  j2 4  j I    j5     2  

  -3  j5 , I  I1  I 2 

 1  -5  j40 ,

 2  -15  j85

 1   2 10  j45   - 3  j5  7.90643.49 A

Chapter 10, Solution 34. The circuit is shown below.

Io I2

5

3A 20 

8

4090 V

+ 

-j2 

I3

10  I1

j15 

j4 

For mesh 1, - j40  (18  j2) I 1  (8  j2) I 2  (10  j4) I 3  0 For the supermesh, (13  j2) I 2  (30  j19) I 3  (18  j2) I 1  0

(1)

(2)

Also, I2  I3  3 Adding (1) and (2) and incorporating (3), - j40  5 (I 3  3)  (20  j15) I 3  0 3  j8 I3   1.46538.48 5  j3 I o  I 3  1.46538.48 A

(3)


4

j2 

Consider the circuit shown below. I3 8

1

-j3 

10  20 V

+ 

I1

-j4 A

I2 -j5 

For the supermesh, - 20  8 I 1  (11  j8) I 2  (9  j3) I 3  0

(1)

Also, I 1  I 2  j4

(2)

(13  j) I 3  8 I 1  (1  j3) I 2  0

(3)

For mesh 3,

Substituting (2) into (1), (19  j8) I 2  (9  j3) I 3  20  j32

(4)

Substituting (2) into (3), - (9  j3) I 2  (13  j) I 3  j32

(5)

From (4) and (5),  19  j8 - (9  j3)  I 2   20  j32   - (9  j3) 13  j  I    j32    3   

  167  j69 , I2 

 2  324  j148

 2 324  j148 356.2  - 24.55    167  j69 180.69  - 22.45 I 2  1.971–2.1 A

Chapter 10, Solution 36. Consider the circuit below. j4 

-j3  +

490 A

I1

2

Vo

I2

+ 

 2

2 I3

20 A Clearly, I 1  4 90  j4

and

I 3  -2

For mesh 2, (4  j3) I 2  2 I 1  2 I 3  12  0 (4  j3) I 2  j8  4  12  0 - 16  j8 I2   -3.52  j0.64 4  j3 Thus, Vo  2 (I 1  I 2 )  (2)(3.52  j4.64)  7.04  j9.28 Vo  11.64852.82 V

120 V


I1 + 120  90 o V -

Ix

Z

Z=80-j35 

I2 Iz

o 120  30 V +

Iy

Z I3

For mesh x, ZI x  ZI z   j120

(1)

ZI y  ZI z  12030 o  103.92  j60

(2)

 ZI x  ZI y  3ZI z  0

(3)

For mesh y,

For mesh z,

Putting (1) to (3) together leads to the following matrix equation: 0 (80  j35)  I x    j120   (80  j35)      0 (80  j35) (80  j35)  I y     103.92  j60     (80  j35) (80  j35) (240  j105)  I   0   z   

Using MATLAB, we obtain  - 0.2641  j2.366    I  inv(A) * B   - 2.181 - j0.954   - 0.815  j1.1066   

I 1  I x  0.2641  j 2.366  2.38  96.37 o A I 2  I y  I x  1.9167  j1.4116  2.38143.63 o A

I 3   I y  2.181  j 0.954  2.38 23.63 o A

 

AI  B


Io I1 j2 

1

I2

2

20 A

+ 

-j4 

40 A

I3

I4

1090 V

1

A Clearly, I1  2

(1)

(2  j4) I 2  2 I 1  j4 I 4  10 90  0

(2)

For mesh 2,

Substitute (1) into (2) to get (1  j2) I 2  j2 I 4  2  j5 For the supermesh, (1  j2) I 3  j2 I 1  (1  j4) I 4  j4 I 2  0 j4 I 2  (1  j2) I 3  (1  j4) I 4  j4 At node A, I3  I4  4 Substituting (4) into (3) gives j2 I 2  (1  j) I 4  2 (1  j3) From (2) and (5), 1  j2 j2  I 2   2  j5  j2 1  j I    2  j6    4  

  3  j3 ,

 1  9  j11

-  1 - (9  j11) 1  (-10  j)  3  j3 3  I o  3.35174.3 A

Io  -I2 

(3) (4) (5)


For mesh 1,

(28  j15)I1  8I 2  j15I 3  1264 o

(1)

 8I1  (8  j9)I 2  j16I 3  0

(2)

j15I1  j16I 2  (10  j)I 3  0

(3)

For mesh 2,

For mesh 3,

In matrix form, (1) to (3) can be cast as

8 j15  I1  1264 o   (28  j15)    (8  j9)  j16  I 2    0   8    j15  j16 (10  j)  I 3   0    

or

AI  B

Using MATLAB, I = inv(A)*B I 1  0.128  j 0.3593  381.4109.6° mA I 2  0.1946  j 0.2841  344.3124.4° mA I 3  0.0718  j 0.1265  145.5–60.42° mA

I x  I 1  I 2  0.0666  j 0.0752  100.548.5° mA 381.4109.6° mA, 344.3124.4° mA, 145.5–60.42° mA, 100.548.5° mA

Chapter 10, Solution 40. Let I o = I o1 + I o2 , where I o1 is due to the dc source and I o2 is due to the ac source. For I o1 , consider the circuit in Fig. (a). Clearly,

4

2 I o1

+ 

8V

(a) I o1 = 8/2 = 4 A For I o2 , consider the circuit in Fig. (b).

4

2 I o2

100 V

+ 

j4 

(b) If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2  4 3  .

I o2 2.50 A

4

2

j4 

(c) By the current division principle, 43 Io2  (2.50) 4 3  j4 I o 2  0.25  j 0.75  0.79 - 71.56 Thus, I o 2  0.79 cos(4t  71.56) A Therefore, I o = I o1 + I o2 = [4 + 0.79cos(4t–71.56)] A

Chapter 10, Solution 41. We apply superposition principle. We let vo = v1 + v2 where v 1 and v 2 are due to the sources 6cos2t and 4sin4t respectively. To find v 1 , consider the circuit below. -j2

+ + _

60o

2

V1 –

1/ 4F

 

1

j C



1   j2 j2 x1/ 4

(6) = 3+j3 = 4.243 45°

Thus, v 1 (t) = 4.243cos(2t+45°) volts. To get v 2 (t), consider the circuit below, –j

+ 40o

+ _

2

V2 –

1/ 4 F

 

1

j C



1   j1 j 4 x1/ 4

or v 2 (t) = 3.578sin(4t+25.56°) volts. Hence, v o = [4.243cos(2t+45˚) + 3.578sin(4t+25.56˚)] volts.

Chapter 10, Solution 42. Using Fig. 10.87, design a problem to help other students to better understand the superposition theorem. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Solve for I o in the circuit of Fig. 10.87. j10 

o

200 V

+ _

50 

Io

60 

–j40 

+ _

3045o V

Figure 10.87 For Prob. 10.42. Solution Let I o  I1  I 2 where I 1 and I 2 are due to 20<0o and 30<45o sources respectively. To get I 1 , we use the circuit below. I1 j10  60 

o

200 V

+ _

50 

–j40 

Let Z 1 = -j40//60 = 18.4615 –j27.6927, Z 2 = j10//50=1.9231 + j9.615 Transforming the voltage source to a current source leads to the circuit below.

I1 Z2

Z1

–j2

Using current division, Z2 I1  ( j 2)  0.6217  j 0.3626 Z1  Z 2 To get I 2 , we use the circuit below.

j10 

I2

50 

60 

–j40 

+ _

After transforming the voltage source, we obtain the circuit below. I2

Z2

Z1 0.545o

Using current division,  Z1 I2  (0.5  45o )  0.5275  j 0.3077 Z1  Z 2

3045o V

Hence, I o = I 1 + I 2 = 0.0942+j0.0509 = 109 30° mA.

Chapter 10, Solution 43. Let I x  I 1  I 2 , where I 1 is due to the voltage source and I 2 is due to the current source.

 2 5 cos(2t  10)   510 10 cos(2t  60)   10  - 60 4H   1  F  8

jL  j8 1 1   -j4 jC j (2)(1 / 8)

For I 1 , consider the circuit in Fig. (a). -j4 

3

I1 + 

j8 

10-60 V

(a) I1 

10 - 60 10  - 60  3  j8  j4 3  j4

For I 2 , consider the circuit in Fig. (b). -j4 

510 A

3

j8  (b)

I2 

- j8 - j40 10 (510)  3  j8  j4 3  j4

1 (10 - 60  j4010) 3  j4 49.51 - 76.04 Ix   9.902 - 129.17 553.13

I x  I1  I 2 

Therefore,

i x  9.902 cos(2t – 129.17) A

I2

Chapter 10, Solution 44. Let v x  v1  v 2 , where v 1 and v 2 respectively. For v 1 ,   6 , 5 H

 

are due to the current source and voltage source

jL  j30

The frequency-domain circuit is shown below.

20 

16 

Is

Let Z  16 //(20  j30) 

+ V1 -

16(20  j30)  11.8  j3.497  12.3116.5 o 36  j30

V1  I s Z  (1210 o )(12.3116.5 o )  147.726.5 o

For v 2 ,   2 , 5 H

j30

 

 

v1  147.7 cos(6 t  26.5 o ) V

jL  j10

The frequency-domain circuit is shown below. 20 

16 

j10

+ V2

+

Vs -

-

Using voltage division,

-

V2 

16(500 o ) 16  21.41  15.52 o Vs  36  j10 16  20  j10

 

v 2  21.41sin(2t  15.52 o ) V

Thus, v x  [147.7cos(6t+26.5°)+21.41sin(2t–15.52°)] V

Chapter 10, Solution 45. Let i  i1  i2 , where i 1 and i 2 are due to 16cos(10t +30o) and 6sin4t sources respectively. To find i 1 , consider the circuit below.

I1

20 

+ _ 16 30o V

jX

X   L  10 x300 x103  3 Type equation here.

= 0.7913 21.47°

i 1 (t) = 791.1cos(10t+21.47°) mA.

To find i 2 (t), consider the circuit below,

I2

20 

+ _

60o V jX

X   L  4 x300 x103  1.2

= 0.2995 176.57° or

i 2 (t) = 299.5sin(4t+176.57°) mA. Thus, i(t) = i 1 (t) + i 2 (t) = [791.1cos(10t+21.47°) + 299.5sin(4t+176.57°)] mA.

Chapter 10, Solution 46. Let v o  v1  v 2  v 3 , where v1 , v 2 , and v 3 are respectively due to the 10-V dc source, the ac current source, and the ac voltage source. For v1 consider the circuit in Fig. (a).

6

2H +

1/12 F

+ 

v1

10 V

 (a) The capacitor is open to dc, while the inductor is a short circuit. Hence, v1  10 V For v 2 , consider the circuit in Fig. (b).  2 2H   jL  j4 1 1 1 F     - j6 jC j (2)(1 / 12) 12

+ 6

-j6 

40 A

V2  (b)

Applying nodal analysis, V V V 1 j j  4  2  2  2      V2 6 - j6 j4  6 6 4  V2  Hence,

24  21.4526.56 1  j0.5 v 2  21.45 sin( 2 t  26.56) V

For v 3 , consider the circuit in Fig. (c). 3

j4 

2H   jL  j6 1 1 1    - j4 F  12 jC j (3)(1 / 12)

6

j6  +

120 V

+ 

-j4 

V3  (c)

At the non-reference node, 12  V3 V3 V3   6 - j4 j6 12 V3   10.73 - 26.56 1  j0.5 Hence, v 3  10.73 cos(3t  26.56) V Therefore, v o  [10+21.45sin(2t+26.56)+10.73cos(3t–26.56)] V

Chapter 10, Solution 47. Let i o  i1  i 2  i 3 , where i1 , i 2 , and i 3 are respectively due to the 24-V dc source, the ac voltage source, and the ac current source. For i1 , consider the circuit in Fig. (a). 1

24 V

1/6 F

2H

 +

i1

2

4

(a) Since the capacitor is an open circuit to dc, 24 4A i1  42

For i 2 , consider the circuit in Fig. (b). 1 2H   jL  j2 1 1   - j6 F  6 jC 1

j2 

-j6 

I2 10-30 V

+ 

2

I1

I2

4

(b)

For mesh 1, - 10  - 30  (3  j6) I 1  2 I 2  0 10  - 30  3 (1  2 j) I 1  2 I 2

(1)

0  -2 I 1  (6  j2) I 2 I 1  (3  j) I 2

(2)

For mesh 2,

Substituting (2) into (1)

Hence,

10  - 30  13  j15 I 2 I 2  0.504 19.1 i 2  0.504 sin( t  19.1) A

For i 3 , consider the circuit in Fig. (c). 3 2H   jL  j6 1 1 1 F     - j2 jC j (3)(1 / 6) 6 1

j6 

-j2 

I3 2

20 A

4

(c)

2 || (1  j2) 

2 (1  j2) 3  j2

Using current division, 2 (1  j2)  (20) 2 (1  j2) 3  j2 I3   2 (1  j2) 13  j3 4  j6  3  j2 I 3  0.3352  - 76.43 Hence i 3  0.3352 cos(3t  76.43) A Therefore,

i o  [4 + 0.504 sin(t + 19.1) + 0.3352 cos(3t – 76.43)] A

Chapter 10, Solution 48. Let i O  i O1  i O 2  i O 3 , where i O1 is due to the ac voltage source, i O 2 is due to the dc voltage source, and i O3 is due to the ac current source. For i O1 , consider the circuit in Fig. (a).   2000 50 cos(2000t )   500

40 mH  

jL  j (2000)(40  10 -3 )  j80 1 1   - j25 jC j (2000)(20  10 -6 )

 20 F 

I

500 V

-j25 

I O1

+ 

80 

j80 

100 

60  (a)

80 || (60  100)  160 3 50 30 I  160 3  j80  j25 32  j33

Using current division,

- 80 I -1 10180  I 80  160 3 4645.9  0.217 134.1 i O1  0.217 cos(2000 t  134.1) A

I O1  I O1 Hence,

For i O 2 , consider the circuit in Fig. (b). i O2 80 

100 

60 

+ 

(b)

24 V

i O2 

24  0.1 A 80  60  100

For i O3 , consider the circuit in Fig. (c).   4000 2 cos(4000t )   20 40 mH  

20 F  

jL  j (4000)(40  10 -3 )  j160 1 1   - j12.5 jC j (4000)(20  10 -6 )

-j12.5 

I2

I O3

80  j160  I3 20 A

60 

I1

(c) ` For mesh 1, I1  2

(1) For mesh 2, (80  j160  j12.5) I 2  j160 I 1  80 I 3  0 Simplifying and substituting (1) into this equation yields (8  j14.75) I 2  8 I 3  j32 (2) For mesh 3, 240 I 3  60 I 1  80 I 2  0 Simplifying and substituting (1) into this equation yields I 2  3 I 3  1.5 (3)

100 

Substituting (3) into (2) yields (16  j44.25) I 3  12  j54.125 12  j54.125 I3   1.17827.38 16  j44.25

Hence,

I O 3  - I 3  -1.17827.38 i O 3  -1.1782 sin( 4000t  7.38) A

Therefore, i O  {0.1 + 0.217 cos(2000t + 134.1) – 1.1782 sin(4000t + 7.38)} A


8 sin( 200t  30)   830,   200

5 mH    1 mF 

jL  j (200)(5  10 -3 )  j 1 1   - j5 jC j (200)(1  10 -3 )

After transforming the current source, the circuit becomes that shown in the figure below. 5

4030 V

I

3

I j

+ 

-j5 

40 30 40 30   4.47256.56 5  3  j  j5 8  j4

i  [4.472sin(200t+56.56)] A

Chapter 10, Solution 50. Using Fig. 10.95, design a problem to help other students to better understand source transformation. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Use source transformation to find v o in the circuit in Fig. 10.95.


5 cos(10 5 t )

 50,   10 5

0.4 mH    0.2 F 

jL  j (10 5 )(0.4  10 -3 )  j40 1 1   - j50 5 jC j (10 )(0.2  10 -6 )

After transforming the voltage source, we get the circuit in Fig. (a). j40  + 20 

0.250

-j50 

80 

Vo 

(a)

Let and

Z  20 || - j 50 

- j100 2  j5

Vs  (0.250) Z 

- j25 2  j5

With these, the current source is transformed to obtain the circuit in Fig.(b). j40 

Z

+ Vs

+ 

80 

Vo 

(b)

By voltage division, 80 80 - j25  Vs  - j100 2  j5 Z  80  j40  80  j40 2  j5 8 (- j25) Vo   3.615 - 40.6 36  j42 v o  3.615 cos(105 t – 40.6) V Vo 

Therefore,

Chapter 10, Solution 51. There are many ways to create this problem, here is one possible solution. Let V 1 = 40 30° V, X L = 10 Ω, X C = 20 Ω, R 1 = R 2 = 80 Ω, and V 2 = 50 V.

If we let the voltage across the capacitor be equal to V x , then I o = [V x /(–j20)] + [(V x –50)/80] = (0.0125+j0.05)V x – 0.625 = (0.051539 75.96°)V x – 0.625. The following circuit is obtained by transforming the voltage sources. Vx

4-60 V

j10 

-j20 

40 

V x = (4-60+1.25)/(–j0.1+j0.05+0.025) = (2–j3.4641+1.25)/(0.025–j0.05) = (3.25–j3.4641)/( 0.025–j0.05) = (4.75 –46.826°)/(0.055902 –63.435°)

= 84.97 16.609° V.

Therefore,

1.250 A

I o = (0.051539 75.96°)(84.97 16.609°) – 0.625 = 4.3793 92.569° – 0.625

= –0.196291+j4.3749 – 0.625 = –0.821291+j4.3749 = 4.451 100.63° A.

Chapter 10, Solution 52. We transform the voltage source to a current source. 60 0 Is   6  j12 2  j4 The new circuit is shown in Fig. (a). -j2  Ix 2

4

6

I s = 6 – j12

590 A

j4 

-j3  (a)

Let

6 (2  j4)  2.4  j1.8 8  j4 Vs  I s Z s  (6  j12)(2.4  j1.8)  36  j18  18 (2  j) Z s  6 || (2  j4) 

With these, we transform the current source on the left hand side of the circuit to a voltage source. We obtain the circuit in Fig. (b). Zs

-j2  Ix

Vs

4

+ 

j5 A -j3  (b)

Let

Z o  Z s  j2  2.4  j0.2  0.2 (12  j) Vs 18 (2  j) Io    15.517  j6.207 Z o 0.2 (12  j)

With these, we transform the voltage source in Fig. (b) to a current source. We obtain the circuit in Fig. (c). Ix 4 Io

j5 A

Zo -j3  (c)

Using current division, Zo 2.4  j0.2 Ix  (I o  j5)  (15.517  j1.207) Z o  4  j3 6.4  j3.2 I x  5  j1.5625  5.23817.35 A

Chapter 10, Solution 53. We transform the voltage source to a current source to obtain the circuit in Fig. (a). -j3 

j4  +

4

50 A

j2 

2

Vo

-j2 

 (a) Let

j8  0.8  j1.6 4  j2 Vs  (50) Z s  (5)(0.8  j1.6)  4  j8

Z s  4 || j2 

With these, the current source is transformed so that the circuit becomes that shown in Fig. (b). Zs

-j3 

j4  +

Vs

+ 

2

-j2 

Vo 

(b) Let

Z x  Z s  j3  0.8  j1.4 V 4  j8 Ix  s   3.0769  j4.6154 Z s 0.8  j1.4

With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c). j4  + Ix

Zx

2

-j2 

Vo 

(c) Let

Z y  2 || Z x 

1.6  j2.8  0.8571  j0.5714 2.8  j1.4

Vy  I x Z y  ( 3.0769  j4.6154)  (0.8571  j0.5714)  j5.7143 With these, we transform the current source to obtain the circuit in Fig. (d). j4 

Zy

+ Vy

+ 

-j2 

Vo 

(d) Using current division, Vo 

- j2 ( j5.7143) - j2 Vy   (3.529 – j5.883) V Z y  j4  j2 0.8571  j0.5714  j4  j2

Chapter 10, Solution 54. 50 x( j 30)  13.24  j 22.059 50  j 30 We convert the current source to voltage source and obtain the circuit below. 50 //(  j 30) 

40 

+ V s =115.91 –j31.06V

13.24 – j22.059 

j20 

+

-

I

134.95-j74.912 V

V -

+ -

Applying KVL gives -115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0 or I 

 250.86  j105.97  4.7817  j1.8055 53.24  j 2.059

But  Vs  (40  j20)I  V  0

 

V  Vs  (40  j20)I

V  115.91  j 31.05  (40  j 20)( 4.7817  j1.8055)  124.06  154 o V which agrees with the result in Prob. 10.7.


To find Z th , consider the circuit in Fig. (a). j20 

10 

Z th

-j10 

(a) ( j20)(- j10) j20  j10  10  j20  22.36-63.43 

Z N  Z th  10  j20 || (- j10)  10 

To find Vth , consider the circuit in Fig. (b). j20 

10  +

5030 V

+ 

-j10 

V th  (b)

Vth 

IN 

(b)

- j10 (50 30)  -5030 V j20  j10 Vth - 50 30  2.236273.4 A  Z th 22.36  - 63.43

To find Z th , consider the circuit in Fig. (c). -j5 

8

j10 

(c)

Z th

Z N  Z th  j10 || (8  j5) 

( j10)(8  j5)  1026  j10  8  j5

To obtain Vth , consider the circuit in Fig. (d). -j5  Io 40 A

8

j10 

+ V th 

(d) By current division, 8 32 Io  (4 0)  8  j10  j5 8  j5 Vth  j10 I o 

IN 

j320  33.9258 V 8  j5

Vth 33.92 58   3.39232 A 10 26 Z th


To find Z th , consider the circuit in Fig. (a). j4  6

-j2 

Z th

(a) ( j4)(- j2)  6  j4 j4  j2 = 7.211-33.69 

Z N  Z th  6  j4 || (- j2)  6 

By placing short circuit at terminals a-b, we obtain, I N  20 A Vth  Z th I th  (7.211 - 33.69) (2 0)  14.422-33.69 V (b)

To find Z th , consider the circuit in Fig. (b). j10 

30 

60 

-j5 

(b)

30 || 60  20 (- j5)(20  j10) 20  j5 = 5.423-77.47 

Z N  Z th  - j5 || (20  j10) 

Z th

To find Vth and I N , we transform the voltage source and combine the 30  and 60  resistors. The result is shown in Fig. (c). j10 

445 A

20 

a IN

-j5 

(c) 20 2 (4 45)  (2  j)(445) 20  j10 5 = 3.57818.43 A

IN 

Vth  Z th I N  (5.423 - 77.47) (3.57818.43) = 19.4-59 V

b

Chapter 10, Solution 57. Using Fig. 10.100, design a problem to help other students to better understand Thevenin and Norton equivalent circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the Thevenin and Norton equivalent circuits for the circuit shown in Fig. 10.100.


To find Z th , consider the circuit in Fig. (a). 5

-j10 

2

Z th

j20 

(a) ( j20)(5  j10) 5  j10  18  j12  21.633-33.7 

Z N  Z th  2  j20 || (5  j10)  2 

To find Vth , consider the circuit in Fig. (b). 5

-j10 

2 +

60120 V

+ 

j20 

V th

 (b)

j20 j4 (60 120)  (60120) 5  j10  j20 1  j2 = 107.3146.56 V

Vth 

IN 

Vth 107.3146.56   4.961-179.7 A Z th 21.633 - 33.7

Chapter 10, Solution 58. Consider the circuit in Fig. (a) to find Z eq .

8

Z eq

j10 

-j6  (a) Z eq  j10 || (8  j 6) 

( j10)(8  j 6)  5 (2  j ) 8  j4 = 11.1826.56 

Consider the circuit in Fig. (b) to find V Thev .

Io

+

8 j10 

545 A -j6  (b)

Io 

4  j3 8  j6 (545) (545)  4  j2 8  j6  j10

VThev  j10 I o 

( j10)(4  j 3)(545) (2)(2  j ) = 55.971.56 V

V Thev

Chapter 10, Solution 59. Calculate the output impedance of the circuit shown in Fig. 10.102. –j2 Ω + 0.2V o

Vo

10  

j40 Ω

Figure 10.102 For Prob. 10.59. Solution Since there are no independent sources, we need to inject a current, best value is to make it 1 amp, into the terminals on the right and then to determine the voltage at the terminals. –j2 Ω V1 + 0.2V o

Vo

10  

a 1A

j40 Ω b

Clearly V o = –(–j2) = j2 and V 1 = (0.2V o + 1)j40 = (1+j0.4)j40 = –16+j40 V. Next, V ab = 10 – j2 – 16 + j40 = –6+j38 = 38.47 98.97° V or

Z eq = (–6+j38) Ω.


To find Z eq , consider the circuit in Fig. (a).

10 

-j4  a

j5 

Z eq

4

b

(a)

Z eq  4 || (- j 4  10 || j 5)  4 || (- j 4  2  j 4) Z eq  4 || 2 = 1.333  To find VThev , consider the circuit in Fig. (b). 10 

-j4 

V1

V2 +

200 V

+ 

j5 

40 A

4

V Thev 

(b)

At node 1, 20  V1 V1 V1  V2   10 j5 - j4 (1  j0.5) V1  j2.5 V2  20

(1)

At node 2, V1  V2 V2  - j4 4 V1  (1  j) V2  j16 4

Substituting (2) into (1) leads to 28  j16  (1.5  j3) V2 28  j16  8  j5.333 V2  1.5  j3

(2)

Therefore, VThev  V2  9.61533.69 V (b)

To find Z eq , consider the circuit in Fig. (c).

Z eq c

d

10 

-j4 

j5 

4

(c)  j10   Z eq  - j 4 || (4  10 || j 5)  - j 4 ||  4  2  j  

Z eq  - j 4 || (6  j 4) 

- j4 (6  j 4)  (2.667 – j4)  6

To find VThev ,we will make use of the result in part (a). V2  8  j5.333  (8 3 ) (3  j2) V1  (1  j) V2  j16  j16  (8 3) (5  j) VThev  V1  V2  16 3  j8  9.61456.31 V

Chapter 10, Solution 61. Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 10.104. V 1 and V 2

4 a

Ix -j3 

o

20 A

+ 1.5I x

V oc

I sc

 b Figure 10.104 For Prob. 10.61. Solution Step 1.

First we solve for the open circuit voltage using the above circuit and writing two node equations. Then we solve for the short circuit current which only need one node equation. For being able to solve for V oc , we need to solve these three equations, –2 + [(V 1 –0)/(–j3)] + [(V 1 –V oc )/4] = 0 and [(V oc –V 1 )/4] – 1.5I x = 0 where I x = [(V 1 –0)/(–j3)]. To solve for I sc , all we need to do is to solve these three equations, –2 + [(V 2 –0)/(–j3)] + [(V 2 –0)/4] = 0, I sc = [V 2 /4] + 1.5I x , and I x = [V 2 /–j3]. Finally, V Thev = V oc and Z eq = V oc /I sc .

Step 2.

Now all we need to do is to solve for the unknowns. For V oc , I x = j0.33333V 1 and (0.25+(1.5)(j0.33333))V 1 = 0.25V oc or (0.25+j0.5)V 1 = (0.55902 63.43°)V 1 = 0.25V oc or

V 1 = (0.44721 –63.43°)V oc which leads to,

(0.25+j0.33333)V 1 – 0.25V oc = 2 = (0.41666

53.13°)(0.44721 –63.43°)V oc – 0.25V oc

= (0.186335 –10.3°)V oc – 0.25V oc = (0.183333–0.25–j0.033333)V oc

= (–0.066667–j0.033333)V oc = (0.074536 –153.435°)V oc = 2 or

V oc = V Thev = 26.83 153.44° V = (–24+j12) V.

Now for I sc , I sc = [V 2 /4] + 1.5I x = (0.25+(1.5)(j0.33333))V 2 = (0.25+j0.5)V 2 . [(V 2 –0)/(–j3)] + [(V 2 –0)/4] = 2 = (0.25+j0.3333)V 2 = (0.41667 53.13°)V 2 = 2 or V 2 = 4.8 –53.13°

I sc = (0.25+j0.5)V 2 = (0.55901 63.435°)(4.8 –53.13°)

= 2.6832 10.305° A

Finally,

Z eq = V oc /I sc = 26.833 153.435°/2.6832 10.305°

= 10 143.13° Ω or = (–8+j6) Ω.

Chapter 10, Solution 62. First, we transform the circuit to the frequency domain. 12 cos( t )   120,   1

2H   1 F   4 1 F   8

jL  j2 1  - j4 jC 1  - j8 jC

To find Z eq , consider the circuit in Fig. (a). 3 Io

Io

4

Vx

j2 

1

Ix 2

-j4 

-j8 

+ 

1V

(a)

At node 1, Vx Vx 1  Vx   3Io  , j2 4 - j4 Thus,

where I o 

Vx 2 Vx 1  Vx   - j4 4 j2 Vx  0.4  j0.8

At node 2, I x  3Io 

1 1  Vx  - j8 j2

I x  (0.75  j0.5) Vx  j

3 8

I x  -0.1  j0.425

Z eq 

1  -0.5246  j 2.229  2.29 - 103.24  Ix

- Vx 4

To find VThev , consider the circuit in Fig. (b). 3 Io

Io

4

j2 

V1 1

120 V

+ 

V2 2

-j4 

-j8 

+ V Thev 

(b) At node 1, 12  V1 V V  V2 12  V1  3Io  1  1 , where I o  4 4 - j4 j2 24  (2  j) V1  j2 V2

(1)

V1  V2 V2  3Io  j2 - j8 72  (6  j4) V1  j3 V2

(2)

At node 2,

From (1) and (2),  24  2  j - j2  V1   72    6  j4 - j3  V   2   

  -5  j6 , Vth  V2 

Thus,

 2  - j24 2  3.073 - 219.8 

2 (2)(3.073 - 219.8) Vth  2  Z th 1.4754  j2.229 6.146 - 219.8 Vo   2.3 - 163.3 2.673 - 56.5 Vo 

Therefore, v o  2.3cos(t–163.3) V

Chapter 10, Solution 63. Transform the circuit to the frequency domain.

4 cos(200t  30)   4 30,   200

10 H    5 F 

jL  j (200)(10)  j2 k 1 1   - j k jC j (200)(5  10 -6 )

Z N is found using the circuit in Fig. (a). -j k

j2 k

ZN

2 k

(a) Z N  - j  2 || j2  - j  1  j  1 k

We find I N using the circuit in Fig. (b). -j k

430 A

j2 k

2 k

(b)

j2 || 2  1  j By the current division principle, 1 j IN  (4 30)  5.657 75 1 j  j Therefore, i N (t) = 5.657 cos(200t + 75) A Z N  1 k

IN

Chapter 10, Solution 64. Z N is obtained from the circuit in Fig. (a).

60 

40 

ZN

-j30 

j80  (a) Z N  (60  40) || ( j80  j30)  100 || j50 

(100)( j50) 100  j50

Z N  20  j40  44.7263.43  To find I N , consider the circuit in Fig. (b).

60 

I1

40 

I2

-j30 

IN 360 A

Is j80 

(b) I s  360 For mesh 1, 100 I 1  60 I s  0 I 1  1.860 For mesh 2, ( j80  j30) I 2  j80 I s  0 I 2  4.860 I N = I 2 – I 1 = 360 A

Chapter 10, Solution 65. Using Fig. 10.108, design a problem to help other students to better understand Norton’s theorem. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Compute i o in Fig. 10.108 using Norton's theorem.


5 cos(2 t )   50,   2

4H   1 F   4 1 F   2

jL  j (2)(4)  j8 1 1   - j2 jC j (2)(1 / 4) 1 1   -j jC j (2)(1 / 2)

To find Z N , consider the circuit in Fig. (a). 2 ZN -j2 

-j 

(a)

Z N  - j || (2  j2) 

- j (2  j2) 1  (2  j10) 2  j3 13

To find I N , consider the circuit in Fig. (b). 50 V

2

+ 

-j2 

-j 

IN (b)

IN 

50  j5 -j

The Norton equivalent of the circuit is shown in Fig. (c). Io ZN

IN

j8 

(c)

Using current division, ZN (1 13)(2  j10)( j5) 50  j10  IN  Io  (1 13)(2  j10)  j8 2  j94 Z N  j8 I o  0.1176  j0.5294  0542 - 77.47 Therefore, i o  542 cos(2t – 77.47) mA

Chapter 10, Solution 66.   10 0.5 H  

jL  j (10)(0.5)  j5 1 1    - j10 10 mF  jC j (10)(10  10 -3 ) To find Z th , consider the circuit in Fig. (a).

-j10 

Vx

+ 10 

j5 

Vo

2 Vo

1A



(a) Vx Vx  , j5 10  j10 19 Vx V - 10  j10 1  x   Vx  10  j10 j5 21  j2 1  2 Vo 

Z N  Z th 

where Vo 

10Vx 10  j10

Vx 14.142 135   670129.56 m 1 21.0955.44

To find Vth and I N , consider the circuit in Fig. (b). 120 V

-j10 

 +

+

+ -j2 A

10 

Vo

j5 

I

V th 



(b)

where

2 Vo

(10  j10  j5) I  (10)(- j2)  j5 (2 Vo )  12  0 Vo  (10)(- j2  I )

Thus,

(10  j105) I  -188  j20 188  j20 I - 10  j105 Vth  j5 (I  2 Vo )  j5 (19I  j40)   j95 I  200  j 95 (188  j 20) (95  90)(189.066.07) Vth   200   200 - 10  j105 105.4895.44  170.28  179.37  200  170.27  j1.8723  200  29.73  j1.8723

Vth  29.79–3.6 V IN 

Vth 29.79  3.6   44.46–133.16 A Z th 0.67129.56


Z N  Z Th  10 //(13  j 5)  12 //( 8  j 6) 

Va 

10 (6045 o )  13.78  j21.44, 23  j5

10(13  j 5) 12(8  j 6)   11.243  j1.079 23  j 5 20  j 6 Vb 

(8  j6) (6045 o )  12.069  j26.08 20  j6

VTh  Va  Vb  1.711  j 4.64  4.945  69.76 o V, IN 

VTh 4.945  69.76   437.8  75.24 o mA 11.295 5.48 Z Th


1H   jL  j10x1  j10 1 1 1      j2 F 1 20 j C j10x 20 We obtain V Th using the circuit below. Io

4 a

+

+

6<0o -

V o /3

+ -

-j2 j10

b

j10( j2)   j2.5 j10  j2 Vo  4I o x ( j2.5)   j10I o 1  6  4I o  Vo  0 3 j10 //( j2) 

(1) (2)

Combining (1) and (2) gives Io 

6 , 4  j10 / 3

Vo

4I o

VTh  Vo   j10I o 

 j60  11.52  50.19 o 4  j10 / 3

v Th  11.52 sin(10t  50.19 o )

To find R Th, we insert a 1-A source at terminals a-b, as shown below. Io

4 a + + -

V o /3

-j2 j10

Vo

4I o -

1 4I o  Vo  0 3



V Io   o 12

Vo Vo   j2 j10 Combining the two equations leads to 1  4I o 

Vo 

1  1.2293  j1.4766 0.333  j0.4 V Z Th  o  1.2293  1.477 1

1<0o

Chapter 10, Solution 69. This is an inverting op amp so that Vo - Z f -R    –jRC 1 jC Vs Zi When Vs  Vm and   1 RC , 1 Vo  - j   RC  Vm  - j Vm  Vm  - 90 RC

Therefore, v o ( t )  Vm sin(t  90)  - V m cos(t)

Chapter 10, Solution 70. Using Fig. 10.113, design a problem to help other students to better understand op amps in AC circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem The circuit in Fig. 10.113 is an integrator with a feedback resistor. Calculate v o (t) if v s = 2 cos 4  104t V.

Figure 10.113 Solution This may also be regarded as an inverting amplifier. 2 cos(4  10 4 t )   2 0,   4  10 4 1 1 10 nF     - j2.5 k 4 jC j (4  10 )(10  10 -9 ) Vo - Z f  Vs Zi

where Z i  50 k and Z f  100k || (- j2.5k )  Thus, If Vs  2 0 ,

Vo - (- j2)  Vs 40  j

- j100 k . 40  j

Vo 

j4 490   0.191.43 40  j 40.01 - 1.43

Therefore, v o ( t )  100 cos(4x104 t + 91.43) mV


8 cos(2t  30 o )

 830 o 1 1 0. 5F      j1M jC j2x 0.5x10  6 At the inverting terminal, Vo  830 o Vo  830 o 830 o    j1000k 10k 2k

 

Vo (1  j100)  830  800  60  4000   60 Vo 

6.928  j4  2400  j4157 4800  59.9   4829.53o 1  j100 100  89.43 v o (t) = 48cos(2t+29.53o) V


4 cos(10 4 t )   4 0,   10 4 1 1 1 nF     - j100 k 4 jC j (10 )(10 -9 )

Consider the circuit as shown below. 50 k

Vo + 

40 V

+ 

-j100 k

At the noninverting node, 4  Vo Vo  50 - j100 Io 

  Vo 

Vo Io 100 k

4 1  j0.5

Vo 4  mA  35.78 - 26.56 A 100k (100)(1  j0.5)

Therefore, i o ( t )  35.78cos(104t–26.56) A

Chapter 10, Solution 73. As a voltage follower, V2  Vo

1 1   -j20 k 3 jC1 j (5  10 )(10  10 -9 ) 1 1 C 2  20 nF     -j10 k 3 jC 2 j (5  10 )(20  10 -9 )  C1  10 nF 

Consider the circuit in the frequency domain as shown below.

-j20 k

I s 10 k

20 k

+ 

V1 VS

+ 

-j10 k

Z in At node 1, Vs  V1 V1  Vo V1  Vo   10 - j20 20 2 Vs  (3  j)V1  (1  j)Vo (1)

At node 2, V1  Vo Vo  0  20 - j10 V1  (1  j2)Vo

(2) Substituting (2) into (1) gives 2 Vs  j6Vo

or

V2

1 Vo  -j Vs 3

2 1 V1  (1  j2)Vo    j  Vs 3 3

Io Vo

Vs  V1 (1 3)(1  j)  Vs 10k 10k 1 j  30k

Is 

Is Vs

Vs 30k   15 (1  j) k Is 1  j Z in  21.21–45 k

Z in 


Zi  R1 

1 , jC1

Zf  R 2 

1 jC 2

1  C   1  j R 2 C 2  V - Zf jC 2      1   Av  o  1 Vs Zi C 2   1  jR 1C1   R1  jC1 R2 

Av  –

At   0 ,

As    ,

Av  –

C1 C2

R2 R1

At  

1 , R 1 C1

 C   1  j R 2 C 2 R 1C1   A v  – 1   1 j   C2  

At  

1 , R 2C2

C   1 j  A v  – 1    C 2   1  j R 1C1 R 2 C 2 


  2  10 3

C1  C 2  1 nF  

1 1   -j500 k 3 jC1 j (2  10 )(1  10 -9 )

Consider the circuit shown below. 100 k

Let V s = 10V. -j500 k

-j500 k

V2 + 

V1

40 k VS

+ 

100 k

Vb

+ Vo

20 k



At node 1, [(V 1 –10)/(–j500k)] + [(V 1 –V o )/105] + [(V 1 –V 2 )/(–j500k)] = 0 or (1+j0.4)V 1 – j0.2V 2 – V o = j2

(1)

[(V 2 –V 1 )/(–j500k)] + [(V 2 –0)/100k] + 0 = 0 or –j0.2V 1 + (1+j0.2)V 2 = 0 or V 1 = [–(1+j0.2)/(–j0.2)]V 2 = (1–j5)V 2

(2)

At node 2,

At node b, Vb =

R3 V Vo  o  V 2 3 R3  R4

From (2) and (3), V 1 = (0.3333–j1.6667)V o Substituting (3) and (4) into (1), (1+j0.4)(0.3333–j1.6667)V o – j0.06667V o – V o = j2 (1+j0.4)(0.3333–j1.6667) = (1.07721.8˚)(1.6997–78.69˚) = 1.8306–56.89˚ = 1–j1.5334

(3)

(4)

(1–1+j(–1.5334–0.06667))V o = (–j1.6001)V o = 1.6001–90˚ Therefore, V o = 290˚/(1.6001–90˚) = 1.2499180˚ Since V s = 10, V o /V s = 0.12499180˚.

Chapter 10, Solution 76. Let the voltage between the -jk  capacitor and the 10k  resistor be V 1. 230 o  V1 V1  Vo V1  Vo    j4k 10k 20k

 

(1)

230 o  (1  j0.6)V1  j0.6Vo = 1.7321+j1 Also,

V1  Vo V  o  j2k 10k



V1  (1  j5)Vo

(2)

Solving (2) into (1) yields 230  (1  j0.6)(1  j5)Vo  j0.6Vo  (1  3  j0.6  j5  j6)Vo = (4+j5)V o 230 Vo   0.3124  21.34 o V 6.40351.34

= 312.4–21.34˚ mV I o = (V 1 –V o )/20k = V o /(–j4k) = (0.3124/4k)(–21.43+90)˚

= 78.168.57˚ µA We can easily check this answer using MATLAM. Using equations (1) and (2) we can identify the following matrix equations: YV = I where

>> Y=[1-0.6i,0.6i;1,-1-0.5i] Y= 1.0000 - 0.6000i 0 + 0.6000i 1.0000 -1.0000 - 5.0000i >> I=[1.7321+1i;0] I= 1.7321 + 1.0000i 0

>> V=inv(Y)*I V= 0.8593 + 1.3410i 0.2909 - 0.1137i = V o = 312.3–21.35˚ mV. The answer checks.


R3 2 R1 1 + 

VS

At node 1,

V1

C2

R2

 +

V1 C1

+ Vo 

Vs  V1  jC V1 R1 Vs  (1  jR 1C1 ) V1

(1)

At node 2, 0  V1 V1  Vo   jC 2 (V1  Vo ) R3 R2

 R3   jC 2 R 3  V1  (Vo  V1 )  R2    1  V1 Vo  1   (R 3 R 2 )  jC 2 R 3 

From (1) and (2), Vo 

  Vs R2  1  1  jR 1C1  R 3  jC 2 R 2 R 3  Vo R 2  R 3  jC 2 R 2 R 3  Vs (1  jR 1C 1 ) ( R 3  jC 2 R 2 R 3 )

(2)

Chapter 10, Solution 78. 2 sin(400t )   20,   400 1 1 0.5 F     - j5 k jC j (400)(0.5  10 -6 ) 1 1    - j10 k 0.25 F  jC j (400)(0.25  10 -6 ) Consider the circuit as shown below.

20 k 10 k V 1

-j5 k

V2 + 

Vo 40 k

20 V

+ 

-j10 k

10 k 20 k

At node 1,

V V  V2 V1  Vo 2  V1  1  1  10 - j10 - j5 20 4  (3  j6) V1  j4 V2  Vo

(1)

V1  V2 V2  10  j5 V1  (1  j0.5) V2

(2)

At node 2,

But

20 1 Vo  Vo 20  40 3 From (2) and (3), 1 V1   (1  j0.5) Vo 3 Substituting (3) and (4) into (1) gives 1 4 1  4  (3  j6)   (1  j0.5) Vo  j Vo  Vo  1  j  Vo 3 3 6  24 Vo   3.945  9.46 6 j Therefore, v o ( t )  3.945sin(400t–9.46) V V2 

(3)

(4)

Chapter 10, Solution 79. 0.5 cos(1000t )   0.50,   1000 1 1    - j10 k 0.1 F  jC j (1000)(0.1  10 -6 ) 1 1 0.2 F     - j5 k jC j (1000)(0.2  10 -6 )

Consider the circuit shown below. 20 k

-j10 k

40 k

10 k

V s = 0.50

 +

+ 

V1

 +

-j5 k

Since each stage is an inverter, we apply Vo 

- Zf V to each stage. Zi i

Vo 

- 40 V1 - j5

(1)

V1 

- 20 || (- j10) Vs 10

(2)

and

From (1) and (2),  - j8  - ( 20)(-j10)  Vo    0.50   10  20  j10  Vo  1.6 ( 2  j)  35.7826.56 Therefore,

v o ( t )  3.578cos(1000t + 26.56) V

+ Vo 


4 cos(1000t  60)   4  - 60,   1000 1 1 0.1 F     - j10 k jC j (1000)(0.1  10 -6 ) 1 1    - j5 k 0.2 F  jC j (1000)(0.2  10 -6 ) The two stages are inverters so that  20 20  - j5   Vo   V   (4 - 60)  50 o  10   - j10 -j -j 2   ( j2)  (4 - 60)   Vo 2 2 5 (1  j 5) Vo  4 - 60 4 - 60 Vo   3.922  - 71.31 1 j 5 Therefore, v o ( t )  3.922 cos(1000t – 71.31) V

1 Chapter 10, Solution 81. We need to get the capacitance and inductance corresponding to –j2  and j4 . 1 1  j2   C   0.5 F  X c 1x 2 X j4   L  L  4H  The schematic is shown below.

When the circuit is simulated, we obtain the following from the output file.

FREQ VM(5) VP(5) 1.592E-01 1.127E+01 -1.281E+02 From this, we obtain V o = 11.27128.1o V.

Chapter 10, Solution 82. The schematic is shown below. We insert PRINT to print V o in the output file. For AC Sweep, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we print out the output file which includes:

FREQ 1.592 E-01 which means that

VM($N_0001) 7.684 E+00 V o = 7.68450.19o V

VP($N_0001) 5.019 E+01

Chapter 10, Solution 83. The schematic is shown below. The frequency is f   / 2 

1000  159.15 2

When the circuit is saved and simulated, we obtain from the output file FREQ 1.592E+02

VM(1) 6.611E+00

VP(1) -1.592E+02

Thus, v o = 6.611cos(1000t – 159.2o) V

Chapter 10, Solution 84. The schematic is shown below. We set PRINT to print V o in the output file. In AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes: FREQ

VM($N_0003)

1.592 E-01

1.664 E+00

VP($N_0003) E+02

Namely,

V o = 1.664–146.4o V

–1.646

Chapter 10, Solution 85. Using Fig. 10.127, design a problem to help other students to better understand performing AC analysis with PSpice. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Use PSpice to find V o in the circuit of Fig. 10.127. Let R 1 = 2 Ω, R 2 = 1 Ω, R 3 = 1 Ω, R 4 = 2 Ω, I s = 20˚ A, X L = 1 Ω, and X C = 1 Ω. Solution The schematic is shown below. We let   1 rad/s so that L=1H and C=1F.

When the circuit is saved and simulated, we obtain from the output file FREQ 1.591E-01

VM(1) 2.228E+00

VP(1) -1.675E+02

From this, we conclude that V o = 2.228–167.5° V.

Chapter 10, Solution 86. The schematic is shown below. We insert three pseudocomponent PRINTs at nodes 1, 2, and 3 to print V 1 , V 2 , and V 3 , into the output file. Assume that w = 1, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After saving and simulating the circuit, we obtain the output file which includes:

FREQ

VM($N_0002)

1.592 E-01

6.000 E+01

FREQ

VM($N_0003)

1.592 E-01

2.367 E+02

FREQ

VM($N_0001)

1.592 E-01

1.082 E+02

VP($N_0002) 3.000

E+01

VP($N_0003) -8.483

E+01

VP($N_0001) E+02

Therefore, V 1 = 6030o V V 2 = 236.7-84.83o V V 3 = 108.2125.4o V

1.254

Chapter 10, Solution 87. The schematic is shown below. We insert three PRINTs at nodes 1, 2, and 3. We set Total Pts = 1, Start Freq = 0.1592, End Freq = 0.1592 in the AC Sweep box. After simulation, the output file includes:

FREQ

VM($N_0004)

1.592 E-01

1.591 E+01

FREQ

VM($N_0001)

1.592 E-01

5.172 E+00

FREQ

VM($N_0003)

1.592 E-01

2.270 E+00

VP($N_0004) 1.696

E+02

VP($N_0001) -1.386

E+02

VP($N_0003) E+02

Therefore, V 1 = 15.91169.6o V V 2 = 5.172-138.6o V V 3 = 2.27-152.4o V

-1.524

Chapter 10, Solution 88. The schematic is shown below. We insert IPRINT and PRINT to print I o and V o in the output file. Since w = 4, f = w/2 = 0.6366, we set Total Pts = 1, Start Freq = 0.6366, and End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes:

FREQ

VM($N_0002)

6.366 E-01

3.496 E+01

1.261

FREQ

IM(V_PRINT2)

IP

6.366 E-01

8.912 E-01

VP($N_0002) E+01

(V_PRINT2) -8.870 E+01

Therefore,

V o = 34.9612.6o V, v o = 34.96 cos(4t + 12.6o)V,

I o = 0.8912-88.7o A i o = 0.8912cos(4t – 88.7o )A

Chapter 10, Solution 89. Consider the circuit below. R1

R2

V in

2

R3

1

V in

4

R4

3  +

0  Vin Vin  V2  R1 R2 R - Vin  V2  2 Vin R1

(1) At node 3, V2  Vin Vin  V4  R3 1 jC Vin  V2 - Vin  V4  jCR 3 (2)

From (1) and (2), - Vin  V4 

- R2 V jCR 3 R 1 in

Thus, I in 

Vin  V4 R2  V R4 jCR 3 R 1 R 4 in

Z in 

Vin jCR 1R 3 R 4   jL eq R2 I in L eq 

I in  +

At node 1,

where

C

R 1R 3 R 4C R2

+ 

V in


Z 4  R ||

1 R  jC 1  jRC 1 1  jRC Z3  R   jC jC

Consider the circuit shown below.

Z3

Vi

+ 

+

Z4

Vo 

R1

Vo R2

R2 Z4 Vi  V R1  R 2 i Z3  Z 4

R Vo R2 1  jC   R 1  jRC R 1  R 2 Vi  1  jC jC 

jRC R2  2 jRC  (1  jRC) R1  R 2

Vo R2 jRC   2 2 2 Vi 1   R C  j3RC R 1  R 2

For Vo and Vi to be in phase,

Vo must be purely real. This happens when Vi

1  2 R 2 C 2  0 1   2f RC

or

f

1 2RC

At this frequency, Vo 1 R2 Av    Vi 3 R 1  R 2


(a)

Let

V2  voltage at the noninverting terminal of the op amp Vo  output voltage of the op amp Z p  10 k  R o 1 jC

Z s  R  jL  As in Section 10.9, Zp V2   Vo Z s  Z p

Ro R  R o  jL 

j C

CR o V2  Vo C (R  R o )  j ( 2 LC  1)

For this to be purely real, o2 LC  1  0   o  fo 

1 2 LC



1 LC

1 2 (0.4  10 -3 )(2  10 -9 ) f o  180 kHz

(b)

At oscillation, o CR o Ro V2   Vo o C (R  R o ) R  R o This must be compensated for by Vo 80 Av   1 5 V2 20 Ro 1  R  Ro 5

  R  4R o  40 k


V2  voltage at the noninverting terminal of the op amp Vo  output voltage of the op amp Zs  R o RL 1 1  Z p  jL || || R  1 1 L  jR ( 2 LC  1) jC  jC  R jL

As in Section 10.9,

RL V2 L  jR (2 LC  1)   RL Vo Z s  Z p Ro  L  jR (2 LC  1) V2 RL  Vo RL  R o L  jR o R (2 LC  1) Zp

For this to be purely real, o2 LC  1   f o 

(a)

1 2 LC

At   o , o RL V2 R   Vo o RL  o R o L R  R o This must be compensated for by Vo Rf 1000k Av   1  1  11 Ro V2 100k Hence, R 1    R o  10R  100 k R  R o 11

(b)

fo 

1 2 (10  10 -6 )(2  10 -9 ) f o  1.125 MHz

Chapter 10, Solution 93. As shown below, the impedance of the feedback is jL

1 jC2

ZT 

1 jC1

ZT

 1 1   ||  jL  jC1  jC 2 

-j  -j  1  jL    LC 2 C1  C 2   ZT   -j -j j (C1  C 2   2 LC1C 2 )  jL  C1 C 2

In order for Z T to be real, the imaginary term must be zero; i.e. C1  C 2  o2 LC1 C 2  0 C  C2 1 o2  1  LC1C 2 LC T 1 fo  2 LC T

Chapter 10, Solution 94. If we select C1  C 2  20 nF C1 C 2 C1 CT    10 nF C1  C 2 2 Since f o 

1 2 LC T L

,

1 1   10.13 mH 2 2 (2f ) C T (4 )(2500  10 6 )(10  10 -9 )

Xc 

1 1   159  C 2 ( 2 )(50  10 3 )(20  10 -9 )

We may select R i  20 k and R f  R i , say R f  20 k . Thus,

C1  C 2  20 nF,

L  10.13 mH

R f  R i  20 k

Chapter 10, Solution 95. First, we find the feedback impedance. C ZT L2

L1

 1   Z T  jL1 ||  jL 2  jC  

 j   jL1  jL 2   2 L1C (1  L 2 ) C  ZT   j j (2 C (L1  L 2 )  1) jL1  jL 2  C

In order for Z T to be real, the imaginary term must be zero; i.e. o2 C (L 1  L 2 )  1  0 1  o  2 f o  C (L1  L 2 ) 1 fo  2 C ( L1  L2 )


Consider the feedback portion of the circuit, as shown below. jL

Vo

V2 

+ 

V1

R

V2

R

jL V R  jL 1

jL

  V1 

R  jL V2 jL

Applying KCL at node 1, Vo  V1 V1 V1   jL R R  jL  1 1  Vo  V1  jL V1    R R  jL 

 j2RL  2 L2   Vo  V1 1  R (R  jL)   (2)

From (1) and (2),  R  jL  j2RL  2 L2  V 1  Vo   R (R  jL)  2  jL  Vo R 2  jRL  j2RL  2 L2  V2 jRL

V2  Vo

1 R  2 L2 3 jRL 2

V2 1  Vo 3  j (L R  R L )

(1)

(b)

V2 must be real, Vo

Since the ratio

o L R  0 R o L o L 

R2 o L

 o  2 f o 

R L

fo 

(c)

R 2 L

When   o V2 1  Vo 3 This must be compensated for by A v  3 . But Av  1

R2 3 R1

R 2  2 R1


v( t )  160 cos(50t ) i(t) = –33sin(50t–30˚) = 33cos(50t–30˚+180˚–90˚) = 33cos(50t+60˚) p(t) = v(t)i(t) = 160x33cos(50t)cos(50t+60˚) = 5280(1/2)[cos(100t+60˚)+cos(60˚)] = [1.320+2.640cos(100t+60˚)] kW. P = [V m I m /2]cos(0–60˚) = 0.5x160x33x0.5 = 1.320 kW.

Chapter 11, Solution 2. Using current division,

j1 Ω

I1

I2

Vo -j4 Ω

I1 

j1  j 4  j6 (2)  5  j1  j 4 5  j3

I2 

5 10 (2)  5  j1  j 4 5  j3

20o A

5Ω

.

For the inductor and capacitor, the average power is zero. For the resistor, 1 1 P  | I1 |2 R  (1.029) 2 (5)  2.647 W 2 2 Vo  5I1  2.6471  j 4.4118 1 1 S  Vo I *  (2.6471  j 4.4118) x 2  2.6471  j 4.4118 2 2 Hence the average power supplied by the current source is 2.647 W.


I + –

90  F

C 1600˚

 

R

1 1    j 5.5556 6 j C j 90 x10 x 2 x103

I = 160/60 = 2.667A The average power delivered to the load is the same as the average power absorbed by the resistor which is P avg = 0.5|I|260 = 213.4 W.

Chapter 11, Solution 4. Using Fig. 11.36, design a problem to help other students better understand instantaneous and average power. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the average power dissipated by the resistances in the circuit of Fig. 11.36. Additionally, verify the conservation of power. Note, we do not talk about rms values of voltages and currents until Section 11.4, all voltages and currents are peak values. 5Ω

2030o V

j4 Ω

+ –

8Ω

–j6 Ω


Solution We apply nodal analysis. At the main node, I1

2030o V

+ –

5 ΩI 2

Vo

j4 Ω

8Ω

–j6 Ω

20  30o  Vo Vo V   o   Vo  5.152  j10.639 = 11.82164.16˚ 5 j 4 8  j6 For the 5-Ω resistor, 20  30o  Vo I1   2.438  3.0661o A 5 The average power dissipated by the resistor is 1 1 P1  | I1 |2 R1  x 2.4382 x5  14.86 W 2 2

For the 8-Ω resistor, I 2 = V o /(8–j6) = (11.812/10)(64.16+36.87)˚ = 1.1812101.03˚ A

The average power dissipated by the resistor is P 2 = 0.5|I 2 |2R 2 = 0.5(1.1812)28 = 5.581 W The complex power supplied is S = 0.5(V s )(I 1 )* = 0.5(2030˚)(2.4383.07˚) = 24.3833.07˚

= (20.43+13.303) VA Adding P 1 and P 2 gives the real part of S, showing the conservation of power. P = 14.86+5.581 = 20.44 W which checks nicely.


Converting the circuit into the frequency domain, we get: 2

1

8–40˚

I1 

+ 

j6

8  40  1.6828  25.38 j6(2  j2) 1 j6  2  j2

1.6828 2 1  1.4159 W P1  2

P 1Ω = 1.4159 W P 3H = P 0.25F = 0 W

I 2 

j6 1.6828  25.38  2.258 j6  2  j2

2.258 2 P2  2  5.097 W 2 P 2Ω = 5.097 W

–j2


  j L  j103 x 20 x103  j 20 1 1 40F     j25 jC j10 3 x 40x10  6 20 mH

We apply nodal analysis to the circuit below. Vo

+

20I x –

Ix j20 50

60o

–j25

10

V  20I x V 0 6 o  o 0 10  j20 50  j25 Vo But I x  . Substituting this and solving for V o leads 50  j25   1 20 1 1  Vo  6    10  j20 (10  j20) (50  j25) 50  j25    1 20 1  Vo  6    22.3663.43 (22.3663.43)(55.9  26.57) 55.9  26.57  0.02  j0.04  0.012802  j0.009598  0.016  j0.008Vo  6 (0.0232 – j0.0224)V o = 6 or V o = 6/(0.03225–43.99˚) = 186.0543.99˚ volts. |I x | = 186.05/55.9 = 3.328 We can now calculate the average power absorbed by the 50-Ω resistor. P avg = [(3.328)2/2]x50 = 276.8 W.


Applying KVL to the left-hand side of the circuit, 820  4 I o  0.1Vo

(1)

Applying KCL to the right side of the circuit, V V1 8Io  1  0 j5 10  j5 10 10  j5 But,  V1  Vo  V1  Vo 10  j5 10 Vo 10  j5 Hence, 8Io  Vo  0 j50 10 I o  j0.025 Vo

(2)

Substituting (2) into (1), 820  0.1 Vo (1  j) 8020 Vo  1 j I1 

Vo 8   - 25 10 2

P

1  1  64  2 I1 R    (10)  160W 2  2  2 

Chapter 11, Solution 8. We apply nodal analysis to the following circuit.

V 1 I o -j20 

V2 I2

60 A

j10 

0.5 I o

40 

At node 1,

6

V1 V1  V2 V1  j120  V2  j10 - j20

(1)

At node 2, 0.5 I o  I o 

But, Hence,

V2 40

V1  V2 - j20 1.5 (V1  V2 ) V2  - j20 40 3V1  (3  j) V2 Io 

(2)

Substituting (1) into (2), j360  3V2  3V2  j V2  0 j360 360 V2   (-1  j6) 6  j 37 I2 

V2 9  (-1  j6) 40 37

1 1 9  2  (40)  43.78 W P  I2 R   2 2  37  2

Chapter 11, Solution 9. This is a non-inverting op amp circuit. At the output of the op amp,  Z   (10  j 6) x103  Vo  1  2  Vs   1   (8.66  j 5)  20.712  j 28.124 (2  j 4) x103    Z1  The current through the 20-k resistor is

Io 

Vo  0.1411  j1.491 mA or |I o | = 1.4975 A 20k  j12k

P = [|I o |2/2]R = [1.48752/2]10–6x20x103 = 22.42 mW

Chapter 11, Solution 10. No current flows through each of the resistors. Hence, for each resistor, P  0 W . It should be noted that the input voltage will appear at the output of each of the op amps.

Chapter 11, Solution 11.   377 , R  10 4 , C  200  10 -9 RC  (377)(10 4 )(200  10 -9 )  0.754

tan -1 (RC)  37.02

Z ab 

10k 1  (0.754) 2

 - 37.02  7.985 - 37.02 k

i (t )  33 sin(377t  22)  33 cos(377t  68) mA I = 33–68˚ mA 2 I 2 Z ab 33 x10 3 (7.985 - 37.02)  103 S  2 2 S = 4.348–37.02˚ VA





P  S cos(37.02)  3.472 W

Chapter 11, Solution 12. We find the Thevenin impedance using the circuit below.

j2 Ω

4Ω

-j3 Ω

5Ω

We note that the inductor is in parallel with the 5-Ω resistor and the combination is in series with the capacitor. That whole combination is in parallel with the 4-Ω resistor. Thus,  5xj2   4  j3  5  j2  4(0.6896  j1.2758) 4(1.4502  61.61)  Z Thev    5xj2 4.69  j1.2758 4.86  15.22 4  j3  5  j2  1.1936  46.39 Z Thev = 0.8233 – j0.8642 or Z L = [823.3 + j864.2] mΩ. We obtain V Th using the circuit below. We apply nodal analysis.

j2 Ω

I

4Ω

–j3 Ω

V2

+

o

1650 V + –

V Th –

5Ω

V2  165 V2  165 V2  0 0   4  j3 5 j2 (0.16  j 0.12  j 0.5  0.2)V2  (0.16  j 0.12  j 0.5)165 4.125 (0.5235  46.55)V2  (0.4123  67.17)165 Thus, V 2 = 129.94–20.62˚V = 121.62–j45.76 I = (165 – V 2 )/(4 – j3) = (165 – 121.62 + j45.76)/(4 – j3) = (63.0646.52˚)/(5–36.87˚) = 12.61383.39˚ = 1.4519+j12.529 V Thev = 165 – 4I = 165 – 5.808 – j50.12 = [159.19 – j50.12] V = 166.89–17.48˚V We can check our value of V Thev by letting V 1 = V Thev . Now we can use nodal analysis to solve for V 1 . At node 1, V1  165 V1  V2 V2  0    0  (0.25  j 0.3333)V1  (0.2  j 0.3333)V2  41.25 4  j3 5

At node 2, V2  V1 V2  165   0   j 0.3333V1  ( j 0.1667)V2   j82.5 j2  j3 >> Y=[(0.25+0.3333i),-0.3333i;-0.3333i,(0.2-0.1667i)] Y= 0.2500 + 0.3333i 0 - 0.3333i 0 - 0.3333i 0.2000 - 0.1667i >> I=[41.25;–82.5i] I= 41.2500 0 -20.0000i >> V=inv(Y)*I V=

159.2221 – 50.1018i 121.6421–45.7677i Please note, these values check with the ones obtained above. To calculate the maximum power to the load, |I L | = (166.89/(2x0.8233)) = 101.34 A P avg = [(|I L | rms )20.8233]/2 = 4.228 mW.

Chapter 11, Solution 13. For maximum power transfer to the load, Z L = [120 – j60] Ω. I L = 165/(240) = 0.6875 A P avg = [|I L |2120]/2 = 28.36 W.

Chapter 11, Solution 14. Using Fig. 11.45, design a problem to help other students better understand maximum average power transfer. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem It is desired to transfer maximum power to the load Z in the circuit of Fig. 11.45. Find Z and the maximum power. Let is  5cos 40t A. 40 mF

is

7.5 mH

8Ω

12 Ω

Z

Figure 11.45 For Prob. 11.14. Solution We find the Thevenin equivalent at the terminals of Z. 40 mF 7.5 mH

1 1   j 0.625 jC j 40 x 40 x103   j L  j 40 x7.5 x103  j 0.3

 

To find Z Th , consider the circuit below.

j0.3

-j0.625

12 Ω

8Ω

Z Th

ZTh  8  j 0.625  12 // j 0.3  8  j 0.625 

12 x0.3  8.0075  j 0.3252 12  0.3

Z L = (Z Thev )* = [8.008 + j0.3252] Ω. To find V Th , consider the circuit below. -j0.625

8Ω

I1 50o

j0.3

12 Ω

+ V Th –

By current division, I 1 = 5(j0.3)/(12+j0.3) = 1.590˚/12.0041.43˚ = 0.1249688.57˚ = 0.003118 + j0.12492A V Thev rms = 12I 1 / 2 = 1.060388.57˚V I Lrms = 1.060388.57˚/2(8.008) = 66.288.57˚mA P avg = |I Lrms |28.008 = 35.09 mW.

Chapter 11, Solution 15. To find Z eq , insert a 1-A current source at the load terminals as shown in Fig. (a).

1

-j 

1

2

+ j

Vo

2 Vo

1A



(a) At node 1,

Vo Vo V2  Vo   1 j -j

  Vo  j V2

(1)

At node 2, 1  2 Vo 

V2  Vo -j

  1  j V2  (2  j) Vo

(2)

Substituting (1) into (2), 1  j V2  (2  j)( j) V2  (1  j) V2 1 V2  1 j V 1 j Z eq  2   0.5  j 0.5 1 2 Z L  Z *eq  [0.5  j 0.5]  We now obtain VThev from Fig. (b). 1

-j  +

+ 120 V

+ 

Vo

j

2 Vo





(b)  2 Vo 

Vo  12 Vo  0 1 j

V Thev

Vo 

- 12 1 j

– Vo  (- j  2 Vo )  VTh  0 (12)(1  j 2) VThev  (1 - j2)Vo  1 j 2

Pmax

  VThev  0.5  j 0.5  0.5  j 0.5     0.5  2

2

 12 5     2    0.5 2 22 x0.5

= 90 W

Chapter 11, Solution 16. 1 1    j5 jC j 4 x1 / 20 We find the Thevenin equivalent at the terminals of Z L . To find V Thev , we use the circuit shown below. 0.5V o

  4,

1H

 

jL  j 4,

 

1 / 20F

2

4

V1

V2 +

+ 10<0o -

+ Vo -

-j5

j4

V Thev -

At node 1, 10  V1 V V  V2  1  0.5V1  1  j5 2 4



5  V1 (1.25  j0.2)  0.25V2

(1)

At node 2,

V1  V2 V  0.25V1  2 4 j4

 

0  0.5V1  V2 (0.25  j 0.25)

Solving (1) and (2) leads to

VThev  V2  6.1947  j 7.0796  9.407248.81o

(2)

To obtain R eq , consider the circuit shown below. We replace Z L by a 1-A current source. 0.5V 1

2

4

V1

-j5

V2

j4

1A

At node 1, V1 V V  V2  1  0.25V1  1 0   2  j5 4 At node 2,

0  V1 (1  j 0.2)  0.25V2

(3)

V1  V2 V  0.25V1  2    1  0.5V1  V2 (0.25  j 0.25) (4) 4 j4 Solving (1) and (2) gives V Z eq  2  1.9115  j 3.3274  3.83760.12 o and Z L = 3.837–60.12° Ω 1 1

Pmax 

| VTh | 2 2 Z eq  Z L

1.9115  2

9.4072 2  5.787 W 2 x 4 x1.9115

Chapter 11, Solution 17. We find Z eq at terminals a-b following Fig. (a).

-j10 

30  a

b

40 

j20  (a)

Z eq   j10  30 ||  j 20  40  

(30  j10)(40  j 20) = 20 Ω = Z L 70  j10

We obtain VThev from Fig. (b). I1

I2

-j10 

30  j5 A

+ V Thev  j20 

40  (b)

Using current division, 30  j20 I1  ( j5)  -1.1  j2.3 70  j10 40  j10 I2  ( j5)  1.1  j2.7 70  j10 VTh  30 I 2  j10 I 1  10  j70 Pmax 

VTh

2

2 Z eq  Z L 

2

ZL 

5000 20  31.25 W (2)(2 x 20) 2

Chapter 11, Solution 18. We find Z Th at terminals a-b as shown in the figure below.

40 

40 

j30 

-j10 

80  a

Z th

b

Z Th  j 30  40 || 40  80 || (-j10)  j30  20 

(80)(-j10) 80  j10

Z Th  21.23  j 20.154 Z L  Z *Th  [21.23–j20.15] Ω

Chapter 11, Solution 19. At the load terminals, Z Th  - j2  6 || (3  j)  -j2 

(6)(3  j) 9 j

Z Th  2.049  j1.561

R L  Z Th  2.576 Ω To get VTh , let Z  6 || (3  j)  2.049  j0.439 .

By transforming the current sources, we obtain VTh  (330) Z  67.62  j14.487 = 69.1612.09˚ 2

P max

69.16 2.576 = = 258.5 W. 2.049  j1.561  2.576 2

Chapter 11, Solution 20. Combine j20  and -j10  to get j20 || -j10  -j20 . To find Z Th , insert a 1-A current source at the terminals of R L , as shown in Fig. (a).

Io

4 Io

40 

V1

V2

+ 

-j20 

-j10 

1A

(a) At the supernode,

V1 V V  1  2 40 - j20 - j10 40  (1  j2) V1  j4 V2

1

Also,

V1  V2  4 I o , 1.1 V1  V2

  V1 

Substituting (2) into (1), V  40  (1  j2)  2   j4 V2  1.1  44 V2  1  j6.4 V2  1.05  j6.71  1 R L  Z Th  6.792 

Z Th 

(1) where I o  V2 1.1

- V1 40

(2)

To find VTh , consider the circuit in Fig. (b).

Io

4 Io

40 

V1

+ 

V2 +

1650 V

+ 

-j20 

-j10 

V th 

(b) At the supernode, 165  V1 V V  1  2 40 - j20 - j10 165  (1  j 2) V1  j 4 V2 Also,

V1  V2  4 I o , where I o 

V1 

V2  16.5 1.1

(3) 165  V1 40

(4)

Substituting (4) into (3), 150–j30  (0.9091  j 5.818) V2 VTh  V2 

150  j 30 152.97  11.31   25.98  92.43 0.9091  j 5.818 5.88981.12 2

P max

25.98 6.792 = = 21.51 W 1.05  j 6.71  6.792 2

Chapter 11, Solution 21. We find Z Th at terminals a-b, as shown in the figure below.

100 

-j10  a

40  50 

Z th

j30  b Z Th  50 || [ - j10  100 || (40  j30) ]

where 100 || (40  j30) 

(100)(40  j30)  31.707  j14.634 140  j30

Z Th  50 || (31.707  j4.634)  Z Th  19.5  j1.73

R L  Z Th  19.58 

(50)(31.707  j4.634) 81.707  j4.634


i(t) = [2–2cos(2t)] amps


Using Fig. 11.54, design a problem to help other students to better understand how to find the rms value of a waveshape. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine the rms value of the voltage shown in Fig. 11.54.

v(t) (V)

10

0

1

2

3

4

t (s)


Solution

T

2 rms

V

1

1 1 100   v 2 (t )dt   102 dt  T 0 30 3 V rms = 5.7735 V

Chapter 11, Solution 24.  5, 0  t  1 v( t )   - 5, 1  t  2

T  2,

2  Vrms

1 2

 5 1

0

2



dt  1 (-5) 2 dt  2

25 [1  1]  25 2 Vrms  5 V




1 T 2 1 1 2 3 f ( t )dt   0 (4) 2 dt   1 0dt  2 4 2 dt  0 T 3 1 32  [16  0  16]  3 3

2 f rms 

f rms 

32  3.266 3 f rms = 3.266



Chapter 11, Solution 26. 5 0t 2 v(t )   20 2  t  4

T  4,

2 Vrms 

4 1 2 2 1 10 dt (20) 2 dt   [200  800 ]  250      0 2  4 4

Vrms  15.811 V.

Chapter 11, Solution 27. i( t )  t , 0  t  5

T  5,

1 5 2 1 t3 t dt    5 0 5 3  2.887 A

I 2rms 

I rms

5 0



125  8.333 15


2 Vrms





5 1 2 2 ( 4 t ) dt  0 2 dt   2 5 0 1 16 t 3 2 16    (8)  8.533 5 3 0 15

2  Vrms

Vrms  2.92 V 2 Vrms 8.533 P   4.267 W R 2

Chapter 11, Solution 29. 5  t  15  60  6t i (t )   - 120  6t 15  t  25

T  20 ,

25 1  15 2 ( 60 6 t ) dt (-120  6t) 2 dt        5 15 20  25 1 15    (900  180t  9t 2 ) dt   (9t 2  360t  3600) dt   15 5  5 1 3 2 25  900t  90t 2  3t 3 15 5  3t  180t  3600t 15 5 1  [750  750]  300 5

I eff2 

I eff2 I eff2 I eff2







 

I eff  17.321 A 2 P  I eff R  (17.321)2x12 = 3.6 kW.


t 0t2 v( t )   - 1 2  t  4 2 Vrms 

1 4

 t 2

0

2



4  1 8 dt  2 (-1) 2 dt    2   1.1667  43

Vrms  1.08 V


V

2

rms

2 1 2  1 4 1 1  2   v(t )dt    (2t ) dt   (4) 2 dt     16  8.6667 2 0 20  1  2 3

Vrms  2.944 V

Chapter 11, Solution 32. 2 1 1 (10t 2 ) 2 dt   0 dt    1 2  0 5 1 t  50 0 t 4 dt  50  10  10 5

I 2rms  I 2rms

I rms  3.162 A

Chapter 11, Solution 33. I

2 rms

T 3 4  1 2 1 1 2   i (t )dt    25t dt   25dt   (5t  20) 2 dt  6 0 T 0 1 3 

2 I rms 

4 1  t3 1 t3 25 25(3 1) (25     100t 2  400t )   11.1056  3 6 3 0 3

I rms = 3.332 A


1 T



1  9t 3   3 3 

2

2  f rms

T 0

0

f 2 (t )dt 

3 1 2 2  ( 3 t ) dt 6 2 dt      0 2 3

  36  20 

f rms  20  4.472

f rms = 4.472


2 Vrms



6 5 4 2 1 1 2 2 2 2 10 dt  20 dt  30 dt  20 dt  10 2 dt      5 4 2 1 6 0 1  [100  400  1800  400  100 ]  466.67 6

2  Vrms

Vrms  21.6 V




(a) I rms = 10 A 2

 3   (b) V rms  4      2 36  9.055 A (c) I rms  64  2 2

(d)

2

V rms 

25 16   4.528 V 2 2

Vrms  16 

9  4.528 V (checked) 2

Chapter 11, Solution 37. Design a problem to help other students to better understand how to determine the rms value of the sum of multiple currents. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Calculate the rms value of the sum of these three currents: i 1 = 8, i 2 = 4 sin(t + 10),

i 3 = 6 cos(2t + 30) A

Solution

i  i1  i2  i3  8  4 sin(t  10 o )  6 cos(2t  30 o ) I rms 

I 2 1rms  I 2 2 rms  I 2 3 rms  64 

16 36   90  9.487 A 2 2

Chapter 11, Solution 38. V 2 2202 S1  *   390.32 Z1 124 S2 

V2 2202   944.4  j1180.5 Z 2* 20  j 25

S3 

V2 2202   300  j 267.03 Z 3* 90  j80

S  S1  S2  S3  1634.7  j 913.47  1872.6  29.196o VA (a) P = Re(S) = 1634.7 W (b) Q = Im (S) = 913.47 VA (leading) (c ) pf = cos (29.196o) = 0.8732

Chapter 11, Solution 39. (a) Z L = 4.2 + j3.6 = 5.5317 40.6o pf = cos 40.6 = 0.7592 2 Vrms 2202 S *   6.643  j 5.694 kVA Z 5.5317  40.6o

P = 6.643 kW Q = 5.695 kVAR (b) C 

P(tan 1  tan  2 ) 6.643 x103 (tan 40.6o  tan 0o )   312  F , 2 Vrms 2 x60 x 2202

{It is important to note that this capacitor will see a peak voltage of 220 2 = 311.08V, this means that the specifications on the capacitor must be at least this or greater!}

Chapter 11, Solution 40. Design a problem to help other students to better understand apparent power and power factor. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem A load consisting of induction motors is drawing 80 kW from a 220-V, 60 Hz power line at a pf of 0.72 lagging. Find the capacitance of a capacitor required to raise the pf to 0.92. Solution

pf 1  0.72  cos  1

  1  43.940

pf 2  0.92  cos  2

   2  23.07 0

C

P(tan 1  tan  2 ) 80 x103 (0.9637  0.4259)   2.4 mF , 2 2 x60 x(220) 2 Vrms

{Again, we need to note that this capacitor will be exposed to a peak voltage of 311.08V and must be rated to at least this level, preferably higher!}


- j2 || ( j5  j2)  -j2 || -j3 

(-j2)(-j3)  -j6 j

Z T  4  j6  7.211 - 56.31

pf  cos(-56.31)  0.5547 (leading) (b)

j2 || (4  j) 

( j2)(4  j)  0.64  j1.52 4  j3

Z  1 || (0.64  j1.52  j) 

0.64  j0.44  0.479321.5 1.64  j0.44

pf  cos(21.5)  0.9304 (lagging)

Chapter 11, Solution 42. pf  0.707  cos      45o (a) S=120, S  S cos   jS sin   84.84  j84.84 VA

(b) S  Vrms I rms

2 (c) S  I rms Z

  I rms 

  Z

(d) If Z = R + jL, then

 L  2 fL  71.278

S 2 I rms

S 120   1.091 A rms Vrms 110  71.278  j 71.278 

R = 71.278 Ω

  L

71.278  0.1891 H = 189.1 mH. 2 x60

Chapter 11, Solution 43. Design a problem to help other students to better understand complex power. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem The voltage applied to a 10-ohm resistor is v(t) = 5 + 3 cos(t + 10) + cos(2t + 30) V (a) Calculate the rms value of the voltage. (b) Determine the average power dissipated in the resistor.

Solution (a) Vrms  V 2 1rms  V 2 2 rms  V 2 3 rms  25  (b) P 

V 2 rms  30 / 10  3 W R

9 1   30  5.477 V 2 2

Chapter 11, Solution 44. 40  F

 

1 1    j12.5 jC j 2000 x 40 x106

60mH

 

j L  j 2000 x60 x103  j120

We apply nodal analysis to the circuit shown below.

But

100  Vo 4I  V V  x o  o 30  j12.5 20 j120 V I x  o . Solving for V o leads to j120 Vo  2.9563  j1.126 Io

1000o

Io 

30 Ω

-j12.5

20 Ω Vo

Ix j120

+ –

100  Vo  2.7696  j1.1165 30  j12.5

1 1 S  Vs I o*  (100)(2.7696  j.1165)  138.48  j 55.825 VA 2 2 S = (138.48 – j55.82) VA

+ –

4I x


(a) V 2 rms  20 2 

60 2  2200 2

I rms  1 2 

 

Vrms  46.9 V

0.5 2  1.125  1.061 A 2

(b) p(t) = v(t)i(t) = 20 + 60cos100t – 10sin100t – 30(sin100t)(cos100t); clearly the average power = 20W.


S  V I *  (220 30)(0.5 - 60)  110  - 30 S  [95.26  j 55] VA Apparent power = 110 VA Real power = 95.26 W Reactive power = 55 VAR pf is leading because current leads voltage

(b)

S  V I *  (250 - 10)(6.2 25)  155015 S  [497.2  j 401.2] VA Apparent power = 1550 VA Real power = 1497.2 W Reactive power = 401.2 VAR pf is lagging because current lags voltage

(c)

S  V I *  (1200)(2.415)  28815 S  [278.2  j 74.54] VA Apparent power = 288 VA Real power = 278.2 W Reactive power = 74.54 VAR pf is lagging because current lags voltage

(d)

S  V I *  (16045)(8.5 - 90)  1360 - 45 S  [961.7 – j961.7] VA Apparent power = 1360 VA Real power = 961.7 W Reactive power = - 961.7 VAR pf is leading because current leads voltage


(b)

V  112 10 , I  4 - 50 1 S  V I *  22460  [112  j194] VA 2 Average power = 112 W Reactive power = 194 VAR V  160 0 , I  445 1 S  V I *  320 - 45  226.3 – j226.3 2

Average power = 226.3 W Reactive power = –226.3 VAR

(c)

S

2

V Z*



(80) 2  12830  110.85 + j64 50 - 30

Average power = 110.85 W Reactive power = 64 VAR (d)

2

S  I Z  (100)(10045)  [7.071  j 7.071] kVA Average power = 7.071 kW Reactive power = 7.071 kVAR


S  P  jQ  [269  j150] VA

(b)

pf  cos   0.9     25.84 Q  S sin    S 

Q 2000   4588.31 sin  sin(25.84)

P  S cos   4129.48 S  [4.129  j 2] kVA

(c)

Q 450   0.75 S 600 pf  0.6614

Q  S sin    sin     48.59 ,

P  S cos   (600)(0.6614)  396.86 S  [396.9  j 450] VA

(d)

2

(220) 2   1210 S 40 Z

V

P  S cos    cos     34.26

Q  S sin   681.25 S  [1  j 0.6812] kVA

P 1000   0.8264 S 1210


(b)

4 sin(cos -1 (0.86)) kVA 0.86 S  [4  j 2.373] kVA S  4 j

pf 

P 1.6 0.8  cos     sin   0.6 S 2

S  1.6  j2 sin   [1.6  j1.2] kVA (c)

S  Vrms I *rms  (20820)(6.550) VA S  1.352 70  [0.4624  j1.2705] kVA 2

(d)

V (120) 2 14400  S *  Z 40  j60 72.11 - 56.31 S  199.7 56.31  [110.77  j166.16] VA


S  P  jQ  1000  j

S  1000  j750

But,

Vrms

S

Vrms

1000 sin(cos -1 (0.8)) 0.8

2

Z* 2

(220) 2 Z    30.98  j23.23 S 1000  j750 Z  [30.98  j 23.23]  *

(b)

2

S  I rms Z Z

S I rms

(c)

2

Vrms



1500  j2000  [10.42  j13.89]  (12) 2

2

2

(120) 2  1.6  - 60 2S (2)(4500 60) S Z  1.660  [0.8  j1.386] 

Z  *



V




Z T  2  (10  j5) || (8  j6) (10  j5)(8  j6) 110  j20 ZT  2   2 18  j 18  j Z T  8.152  j0.768  8.1885.382

pf  cos(5.382)  0.9956 (lagging)

(b)

S  VI  *

V

2



Z* S  31.265.382

(16) 2 (8.188 - 5.382)

P  S cos   31.12 W

(c)

Q  S sin   2.932 VAR

(d)

S  S  31.26 VA

(e)

S  31.265.382  (31.12+j2.932) VA

(a) 0.9956 (lagging, (b) 31.12 W, (c) 2.932 VAR, (d) 31.26 VA, (e) [31.12+j2.932] VA


2000 0.6  2000  j1500 0.8 S B  3000 x 0.4  j3000 x 0.9165  1200  j2749 S A  2000  j

SC  1000  j500 S  S A  S B  SC  4200  j749

4200

(a)

pf 

(b)

S  Vrms I rms   I rms 

4200 2  749 2

 0.9845 leading

I rms = 35.5555.11˚ A.

4200  j749  35.55  55.11 12045

Chapter 11, Solution 53. S = S A + S B + S C = 4000(0.8–j0.6) + 2400(0.6+j0.8) + 1000 + j500 = 5640 + j20 = 56400.2˚

(a)

 I rms 

S B S A  SC S 56400.2  47  29.8    Vrms Vrms Vrms 120 30

I  47 29.8  47 29.8 A (b)

pf = cos(0.2˚) ≈ 1.0 lagging.

Chapter 11, Solution 54. Consider the circuit shown below. 8 - 20  1.6 16.87 4  j3 8 - 20 I2   1.6  - 110 j5 I1 

I  I 1  I 2  (-0.5472  j1.504)  (1.531  j0.4643) I  0.9839  j1.04  1.432 - 46.58

For the source, S  V I *  (8 - 20)(1.43246.58) S  11.45626.58  (10.24+j3.12) VA For the capacitor, S  I1

2

Z c  (1.6) 2 (-j3)  –j7.68 VA

For the resistor, S  I1

2

Z R  (1.6) 2 (4)  10.24 VA

For the inductor, S  I2

2

Z L  (1.6) 2 ( j 5)  j12.8 VA

Chapter 11, Solution 55. Using Fig. 11.74, design a problem to help other students to better understand the conservation of AC power. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the complex power absorbed by each of the five elements in the circuit of Fig. 11.74.


We apply mesh analysis to the following circuit. -j20 

j10  I3

400 V rms

+ 

I1

20 

I2

+ 

5090 V rms

For mesh 1, 40  (20  j20) I1  20 I 2 2  (1  j) I1  I 2

(1)

For mesh 2, - j50  (20  j10) I 2  20 I1 - j5  -2 I1  (2  j) I 2 Putting (1) and (2) in matrix form,  2  1  j - 1  I1   - j5   - 2 2  j I      2 

(2)

  1 j ,

 1  4  j3 ,

 2  -1  j5

 4  j3 1 I1  1   (7  j)  3.5358.13  1 j 2  2 - 1  j5  2  j3  3.605 - 56.31  I2   1 j I 3  I1  I 2  (3.5  j0.5)  (2  j3)  1.5  j3.5  3.80866.8 For the 40-V source, 1  S  -V I 1*  -(40)   (7  j)   [-140  j20] VA 2  For the capacitor,

S  I1 For the resistor,

2

Z c  - j250 VA

S  I3

2

R  290 VA

For the inductor, 2

S  I 2 Z L  j130 VA For the j50-V source, S  V I *2  ( j50)(2  j3)  [-150  j100] VA

Chapter 11, Solution 56. (6)(- j 2) 12  90   1.897365  71.565  0.6  j1.8 6  j2 6.32456  18.435 3  j 4  [(-j2) || 6]  3.6  j2.2 - j 2 || 6 

The circuit is reduced to that shown below.

Io + 230 A

Vo

5

3.6 + j2.2 

 Io 

3.6  j 2.2 4.21931.4296 (230)  (230)  0.9505547.08 8.6  j 2.2 8.8769414.3493

Vo  5 I o  4.7527547.08

S  Vo I*s  (4.7527547.08)(2  30) S  9.505517.08  (9.086+j2.792) VA

Chapter 11, Solution 57. Consider the circuit as shown below.

4

Vo

-j1 

V1

2 +

240 V

+ 

1

j2 

V2

2 Vo

 At node o,

24  Vo Vo Vo  V1   4 1 -j 24  (5  j4) Vo  j4 V1

(1)

Vo  V1 V1  2 Vo  -j j2 V1  (2  j4) Vo

(2)

At node 1,

Substituting (2) into (1), 24  (5  j4  j8  16) Vo - 24 (-24)(2 - j4) Vo  , V1  11  j4 11  j4 The voltage across the dependent source is V2  V1  (2)(2 Vo )  V1  4 Vo (-24)(6  j4) - 24 V2   (2  j4  4)  11  j4 11  j4

S  V2 I *  V2 ( 2 Vo* ) S

(-24)(6  j4) - 48  1152    (6  j4) 11  j4 11 - j4  137  S  (50.45–j33.64) VA

Chapter 11, Solution 58. From the left portion of the circuit, 0.2 Io   0.4 mA 500

20 I o  8 mA which then leads to the following circuit, I x -j3 k

j1 k

4 k

8 mA

10 k

From the right portion of the circuit, 16 4 Ix  mA (8 mA)  7 j 4  10  j  j3 S  Ix

2

R

(16  10 -3 ) 2  (10  10 3 ) 50 S  51.2 mVA

It should be noted that the complex power delivered to a resistor is always watts.

Chapter 11, Solution 59. Let V o represent the voltage across the current source and then apply nodal analysis to the circuit and we get:

Vo Vo 240  Vo   50 - j20 40  j30 88  (0.36  j0.38) Vo 88 Vo   168.13 - 46.55 0.36  j0.38 4

Vo  8.4143.45 - j20 Vo I2   3.363 - 83.42 40  j30

I1 

Reactive power in the inductor is 2 S  I 2 Z L  (3.363) 2 ( j30)  j339.3 VAR Reactive power in the capacitor is 2 S  I1 Z c  (8.41) 2 (- j20)  –j1.4146 kVAR


20 sin(cos -1 (0.8))  20  j15 0.8 16 S 2  16  j sin(cos -1 (0.9))  16  j7.749 0.9

S1  20  j

S  S1  S 2  36  j22.749  42.58532.29

But

S  Vo I *  6 Vo Vo 

S  7.098  32.29 6

pf  cos(32.29)  0.8454 (lagging)

Chapter 11, Solution 61. Consider the network shown below.

Io

I2

+

I1

S2

Vo

So

S1

S3

 S 2  1.2  j0.8 kVA

S3  4  j

4 sin(cos -1 (0.9))  4  j1.937 kVA 0.9

Let

S 4  S 2  S 3  5.2  j1.137 kVA

But

S 4  Vo I *2 S 4 (5.2  j1.137)  10 3   11.37  j 52 10090 Vo I 2  11.37  j 52 I *2 

Similarly,

S1  2  j

But

S1  Vo I 1*

2 sin(cos -1 (0.707))  2 (1  j) kVA 0.707

S1 (1.4142  j1.4142)  10 3   -14.142  j14.142 Vo j100 I1  – 14.142 + j14.142

I 1* 

I o  I 1  I 2  - 2.772  j66.14  66.292.4° A S o  Vo I o* S o  (10090)(66.2 - 92.4) VA S o  6.62–2.4° kVA 66.292.4° A, 6.62–2.4° kVA


0.2 + j0.04 

I

I2

0.3 + j0.15 

I1 Vs

+ 

S 2  15  j But

+

+

V1

V2





15 sin(cos -1 (0.8))  15  j11.25 0.8

S 2  V2 I *2 S 2 15  j11.25  V2 120 I 2  0.125  j0.09375

I *2 

V1  V2  I 2 (0.3  j0.15) V1  120  (0.125  j0.09375)(0.3  j0.15) V1  120.02  j0.0469

S1  10  j But

10 sin(cos -1 (0.9))  10  j4.843 0.9

S1  V1 I 1* S 1 11.11125.84  V1 120.02 0.02 I 1  0.093 - 25.82  0.0837  j0.0405

I 1* 

I  I 1  I 2  0.2087  j0.053

Vs  V1  I (0.2  j0.04) Vs  (120.02  j0.0469)  (0.2087  j0.053)(0.2  j0.04) Vs  120.06  j0.0658 Vs  120.060.03 V


S  S1  S 2  S 3 .

S1  12  j

12 sin(cos -1 (0.866))  12  j6.929 0.866

S 2  16  j

16 sin(cos -1 (0.85))  16  j9.916 0.85

S3 

(20)(0.6)  j 20  15  j 20 sin(cos -1 (0.6))

S  43  j 22.987  V I *o I *o 

S (43  j 22.99) x10 3   195.45  j104.5  221.628.13 V 220 I o  221.6–28.13˚ A

Chapter 11, Solution 64. I2 I1 8 + 

Is

1200º V

j12

I s + I 2 = I 1 or I s = I 1 – I 2 I1  But,

120 8  j12

 4.615  j6.923

2500  j400 S  I 2    20.83  j3.333 S  VI 2  V 120 or I 2  20.83  j3.333 I s = I 1 – I 2 = –16.22 – j10.256 = 19.19–147.69˚ A.

Chapter 11, Solution 65. C  1 nF  

1 -j  4  -j100 k jC 10  10 -9

At the noninverting terminal, Vo 40  Vo 4    Vo  100 - j100 1 j 4 Vo   - 45 2 4 v o (t)  cos(10 4 t  45) 2 2  4 1   1  Vrms   W   P R  2 2   50  10 3  2

P  80 W

Chapter 11, Solution 66. As an inverter, - Zf - (2  j4) Vo  Vs   (4 45) Zi 4  j3

Io 

Vo - (2  j4)(445) mA mA  (6 - j2)(4  j3) 6  j2

The power absorbed by the 6-k resistor is 2

P  Io

2

 20  4    10 -6  6  10 3 R     40  5 

P  1.92 mW


  2,

3H

 

jL  j 6,

0.1F

 

1 1    j5 jC j 2 x0.1

 j 50  2  j4 10  j 5 The frequency-domain version of the circuit is shown below. Z 2 =2-j4  10 //(  j 5) 

Z 1 =8+j6 

I1 +

Io

+

+

0.620 o V

Z 3  12

Vo

-

(a) I 1 

0.620 o  0 0.5638  j 0.2052   0.06  16.87 o 8  j6 8  j6

1 S  Vs I *1  (0.320 o )(0.06  16.87 o )  14.4  j10.8 mVA  1836.86 o mVA 2 S = (14.4+j10.8) mVA = 1836.86° mVA V Z2 (2  j 4) (0.620 o )  0.022499.7 o Vs , Io  o   12(8  j 6) Z1 Z3 1 P  | I o | 2 R  0.5(0.0224) 2 (12)  2.904 mW 2

(b) Vo  

P = 2.904 mW (a) 1836.86° mVA, (b) 2.904 mW

Chapter 11, Solution 68. Let where

S  SR  SL  Sc

1 2 I R  j0 2 o 1 S L  PL  jQ L  0  j I o2 L 2 1 2 1 S c  Pc  jQ c  0  j I o  2 C S R  PR  jQ R 

Hence, S

 1  1 2  I o  R  jL   C  2 


(b)

(c)

Given that Z  10  j12 12     50.19 tan   10 pf  cos   0.6402 2

(120) 2  295.12  j354.09 2 Z * (2)(10  j12) The average power absorbed  P  Re(S)  295.1 W S

V



For unity power factor, 1  0 , which implies that the reactive power due to the capacitor is Q c  354.09 But C

Qc 

V2 1  C V 2 2 Xc 2

2 Qc (2)(354.09)  130.4 F 2  V (2 )(60)(120) 2

Chapter 11, Solution 70. Design a problem to help other students to better understand power factor correction. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem An 880-VA, 220-V, 50-Hz load has a power factor of 0.8 lagging. What value of parallel capacitance will correct the load power factor to unity? Solution

pf  cos   0.8   sin   0.6 Q  S sin   (880)(0.6)  528 If the power factor is to be unity, the reactive power due to the capacitor is Q c  Q  528 VAR 2 Vrms Qc  C V 2   C  Xc V2 (528)  34.72 µF C (2)(50)(220) 2

But

Q

Chapter 11, Solution 71. (a) For load 1, Q 1 = 60 kVAR, pf = 0.85 or θ 1 = 31.79˚ Q 1 = S 1 sinθ 1 = 60k or S 1 = 113.89k and P 1 = 113.89cos(31.79) = 96.8kW S 1 = 96.8 + j60 kVA For load 2, S 2 = 90 – j50 kVA For load 3, S 3 =100 kVA Hence, S = S 1 + S 2 + S 3 = 286.8 + j10kVA = 2872˚kVA But Thus,

S = (V rms )2/Z* or Z* = 1202/2872˚k = 0.05017–2˚ Z = 0.050172˚Ω or [50.14 + j1.7509] mΩ.

(b) From above, pf = cos2˚ = 0.9994. (c) I rms = V rms /Z = 120/0.050172˚ = 2.392–2˚ kA or [2.391 – j0.08348] kA.

Chapter 11, Solution 72. (a) P  S cos 1

  S

P 2.4   3.0 kVA cos 1 0.8

pf  0.8  cos 1   1  36.87 o Q  S sin 1  3.0sin 36.87 o  1.8 kVAR Hence, S = 2.4 + j1.8 kVA P 1.5 S1  1   2.122 kVA cos  0.707 pf  0.707  cos      45o

Q1  P1  1.5 kVAR   S1  1.5  j1.5 kVA Since, S  S1  S 2   S2  S  S1  (2.4  j1.8)  (1.5  j1.5)  0.9  j 0.3 kVA S 2  0.9497  18.43o pf = cos 18.43o = 0.9487    2  25.84o (b) pf  0.9  cos  2 P(tan 1  tan  2 ) 2400(tan 36.87  tan 25.84) C   117.5  F 2 2 x60 x(120) 2 Vrms


S  10  j15  j22  10  j7 kVA S  S  10 2  7 2  12.21 kVA

(b)

S  V I*

  I * 

S 10,000  j7,000  240 V

I  41.667  j29.167  50.86 - 35 A (c)

7 1  tan -1    35 , 10 

 2  cos -1 (0.96)  16.26

Q c  P1 [ tan 1  tan  2 ]  10 [ tan(35) - tan(16.26) ] Q c  4.083 kVAR C

(d)

Qc 4083   188.03 F 2  Vrms (2 )(60)(240) 2

S 2  P2  jQ 2 ,

P2  P1  10 kW

Q 2  Q1  Q c  7  4.083  2.917 kVAR S 2  10  j2.917 kVA

But

S 2  V I *2

S 2 10,000  j2917  240 V I 2  41.667  j12.154  43.4 - 16.26 A I *2 


1  cos -1 (0.8)  36.87 P1 24 S1    30 kVA cos 1 0.8 Q1  S1 sin 1  (30)(0.6)  18 kVAR S1  24  j18 kVA  2  cos -1 (0.95)  18.19 P2 40 S2    42.105 kVA cos  2 0.95 Q 2  S 2 sin  2  13.144 kVAR S 2  40  j13.144 kVA S  S1  S 2  64  j31.144 kVA  31.144    25.95   tan -1   64  pf  cos   0.8992

(b)

 2  25.95 ,

1  0

Q c  P [ tan  2  tan 1 ]  64 [ tan(25.95)  0 ]  31.144 kVAR C

Qc 31,144   5.74 mF 2  Vrms (2 )(60)(120) 2


(a)

S1 

V Z1*

2



(240) 2 5760   517.75  j323.59 VA 80  j50 8  j5

S2 

(240) 2 5760   358.13  j208.91 VA 120  j70 12  j7

S3 

(240) 2  960 VA 60

S  S1  S 2  S 3  [1.8359  j 0.11468] kVA (b)

(c)

 114.68    3.574   tan -1  1835.88  pf  cos   0.998 {leading}

Since the circuit already has a leading power factor, near unity, no compensation is necessary.

Chapter 11, Solution 76. The wattmeter reads the real power supplied by the current source. Consider the circuit below.

4

120 V

-j3 

+ 

Vo

j2 

8

12  Vo Vo Vo   4  j3 j2 8 36.14  j23.52 Vo   0.7547  j11.322  11.347 86.19 2.28  j3.04

330 

S  Vo I *o  (11.34786.19)(3 - 30) S  34.0456.19 VA

P  Re(S)  18.942 W

330 A

Chapter 11, Solution 77. The wattmeter measures the power absorbed by the parallel combination of 0.1 F and 150 . 120 cos(2 t )   1200 ,  2

4H  

jL  j8 1 0.1 F    -j5 jC Consider the following circuit. 6

1200 V

Z  15 || (-j5) 

I

j8 

I

+ 

(15)(-j5)  1.5  j4.5 15  j5

120  14.5 - 25.02 (6  j8)  (1.5  j4.5)

1 1 2 1 V I *  I Z   (14.5) 2 (1.5  j4.5) 2 2 2 S  157.69  j473.06 VA S

The wattmeter reads P  Re(S)  157.69 W

Z

Chapter 11, Solution 78. The wattmeter reads the power absorbed by the element to its right side. 2 cos(4 t )   20 ,  4

1H   jL  j4 1 1 F    -j3 12 jC Consider the following circuit. 10 

200 V

I

+ 

Z  5  j4  4 || - j3  5  j4 

Z

(4)(- j3) 4  j3

Z  6.44  j2.08 I

20  1.207  - 7.21 16.44  j2.08

S

1 2 1 I Z   (1.207) 2 (6.44  j2.08) 2 2

P  Re(S)  4.691 W

Chapter 11, Solution 79. The wattmeter reads the power supplied by the source and partly absorbed by the 40-  resistor.   100, 10 mH

j100 x10x10  3  j,

 

500F

 

1 1    j20 jC j100 x 500 x10  6

The frequency-domain circuit is shown below.

20

I

40

Io

j V1

V2

+1 2 Io

10<0o

-j20 -

At node 1,

V  V2 V1  V2 3(V1  V2 ) V1  V2 10  V1  2I o  1    j 20 20 j 40 10  (7  j40)V1  (6  j40)V2



(1)

At node 2, V1  V 2 V1  V 2 V   2 j 20  j 20

 

0  (20  j )V1  (19  j )V 2

Solving (1) and (2) yields V 1 = 1.5568 –j4.1405 I

10  V1  0.2111  j0.1035, 40

S

1 V1I   0.04993  j0.5176 2

P = Re(S) = 50 mW.

(2)


(b)

I

V 110   17.19 A Z 6.4

S

V Z

2



(110) 2  1890.6 6.4

cos   pf  0.825     34.41 P  S cos   1559.8  1.6 kW

Chapter 11, Solution 81. Design a problem to help other students to better understand how to correct power factor to values other than unity. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem A 120-V rms, 60-Hz electric hair dryer consumes 600 W at a lagging pf of 0.92. Calculate the rms-valued current drawn by the dryer. How would you power factor correct this to a value of 0.95? Solution

    23.074o P P  S cos    S  652.17 VA 0.92 S = P+jQ = 600 + j652.17sin23.09o = 600 +j255.6 * S  Vrms I rms But . S 600  j 255.6 * I rms   Vrms 120 P = 600 W,

pf  0.92

I rms = 5 – j2.13 = 5.435–23.07˚A. To correct this to a pf = 0.95, I would add a capacitor in parallel with the hair dryer (remember, series compensation will increase the power delivered to the load and probably burn out the hair dryer. pf = 0.95 = 600/S or S = 631.6 VA and θ = 18.19˚ and VARs = 197.17 Thus, VARs cap = 255.6 – 197.17 = 58.43 = 120xI C or I C = 58.43/120 = 0.4869A Next, X C = 120/0.4869 = 246.46 = 1/(377xC) or C = 10.762 µF

Chapter 11, Solution 82. (a) P1  5,000,

Q1  0

P2  30,000 x0.82  24,600,

Q2  30,000 sin(cos 1 0.82)  17,171

S  S1  S 2  (P1  P2 )  j(Q1  Q 2 )  29,600  j17,171 S | S | 34.22 kVA Q = 17.171 kVAR

(b) (c ) pf 

P 29,600   0.865 S 34,220

Qc  P (tan  1  tan  2 )





 29,600 tan(cos 1 0.865)  tan(cos 1 0.9)  2833 VAR (c)

C

Qc 2833   130.46  F 2 V rms 2x 60 x 240 2


1 1 (a) S  VI   (21060 o )(8  25 o )  84035 o 2 2 P  S cos   840 cos 35 o  688.1 W (b) S = 840 VA (c) Q  S sin   840 sin 35 o  481.8 VAR (d) pf  P / S  cos 35 o  0.8191 (lagging)


Maximum demand charge  2,400  30  $72,000 Energy cost  $0.04  1,200  10 3  $48,000 Total charge = $120,000

(b)

To obtain $120,000 from 1,200 MWh will require a flat rate of $120,000 per kWh  $0.10 per kWh 1,200  10 3

Chapter 11, Solution 85.   j 2x60 x15 x10 3  j 5.655 (a) 15 mH We apply mesh analysis as shown below. I1 + Ix

o

120<0 V -

10 

In

30  Iz 10 

+ 120<0o V

Iy j5.655  I2

For mesh x, (1) 120 = 10 I x - 10 I z For mesh y, 120 = (10+j5.655) I y - (10+j5.655) I z (2) For mesh z, 0 = -10 I x –(10+j5.655) I y + (50+j5.655) I z (3) Solving (1) to (3) gives I x =20, I y =17.09-j5.142, I z =8 Thus, I 1 =I x =20 A I 2 =-I y =-17.09+j5.142 = 17.85163.26 o A I n =I y - I x = –2.91 –j5.142 = 5.907  119.5 o A (b) S1  (120)I  x  120x 20  2400,

S 2  (120)I  y  2051  j617

S  S 1  S 2  [4.451  j 0.617] kVA

(c ) pf = P/S = 4451/4494 = 0.9904 (lagging)

Chapter 11, Solution 86. For maximum power transfer Z L  Z *Th   Z i  Z Th  Z *L

Z L  R  jL  75  j (2)(4.12  10 6 )(4  10 -6 ) Z L  75  j103.55  Z i  [75  j103.55] 


Z  R  jX VR  I R

Z

2

  R 

 R 2  X2

VR 80   1.6 k 50  10 -3 I

  X 2  Z

2

 R 2  (3) 2  (1.6) 2

X  2.5377 k

X  2.5377    57.77   tan -1    tan -1  R   1.6  pf  cos   0.5333


S  (110)(2 55)  22055 P  S cos   220 cos(55)  126.2 W

(b)

S  S  220 VA


Apparent power  S  12 kVA

P  S cos   (12)(0.78)  9.36 kW Q  S sin   12 sin(cos -1 (0.78))  7.51 kVAR

S  P  jQ  [9.36  j 7.51] kVA

(b)

S

V Z*

2

  Z 

Z  [2.866 + j2.3] Ω

*

V S

2

(210) 2  = 2.866 – j2.3 (9.36  j7.51)  10 3

Chapter 11, Solution 90 Original load : P1  2000 kW , cos 1  0.85   1  31.79 P1 S1   2352.94 kVA cos 1 Q1  S1 sin 1  1239.5 kVAR

Additional load : P2  300 kW , cos  2  0.8    2  36.87 P2 S2   375 kVA cos  2 Q 2  S 2 sin  2  225 kVAR Total load : S  S1  S 2  (P1  P2 )  j (Q1  Q 2 )  P  jQ

P  2000  300  2300 kW Q  1239.5  225  1464.5 kVAR

The minimum operating pf for a 2300 kW load and not exceeding the kVA rating of the generator is P 2300   0.9775 cos   S1 2352.94   12.177 or The maximum load kVAR for this condition is Q m  S1 sin   2352.94 sin(12.177) Q m  496.313 kVAR The capacitor must supply the difference between the total load kVAR ( i.e. Q ) and the permissible generator kVAR ( i.e. Q m ). Thus, Q c  Q  Q m  968.2 kVAR

Chapter 11, Solution 91 The nameplate of an electric motor has the following information: Line voltage: 220 V rms Line current: 15 A rms Line frequency: 60 Hz Power: 2700 W Determine the power factor (lagging) of the motor. Find the value of the capacitance C that must be connected across the motor to raise the pf to unity. Solution

I = V/Z which leads to Z = [220/15] θ = 14.6667 θ, S = (220)(15) θ = 3.3 θ kVA, where cos–1(2700/3300) = cos–1(0.818182) = 35.097°, and X L = 3300sin(35.097°) = 1897.38 = X C . This leads to C = 1/[377(1897.38)] = 1.398 µF. pf = 0.8182 (lagging) C = 1.398 µF 0.8182 (lagging), 1.398 µF


Apparent power drawn by the motor is P 60   80 kVA Sm  cos  0.75 Q m  S 2  P 2  (80) 2  (60) 2  52.915 kVAR

Total real power P  Pm  Pc  PL  60  0  20  80 kW Total reactive power Q  Q m  Q c  Q L  52.915  20  0  32.91 kVAR Total apparent power S  P 2  Q 2  86.51 kVA (b)

pf 

(c)

I

P 80   0.9248 S 86.51

S 86510   157.3 A V 550


P1  (5)(0.7457)  3.7285 kW P1 3.7285   4.661 kVA S1  pf 0.8 Q1  S1 sin(cos -1 (0.8))  2.796 kVAR S1  3.7285  j2.796 kVA P2  1.2 kW , S 2  1.2  j0 kVA

Q 2  0 VAR

P3  (10)(120)  1.2 kW , S 3  1.2  j0 kVA

Q 3  0 VAR

Q 4  1.6 kVAR , cos  4  0.6   sin  4  0.8 Q4 S4   2 kVA sin  4 P4  S 4 cos  4  (2)(0.6)  1.2 kW S 4  1.2  j1.6 kVA

S  S1  S 2  S 3  S 4 S  7.3285  j1.196 kVA

Total real power = 7.3285 kW Total reactive power = 1.196 kVAR (b)

 1.196    9.27   tan -1   7.3285 

pf  cos   0.987


cos 1  0.7   1  45.57 S1  1 MVA  1000 kVA P1  S1 cos 1  700 kW Q1  S1 sin 1  714.14 kVAR For improved pf, cos  2  0.95    2  18.19 P2  P1  700 kW P2 700 S2    736.84 kVA cos  2 0.95 Q 2  S 2 sin  2  230.08 kVAR

P 1 = P 2 = 700 kW 1

2

Q2 S2 S1 Q1

(a)

Qc

Reactive power across the capacitor Q c  Q1  Q 2  714.14  230.08  484.06 kVAR Cost of installing capacitors  $30  484.06  $14,521.80

(b)

Substation capacity released  S1  S 2  1000  736.84  263.16 kVA Saving in cost of substation and distribution facilities  $120  263.16  $31,579.20

(c)

Yes, because (a) is greater than (b). Additional system capacity obtained by using capacitors costs only 46% as much as new substation and distribution facilities.


Source impedance Load impedance

Zs  R s  jXc ZL  R L  jX 2

For maximum load transfer Z L  Z *s   R s  R L , X c  X L 1 Xc  XL    L C 1 or   2 f LC

f

1 2 LC



1 2 (80  10 -3 )(40  10 -9 ) 2

(b)

 2.814 kHz

2  V  s  4   4.6  4  431.8 mW (since V is in rms) P s  (10  4)   14   


Z Th

+ 

V Th

(a)

VTh  146 V, 300 Hz Z Th  40  j8 

Z L  Z *Th  [40  j 8] 

(b)

P

VTh

2

8 R Th

(146) 2   66.61 W (8)(40)

ZL

Chapter 11, Solution 97 Z T  (2)(0.1  j)  (100  j20)  100.2  j22  Vs 240  I Z T 100.2  j22 2

P  I R L  100 I

2



(100)(240) 2  547.3 W (100.2) 2  (22) 2


If Vab  400 , then 400 Van   - 30  231 - 30 V 3 Vbn  231 - 150 V Vcn  231 - 270 V

(b)

For the acb sequence, Vab  Van  Vbn  Vp 0  Vp 120

 1 3 Vab  Vp 1   j   Vp 3 - 30 2   2 i.e. in the acb sequence, Vab lags Van by 30. Hence, if Vab  400 , then 400 Van  30  23130 V 3 Vbn  231150 V Vcn  231 - 90 V

Chapter 12, Solution 2. Since phase c lags phase a by 120, this is an acb sequence. Vbn  120(30  120)  120150 V

Chapter 12, Solution 3. Since Vbn leads Vcn by 120, this is an abc sequence. Van  440(130  120)  440–110˚ V.

Chapter 12, Solution 4. Knowing the line-to-line voltages we can calculate the wye voltages and can let the value of V a be a reference with a phase shift of zero degrees.

V L = 440 =

V p or V p = 440/1.7321 = 254 V or V an = 254 0° V which

determines, using abc roataion, both V bn = 254 –120° and V cn = 254 120°.

I a = V an /Z Y = 254/(40 30°) = 6.35–30˚ A

I b = I a –120˚ = 6.35–150˚ A I c = I a +120˚ = 6.3590˚ A

Chapter 12, Solution 5. V AB = 1.7321xV AN +30˚ = 207.8(32˚+30˚) = 207.862˚ V or v AB = 207.8cos(ωt+62˚) V which also leads to, v BC = 207.8cos(ωt–58˚) V and v CA = 207.8cos(ωt+182˚) V

207.8cos(ωt+62˚) V, 207.8cos(ωt–58˚) V, 207.8cos(ωt+182˚) V

Chapter 12, Solution 6. Using Fig. 12.41, design a problem to help other students to better understand balanced wye-wye connected circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the Y-Y circuit of Fig. 12.41, find the line currents, the line\ voltages, and the load voltages.


Z Y  10  j5  11.1826.56

The line currents are Van 220 0  Ia   19.68 - 26.56 A Z Y 11.1826.56 I b  I a  - 120  19.68 - 146.56 A I c  I a 120  19.6893.44 A The line voltages are Vab  220 3 30  38130 V Vbc  381 - 90 V Vca  381 - 210 V The load voltages are VAN  I a Z Y  Van  2200 V VBN  Vbn  220 - 120 V VCN  Vcn  220120 V

Chapter 12, Solution 7. This is a balanced Y-Y system.

4400 V

+ 

Z Y = 6  j8 

Using the per-phase circuit shown above, 4400 Ia   4453.13 A 6  j8 I b  I a  - 120  44 - 66.87 A

I c  I a 120  44173.13 A

Chapter 12, Solution 8. Consider the per phase equivalent circuit shown below. Zl

V an

+ _

ZL 5.396–35.1˚ A

I a = V an /( Z l + Z L ) = (100 20°)/(10.6+j15.2) = (100 20°)/(18.531 55.11°)

= 5.396 –35.11° amps.

I b = I a –120° = 5.396 –155.11° amps.

Ic = Ia

120° = 5.396 84.89° amps.

V La = I a Z L = (4.414–j3.103)(10+j14) = (5.396 –35.11° 17.205 54.46°)

= 92.84 19.35° volts.

V Lb = V La –120° = 94.84 –100.65° volts.

V Lc = V La

120° = 94.84 139.35° volts.

Chapter 12, Solution 9. Ia 

Van 1200   4.8 - 36.87 A Z L  Z Y 20  j15

I b  I a  - 120  4.8 - 156.87 A

I c  I a 120  4.883.13 A As a balanced system, I n  0 A

Chapter 12, Solution 10. Since the neutral line is present, we can solve this problem on a per-phase basis. For phase a, Ia 

Van 4400 440    15.28320.32 Z A  2 27  j10 28.79  20.32

Ib 

Vbn 440 - 120   20 - 120 ZB  2 22

Ic 

Vcn 440120 440120    33.8597.38 12  j 5 1322.62 ZC  2

For phase b,

For phase c,

The current in the neutral line is I n  -(I a  I b  I c ) or - I n  I a  I b  I c

- I n  (14.332  j 5.308)  (10  j17.321)  (4.346  j33.57) I n  0.014  j 21.56  21.56–89.96°A

Chapter 12, Solution 11. Given that V p = 240 and that the system is balanced, V L = 1.7321V p = 415.7 V. I p = V L /|2–j3| = 415.7/3.606 = 115.29 A and I L = 1.7321x115.29 = 199.69 A.

Chapter 12, Solution 12. Using Fig. 12.45, design a problem to help other students to better understand wye-delta connected circuits. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Solve for the line currents in the Y- circuit of Fig. 12.45. Take Z  = 6045.

Figure 12.45 Solution Convert the delta-load to a wye-load and apply per-phase analysis.

Ia 1100 V

ZY 

+ 

Z  20 45  3

110 0  5.5 - 45 A 2045 I b  I a  - 120  5.5 - 165 A I c  I a 120  5.575 A Ia 

ZY

Chapter 12, Solution 13. Convert the delta load to wye as shown below.

1100o V rms 2

–+ 110–120o V rms

ZY 2 ZY

–+ 110120o V rms

ZY

2

–+ 1 ZY  Z   3  j 2  3

We consider the single phase equivalent shown below. 2

1100˚ V rms

+ _

3 – j2 

I a = 110/(2 + 3 – j2) = 20.4321.8° A

I L = |I a | = 20.43 A S = 3|I a |2Z Y = 3(20.43)2(3–j2) = 4514–33.96˚ = 3744 – j2522 P = Re(S) = 3.744 kW.

Chapter 12, Solution 14. We apply mesh analysis with Z L = (12+j12) Ω.

a

Ia 1+j2 Ω

+ 

1000

I1

A

n 100120 + c





100–120 +

I2

ZL

I3

ZL

Ib b

1+j2 Ω

B

C

ZL

1+j2 Ω

Ic For mesh 1,  100  100  120o  I1 (14  j16)  (1  j 2) I 2  (12  j12) I 3  0 or (14  j16) I1  (1  j 2) I 2  (12  j12) I 3  100  50  j86.6  150  j86.6

(1)

For mesh 2, 100120 o  100  120 o  I 1 (1  j 2)  (12  j12) I 3  (14  j16) I 2  0 or  (1  j 2) I 1  (14  j16) I 2  (12  j12) I 3  50  j86.6  50  j86.6   j173.2

(2)

For mesh 3,  (12  j12) I 1  (12  j12) I 2  (36  j36) I 3  0 or I 3 = I 1 + I 2

(3)

Solving for I 1 and I 2 using (1) to (3) gives I 1 = 12.804–50.19° A = (8.198 – j9.836) A and I 2 = 12.804–110.19° A = (–4.419 – j12.018) A I a = I 1 = 12.804–50.19° A I b = I 2 – I 1 = 12.804–170.19° A I c = –I 2 = 12.80469.81° A

As a check we can convert the delta into a wye circuit. Thus,

Z Y = (12+j12)/3 = 4+j4 and I a = 100/(1+j2+4+j4) = 100/(5+j6)

= 100/(7.8102 50.19°) =

12.804 –50.19° A.

So, the answer does check.

Chapter 12, Solution 15. Convert the delta load, Z  , to its equivalent wye load. Z  8  j10 Z Ye  3 (12  j5)(8  j10)  8.076  - 14.68 20  j5 Z p  7.812  j2.047

Z p  Z Y || Z Ye 

Z T  Z p  Z L  8.812  j1.047 Z T  8.874  - 6.78

We now use the per-phase equivalent circuit. Vp 210 Ia  , where Vp  Zp  ZL 3 Ia 

210 3 (8.874  - 6.78)

I L  I a  13.66 A

 13.66 6.78


I CA  - I AC  5(-30  180)  5150 This implies that I AB  530 I BC  5  90 I a  I AB 3  - 30  8.660˚ A

I b  8.66–120˚ A I c  8.66120˚ A (b)

Z 

VAB 1100   22–30˚ Ω. I AB 530

Chapter 12, Solution 17. I a = 1.7321xI AB –30˚ or I AB = I a /(1.7321–30˚) = 2.887(–25˚+30˚) = 2.8875˚ A I BC = I AB –120˚ = 2.887–115˚ A I CA = I AB +120˚ = 2.887125˚ A 2.8875˚ A, 2.887–115˚ A, 2.887125˚ A

Chapter 12, Solution 18. VAB  Van 3 30  (22060)( 3 30)  381.190 Z   12  j9  1536.87

I AB 

VAB 381.190   25.453.13˚ A Z  1536.87

I BC  I AB  - 120  25.4–66.87˚ A I CA  I AB 120  25.4173.13˚ A

Chapter 12, Solution 19. Z   30  j10  31.62 18.43

The phase currents are Vab 1730  5.47  - 18.43 A I AB   Z  31.62 18.43 I BC  I AB  - 120  5.47  - 138.43 A I CA  I AB 120  5.47 101.57 A The line currents are I a  I AB  I CA  I AB 3  - 30 I a  5.47 3  - 48.43  9.474 - 48.43 A

I b  I a  - 120  9.474 - 168.43 A I c  I a 120  9.47471.57 A

Chapter 12, Solution 20. Using Fig. 12.51, design a problem to help other students to better understand balanced deltadelta connected circuits. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Refer to the - circuit in Fig. 12.51. Find the line and phase currents. Assume that the load impedance is 12 + j9 per phase.

Figure 12.51 Solution Z   12  j9  1536.87

The phase currents are

2100  14 - 36.87 A 1536.87  I AB  - 120  14 - 156.87 A  I AB 120  1483.13 A

I AB 

I BC I CA

The line currents are I a  I AB 3  - 30  24.25 - 66.87 A I b  I a  - 120  24.25 - 186.87 A I c  I a 120  24.2553.13 A


I AC 

 230120  230120   17.96  98.66 A 10  j8 12.80638.66 I AC = 17.96–98.66˚ A

230  120 2300  10  j8 10  j8  17.96  158.66  17.96  38.66  16.729  j 6.536  14.024  j11.220  30.75  j 4.684

I bB  I BC  I BA  I BC  I AB  (b)

I bB = 31.1171.34˚ A.

Chapter 12, Solution 22. Convert the -connected source to a Y-connected source. Vp 440 Van   - 30   - 30  254 - 30 3 3 Convert the -connected load to a Y-connected load. Z (4  j6)(4  j5) Z  Z Y ||  (4  j6) || (4  j5)  8 j 3 Z  5.723  j0.2153

ZL

V an

Ia

+ 

Z

Van 254  30   32.88–28.4˚ A Z L  Z 7.723  j 0.2153 I b  I a  - 120  32.88–148.4˚ A

Ia 

I c  I a 120  32.8891.6˚ A


(a)

I AB 

I a  I AB

V AB 202  Z 2560 o

202 3  30 o 3  30   13.995  90 o o 25 60 o

I L | I a | 13.995 A (b)

 202 3   cos 60 o P  P1  P2  3V L I L cos   3 ( 202)  25   = 2.448 kW

Chapter 12, Solution 24. Convert both the source and the load to their wye equivalents. Z  20 30  17.32  j10 ZY  3 Vab Van   - 30  240.20 3

We now use per-phase analysis. 1+j

V an

Ia 

+ 

Ia

2030 

Van 240.2   11.24 - 31 A (1  j)  (17.32  j10) 21.37 31

I b  I a  - 120  11.24 - 151 A

I c  I a 120  11.2489 A But I AB 

I a  I AB 3  - 30

11.24  - 31 3  - 30

 6.489 - 1 A

I BC  I AB  - 120  6.489 - 121 A I CA  I AB 120  6.489119 A

Chapter 12, Solution 25. Convert the delta-connected source to an equivalent wye-connected source and consider the single-phase equivalent. Ia 

where

440 (10  30)

3 ZY ZY  3  j 2  10  j8  13  j 6  14.318  24.78

Ia 

440  20 \= 17.742 4.78° amps. 3 (14.318  24.78)

I b = I a –120° = 17.742 –115.22° amps.

I c = I a +120° = 17.742 124.78° amps.

Chapter 12, Solution 26. Using Fig. 12.55, design a problem to help other students to better understand balanced delta connected sources delivering power to balanced wye connected loads. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the balanced circuit in Fig. 12.55, V ab = 1250 V. Find the line currents I aA , I bB , and I cC .


Transform the source to its wye equivalent. Vp Van   - 30  72.17  - 30 3 Now, use the per-phase equivalent circuit. Van I aA  , Z  24  j15  28.3 - 32 Z

I aA 

72.17  - 30  2.55 2 A 28.3 - 32

I bB  I aA  - 120  2.55 - 118 A I cC  I aA 120  2.55122 A

Chapter 12, Solution 27. Since Z L and Z  are in series, we can lump them together so that ZY  2  j  6  j 4  8  j 5 VP  30o 208  30o  Ia  3 ZY 3(8  j 5) 208(0.866  j 0.5)(6  j 4) VL  (6  j 4) I a   80.81  j 43.54 3(8  j 5)

|V L | = 91.79 V

Chapter 12, Solution 28. VL  Vab  440  3VP or V P = 440/1.7321 = 254 For reference, let V AN = 2540˚ V which leads to V BN = 254–120˚ V and V CN = 254120˚ V. The line currents are found as follows, I a = V AN /Z Y = 254/2530˚ = 10.16–30˚ A. This leads to, I b = 10.16–150˚ A and I c = 10.1690˚ A.

Chapter 12, Solution 29. We can replace the delta load with a wye load, Z Y = Z Δ /3 = 17+j15Ω. The per-phase equivalent circuit is shown below. Zl

V an

+ _

ZY

I a = V an /|Z Y + Z l | = 240/|17+j15+0.4+j1.2| = 240/|17.4+j16.2| = 240/23.77 = 10.095 S = 3[(I a )2(17+j15)] = 3x101.91(17+j15) = [5.197+j4.586] kVA.

Chapter 12, Solution 30. Since this a balanced system, we can replace it by a per-phase equivalent, as shown below.

+

ZL

Vp 3V 2 p S  3S p  * , Z p

Vp 

VL 3

V 2L (208) 2 S *   1.442145 o kVA o Z p 30  45

P  S cos   1.02 kW


(a) Pp  6,000,

cos  0.8,

Q p  S P sin   4.5 kVAR

Sp 

PP  6 / 0.8  7.5 kVA cos 

S  3S p  3(6  j 4.5)  18  j13.5 kVA For delta-connected load, V p = V L = 240 (rms). But S

(b)

3V 2 p Z*p

 

Z*p 

Pp  3V L I L cos 

3V 2 p 3( 240) 2  , S (18  j13.5) x10 3

 

IL 

6000 3 x 240 x 0.8

Z P  [6.144  j 4.608] 

 18.04 A

(c ) We find C to bring the power factor to unity Qc  Q p  4.5 kVA

 

C

Qc 4500   207.2 F 2 V rms 2x 60 x 240 2

Chapter 12, Solution 32. Design a problem to help other students to better understand power in a balanced threephase system. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A balanced wye load is connected to a 60-Hz three-phase source with V ab = 2400˚V. The load has lagging pf =0.5 and each phase draws 5 kW. (a) Determine the load impedance Z Y . (b) Find I a , I b , and I c . Solution (a) | Vab | 3V p  240

  Vp 

240  138.56 3

Van  V p  30o

pf  0.5  cos 

    60o P 5   S   10 kVA P  S cos  cos  0.5 Q  S sin   10sin 60  8.66 S p  5  j8.66 kVA But SP 

V p2 Z *p

V p2

138.562   Z    0.96  j1.663 S p (5  j8.66) x103 * p

Z p = [0.96 + j1.663] 

(b)

Van 138.56  30o Ia    72.17  90o A = 72.17 –90° A ZY 0.96  j1.6627

I b  I a  120o  72.17  210o A = 72.17 150° A

I c  I a  120o  72.17  30o A = 72.17 30° A


S  3 VL I L  S  S  3 VL I L For a Y-connected load, IL  Ip , VL  3 Vp

S  3 Vp I p IL  Ip 

S 4800   7.69 A 3 Vp (3)(208)

VL  3 Vp  3  208  360.3 V


Vp 

Ia 

VL 3 Vp ZY





220 3 220 3 (10  j16)



127.02  6.73258 18.868  58

I L  I p  6.732A

S  3 VL I L   3  220  6.732 - 58  2565  58 S = [1.3592–j2.175] kVA


(a) This is a balanced three-phase system and we can use per phase equivalent circuit. The delta-connected load is converted to its wye-connected equivalent

Z '' y 

1 Z   (60  j 30) / 3  20  j10 3 IL + 230 V -

Z y  Z ' y // Z '' y  (40  j10) //( 20  j10)  13.5  j 5.5

IL 

230  [14.61  j5.953] A 13.5  j5.5

(b) S  3Vs I * L  [10.081  j4.108] kVA (c ) pf = P/S = 0.9261

Z’y

Z’’y


(a)

(b) S  3V p I * p

S = 1 [0.75 + sin(cos-10.75) ] = 0.75 + j0.6614 MVA

 

I*p 

S (0.75  j 0.6614) x10 6   59.52  j 52.49 3V p 3x 4200

PL | I p | 2 Rl  (79.36) 2 (4)  25.19 kW

(c) V s  V L  I p (4  j )  4.4381  j 0.21 kV  4.443 - 2.709 o kV

Chapter 12, Solution 37. S

P 12   20 pf 0.6

S  S  20  12  j16 kVA But

S  3 VL I L  IL 

S  3 Ip

20  10 3 3  208

 55.51 A

2

Zp

For a Y-connected load, I L  I p .

Zp 

S 3 IL

2



(12  j16)  10 3 (3)(55.51) 2 Z p  [1.298  j1.731] 

Chapter 12, Solution 38. As a balanced three-phase system, we can use the per-phase equivalent shown below. Ia 

110 0 110 0  (1  j2)  (9  j12) 10  j14

S p  Ia

2

ZY 

(110) 2  (9  j12) (10 2  14 2 )

The complex power is (110) 2 S  3S p  3  (9  j12) 296 S  (1.1037+j1.4716) kVA

Chapter 12, Solution 39. Consider the system shown below. 5

a

A 4

-j6  100120

c

+



+



 + 100-120

1000

I1 5

b

8 B

I2

j3 

I3

C 10 

5

For mesh 1, 100  (18  j6) I 1  5 I 2  (8  j6) I 3

(1)

100  - 120  20 I 2  5 I 1  10 I 3 20 - 120  - I 1  4 I 2  2 I 3

(2)

0  - (8  j6) I 1  10 I 2  (22  j3) I 3

(3)

For mesh 2,

For mesh 3,

To eliminate I 2 , start by multiplying (1) by 2, 200  (36  j12) I 1  10 I 2  (16  j12) I 3

(4)

Subtracting (3) from (4), 200  (44  j18) I 1  (38  j15) I 3

(5)

Multiplying (2) by 5 4 , 25 - 120  -1.25 I 1  5 I 2  2.5 I 3

(6)

Adding (1) and (6), 87.5  j21.65  (16.75  j6) I 1  (10.5  j6) I 3

(7)

In matrix form, (5) and (7) become

   44  j18 - 38  j15  I 1  200 87.5  j12.65  16.75  j6 - 10.5  j6  I      3 

  192.5  j26.25 ,

 1  900.25  j935.2 ,

 3  110.3  j1327.6

I1 

 1 1298.1 - 46.09   6.682  - 38.33  5.242  j4.144  194.28 - 7.76

I3 

 3 1332.2 - 85.25   6.857 - 77.49  1.485  j6.694  194.28 - 7.76

We obtain I 2 from (6), 1 1 I 2  5 - 120  I 1  I 3 4 2 I 2  (-2.5  j4.33)  (1.3104  j1.0359)  (0.7425  j3.347) I 2  -0.4471  j8.713

The average power absorbed by the 8- resistor is 2

2

P1  I 1  I 3 (8)  3.756  j2.551 (8)  164.89 W The average power absorbed by the 4- resistor is 2

P2  I 3 (4)  (6.8571) 2 (4)  188.1 W The average power absorbed by the 10- resistor is 2 2 P3  I 2  I 3 (10)  - 1.9321  j2.019 (10)  78.12 W

Thus, the total real power absorbed by the load is P  P1  P2  P3  431.1 W

Chapter 12, Solution 40. Transform the delta-connected load to its wye equivalent. Z  7  j8 ZY  3 Using the per-phase equivalent circuit above, 100 0 Ia   8.567  - 46.75 (1  j0.5)  (7  j8) For a wye-connected load, I p  I a  I a  8.567 S  3 Ip

2

Z p  (3)(8.567) 2 (7  j8)

P  Re(S)  (3)(8.567) 2 (7)  1.541 kW

Chapter 12, Solution 41. S

But

P 5 kW   6.25 kVA pf 0.8

S  3 VL I L IL 

S 3 VL



6.25  10 3 3  400

 9.021 A

Chapter 12, Solution 42. The load determines the power factor. 40 tan    1.333     53.13 30

pf  cos   0.6 (leading)  7.2  S  7.2  j (0.8)  7.2  j9.6 kVA  0.6  But

S  3 Ip 2

Ip



2

Zp

S (7.2  j9.6)  10 3   80 3Zp (3)(30  j40)

I p  8.944 A I L  I p  8.944 A VL 

S 3 IL



12  10 3 3 (8.944)

 774.6 V

Chapter 12, Solution 43. S  3 Ip

2

Zp ,

I p  I L for Y-connected loads

S  (3)(13.66) 2 (7.812  j2.047) S  [4.373  j1.145] kVA

Chapter 12, Solution 44. For a -connected load, Vp  VL ,

IL  3 Ip

S  3 VL I L

IL 

S 3 VL



(12 2  5 2 )  10 3 3 (240)

 31.273

At the source, VL'  VL  I L Z l + I L Z l VL'  2400  2(31.273)(1  j 3) = 240+62.546+j187.638 VL'  302.546+j187.638 = 356 31.81° VL'  356 V

Also, at the source, S’ = 3(31.273)2(1+j3) + (12,000+j5,000) = 2,934+12,000+j(8,802+5,000) = 14,934+j13,802 = 20,335 42.744° thus, θ = 42.744°.

pf = cos(42.744°) = 0.7344 Checking, V Y = 240/1.73205 = 138.564, S = 3(138.564)2/(Z Y )* = 12,000+l5,000, and Z Y = 57,600/(12,000–j5,000) = 57.6/(13 –22.62°) = 4.4308 22.62° = 4.09+j1.70416. The

total load seen by the source is 1+j3+4.09+j1.70416 = 5.09+j4.70416 = 6.9309 42.74° per phase. This leads to θ = Tan-1(4.70416/5.09) = Tan-1(0.9242) = 42.744°. Clearly, the answer checks. I l = 138.564/4.4308 = 31.273 A. Again the answer checks. Finally,

3(31.273)2(5.09+j4.70416) = 2,934(6.9309 42.74°) = 20,335 42.74°, the same as we calculated above.


S  3 VL I L  IL 

IL 

S - 3 VL

,

(635.6)  -  3  440

P 450  10 3 S    635.6 kVA pf 0.708  834  - 45 A

At the source, VL  440 0  I L (0.5  j2) VL  440  (834  - 45)(2.062 76) VL  440  1719.7 31 VL  1914.1  j885.7 VL  2.109 24.83 V

Chapter 12, Solution 46. For the wye-connected load, IL  Ip , VL  3 Vp

I p  Vp Z

2

S  3V I  * p p

S

VL

2

Z*

3 Vp



Z*

3 VL

(110) 2   121 W 100

2

S  3V I  S

2

Z*

For the delta-connected load, Vp  VL , IL  3 Ip , * p p

3

3 Vp Z*



3 VL

I p  Vp Z 2

Z*

(3)(110) 2  363 W 100

This shows that the delta-connected load will absorb three times more average power than the wye-connected load using the same elements.. This is also evident Z from Z Y   . 3


pf  0.8 (lagging)     cos -1 (0.8)  36.87 S1  250 36.87  200  j150 kVA pf  0.95 (leading)     cos -1 (0.95)  -18.19 S 2  300  - 18.19  285  j93.65 kVA pf  1.0     cos -1 (1)  0 S 3  450 kVA S T  S1  S 2  S 3  935  j56.35  936.7 3.45 kVA S T  3 VL I L

IL 

936.7  10 3 3 (13.8  10 3 )

 39.19 A rms

pf  cos   cos(3.45)  0.9982 (lagging)

Chapter 12, Solution 48. (a) We first convert the delta load to its equivalent wye load, as shown below. A A 18-j12 

ZA

40+j15 

ZB ZC C

B

C

B

60 

(40  j15)(18  j12)  7.577  j1.923 118  j 3 60(40  j15). ZB   20.52  j7.105 118  j3 60(18  j12) ZC   8.992  j 6.3303 118  j 3 The system becomes that shown below. ZA 

`

a

2+j3

A

+ 240<0o -

ZA I1

240<120o + c

-

I2

240<-120o + b

ZB

ZC

2+j3 B

2+j3 We apply KVL to the loops. For mesh 1,

C

 240  240  120 o  I 1 (2Z l  Z A  Z B )  I 2 ( Z B  Z l )  0 or (32.097  j11.13) I 1  (22.52  j10.105) I 2  360  j 207.85 For mesh 2, 240120 o  240  120 o  I 1 ( Z B  Z l )  I 2 (2Z l  Z B  Z C )  0 or

(1)

 (22.52  j10.105) I 1  (33.51  j 6.775) I 2   j 415.69 Solving (1) and (2) gives I 1  23.75  j 5.328, I 2  15.165  j11.89

(2)

I aA  I 1  24.34  12.64o A,

I bB  I 2  I 1  10.81  142.6o A

I cC   I 2  19.27141.9o A (b) S a  ( 2400o )( 24.3412.64o )  5841.612.64o

S b  ( 240  120o )(10.81142.6o )  2594.4 22.6o S c  ( 240120o )(19.27  141.9o )  4624.8  21.9o S  S a  S b  S c  12.386  j0.55 kVA  12.4 2.54o kVA

Chapter 12, Solution 49. (a) For the delta-connected load, Z p  20  j10, S

V p  VL  220 (rms) ,

3V 2 p 3 x 220 2   5808  j 2904  6.94326.56 o kVA * (20  j10) Z p

P = 5.808 kW (b) For the wye-connected load, Z p  20  j10, S

3V 2 p 3 x 220 2   2.16426.56 o kVA 3(20  j10) Z*p

P = 1.9356 kW

V p  VL / 3 ,


S  S 1  S 2  8(0.6  j 0.8)  4.8  j 6.4 kVA, Hence,

S 1  3 kVA

S 2  S  S 1  1.8  j 6.4 kVA But S 2 

3V 2 p , Z*p

Z

*

p



Vp 

V *L S2

VL 3

 

240 2  (1.8  j 6.4) x10 3

S2 

.V 2 L Z*p

 

Z p  2.346  j 8.34

Chapter 12, Solution 51. This is an unbalanced system. 240  0o 240  0o I AB    19.2-j14.4 A Z1 8  j6

I BC 

240120 240120   50.62147.65˚ = [–42.76+j27.09] A Z2 4.7413  27.65

I CA 

240  120 240  120   [–12–j20.78] A Z3 10

At node A, I aA  I AB  I CA  (19.2  j14.4)  (12  j 20.78)  31.2  j 6.38 A

I bBI b  I BC  I AB  (42.76  j 27.08)  (19.2  j14.4)  61.96  j 41.48 A I cCI c  I CA  I

BC

 (12  j 20.78)  (42.76  j 27.08)  30.76  j 47.86 A

Chapter 12, Solution 52. Since the neutral line is present, we can solve this problem on a per-phase basis. Van 120 120  Ia   6 60 20 60 Z AN Vbn 120 0  4 0 Ib   Z BN 30 0 Vcn 120  - 120  3 - 150 Ic   Z CN 40 30 Thus, - In - In - In - In

 Ia  Ib  Ic  6 60  4 0  3 - 150  (3  j5.196)  (4)  (-2.598  j1.5)  4.405  j3.696  5.7540

I n  5.75220 A

Chapter 12, Solution 53. Using Fig. 12.61, design a problem that will help other students to better understand unbalanced three-phase systems. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem In the wye-wye system shown in Fig. 12.61, loads connected to the source are unbalanced. (a) Calculate I a , I b , and I c . (b) Find the total power delivered to the load. Take V P = 240 V rms. Ia

+ _

V P 0˚

V P 120˚ + –

100

V P –120˚ –+

80 Ib

60 

Ic Figure 12.61

For Prob. 12.53.

Solution Applying mesh analysis as shown below, we get. Ia

+ _

V P 0˚

100 I1

V P 120˚ + –

V P –120˚ – +

80 Ib

60  I2

Ic

240–120˚ – 240 + 160I 1 – 60I 2 = 0 or 160I 1 – 60I 2 = 360+j207.84

(1)

240120˚ – 240–120˚ – 60I 1 + 140I 2 = 0 or – 60I 1 + 140I 2 = –j415.7

(2)

In matrix form, (1) and (2) become 160 60   I1  360  j 207.84   60 140   I     j 415.7    2  

Using MATLAB, we get, >> Z=[160,-60;-60,140] Z= 160 -60 -60 140 >> V=[(360+207.8i);-415.7i] V= 1.0e+002 * 3.6000 + 2.0780i 0 - 4.1570i >> I=inv(Z)*V I= 2.6809 + 0.2207i 1.1489 - 2.8747i

I 1 = 2.681+j0.2207 and I 2 = 1.1489–j2.875 I a = I 1 = 2.694.71˚ A I b = I 2 – I 1 = –1.5321–j3.096 = 3.454–116.33˚ A I c = –I 2 = 3.096111.78˚ A

S a | I a |2 Z a  (2.69) 2 x100  723.61

Sb | I b |2 Z b  (3.454)2 x60  715.81 Sc | I c |2 Z c  (3.0957) 2 x80  766.67 S  S a  Sb  Sc  2.205 kVA

Chapter 12, Solution 54. Consider the load as shown below.

Ia

A

N

Ib

B

Ic

Ia 

210  0o  2.625 A 80

Ib 

2100 210   1.9414–56.31˚ A 60  j90 108.1756.31

210  0o  2.625  90o A j80 * S a  VI a  210 x 2.625  551.25 Ic 

Sb  VI b* 

| V |2 2102   226.15  j 339.2 Z b* 60  j 90

| V |2 2102   j 551.25 Z c*  j80 S  S a  Sb  Sc  777.4  j890.45 VA

Sc 

C

1 Chapter 12, Solution 55. The phase currents are: I AB = 240/j25 = 9.6–90˚ A I CA = 240120˚/40 = 6120˚ A I BC = 240–120˚/3030˚ = 8–150˚ A The complex power in each phase is: S AB | I AB |2 Z AB  (9.6) 2 j 25  j 2304

S AC | I AC |2 Z AC  (6) 2 40  0o  1440

S BC | I BC |2 Z BC  (8) 2 30  30o  1662.77  j960 The total complex power is S  S AB  S AC  S BC  3102.77  j 3264 VA

= [3.103+j3.264] kVA

Chapter 12, Solution 56. Using Fig. 12.63, design a problem to help other students to better understand unbalanced threephase systems. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Refer to the unbalanced circuit of Fig. 12.63. Calculate: (a) the line currents (b) the real power absorbed by the load (c) the total complex power supplied by the source


(a)

Consider the circuit below. a

A

4400 +  b 440120

+



j10 

I1 B

 + 440-120

I2

c For mesh 1, 440 - 120  440 0  j10 (I 1  I 3 )  0

I3

-j5 

20  C

I1  I 3 

(440)(1.5  j0.866)  76.21 - 60 j10

For mesh 2, 440120  440 - 120  20 (I 2  I 3 )  0 (440)( j1.732) I3  I2   j38.1 20 For mesh 3, j10 (I 3  I 1 )  20 (I 3  I 2 )  j5 I 3  0

Substituting (1) and (2) into the equation for mesh 3 gives, (440)(-1.5  j0.866) I3   152.4260 j5 From (1), I 1  I 3  76.21 - 60  114.315  j66  13230 From (2), I 2  I 3  j38.1  76.21  j93.9  120.9350.94 I a  I 1  13230 A I b  I 2  I 1  -38.105  j27.9  47.23143.8 A I c  - I 2  120.9230.9 A (b)

2

S AB  I 1  I 3 ( j10)  j58.08 kVA 2

S BC  I 2  I 3 (20)  29.04 kVA 2

S CA  I 3 (-j5)  (152.42) 2 (-j5)  -j116.16 kVA S  S AB  S BC  S CA  29.04  j58.08 kVA Real power absorbed = 29.04 kW (c)

Total complex supplied by the source is S  29.04  j58.08 kVA

(1)

(2)

(3)

Chapter 12, Solution 57. We apply mesh analysis to the circuit shown below. Ia + Va –

80  j 50 I1

–

20  j 30

–

Vc +

60  j 40

Vb +

Ib

I2 Ic

(100  j80) I 1  (20  j 30) I 2  Va  Vb  165  j 95.263  (20  j 30) I 1  (80  j10) I 2  Vb  Vc   j190.53 Solving (1) and (2) gives I 1  1.8616  j 0.6084, I 2  0.9088  j1.722 . I a  I 1  1.9585  18.1o A,

(1) (2)

I b  I 2  I 1  0.528  j1.1136  1.4656  130.55 o A I c   I 2  1.947117.8 o A

Chapter 12, Solution 58. The schematic is shown below. IPRINT is inserted in the neutral line to measure the current through the line. In the AC Sweep box, we select Total Ptss = 1, Start Freq. = 0.1592, and End Freq. = 0.1592. After simulation, the output file includes FREQ

IM(V_PRINT4)

IP(V_PRINT4)

1.592 E–01

2.156 E+01

–8.997 E+01

i.e.

I n = 21.56–89.97 A

ACMAG=440V ACMAG=440V

ACMAG=440V

Chapter 12, Solution 59. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, we obtain an output file which includes

i.e.

FREQ

VM(1)

VP(1)

6.000 E+01

2.206 E+02

–3.456 E+01

FREQ

VM(2)

VP(2)

6.000 E+01

2.141 E+02

–8.149 E+01

FREQ

VM(3)

VP(3)

6.000 E+01

4.991 E+01

–5.059 E+01

V AN = 220.6–34.56, V BN = 214.1–81.49, V CN = 49.91–50.59 V

Chapter 12, Solution 60. The schematic is shown below. IPRINT is inserted to give I o . We select Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. Upon simulation, the output file includes

from which,

FREQ

IM(V_PRINT4)

IP(V_PRINT4)

1.592 E–01

1.953 E+01

–1.517 E+01

I o = 19.53–15.17 A

+

–

Chapter 12, Solution 61. The schematic is shown below. Pseudocomponents IPRINT and PRINT are inserted to measure I aA and V BN . In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. Once the circuit is simulated, we get an output file which includes FREQ

VM(2)

VP(2)

1.592 E–01

2.308 E+02

–1.334 E+02

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

1.115 E+01

3.699 E+01

from which I aA = 11.1537 A, V BN = 230.8–133.4 V

Chapter 12, Solution 62. Using Fig. 12.68, design a problem to help other students to better understand how to use PSpice to analyze three-phase circuits. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem The circuit in Fig. 12.68 operates at 60 Hz. Use PSpice to find the source current I ab and the line current I bB .

Figure 12.68 Solution Because of the delta-connected source involved, we follow Example 12.12. In the AC Sweep box, we type Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, the output file includes FREQ

IM(V_PRINT2)

IP(V_PRINT2)

6.000 E+01

5.960 E+00

–9.141 E+01

FREQ

IM(V_PRINT1)

IP(V_PRINT1)

6.000 E+01

7.333 E+07

1.200 E+02

From which I ab = 3.432-46.31 A, I bB = 10.39–78.4 A

Chapter 12, Solution 63. Let   1 so that L  X/  20 H, and C 

1  0.0333 F X

The schematic is shown below..

. When the file is saved and run, we obtain an output file which includes the following:

FREQ

IM(V_PRINT1)IP(V_PRINT1)

1.592E-01 1.867E+01 1.589E+02

FREQ


1.592E-01 1.238E+01 1.441E+02 From the output file, the required currents are: I aA  18.67158.9 o A, I AC  12.38144.1 o A

Chapter 12, Solution 64. We follow Example 12.12. In the AC Sweep box we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation the output file includes FREQ

IM(V_PRINT1)

IP(V_PRINT1)

1.592 E–01

4.710 E+00

7.138 E+01

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

6.781 E+07

–1.426 E+02

FREQ

IM(V_PRINT3)

IP(V_PRINT3)

1.592 E–01

3.898 E+00

–5.076 E+00

FREQ

IM(V_PRINT4)

IP(V_PRINT4)

1.592 E–01

3.547 E+00

6.157 E+01

FREQ

IM(V_PRINT5)

IP(V_PRINT5)

1.592 E–01

1.357 E+00

9.781 E+01

FREQ

IM(V_PRINT6)

IP(V_PRINT6)

1.592 E–01

3.831 E+00

–1.649 E+02

from this we obtain I aA = 4.7171.38 A, I bB = 6.781–142.6 A, I cC = 3.898–5.08 A I AB = 3.54761.57 A, I AC = 1.35797.81 A, I BC = 3.831–164.9 A

Chapter 12, Solution 65. Due to the delta-connected source, we follow Example 12.12. We type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. The schematic is shown below. After it is saved and simulated, we obtain an output file which includes

Thus,

FREQ

IM(V_PRINT1)

IP(V_PRINT1)

1.592E-01

1.140E+01

8.664E+00

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592E-01

1.140E+01

-1.113E+02

FREQ

IM(V_PRINT3)

IP(V_PRINT3)

1.592E-01

1.140E+01

1.287E+02

I aA = 11.0212 A, I bB = 11.02–108 A, I cC = 11.02132 A

Since this is a balanced circuit, we can perform a quick check. The load resistance is large compared to the line and source impedances so we will ignore them (although it would not be difficult to include them). Converting the sources to a Y configuration we get: V an = 138.56 –20˚ Vrms and Z Y = 10 – j6.667 = 12.019–33.69˚ Now we can calculate, I aA = (138.56 –20˚)/(12.019–33.69˚) = 11.52813.69˚ Clearly, we have a good approximation which is very close to what we really have.


VL

Vp 

(b)

Because the load is unbalanced, we have an unbalanced three-phase system. Assuming an abc sequence, 120 0 I1   2.50 A 48 120 - 120 I2   3 - 120 A 40 120120 I3   2120 A 60

3



208

(a)

3

 120 V

  3 3 - I N  I 1  I 2  I 3  2.5  (3)  - 0.5  j   (2)  - 0.5  j  2  2    IN  j

3  j0.866  0.86690 A 2

Hence, I1  2.5 A ,

I2  3 A ,

(c)

P1  I12 R 1  (2.5) 2 (48)  300 W P2  I 22 R 2  (3) 2 (40)  360 W P3  I 32 R 3  (2) 2 (60)  240 W

(d)

PT  P1  P2  P3  900 W

I3  2 A ,

I N  0.866 A


The power to the motor is PT  S cos   (260)(0.85)  221 kW The motor power per phase is 1 Pp  PT  73.67 kW 3

Hence, the wattmeter readings are as follows: Wa  73.67  24  97.67 kW Wb  73.67  15  88.67 kW Wc  73.67  9  82.67 kW (b)

The motor load is balanced so that I N  0 . For the lighting loads, 24,000  200 A Ia  120 15,000 Ib   125 A 120 9,000 Ic   75 A 120

If we let I a  I a 0  2000 A I b  125 - 120 A I c  75120 A Then, - I N  Ia  Ib  Ic   3 3 - I N  200  (125) - 0.5  j   (75) - 0.5  j  2  2    - I N  100  j43.3 A

I N  108.97 A


S  3 VL I L  3 (330)(8.4)  4801 VA

(b)

P  S cos    pf  cos  

pf 

4500  0.9372 4801.24

(c)

For a wye-connected load, I p  I L  8.4 A

(d)

Vp 

VL 3



330 3

 190.53 V

P S

Chapter 12, Solution 69. For load 1, S 1  S1 cos  1  jS1 sin 1 pf  0.85  cos 1   1  31.79o

S 1  13.6  j8.43 kVA For load 2, S 2  12 x0.6  j12 x0.8  7.2  j 9.6 kVA For load 3, S 3  8  j 0 kVA Therefore, S = S 1 + S 2 + S 3 = [28.8+j18.03] kVA

Although we can solve this using a delta load, it will be easier to assume our load is wye connected. We also need the wye voltages and will assume that the phase angle on V an = 208/1.73205 = 120.089 is –30 degrees.

Since S = 3VI* or I* = S/(3V) = (33,978 32.048°)/[3(120.089) –30°] =

94.31 62.05° A.

I a = 94.31 –62.05° A, I b = 94.31 177.95° A, I c = 94.31 57.95° A

I = 138.46 – j86.68 = 163.35-32˚ A.

Chapter 12, Solution 70. PT  P1  P2  1200  400  800 Q T  P2  P1  -400  1200  -1600

tan  

Q T - 1600   -2     -63.43 PT 800

pf  cos   0.4472 (leading) Zp 

VL 240   40 IL 6

Z p  40 - 63.43 


If Vab  2080 , Vbc  208 - 120 , Vca  208120 , Vab 2080 I AB    10.4 0 20 Z Ab Vbc 208 - 120 I BC    14.708 - 75 Z BC 10 2  - 45 I CA 

Vca 208120  16 97.38  Z CA 1322.62

I aA  I AB  I CA  10.40  16 97.38 I aA  10.4  2.055  j15.867 I aA  20.171 - 51.87 I cC  I CA  I BC  16 97.83  14.708 - 75 I cC  30.64 101.03

P1  Vab I aA cos( Vab  IaA ) P1  (208)(20.171) cos(0  51.87)  2.590 kW

P2  Vcb I cC cos( Vcb   IcC ) But

Vcb  -Vbc  20860

P2  (208)(30.64) cos(60  101.03)  4.808 kW

(b)

PT  P1  P2  7398.17 W

Q T  3 (P2  P1 )  3840.25 VAR S T  PT  jQ T  7398.17  j3840.25 VA

S T  S T  8.335 kVA

Chapter 12, Solution 72. From Problem 12.11, VAB  220 130 V

and

I aA  30 180 A

P1  (220)(30) cos(130  180)  4.242 kW

VCB  -VBC  220190 I cC  30 - 60 P2  (220)(30) cos(190  60)  - 2.257 kW


I1 Ia 240-60 V

+ 

Z Z

240-120 V

 +

Z I2

Ib

Ic

Z  10  j30  31.6271.57

240 - 60  7.59 - 131.57 31.6271.57 240  - 120 Ib   7.59 - 191.57 31.62 71.57 Ia 

I c Z  240 - 60  240  - 120  0 - 240 Ic   7.59108.43 31.6271.57 I 1  I a  I c  13.146 - 101.57 I 2  I b  I c  13.146138.43 P1  Re  V1 I 1*   Re  (240 - 60)(13.146 101.57)   2.360 kW P2  Re  V2 I *2   Re  (240  - 120)(13.146 - 138.43)   - 632.8 W


Z = 60  j30 

2080 V

+ 

I1 Z

208-60 V

 +

I2 Z

For mesh 1, 208  2 Z I 1  Z I 2

For mesh 2, - 208 - 60  - Z I 1  2 Z I 2

In matrix form,    2 Z - Z  I 1  208  - 208 - 60   - Z 2 Z  I   2       3Z 2 ,

 1  (208)(1.5  j0.866) Z ,

 2  (208)( j1.732) Z

I1 

 1 (208)(1.5  j0.866)  1.78956.56   (3)(60  j30)

I2 

 2 (208)( j1.732)   1.79116.56 (3)(60  j30) 

P1  Re  V1 I 1*   Re  (208)(1.789 - 56.56)   208.98 W P2  Re  V2 (- I 2 ) *   Re  (208 - 60))(1.7963.44)   371.65 W


I

V 12   20 mA R 600

(b)

I

V 120   200 mA R 600

Chapter 12, Solution 76. If both appliances have the same power rating, P, P I Vs

For the 120-V appliance,

For the 240-V appliance,  P2 R  2 Power loss = I 2 R   120 2 P R  240 2 Since

P . 120 P . I2  240

I1 

for the 120-V appliance for the 240-V appliance

1 1 , the losses in the 120-V appliance are higher. 2  120 240 2


Pg  PT  Pload  Pline , But

pf  0.85

PT  3600 cos   3600  pf  3060

Pg  3060  2500  (3)(80)  320 W


51  0.85   1  31.79 60 Q1  S1 sin 1  (60)(0.5268)  31.61 kVAR

cos 1 

P2  P1  51 kW

cos  2  0.95    2  18.19 P2 S2   53.68 kVA cos  2 Q 2  S 2 sin  2  16.759 kVAR Q c  Q1  Q 2  3.61  16.759  14.851 kVAR For each load, Qc  4.95 kVAR 3 Q c1 4950  67.82 F C 2  (2 )(60)(440) 2 V

Q c1 

Chapter 12, Solution 79. Consider the per-phase equivalent circuit below.

Ia a

A + 

V an

2

n Ia 

Z Y = 12 + j5  N Van 2550   17.15 - 19.65 A Z Y  2 14  j5

Thus, I b  I a  - 120  17.15 - 139.65 A I c  I a 120  17.15 100.35 A

VAN  I a Z Y  (17.15 - 19.65)(1322.62)  223 2.97 V Thus,

VBN  VAN  - 120  223 - 117.63 V VCN  VAN 120  223122.97 V


S  S1  S 2  S 3  6[0.83  j sin(cos 1 0.83)]  S 2  8(0.7071  j 0.7071) S  10.6368  j 2.31  S 2 kVA

(1)

But S  3VL I L   3 (208)(84.6)(0.8  j 0.6) VA  24.383  j18.287 kVA

(2)

From (1) and (2), S 2  13.746  j 20.6  24.7656.28 kVA Thus, the unknown load is 24.76 kVA at 0.5551 pf lagging.


pf  0.8 (leading)   1  -36.87 S1  150  - 36.87 kVA pf  1.0    2  0 S 2  100 0 kVA pf  0.6 (lagging)   3  53.13

S 3  20053.13 kVA S 4  80  j95 kVA

S  S1  S 2  S 3  S 4 S  420  j165  451.221.45 kVA

S  3 VL I L IL 

S 3 VL



451.2  10 3 3  480

 542.7 A

For the line, S L  3 I 2L Z L  (3)(542.7) 2 (0.02  j0.05) S L  17.67  j44.18 kVA At the source, S T  S  S L  437.7  j209.2 S T  485.125.55 kVA VT 

ST 3 IL



485.1  10 3 3  542.7

 516 V


S 1  400(0.8  j 0.6)  320  j 240 kVA,

S2  3

V 2p Z*p

For the delta-connected load, V L  V p S 2  3x

(2400) 2  1053.7  j842.93 kVA 10  j8

S  S 1  S 2  1.3737  j1.0829 MVA Let I = I 1 + I 2 be the total line current. For I 1 , S1  3V p I *1 ,

I *1 

S1

Vp 

VL

3 (320  j 240) x10 3

 , 3VL 3 (2400) For I 2 , convert the load to wye.

I 1  76.98  j 57.735

2400 3  30 o  273.1  j 289.76 10  j8 I  I 1  I 2  350  j 347.5

I 2  I p 3  30 o 

V s  V L  Vline  2400  I ( 3  j 6)  5.185  j1.405 kV

 

| V s | 5.372 kV

Chapter 12, Solution 83. S1  120 x746 x0.95(0.707  j 0.707)  60.135  j 60.135 kVA,

S 2  80 kVA

S  S1  S 2  140.135  j 60.135 kVA

But | S | 3V L I L

 

IL 

|S| 3V L



152.49 x10 3 3 x 480

 183.42 A

Chapter 12, Solution 84. We first find the magnitude of the various currents. For the motor, S

IL 

3 VL



4000 440 3

 5.248 A

For the capacitor, IC 

Q c 1800   4.091 A VL 440

For the lighting, Vp 

440

I Li 

PLi 800   3.15 A Vp 254

3

 254 V

Consider the figure below.

Ia a

IC

+

-jX C

V ab b



Ib

I2

Ic

I3

c

I Li In

n If Van  Vp 0 ,

Vab  3 Vp 30

Vcn  Vp 120 IC 

I1

Vab  4.091120 -j X C

R

I1 

Vab  4.091(  30) Z

where   cos -1 (0.72)  43.95 I 1  5.249 73.95

I 2  5.249  - 46.05

I 3  5.249193.95 I Li 

Vcn  3.15120 R

Thus, I a  I 1  I C  5.24973.95  4.091120 I a  8.60893.96 A I b  I 2  I C  5.249 - 46.05  4.091120 I b  9.271 - 52.16 A I c  I 3  I Li  5.249193.95  3.15120 I c  6.827 167.6 A I n  - I Li  3.15 - 60 A


ZY  R

Vp 

VL 3



240 3

 138.56 V

Vp2 27  9 kW  P  Vp I p  2 R

Vp2

(138.56) 2 R   2.133  P 9000 Thus,

Z Y  2.133 


1 a

A + 

1200 V rms

24 – j2 

I1 1

n

N I2

+ 

1200 V rms

15 + j4 

1 b

B

For the two meshes, 120  (26  j2) I 1  I 2 120  (17  j4) I 2  I 1

(1) (2)

In matrix form, 120  26  j2 - 1  I 1  120   - 1 17  j4 I 2    

  449  j70 ,

 1  (120)(18  j4) ,

 2  (120)(27  j2)

 1 120  18.44 12.53   4.87 3.67  454.42 8.86  2 120  27.07  - 4.24 I2    7.15 - 13.1 454.42 8.86  I1 

I aA  I 1  4.87  3.67 A

I bB  - I 2  7.15166.9 A I nN  I 2  I 1 

 2  1 

I nN 

(120)(9  j6)  2.856 - 42.55 A 449  j70

Chapter 12, Solution 87.  jL  j (2)(60)(5010 -3 )  j18.85 L  50 mH  Consider the circuit below.

1

115 V

+ 

I1

20 

2 15 + j18.85 

115 V

+ 

I2

30 

1

Applying KVl to the three meshes, we obtain 23 I 1  2 I 2  20 I 3  115 - 2 I 1  33 I 2  30 I 3  115 - 20 I 1  30 I 2  (65  j18.85) I 3  0 In matrix form, - 20   I 1  115  23 - 2  I   115  - 2 33 - 30  2    - 20 - 30 65  j18.85 I 3   0 

  12,775  j14,232 ,  2  (115)(1825  j471.3) ,

(1) (2) (3)

 1  (115)(1975  j659.8)  3  (115)(1450)

 1 115  208218.47   12.52 - 29.62  1921448.09  2 115  1884.9 14.48 I2    11.33 - 33.61 19124 48.09     1 (115)(-150  j188.5) I n  I 2  I1  2   1.448 - 176.6 A  12,775  j14,231.75 I1 

S 1  V1 I *1  (115)(12.52 29.62)  [1.252  j 0.7116] kVA S 2  V2 I *2  (115)(1.3333.61)  [1.085  j 0.7212] kVA

Chapter 13, Solution 1. For coil 1, L 1 – M 12 + M 13 = 12 – 8 + 4 = 8 For coil 2, L 2 – M 21 – M 23 = 16 – 8 – 10 = – 2 For coil 3, L 3 + M 31 – M 32 = 20 + 4 – 10 = 14 L T = 8 – 2 + 14 = 20H or

L T = L 1 + L 2 + L 3 – 2M 12 – 2M 23 + 2M 13

L T = 12 + 16 + 20 – 2x8 – 2x10 + 2x4 = 48 – 16 – 20 + 8 = 20H

Chapter 13, Solution 2. Using Fig. 13.73, design a problem to help other students to better understand mutual inductance. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine the inductance of the three series-connected inductors of Fig. 13.73.

Figure 13.73 Solution L = L 1 + L 2 + L 3 + 2M 12 – 2M 23 –2M 31 = 10 + 12 +8 + 2x6 – 2x6 –2x4 = 22H

Chapter 13, Solution 3. L 1 + L 2 + 2M = 500 mH

(1)

L 1 + L 2 – 2M = 300 mH

(2)

Adding (1) and (2), 2L 1 + 2L 2 = 800 mH But,

L 1 = 3L 2, , or 8L 2 + 400,

and L 2 = 100 mH

L 1 = 3L 2 = 300 mH From (2),

150 + 50 – 2M = 150 leads to M = 50 mH k = M/ L1 L2  50 / 100 x300 = 0.2887

300 mH, 100 mH, 50 mH, 0.2887

Chapter 13, Solution 4. (a) For the series connection shown in Figure (a), the current I enters each coil from its dotted terminal. Therefore, the mutually induced voltages have the same sign as the self-induced voltages. Thus, L eq = L 1 + L 2 + 2M M

Is

L1

I

I1 Vs

M

I2

+ –

L2

L2

L1

L eq (a) (b)

(b)

For the parallel coil, consider Figure (b). Is = I1 + I2

and

Z eq = V s /I s

Applying KVL to each branch gives, V s = jL 1 I 1 + jMI 2

(1)

V s = jMI 1 + j L 2 I 2

(2)

 Vs   jL1  V    jM  s 

or

jM   I1  jL 2  I 2 

 = –2L 1 L 2 + 2M2,  1 = jV s (L 2 – M),  2 = jV s (L 1 – M) I 1 =  1 /, and I 2 =  2 / I s = I 1 + I 2 = ( 1 +  2 )/ = j(L 1 + L 2 – 2M)V s /( –2(L 1 L 2 – M2)) = (L 1 + L 2 – 2M)V s /( j(L 1 L 2 – M2)) Z eq = V s /I s = j(L 1 L 2 – M2)/(L 1 + L 2 – 2M) = jL eq i.e.,

L eq = (L 1 L 2 – M2)/(L 1 + L 2 – 2M)


(a) If the coils are connected in series,

L  L1  L2  2 M  50  120  2(0.5) 50 x120  247.4 mH (b) If they are connected in parallel, L

50 x120  38.72 2 L1 L2  M 2  mH  48.62 mH L1  L2  2M 50  120  2 x38.72

(a) 247.4 mH, (b) 48.62 mH

Chapter 13, Solution 6. M  k L1 L2  0.6 40 x5  8.4853 mH 40mH 5mH

 

3

j L  j 2000 x 40 x10  j80 3

j L  j 2000 x5 x10  j10

 

3

8.4853mH   j M  j 2000 x8.4853 x10  j16.97 We analyze the circuit below.

16.77  I1 +

I2  j80 

+ j10 

V1

V2 

_

_

V1  j80 I1  j16.97 I 2 V2  16.97 I1  j10 I 2

(1) (2)

But, V 1 = 200˚ V and I 2 = 4–90˚ A. Substituting these into (1) produces I 1 = [(V 1 +j16.97I 2 )/j80] = [(20+j16.97(–j4))/j80] = 1.0986–90˚ A or i 1 = 1.0986sin(ωt) A From (2), V 2 = –16.97x(–j1.0986) + j10(–j4) = 40 + j18.643 = 44.1325˚ or v 2 = 44.13cos(ωt+25˚) V.

Chapter 13, Solution 7. We apply mesh analysis to the circuit as shown below. j1 

2

1

–j1 

 12

+ _

j6 

I1

+ j4 

I2

1



For mesh 1, (2+j6)I 1 + jI 2 = 24 For mesh 2, jI 1 + (2–j+j4)I 2 = jI 1 + (2+j3)I 2 = 0 or I 1 = (–3+j2)I 2 Substituting into the first equation results in I 2 = (–0.8762+j0.6328) A.

V o = I 2 x1 = 1.081144.16˚ V.

Vo _

Chapter 13, Solution 8. 2H 1H

   

j L  j 4 x 2  j8 j L  j 4 x1  j 4

Consider the circuit below.

j4

4

2 0o

+ _

I1





j8

j4

+ I2

2  (4  j8) I1  j 4 I 2 0   j 4 I1  (2  j 4) I 2

2

v(t) _

(1) (2)

In matrix form, these equations become  2   4  j8  j 4   I1  0    j 4 2  j 4  I      2 Solving this leads to I 2 = 0.2353 – j0.0588 V = 2I 2 = 0.4851 –14.04o Thus, v(t) = 485.1cos(4t–14.04°) mV


2

830o

+ –

I1

2

j4

j4

I2

-j1

-j2V

+ –

For loop 1, 830 = (2 + j4)I 1 – jI 2 For loop 2,

(j4 + 2 – j)I 2 – jI 1 + (–j2) = 0 or

Substituting (2) into (1),

(1)

I 1 = (3 – j2)i 2 – 2 830 + (2 + j4)2 = (14 + j7)I 2 I 2 = (10.928 + j12)/(14 + j7) = 1.03721.12 V x = 2I 2 = 2.07421.12 V

(2)

Chapter 13, Solution 10.   2H j L  j 2 x 2  j 4 0.5 H   j L  j 2 x0.5  j 1 1 1 F    j jC j 2 x1/ 2 2 Consider the circuit below. j 

 +

24  0

+ _

j4

I1

j4

I2

Vo

–j

_

24  j 4 I1  jI 2 0   jI1  ( j 4  j ) I 2 In matrix form,

  0   I1  3I 2

 24   j 4  j   I1   0    1 3   I      2

Solving this, I 2   j 2.1818,

Vo   jI 2  2.1818 v o (t) = –2.1818cos2t V

(1) (2)

Chapter 13, Solution 11. 3

800mH

 

j L  j 600 x800 x10  j 480

600mH

 

j L  j 600 x600 x10  j 360

3

3

j L  j 600 x1200 x10  j 720 j 1  = –j138.89 12F  jC 600x12x10  6  

1200mH

After transforming the current source to a voltage source, we get the circuit shown below. 200



j480

-j138.89

150

Ix 800  0

+ _

I1

j360

j720

I2

+ _

110  30



For mesh 1, 800  (200  j 480  j 720) I1  j 360 I 2  j 720 I 2 or 800  (200  j1200) I1  j 360 I 2

(1)

For mesh 2, 11030˚ + 150–j138.89+j720)I 2 + j360I 1 = 0 or 95.2628  j 55   j 360 I1  (150  j 581.1) I 2

(2)

In matrix form,  j 360   I1  800    200  j1200   95.2628  j 55   j 360 150  j 581.1  I 2     Solving this using MATLAB leads to: >> Z = [(200+1200i),-360i;-360i,(150+581.1i)] Z= 1.0e+003 * 0.2000 + 1.2000i 0 - 0.3600i

0 - 0.3600i 0.1500 + 0.5811i >> V = [800;(-95.26-55i)] V= 1.0e+002 * 8.0000 -0.9526 - 0.5500i >> I = inv(Z)*V I= 0.1390 - 0.7242i 0.0609 - 0.2690i I x = I 1 – I 2 = 0.0781 – j0.4552 = 0.4619–80.26˚. Hence, i x (t) = 461.9cos(600t–80.26˚) mA.

Chapter 13, Solution 12. Let   1.

j8 j4  +

j12

1V -

j16

I1

j20 I2



Applying KVL to the loops, 1  j16 I1  j8 I 2 0  j8 I1  j 36 I 2

(1) (2)

Solving (1) and (2) gives I 1 = –j0.0703. Thus

Z

1  jLeq I1

 

Leq 

1  14.225 H. jI1

We can also use the equivalent T-section for the transform to find the equivalent inductance.


4

I1

80  0

j5

–j Ω

j5 I2

+ _

j2I 2

+ –

– +

j2I 1

j2 

–80 + (4+j5)I 1 + j2I 2 = 0 or (4+j5)I 1 + j2I 2 = 80

j2I 1 +(4+j6)I 2 = 0 or I 2 = [–j2/(7.2111 56.31°)]I 1 = (0.27735 –146.31°)J 1

[4+j5 + j2(–0.230769–j0.153846)]I 1 = [4+j5+0.307692–j0.461538]I 1 = 80

[4.307692+j4.538462]I 1 = 80 or I 1 = 80/(6.2573 46.494°)

= 12.78507 –46.494° A.

Z in = 80/I 1 = 6.2573 46.494° Ω = (4.308 + j4.538) Ω

An alternate approach would be to use the equation, Z in  4  j(5) 

4 4  4  j5  j5  4  j  j2 7.211156.31

= 4+j5+0.5547 –56.31° = 4+0.30769+j(5–0.46154)

= [4.308+j4.538] Ω.

Chapter 13, Solution 14. To obtain V Th , convert the current source to a voltage source as shown below.

j2 5

j6 

j8 

-j3 

2

a j50 V

+

+ –

I

V Th

40 V

+ –

– b

Note that the two coils are connected series aiding. L = L 1 + L 2 – 2M jL = j6 + j8 – j4 = j10 Thus,

–j50 + (5 + j10 – j3 + 2)I + 40 = 0 I = (– 40 + j50)/ (7 + j7)

But,

–j50 + (5 + j6)I – j2I + V Th = 0 V Th = j50 – (5 + j4)I = j50 – (5 + j4)(–40 + j50)/(7 + j7) V Th = 26.7434.11 V

To obtain Z Th , we set all the sources to zero and insert a 1-A current source at the terminals a–b as shown below.

j2 5

j6 

I1

a

j8 

+ Vo

1A

–

b

-j3 

2

I2

Clearly, we now have only a super mesh to analyze. (5 + j6)I 1 – j2I 2 + (2 + j8 – j3)I 2 – j2I 1 = 0 (5 + j4)I 1 + (2 + j3)I 2 = 0

(1)

But,

I 2 – I 1 = 1 or I 2 = I 1 – 1

(2)


(5 + j4)I 1 +(2 + j3)(1 + I 1 ) = 0 I 1 = –(2 + j3)/(7 + j7)

Now,

((5 + j6)I 1 – j2I 1 + V o = 0 V o = –(5 + j4)I 1 = (5 + j4)(2 + j3)/(7 + j7) = (–2 + j23)/(7 + j7) = 2.33250 Z Th = V o /1 = 2.33250 Ω.

Chapter 13, Solution 15. The first step is to replace the mutually coupled circuits with the equivalent circuits using dependent sources. To obtain I N , short-circuit a–b as shown in Figure (a) and solve for I sc .

j20 Ω

20 

a

+ 

+ –

j5(I 1 –I 2 )

I1

I sc

j10 Ω I2

10030o + –

j5I 2

b (a) Now all we need to do is to write our two mesh equations. Loop 1.

–10060˚ + 20I 1 + j10(I 1 –I 2 ) + j5I 2 = 0 or (20+j10)I 1 – j5I 2 = 10060˚ or (4+j2)I 1 – jI 2 = 2030˚

Loop 2.

–j5I 2 + j10(I 2 –I 1 ) + j20I 2 + j5(I 1 –I 2 ) = 0 or –j5I 1 + j20I 2 = 0 or I 1 = 4I 2

Substituting back into the first equation, we get, (4+j2)4I 2 – jI 2 = 2030˚ or (16+j7)I 2 = 2030˚. Now to solve for I 2 = I sc = I N = (2030˚)/(16+j7) = (2030˚)/(17.46423.63˚) = 1.14526.37˚ A.

j20 Ω

20 

+ 

+ –

j5(I 1 –I 2 )

I1

j10 Ω I2

10030o + –

j5I 2

a + V oc – b

(b) To solve for Z N = Z eq = V oc /I sc , all we need to do is to solve for V oc . In circuit (b) we note that I 2 = 0 and we get the mesh equation, –10030˚ + (20+j10)I 1 = 0 or I 1 = (10030˚)/(22.3626.57˚) = 4.4723.43˚ A. V oc = j10I 1 – j5I 1 (induced voltage due to the mutual coupling) = j5I 1 = 22.3693.43˚ V. Z eq = Z N = (22.3693.43˚)/(1.14526.37˚) = 19.52587.06˚ Ω. or [1.0014+j19.498] Ω.

Chapter 13, Solution 16. To find I N , we short-circuit a-b. 8

j

-j2 

a

  j4 

+ 800 V -

j6 

I2

IN

I1

o

b  80  (8  j2  j4)I1  jI 2  0   j6I 2  jI1  0   I1  6I 2

(8  j2)I1  jI 2  80

(1) (2)

Solving (1) and (2) leads to I N  I2 

80  1.584  j0.362  1.6246 –12.91° A 48  j11

To find Z N , insert a 1-A current source at terminals a-b. Transforming the current source to voltage source gives the circuit below. j 8 -j2  2 a

  j4 

+ j6 

I1

I2

2V -

b

0  (8  j2)I1  jI 2

 

I1 

jI 2 8  j2

= [j/(8.24621 14.036°)]I 2 = 0.121268 75.964°I 2 = (0.0294113+j0.117647)I 2

(3)

2  (2  j 6) I 2  jI 1  0

(4)

Solving (3) and (4) leads to (2+j6)I 2 – j(0.0294113+j0.117647)I 2 = –2 or (2.117647+j5.882353)I 2 = –2 or I 2 = –2/(6.25192 70.201°) = 0.319902 109.8°.

V ab = 2(1+ I 2 ) = 2(1–0.1083629+j0.30099) = (1.78327+j0.601979) V = 1Z eq or

Z eq = (1.78327+j0.601979) = 1.8821 18.65° Ω

An alternate approach would be to calculate the open circuit voltage. –80 + (8+j2)I 1 – jI 2 = 0 or (8+j2)I 1 – jI 2 = 80 (2+j6)I 2 – jI 1 = 0 or I 1 = (2+j6)I 2 /j = (6–j2)I 2

(5) (6)

Substituting (6) into (5) we get,

(8.24621 14.036°)(6.32456 –18.435°)I 2 – jI 2 = 80 or

[(52.1536 –4.399°)–j]I 2 = [52–j5]I 2 = (52.2398 –5.492°)I 2 = 80 or

I 2 = 1.5314 5.492° A and V oc = 2I 2 = 3.0628 5.492° V which leads to,

Z eq = V oc /I sc = (3.0628 5.492°)/(1.6246 –12.91°) = 1.8853 18.4° Ω

This is in good agreement with what we determined before.


j L  j 40

40 40 2667 rad/s   2666.67 L 15 x103

  

M  k L1 L2  0.6 12 x103 x30 x103  62.35 mHmH 11.384 If Then

15 mH

40 Ω

12 mH 30 mH 11.384 mH

32 Ω 80 Ω 30.36 Ω

The circuit becomes that shown below. j30.36 

22 

60 



j32 

j80 

Z L =j40



Z in  22  j 32 

 2M 2

 22  j 32 

(30.36) 2 60  j120

j80  60  j 40 921.7  22  j 32   22  j 32  6.87  63.43  22  j 32  3.073  j 6.144 134.1663.43

= [25.07 + j25.86] Ω.

Chapter 13, Solution 18. Replacing the mutually coupled circuit with the dependent source equivalent we get, –j4 Ω

j2 

j5Ω j5I 2 120V

j20Ω – +

+ –

j5I 1

+ 

I2 I1

4

j6 Ω

Now all we need to do is to find V oc and I sc . To calculate the open circuit voltage, we note that I 2 is equal to zero. Thus,

–120 + (4 + j(–4+5+6))I 1 = 0 or I 1 = 120/(4+j7) = 120/(8.06226 60.255°)

= 14.8842 –60.255°.

V oc = V Thev = j5I 1 + (4+j6)I 1 = (4+j11)I 1 = (11.7047 70.017°)(14.8842 –60.255°) = 174.22 9.76° V

To find the short circuit current (I sc = I 2 ), we need to solve the following mesh equations,

Mesh 1 –120 + (–j4+j5)I 1 – j5I 2 + (4+j6)(I 1 –I 2 ) = 0 or (4+j7)I 1 – (4+j11)I 2 = 120

(1)

Mesh 2 (4+j6)(I 2 –I 1 ) – j5I 1 + j22I 2 = 0 or –(4+j11)I 1 + (4+j28)I 2 = 0 or I 1 = (28.2843 81.87°)I 2 /(11.7047 70.0169°) = (2.4165 11.853°)I 2

Substituting this into equation (1) we get,

(8.06226 60.255°)(2.4165 11.853°)I 2 – (4+j11)I 2 = 120 or

[(19.4825 72.108°) – 4 –j11]I 2 = 120 and [5.9855+j18.5403–4–j11]I 2 = (1.9855+j7.5403)I 2 = 120 or

I 2 = I sc = 120/(7.79733 75.248°) = 15.3899 ─75.248° A

Checking using MATLAB we get, >> Z = [(4+7j) (-4-11j);(-4-11j) (4+28j)] Z= 4.0000 + 7.0000i -4.0000 -11.0000i -4.0000 -11.0000i 4.0000 +28.0000i >> V = [120;0] V= 120 0

>> I = inv(Z)*V I= 16.6551 -33.2525i

(I 1 )

3.9188 -14.8829i

(I 2 = I sc ) = 15.3902 ─75.248° (answer checks)

Finally, Z eq = V Thev /I sc = (174.22 9.76°)/(15.3899 ─75.248°)

= (11.32 85.01°) Ω

Chapter 13, Solution 19. X La = X L1 – (–X M ) = 40 + 25 = 65 Ω and X Lb = X L2 – (–X M ) = 40 + 25 = 55 Ω. Finally, X C = X M thus, the T-section is as shown below. j65 

j55 

–j25 

Chapter 13, Solution 20. Transform the current source to a voltage source as shown below.

k=0.5 4

j10

8

j10 I3

+ –

j12

-j5

I1

I2

200o

+ –

k = M/ L1 L 2 or M = k L1 L 2 M = k L1L 2 = 0.5(10) = 5 For mesh 1,

j12 = (4 + j10 – j5)I 1 + j5I 2 + j5I 2 = (4 + j5)I 1 + j10I 2

For mesh 2,

0 = 20 + (8 + j10 – j5)I 2 + j5I 1 + j5I 1 –20 = +j10I 1 + (8 + j5)I 2

From (1) and (2),

(1)

 j12 4  j5  j10   I1   20     j10 8  j5 I      2   = 107 + j60,  1 = –60 –j296,  2 = 40 – j100 I 1 =  1 / = 2.46272.18 A I 2 =  2 / = 878–97.48 mA I 3 = I 1 – I 2 = 3.32974.89 A i 1 = 2.462 cos(1000t + 72.18) A i 2 = 0.878 cos(1000t – 97.48) A

At t = 2 ms, 1000t = 2 rad = 114.6 i 1 (0.002) = 2.462cos(114.6 + 72.18) = –2.445A

(2)

–2.445 i 2 = 0.878cos(114.6 – 97.48) = –0.8391 The total energy stored in the coupled coils is w = 0.5L 1 i 1 2 + 0.5L 2 i 2 2 + Mi 1 i 2 Since L 1 = 10 and  = 1000, L 1 = L 2 = 10 mH, M = 0.5L 1 = 5mH w = 0.5(0.01)(–2.445)2 + 0.5(0.01)(–0.8391)2 + 0.05(–2.445)(–0.8391) w = 43.67 mJ

Chapter 13, Solution 21. Using Fig. 13.90, design a problem to help other students to better understand energy in a coupled circuit. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find I 1 and I 2 in the circuit of Fig. 13.90. Calculate the power absorbed by the 4- resistor.

Figure 13.90 Solution For mesh 1, 3630 = (7 + j6)I 1 – (2 + j)I 2 For mesh 2,

0 = (6 + j3 – j4)I 2 – 2I 1 – jI 1 = –(2 + j)I 1 + (6 – j)I 2

(1) (2)

3630  7  j6  2  j  I1  Placing (1) and (2) into matrix form,      0   2  j 6  j  I 2 

 = 45 + j25 = 51.4829.05,  1 = (6 – j)3630 = 21920.54  2 = (2 + j)3630 = 80.556.57, I 1 =  1 / = 4.254–8.51 A , I 2 =  2 / = 1.563727.52 A Power absorbed by the 4-ohm resistor, = 0.5(I 2 )24 = 2(1.5637)2 = 4.89 watts

Chapter 13, Solution 22. With more complex mutually coupled circuits, it may be easier to show the effects of the coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure 13.85 then becomes, -j50 Io I3 j20I c

j40

+ 

j10I b

j60

+  Ia j30I c

 +  +

 + Ix

j30I b

j20I a

500 V + 

j80

I1

I2

100 

Ib  +

j10I a

Note the following, Ia = I1 – I3 Ib = I2 – I1 Ic = I3 – I2 and

Io = I3

Now all we need to do is to write the mesh equations and to solve for I o . Loop # 1, -50 + j20(I 3 – I 2 ) j 40(I 1 – I 3 ) + j10(I 2 – I 1 ) – j30(I 3 – I 2 ) + j80(I 1 – I 2 ) – j10(I 1 – I 3 ) =0 j100I 1 – j60I 2 – j40I 3 = 50 Multiplying everything by (1/j10) yields 10I 1 – 6I 2 – 4I 3 = - j5

(1)

Loop # 2, j10(I 1 – I 3 ) + j80(I 2 –I 1 ) + j30(I 3 –I 2 ) – j30(I 2 – I 1 ) + j60(I 2 – I 3 ) – j20(I 1 – I 3 ) + 100I 2 =0 -j60I 1 + (100 + j80)I 2 – j20I 3 = 0

(2)

Loop # 3, -j50I 3 +j20(I 1 –I 3 ) +j60(I 3 –I 2 ) +j30(I 2 –I 1 ) –j10(I 2 –I 1 ) +j40(I 3 –I 1 ) –j20(I 3 –I 2 ) = 0 -j40I 1 – j20I 2 + j10I 3 = 0 Multiplying by (1/j10) yields,

-4I 1 – 2I 2 + I 3 = 0

(3)

Multiplying (2) by (1/j20) yields -3I 1 + (4 – j5)I 2 – I 3 = 0

(4)

Multiplying (3) by (1/4) yields

(5)

-I 1 – 0.5I 2 – 0.25I 3 = 0

Multiplying (4) by (-1/3) yields I 1 – ((4/3) – j(5/3))I 2 + (1/3)I 3 = -j0.5

(7)

Multiplying [(6)+(5)] by 12 yields

(-22 + j20)I 2 + 7I 3 = 0

(8)

Multiplying [(5)+(7)] by 20 yields

-22I 2 – 3I 3 = -j10

(9)

(8) leads to I 2 = -7I 3 /(-22 + j20) = 0.235542.3o = (0.17418+j0.15849)I 3 (9) leads to I 3 = (j10 – 22I 2 )/3, substituting (1) into this equation produces, I 3 = j3.333 + (-1.2273 – j1.1623)I 3 or

I 3 = I o = 1.304063o amp.

(10)

Chapter 13, Solution 23.  = 10 0.5 H converts to jL 1 = j5 ohms 1 H converts to jL 2 = j10 ohms 0.2 H converts to jM = j2 ohms 25 mF converts to 1/(jC) = 1/(10x25x10-3) = –j4 ohms The frequency-domain equivalent circuit is shown below.

j2I 2

j5

j10

+ 

120

For mesh 1,

For mesh 2,

From (1),

+ 

I1

j2I 1 + 

–j4

I2

5

–12 + j5I 1 + j2I 2 + (–j4)(I 1 –I 2 ) = 0 or jI 1 + j6I 2 = 12 or I 1 + 6I 2 = –j12

(1)

(–j4)(I 2 –I 1 ) + j10I 2 + j2I 1 + 5I 2 = 0 or j6I 1 + (5+j6)I 2 = 0

(2)

I 1 = –j12 – 6I 2

Substituting this into (2) produces, j6(–j12–6I 2 ) + (5+j6)I 2 = 0 = 72 + (5+j6–j36)I 2 or (5–j30)I 2 = (30.414 –80.54°)I 2 = –72 or I 2 = 2.367 –99.46° A

I 1 = –j12 – 6( –0.38909–j2.3351) = 2.33454 + j(–12+14.0106) = 2.33454 + j2.0106) = 3.081 40.74° A

Checking using matrices,

I 1 = [(72–j60)/(5–j30)] = (93.723 –39.806°)/(30.414 –80.538°) = 3.082 40.73° A and

I 2 = [–72/(30.414 –80.54)] = 2.367 –99.46° A

Thus, i 1 (t) = 3.081cos(10t + 40.74) A, i 2 (t) = 2.367cos(10t – 99.46) A. At t = 15 ms,

10t = 10x15x10-3 = 0.15 rad = 8.59 i 1 = 3.081cos(49.33) = 2.00789 A i 2 = 2.367cos(–90.87) = –0.03594 A w = 0.5(5)(2.00789)2 + 0.5(1)(–0.03594)2 – (0.2)(2.00789)(–0.03594) = 10.079056 + 0.0006458 + 0.0144327 = 10.094 J.

3.081cos(10t + 40.74) A, 2.367cos(10t – 99.46) A, 10.094 J.


k = M/ L1 L 2 = 1/ 4 x 2 = 0.3535

(b)

 = 4 1/4 F leads to 1/(jC) = –j/(4x0.25) = –j 1||(–j) = –j/(1 – j) = 0.5(1 – j) 1 H produces jM = j4 4 H produces j16 2 H becomes j8 j4 2 j8 120

+ 

I1

I2

0.5(1–j)

j16

12 = (2 + j16)I 1 + j4I 2 or

6 = (1 + j8)I 1 + j2I 2

0 = (j8 + 0.5 – j0.5)I 2 + j4I 1 or I 1 = (0.5 + j7.5)I 2 /(–j4)

(1) (2)

Substituting (2) into (1), 24 = (–11.5 – j51.5)I 2 or I 2 = –24/(11.5 + j51.5) = –0.455–77.41 V o = I 2 (0.5)(1 – j) = 0.321757.59 v o = 321.7cos(4t + 57.6) mV (c)

From (2),

I 1 = (0.5 + j7.5)I 2 /(–j4) = 0.855–81.21

i 1 = 0.885cos(4t – 81.21) A, i 2 = –0.455cos(4t – 77.41) A At t = 2s, 4t = 8 rad = 98.37

i 1 = 0.885cos(98.37 – 81.21) = 0.8169 i 2 = –0.455cos(98.37 – 77.41) = –0.4249 w = 0.5L 1 i 1 2 + 0.5L 2 i 2 2 + Mi 1 i 2 = 0.5(4)(0.8169)2 + 0.5(2)(–.4249)2 + (1)(0.1869)(–0.4249) = 1.168 J

Chapter 13, Solution 25. m = k L1 L 2 = 0.5 H We transform the circuit to frequency domain as shown below. 12sin2t converts to 120,  = 2 0.5 F converts to 1/(jC) = –j 2 H becomes jL = j4 j1 Io 4 

1

a

3

–j1

120

+ 

j2

j2

j4

2 b

Applying the concept of reflected impedance, Z ab = (2 – j)||(1 + j2 + (1)2/(j2 + 3 + j4)) = (2 – j)||(1 + j2 + (3/45) – j6/45) = (2 – j)||(1 + j2 + (3/45) – j6/45) = (2 – j)||(1.0667 + j1.8667) =(2 – j)(1.0667 + j1.8667)/(3.0667 + j0.8667) = 1.508517.9 Ω I o = 120/(Z ab + 4) = 12/(5.4355 + j0.4636) = 2.2–4.88 i o = 2.2sin(2t – 4.88) A

Chapter 13, Solution 26. M = k L1L 2 M = k L1L 2 = 0.601 20x 40 = 17 The frequency-domain equivalent circuit is shown below. j17

50 

20060

For mesh 1,

For mesh 2,

+ 

–j30

I1

Io

j20

j40

I2

10 

–200 60° + (50–j30+j20)I 1 – j17I 2 = 0 or

(50–j10)I 1 – j17I 2 = 200 60°

(1)

(10 + j40)I 2 – j17I 1 = 0 or –j17I 1 + (10+j40)I 2 = 0

(2)

In matrix form,  j17   I1  20060 50  j10       or 0     j17 10  j40 I 2  j17  10  j40  j17 50  j10  I1  20060    I  500  400  289  j100  j2,000   0    2

I 1 = (10+j40)(200 60°)/(1,189+j1,900)

= (41.231 75.964°)(200 60°)/(2,241.4 57.962°) = 3.679 78° A and

I 2 = j17(200 60°)/(2,241.4 57.962°) = 1.5169 92.04° A

I o = I 2 = 1.5169 92.04° A

It should be noted that switching the dot on the winding on the right only reverses the direction of I o .

Chapter 13, Solution 27.   1H j L  j 20 2H   j L  j 40 0.5 H   j L  j10 We apply mesh analysis to the circuit as shown below. 10  j10 

8

I3



40  0

+ _

I1

j20 

 j40 

50 

I2

To make the problem easier to solve, let us have I 3 flow around the outside loop as shown. For mesh 1, –40 + 8I 1 + j20I 1 – j10I 2 = 0 or (8+j20)I 1 – j10I 2 = 40

(1)

For mesh 2, j40I 2 – j10I 1 + 50(I 2 + I 3 ) = 0 or –j10I 1 + (50+j40)I 2 + 50I 3 = 0

(2)

For mesh 3, –40 + 10I 3 + 50(I 3 + I 2 ) = 0 or 50I 2 + 60I 3 = 40

(3)

In matrix form, (1) to (3) become  j10 0  40 8  j20   j10 50  j40 50 I   0       0 50 60 40

>> Z=[(8+20i),-10i,0;-10i,(50+40i),50;0,50,60] Z= 8.0000 +20.0000i 0 -10.0000i 0

0 -10.0000i 50.0000 +40.0000i 50.0000 0 50.0000 60.0000 >> V=[40;0;40] V= 40 0 40 >> I=inv(Z)*V I= 0.6354 - 1.5118i 0.0613 + 0.4682i 0.6156 - 0.3901i Solving this leads to I 50 = I 2 + I 3 = 0.0613+0.6156 + j(0.4682–0.3901) = 0.6769+j0.0781 = 0.6814 6.58° A or I 50rms |I 50rms | = 0.6814/1.4142 = 481.8 mA.

The power delivered to the 50- resistor is P = (I 50rms )2R = (0.4818)250 = 11.608 W.

Chapter 13, Solution 28. We find Z Th by replacing the 20-ohm load with a unit source as shown below. j10 

8

-jX

  j12 

j15 

+ 1V -

I2

I1

For mesh 1,

0  (8  jX  j12) I 1  j10 I 2

For mesh 2, 1  j15 I 2  j10 I 1  0

 

(1)

I 1  1.5 I 2  0.1 j

(2)

Substituting (2) into (1) leads to  1.2  j 0.8  0.1X I2  12  j8  j1.5 X

Z Th 

| Z Th | 20 

1 12  j8  j1.5 X   I 2 1.2  j 0.8  0.1X

12 2  (8  1.5 X ) 2 (1.2  0.1X )  0.8 2

2

 

Solving the quadratic equation yields X = 6.425 Ω

0  1.75 X 2  72 X  624

Chapter 13, Solution 29. 30 mH becomes jL = j30x10-3x103 = j30 50 mH becomes j50 Let X = M Using the concept of reflected impedance, Z in = 10 + j30 + X2/(20 + j50) I 1 = V/Z in = 165/(10 + j30 + X2/(20 + j50)) p = 0.5|I 1 |2(10) = 320 leads to |I 1 |2 = 64 or |I 1 | = 8 8 = |165(20 + j50)/(X2 + (10 + j30)(20 + j50))| = |165(20 + j50)/(X2 – 1300 + j1100)| 64 = 27225(400 + 2500)/((X2 – 1300)2 + 1,210,000)

or

(X2 – 1300)2 + 1,210,000 = 1,233,633 X = 33.86 or 38.13 If X = 38.127 = M M = 38.127 mH k = M/ L1 L 2 = 38.127/ 30x 50 = 0.984 j38.127

10 

1650

+ 

I1

j30

j50

I2

20 

165 = (10 + j30)I 1 – j38.127I 2 0 = (20 + j50)I 2 – j38.127I 1

In matrix form,

(1) (2)

165  10  j30  j38.127  I1   0     j38.127 20  j50   I      2 

 = 154 + j1100 = 1110.7382.03,  1 = 888.568.2,  2 = j6291 I 1 =  1 / = 8–13.81, I 2 =  2 / = 5.6647.97 i 1 = 8cos(1000t – 13.83), i 2 = 5.664cos(1000t + 7.97) At t = 1.5 ms, 1000t = 1.5 rad = 85.94 i 1 = 8cos(85.94 – 13.83) = 2.457 i 2 = 5.664cos(85.94 + 7.97) = –0.3862 w = 0.5L 1 i 1 2 + 0.5L 2 i 2 2 + Mi 1 i 2 = 0.5(30)(2.547)2 + 0.5(50)(–0.3862)2 – 38.127(2.547)(–0.3862) = 130.51 mJ

Chapter 13, Solution 30. Z in = j40 + 25 + j30 + (10)2/(8 + j20 – j6)

(a)

= 25 + j70 + 100/(8 + j14) = (28.08 + j64.62) ohms jL a = j30 – j10 = j20, jL b = j20 – j10 = j10, jL c = j10

(b)

Thus the Thevenin Equivalent of the linear transformer is shown below.

j40

25 

j20

j10

j10

8

–j6

Z in Z in = j40 + 25 + j20 + j10||(8 + j4) = 25 + j60 + j10(8 + j4)/(8 + j14) = (28.08 + j64.62) ohms

Chapter 13, Solution 31. Using Fig. 13.100, design a problem to help other students to better understand linear transformers and how to find T-equivalent and –equivalent circuits. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the circuit in Fig. 13.99, find: (a) the T-equivalent circuit, (b) the -equivalent circuit.


(a)

L a = L 1 – M = 10 H L b = L 2 – M = 15 H Lc = M = 5 H

(b)

L 1 L 2 – M2 = 300 – 25 = 275 L A = (L 1 L 2 – M2)/(L 1 – M) = 275/15 = 18.33 H L B = (L 1 L 2 – M2)/(L 1 – M) = 275/10 = 27.5 H L C = (L 1 L 2 – M2)/M = 275/5 = 55 H

Chapter 13, Solution 32. We first find Z in for the second stage using the concept of reflected impedance.

Lb

R

LB

Z in ’ Z in ’ = jL b + 2M b 2/(R + jL b ) = (jL b R - 2L b 2 + 2M b 2)/(R + jL b ) (1) For the first stage, we have the circuit below.

La

LA

Z in ’

Z in Z in = jL a + 2M a 2/(jL a + Z in ) = (–2L a 2 + 2M a 2 + jL a Z in )/( jL a + Z in )

(2)

Substituting (1) into (2) gives, ( jL b R   2 L2b   2 M 2b ) R  j L b 2 2 jL b R   L b   2 M 2b jL a  R  j L b

  2 L2a   2 M a2  jL a =

–R2L a 2 + 2M a 2R – j3L b L a + j3L b M a 2 + jL a (jL b R – 2L b 2 + 2M b 2) = jRLa –2L a L b + jL b R – 2L a 2 + 2M b 2 2R(L a 2 + L a L b – M a 2) + j3(L a 2L b + L a L b 2 – L a M b 2 – L b M a 2) Z in =

2(L a L b +L b 2 – M b 2) – jR(L a +L b )

Chapter 13, Solution 33. Z in = 10 + j12 + (15)2/(20 + j 40 – j5) = 10 + j12 + 225/(20 + j35) = 10 + j12 + 225(20 – j35)/(400 + 1225) = (12.769 + j7.154) Ω

Chapter 13, Solution 34. Using Fig. 13.103, design a problem to help other students to better understand how to find the input impedance of circuits with transformers. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the input impedance of the circuit in Fig. 13.102.

Figure 13.102 Solution Insert a 1-V voltage source at the input as shown below. j6 

1



+

j12 

o

1<0 V

8 

j10 

I1

j4  I2

-j2 

For loop 1, 1  (1  j10) I 1  j 4 I 2

(1)

For loop 2, 0  (8  j 4  j10  j 2) I 2  j 2 I 1  j 6 I 1

 

0   jI 1  (2  j 3) I 2

(2)

Solving (1) and (2) leads to I 1 =0.019 –j0.1068 Z

1  1.6154  j 9.077  9.219 79.91 o  I1

Alternatively, an easier way to obtain Z is to replace the transformer with its equivalent T circuit and use series/parallel impedance combinations. This leads to exactly the same result.

Chapter 13, Solution 35. For mesh 1, 16  (10  j 4) I 1  j 2 I 2

(1)

For mesh 2,

0  j 2 I 1  (30  j 26) I 2  j12 I 3

(2)

For mesh 3,

0   j12 I 2  (5  j11) I 3

(3)

We may use MATLAB to solve (1) to (3) and obtain I 1  1.3736  j0.5385  1.4754–21.41° A I 2  0.0547  j0.0549  77.5–134.85° mA I 3  0.0268  j0.0721  77–110.41° mA 1.4754–21.41° A, 77.5–134.85° mA, 77–110.41° mA

Chapter 13, Solution 36. Following the two rules in section 13.5, we obtain the following: (a)

V 2 /V 1 = –n,

I 2 /I 1 = –1/n

(b)

V 2 /V 1 = –n,

I 2 /I 1 = –1/n

(c)

V 2 /V 1 = n,

I 2 /I 1 = 1/n

(d)

V 2 /V 1 = n,

I 2 /I 1 = –1/n

(n = V 2 /V 1 )

Chapter 13, Solution 37. (a) n 

V2 2400  5 V1 480

(b) S 1  I 1V1  S 2  I 2V 2  50,000 (c ) I 2 

50,000  20.83 A 2400

 

I1 

50,000  104.17 A 480

Chapter 13, Solution 38. Design a problem to help other students to better understand ideal transformers. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A 4-kVA, 2300/230-V rms transformer has an equivalent impedance of 210 on the primary side. If the transformer is connected to a load with 0.6 power factor leading, calculate the input impedance. Solution Z in = Z p + Z L /n2, n = v 2 /v 1 = 230/2300 = 0.1 v 2 = 230 V, s 2 = v 2 I 2 * I 2 * = s 2 /v 2 = 17.391–53.13 or I 2 = 17.39153.13 A Z L = v 2 /I 2 = 2300/17.39153.13 = 13.235–53.13 Z in = 210 + 1323.5–53.13 = 1.97 + j0.3473 + 794.1 – j1058.8 Z in = 1.324–53.05 kΩ

Chapter 13, Solution 39. Referred to the high-voltage side, Z L = (1200/240)2(0.810) = 2010 Z in = 60–30 + 2010 = 76.4122–20.31 I 1 = 1200/Z in = 1200/76.4122–20.31 = 15.720.31 A Since S = I 1 v 1 = I 2 v 2 , I 2 = I 1 v 1 /v 2 = (1200/240)( 15.720.31) = 78.520.31 A


R Th

I1 1:5

V Th

I2

+ _

200 

We reflect the 200- load to the primary side.

200  108 52 I I2  1  2 108 n

Z p  100 

I1 

10 , 108

P

1 1 2 2 | I 2 |2 RL  ( ) (200)  34.3 mW 2 2 108

Chapter 13, Solution 41. We reflect the 2-ohm resistor to the primary side. Z in = 10 + 2/n2,

n = –1/3

Since both I 1 and I 2 enter the dotted terminals,

Z in = 10 + 18 = 28 ohms

I 1 = 140/28 = 500 mA and I 2 = I 1 /n = 0.5/(–1/3) = –1.5 A

Chapter 13, Solution 42. We apply mesh analysis to the circuit as shown below. 50 

80

+ _

–j1 

I1

j20 

1:2 +

+

V1

V2

_

_

For mesh 1, –80 + (50–j)I 1 + V 1 = 0 For mesh 2, –V 2 + (2+j20)I 2 = 0 At the transformer terminals, V 2 = 2V 1 or 2V 1 – V 2 = 0 I 1 = 2I 2 or I 1 – 2I 2 = 0 From (1) to (4),

Solving this with MATLAB, >> A = [(50-j) 0 1 0;0 (2+20j) 0 -1;0 0 2 -1;1 -2 0 0] A= Columns 1 through 3 50.0000 - 1.0000i 0 1.0000 0 2.0000 +20.0000i 0 0 0 2.0000 1.0000 -2.0000 0 Column 4 0 -1.0000 -1.0000

I2

2

(1) (2) (3) (4)

0 >> B = [80;0;0;0] B= 80 0 0 0 >> C = inv(A)*B C= 1.5743 - 0.1247i 0.7871 - 0.0623i 1.4106 + 7.8091i 2.8212 +15.6181i

(I 1 ) (I 2 ) (V 1 ) (V 2 )

I 2 = (787.1–j62.3) mA or 789.6 –4.53° mA

The power absorbed by the 2- resistor is P = |I 2 |2R = (0.7896)22 = 1.2469 W.

Chapter 13, Solution 43. Transform the two current sources to voltage sources, as shown below. 10  + 20 V

+ –

Using mesh analysis,

I1

12 

1:4

v1 

+ v2 

I2

12V

+ –

–20 + 10I 1 + v 1 = 0 20 = v 1 + 10I 1

12 + 12I 2 – v 2 = 0 or 12 = v 2 – 12I 2 At the transformer terminal, v 2 = nv 1 = 4v 1 I 1 = nI 2 = 4I 2

(1) (2) (3) (4)

Substituting (3) and (4) into (1) and (2), we get,

Solving (5) and (6) gives

20 = v 1 + 40I 2

(5)

12 = 4v 1 – 12I 2

(6)

v 1 = 4.186 V and v 2 = 4v = 16.744 V

Chapter 13, Solution 44. We can apply the superposition theorem. Let i 1 = i 1 ’ + i 1 ” and i 2 = i 2 ’ + i 2 ” where the single prime is due to the DC source and the double prime is due to the AC source. Since we are looking for the steady-state values of i 1 and i 2 , i 1 ’ = i 2 ’ = 0. For the AC source, consider the circuit below. R

1:n +

i1”

v1 

v 2 /v 1 = –n,

+ v2 

i2

+ –

V n 0

I 2 ”/I 1 ” = –1/n

But v 2 = v m , v 1 = –v m /n or I 1 ” = v m /(Rn) I 2 ” = –I 1 ”/n = –v m /(Rn2) Hence,

i 1 (t) = (v m /Rn)cost A, and i 2 (t) = (–v m /(n2R))cost A

Chapter 13, Solution 45. 48 

4–90˚

ZL  8  Z

+ 

Z

j  8  j4 , n = 1/3 C

ZL

 9Z L  72  j36 n2 4  90 4  90 I   0.03193  73.3 48  72  j36 125.28  16.7

We now have some choices, we can go ahead and calculate the current in the second loop and calculate the power delivered to the 8-ohm resistor directly or we can merely say that the power delivered to the equivalent resistor in the primary side must be the same as the power delivered to the 8-ohm resistor. Therefore, P8 

I2 72  0.5098x10  3 72  36.71 mW 2

The student is encouraged to calculate the current in the secondary and calculate the power delivered to the 8-ohm resistor to verify that the above is correct.


Reflecting the secondary circuit to the primary, we have the circuit shown below. Z in

1660

+ 

I1

+ 

1030/(–n) = –530

Z in = 10 + j16 + (1/4)(12 – j8) = 13 + j14 –1660 + Z in I 1 – 530 = 0 or I 1 = (1660 + 530)/(13 + j14) Hence, (b)

I 1 = 1.0725.88 A, and I 2 = –0.5I 1 = 0.536185.88 A

Switching a dot will not affect Z in but will affect I 1 and I 2 . I 1 = (1660 – 530)/(13 + j14) = 0.625 25 A and I 2 = 0.5I 1 = 0.312525 A

Chapter 13, Solution 47.  (1/3) F 1F 

1 1    j1 jC j 3 x1/ 3

Consider the circuit shown below.

–j 1

2  +

4  0º

+ _

I1

I3

1:4 

+

+

V1

V2

_

_

I2

5  v(t) –

For mesh 1, 3I 1 – 2I 3 + V 1 = 4 For mesh 2, 5I 2 – V 2 = 0 For mesh 3, –2I 1 (2–j)I 3 – V 1 + V 2 =0 At the terminals of the transformer, V2  nV1  4V1 I 1 – I 3 = 4(I 2 – I 3 ) In matrix form,

(1) (2) (3) (4) (5)

2 0 1 0   I1  4 3 0 5 0 0  1  I 2  0    2 0 2  j  1 1   I 3   0        4 1   V1  0 0 0 0  1  4 3 0 0  V2  0

Solving this using MATLAB yields >>A = [3,0,-2,1,0; 0,5,0,0,-1; -2,0,(2-j),-1,1 ;0,0,0,-4,1; 1,-4,3,0,0]

A= 3.0000 0 -2.0000 0 1.0000

0 5.0000 0 0 -4.0000

-2.0000 1.0000 0 0 0 -1.0000 2.0000 - 1.0000i -1.0000 1.0000 0 -4.0000 1.0000 3.0000 0 0

>>U = [4;0;0;0;0] >>X = inv(A)*U X= 1.2952 + 0.0196i 0.5287 + 0.0507i 0.2733 + 0.0611i 0.6609 + 0.0634i 2.6437 + 0.2537i

V = 5I 2 = V 2 = 2.6437+j0.2537 = 2.656 5.48° V, therefore,

v(t) = 2.656cos(3t+5.48˚) V

Chapter 13, Solution 48. Using Fig. 13.113, design a problem to help other students to better understand how ideal transformers work. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find I x in the ideal transformer circuit of Fig. 13.112.

Figure 13.112 Solution We apply mesh analysis.

8

10 

2:1 +

+ 

+



V1

V2 -

I1 1000o V -

Ix

j6  I2

-j4 

100  (8  j 4) I 1  j 4 I 2  V1

(1)

0  (10  j 2) I 2  j 4 I 1  V 2

(2)

But V2 1 n V1 2

 

V1  2V2

(3)

I2 1    2 I1 n

 

I 1  0.5 I 2

(4)

Substituting (3) and (4) into (1) and (2), we obtain 100  (4  j 2) I 2  2V2 0  (10  j 4) I 2 V2

(1)a (2)a

Solving (1)a and (2)a leads to I 2 = -3.5503 +j1.4793

I x  I 1  I 2  0.5 I 2  1.923157.4 o A


1 F 20

  2,

 

1   j10 j C Ix

2

-j10

I1

1:3

I2

1 +

2 +  V1

o

12<0 V -

+ V2

-

6



At node 1,

12  V1 V1  V2   I1   12  2 I 1  V1 (1  j 0.2)  j 0.2V2  j10 2 At node 2, V  V2 V2 I2  1    0  6 I 2  j 0.6V1  (1  j 0.6)V2  j10 6 1 At the terminals of the transformer, V2  3V1 , I 2   I1 3 Substituting these in (1) and (2), 12  6 I 2  V1 (1  j 0.8),

0  6 I 2  V1 (3  j 2.4)

Adding these gives V 1 =1.829 –j1.463 and Ix 

V1  V2 4V1   0.93751.34 o  j10  j10 i x (t) = 937cos(2t+51.34˚) mA.

(1)

(2)

Chapter 13, Solution 50. The value of Z in is not effected by the location of the dots since n2 is involved. Z in ’ = (6 – j10)/(n’)2, n’ = 1/4 Z in ’ = 16(6 – j10) = 96 – j160 Z in = 8 + j12 + (Z in ’ + 24)/n2,

n = 5

Z in = 8 + j12 + (120 – j160)/25 = 8 + j12 + 4.8 – j6.4 Z in = (12.8 + j5.6) Ω

Chapter 13, Solution 51. Let Z 3 = 36 +j18, where Z 3 is reflected to the middle circuit. Z R ’ = Z L /n2 = (12 + j2)/4 = 3 + j0.5 Z in = 5 – j2 + Z R ’ = [8 – j1.5] Ω I 1 = 240/Z eq = 240/(8 – j1.5) = 240/8.14–10.62 = 8.9510.62 A [8 – j1.5] Ω, 8.9510.62 A

Chapter 13, Solution 52. For maximum power transfer, 40 = Z L /n2 = 10/n2 or n2 = 10/40 which yields n = 1/2 = 0.5 I = 120/(40 + 40) = 3/2 p = I2R = (9/4)x40 = 90 watts.


The Thevenin equivalent to the left of the transformer is shown below. 8

20 V

+ 

The reflected load impedance is Z L ’ = Z L /n2 = 200/n2. For maximum power transfer, (b)

8 = 200/n2 produces n = 5.

If n = 10, Z L ’ = 200/10 = 2 and I = 20/(8 + 2) = 2 p = I2Z L ’ = (2)2(2) = 8 watts.


Z Th

VS

I1

+ 

Z L /n2

For maximum power transfer, Z Th = Z L /n2, or n2 = Z L /Z Th = 8/128 n = 0.25 (b)

I 1 = V Th /(Z Th + Z L /n2) = 10/(128 + 128) = 39.06 mA

(c)

v 2 = I 2 Z L = 156.24x8 mV = 1.25 V But

v 2 = nv 1 therefore v 1 = v 2 /n = 4(1.25) = 5 V

Chapter 13, Solution 55. We first reflect the 60- resistance to the middle circuit. 60 Z L'  20  2  26.67 3 We now reflect this to the primary side. Z L' 26.67 ZL  2   1.667 1.6669  Ω 4 16

Chapter 13, Solution 56. We apply mesh analysis to the circuit as shown below. 2

1:2 +

+

v1 46V

+ 

v2



I1



I2

10 

5

For mesh 1,

46 = 7I 1 – 5I 2 + v 1

(1)

For mesh 2,

v 2 = 15I 2 – 5I 1

(2)

At the terminals of the transformer, v 2 = nv 1 = 2v 1

(3)

I 1 = nI 2 = 2I 2

(4)

Substituting (3) and (4) into (1) and (2),

Combining (5) and (6),

46 = 9I 2 + v 1

(5)

v 1 = 2.5I 2

(6)

46 = 11.5I 2 or I 2 = 4 P 10 = 0.5|I 2 |2(10) = 80 watts.


Z L = j3||(12 – j6) = j3(12 – j6)/(12 – j3) = (12 + j54)/17 Reflecting this to the primary side gives Z in = 2 + Z L /n2 = 2 + (3 + j13.5)/17 = 2.316820.04 I 1 = v s /Z in = 6090/2.316820.04 = 25.969.96 A(rms) I 2 = I 1 /n = 12.9569.96 A(rms)

(b)

6090 = 2I 1 + v 1 or v 1 = j60 –2I 1 = j60 – 51.869.96 v 1 = 21.06147.44 V(rms) v 2 = nv 1 = 42.12147.44 V(rms) v o = v 2 = 42.12147.44 V(rms)

(c)

S = v s I 1 * = (6090)(25.9–69.96) = 1.55420.04 kVA

Chapter 13, Solution 58. Consider the circuit below. 20 

I3 20 

1:5 +

800

+ –

V1

I1



For mesh1,

V2 

+ I2

vo

100 



–80 + 20I 1 – 20I 3 + V 1 = 0 or 20I 1 – 20I 3 + V 1 = 80

(1)

V 2 = 100I 2 or 100I 2 – V 2 = 0

For mesh 2, (2) For mesh 3,

+

40I 3 – 20I 1 + V 2 – V 1 = 0 which leads to –20 I 1 + 40I 3 – V 1 + V 2 = 0

(3)

At the transformer terminals, V 2 = –nV 1 = –5V 1 or 5V 1 + V 2

(4)

I 1 – I 3 = –n(I 2 – I 3 ) = –5(I 2 – I 3 ) or I 1 + 5I 2 – 6I 3 = 0

(5)

Solving using MATLAB, >>A = [ 20 0 -20 1 0 ; 0 100 0 0 -1; -20 0 40 -1 1; 0 0 0 5 1; 1 5 -6 0 0] A= 20 0 -20 1 0 0 100 0 0 -1 -20 0 40 -1 1 0 0 0 5 1 1 5 -6 0 0

>> B = [ 80 0 0 0 0 ]' B= 80 0 0 0 0 >> Y = inv(A)*B Y= 5.9355 0.5161 1.4194 -10.3226 51.6129 P20,1 = 0.5*( I1 – I3 )^2*20 = 0.5*( 5.9355– 1.4194)^2*20 = 203.95 p 20 (the one between 1 and 3) = 0.5(20)(I 1 – I 3 )2 = 10(5.9355–1.4194)2 = 203.95 watts p 20 (at the top of the circuit) = 0.5(20)I 3 2 = 20.15 watts p 100 = 0.5(100)I 2 2 = 13.318 watts

Chapter 13, Solution 59. We apply mesh analysis to the circuit as shown below.

10  

1:4

 +

+

40  0

+ –

V_2

V1 _ I1

20 

I2 12

For mesh 1, –40 + 22I 1 – 12I 2 + V 1 = 0

(1)

For mesh 2, –12I 1 + 32I 2 – V 2 = 0

(2)

At the transformer terminals, –4V 1 + V 2 = 0 I 1 – 4I 2 = 0 Putting (1), (2), (3), and (4) in matrix form, we obtain 0  40  22  12 1  12 32 0  1  0   I  0 0 4 1   0      4 0 0 0  1 >> A=[22,-12,1,0;-12,32,0,-1;0,0,-4,1;1,-4,0,0] A= 22 -12 1 0 -12 32 0 -1 0 0 -4 1 1 -4 0 0

(3) (4)

>> U=[40;0;0;0] U= 40 0 0 0 >> X=inv(A)*U X= 2.2222 0.5556 -2.2222 -8.8889 For 10- resistor, P 10 = [(2.222)2/2]10 = 24.69 W For 12- resistor, P 12 = [(2.222–0.5556)2/2]12 = 16.661 W For 20- resistor, P 20 = [(0.5556)2/2]20 = 3.087 W. 24.69 W, 16.661 W, 3.087 W


Transferring the 40-ohm load to the middle circuit, Z L ’ = 40/(n’)2 = 10 ohms where n’ = 2 10||(5 + 10) = 6 ohms We transfer this to the primary side. Z in = 4 + 6/n2 = 4 + 0.375 = 4.375 ohms, where n = 4 I 1 = 120/4.375 = 27.43 A and I 2 = I 1 /n = 6.857 A 4

1:4

I1 +

1200

+ –

v1

10 



Using current division, I 2 ’ = (10/25)I 2 = 2.7429 and I 3 = I 2 ’/n’ = 1.3714 A (b)

I2’

+ v2



5

I2

p = 0.5(I 3 )2(40) = 37.62 watts

10 

Chapter 13, Solution 61. We reflect the 160-ohm load to the middle circuit. Z R = Z L /n2 = 160/(4/3)2 = 90 ohms, where n = 4/3 2

1:5

I1 +

240

+ –

v1 

Io

14 

Io’

+ vo 

14 + 60||90 = 14 + 36 = 50 ohms We reflect this to the primary side. Z R ’ = Z L ’/(n’)2 = 50/52 = 2 ohms when n’ = 5 I 1 = 24/(2 + 2) = 6A 24 = 2I 1 + v 1 or v 1 = 24 – 2I 1 = 12 V v o = –nv 1 = –60 V, I o = –I 1 /n 1 = –6/5 = –1.2 I o ‘ = [60/(60 + 90)]I o = –0.48A I 2 = –I o ’/n = 0.48/(4/3) = 360 mA

60 

90 


Reflect the load to the middle circuit. Z L ’ = 8 – j20 + (18 + j45)/32 = 10 – j15

We now reflect this to the primary circuit so that Z in = 6 + j4 + (10 – j15)/n2 = 7.6 + j1.6 = 7.76711.89, where n = 5/2 = 2.5 I 1 = 40/Z in = 40/7.76711.89 = 5.15–11.89 S = v s I 1 * = (400)(5.1511.89) = 20611.89 VA (b)

I 2 = –I 1 /n,

n = 2.5

I 3 = –I 2 /n’, n = 3 I 3 = I 1 /(nn’) = 5.15–11.89/(2.5x3) = 0.6867–11.89 p = |I 2 |2(18) = 18(0.6867)2 = 8.488 watts

Chapter 13, Solution 63. Reflecting the (9 + j18)-ohm load to the middle circuit gives, Z in ’ = 7 – j6 + (9 + j18)/(n’)2 = 7 – j6 + 1 + j2 = 8 – j4 when n’ = 3 Reflecting this to the primary side, Z in = 1 + Z in ’/n2 = 1 + 2 – j = 3 – j, where n = 2 I 1 = 120/(3 – j) = 12/3.162–18.43 = 3.79518.43A I 2 = I 1 /n = 1.897518.43 A I 3 = –I 2 /n2 = 632.5161.57 mA

Chapter 13, Solution 64. The Thevenin equivalent to the left of the transformer is shown below.

8 k

24  0 V

+ _

The reflected load impedance is Z 30k Z L'  2L  2 n n

For maximum power transfer, 30k  8k     n 2  30 / 8  3.75 n2

n =1.9365

Chapter 13, Solution 65. 40 

10 

I1

200 V (rms) -

50 

I2



1 +

1:2

I2

1:3

+

+

+

V1 -

V2 -

V3 -

I3 

2 + V4 -

20 



At node 1, 200  V1 V1  V4   I1 10 40

 

200  1.25V1  0.25V4  10 I 1

(1)

At node 2, V1  V4 V4   I3 40 20

 

V1  3V4  40 I 3

(2)

At the terminals of the first transformer, V2  2   V2  2V1 V1 I2  1 / 2   I 1  2 I 2 I1

(3) (4)

For the middle loop,

 V2  50 I 2  V3  0

 

V3  V2  50 I 2

(5)

At the terminals of the second transformer, V4 3 V3

 

V4  3V3

(6)

I3  1 / 3   I 2  3 I 3 I2 We have seven equations and seven unknowns. Combining (1) and (2) leads to

(7)

200  3.5V4  10 I 1  50 I 3 But from (4) and (7), I 1  2 I 2  2(3I 3 )  6 I 3 . Hence 200  3.5V4  110 I 3

(8)

From (5), (6), (3), and (7), V4  3(V2  50 I 2 )  3V2  150 I 2  6V1  450 I 3

Substituting for V 1 in (2) gives V4  6(3V4  40 I 3 )  450 I 3

 

I3 

19 V4 210

Substituting (9) into (8) yields 200  13.452V4

 

V4  14.87 P

V 24  11.05 W 20

(9)

Chapter 13, Solution 66. Design a problem to help other students to better understand how the ideal autotransformer works. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem An ideal autotransformer with a 1:4 step-up turns ratio has its secondary connected to a 120- load and the primary to a 420-V source. Determine the primary current. Solution v 1 = 420 V

(1)

v 2 = 120I 2

(2)

v 1 /v 2 = 1/4 or v 2 = 4v 1

(3)

I 1 /I 2 = 4 or I 1 = 4 I 2

(4)

Combining (2) and (4), v 2 = 120[(1/4)I 1 ] = 30 I 1 4v 1 = 30I 1 4(420) = 1680 = 30I 1 or I 1 = 56 A


(a)

V1 N 1  N 2 1   V2 N2 0.4

(b)

S 2  I 2V 2  5,000

(c )

S 2  S 1  I 1V1  5,000

 

 

V 2  0.4V1  0.4 x 400  160 V

I2 

 

5000  31.25 A 160

I1 

5000  12.5 A 400

Chapter 13, Solution 68. This is a step-up transformer.

I2 + N1

2 – j6

v2

I1

10 + j40 +

2030

N2 

v1

+ 



For the primary circuit,

2030 = (2 – j6)I 1 + v 1

(1)

For the secondary circuit,

v 2 = (10 + j40)I 2

(2)

At the autotransformer terminals, v 1 /v 2 = N 1 /(N 1 + N 2 ) = 200/280 = 5/7,

Also,

thus v 2 = 7v 1 /5

(3)

I 1 /I 2 = 7/5 or I 2 = 5I 1 /7

(4)

Substituting (3) and (4) into (2),

v 1 = (10 + j40)25I 1 /49

Substituting that into (1) gives

2030 = (7.102 + j14.408)I 1

I 1 = 2030/16.06363.76 = 1.245–33.76 A I 2 = 5I 1 /7 = 889.3–33.76 mA I o = I 1 – I 2 = [(5/7) – 1]I 1 = –2I 1 /7 = 355.7146.2 mA p = |I 2 |2R = (0.8893)2(10) = 7.51 watts

Chapter 13, Solution 69. We can find the Thevenin equivalent. I2

+

+

N2 j125 

75 

v2 V Th

I1 1200

+

N1

v1

+ 





 I1 = I2 = 0 As a step up transformer,

v 1 /v 2 = N 1 /(N 1 + N 2 ) = 600/800 = 3/4

v 2 = 4v 1 /3 = 4(120)/3 = 1600 rms = V Th . To find Z Th , connect a 1-V source at the secondary terminals. We now have a step-down transformer. + j125 

75  I2

I1

v1 10 V

+ v2

+ 



 v 1 = 1V, v 2 =I 2 (75 + j125) But

v 1 /v 2 = (N 1 + N 2 )/N 1 = 800/200 which leads to v 1 = 4v 2 = 1 and v 2 = 0.25 I 1 /I 2 = 200/800 = 1/4 which leads to I 2 = 4I 1

Hence

0.25 = 4I 1 (75 + j125) or I 1 = 1/[16(75 + j125) Z Th = 1/I 1 = 16(75 + j125) Therefore, Z L = Z Th * = (1.2 – j2) k Since V Th is rms, p = (|V Th |/2)2/R L = (80)2/1200 = 5.333 watts

Chapter 13, Solution 70. This is a step-down transformer. 30 + j12

I1 + I2

v1 1200

+ 

+ v2 



20 – j40

I 1 /I 2 = N 2 /(N 1 + N 2 ) = 200/1200 = 1/6, or I 1 = I 2 /6

(1)

v 1 /v 2 = (N 2 + N 2 )/N 2 = 6, or v 1 = 6v 2

(2)

For the primary loop,

120 = (30 + j12)I 1 + v 1

(3)

For the secondary loop,

v 2 = (20 – j40)I 2

(4)

Substituting (1) and (2) into (3), 120 = (30 + j12)( I 2 /6) + 6v 2 and substituting (4) into this yields 120 = (49 – j38)I 2 or I 2 = 1.93537.79 p = |I 2 |2(20) = 74.9 watts.

Chapter 13, Solution 71. Z in = V 1 /I 1 But Hence

V 1 I 1 = V 2 I 2 , or V 2 = I 2 Z L and I 1 /I 2 = N 2 /(N 1 + N 2 ) V 1 = V 2 I 2 /I 1 = Z L (I 2 /I 1 )I 2 = Z L (I 2 /I 1 )2I 1 V 1 /I 1 = Z L [(N 1 + N 2 )/N 2 ] 2 Z in = [1 + (N 1 /N 2 )] 2Z L


Consider just one phase at a time. 1:n A

a B b C c n = V L / 3VLp  7200 /(12470 3 ) = 1/3 (b)

The load carried by each transformer is 60/3 = 20 MVA. Hence

I Lp = 20 MVA/12.47 k = 1604 A I Ls = 20 MVA/7.2 k = 2778 A

(c)

The current in incoming line a, b, c is 3I Lp  3x1603.85 = 2778 A Current in each outgoing line A, B, C is 2778/(n 3 ) = 4812 A

20MVA Load


This is a three-phase -Y transformer.

(b)

V Ls = nv Lp / 3 = 450/(3 3 ) = 86.6 V, where n = 1/3 As a Y-Y system, we can use per phase equivalent circuit. I a = V an /Z Y = 86.60/(8 – j6) = 8.6636.87 I c = I a 120 = 8.66156.87 A I Lp = n 3 I Ls I 1 = (1/3) 3 (8.6636.87) = 536.87 I 2 = I 1 –120 = 5–83.13 A

(c)

p = 3|I a |2(8) = 3(8.66)2(8) = 1.8 kw.


This is a - connection.

(b)

The easy way is to consider just one phase. 1:n = 4:1 or n = 1/4 n = V 2 /V 1 which leads to V 2 = nV 1 = 0.25(2400) = 600 i.e. V Lp = 2400 V and V Ls = 600 V S = p/cos = 120/0.8 kVA = 150 kVA p L = p/3 = 120/3 = 40 kw 4:1 IL

I Ls

V Lp

I ps

I pp

p Ls = V ps I ps

But

For the -load,

IL =

Hence,

I ps = 40,000/600 = 66.67 A I Ls =

(c)

V Ls

3 I ps =

3 I p and V L = V p

3 x66.67 = 115.48 A

Similarly, for the primary side p pp = V pp I pp = p ps or I pp = 40,000/2400 = 16.667 A and

(d)

I Lp =

3 I p = 28.87 A

Since S = 150 kVA therefore S p = S/3 = 50 kVA


n = V Ls /( 3 V Lp ) = 900/(4500 3 ) = 0.11547

(b)

S =

3 V Ls I Ls or I Ls = 120,000/(900 3 ) = 76.98 A

I Ls = I Lp /(n 3 ) = 76.98/(2.887 3 ) = 15.395 A

Chapter 13, Solution 76. Using Fig. 13.138, design a problem to help other students to better understand a wye-delta, three-phase transformer and how they work. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A Y- three-phase transformer is connected to a 60-kVA load with 0.85 power factor (leading) through a feeder whose impedance is 0.05 + j0.1 per phase, as shown in Fig. 13.137 below. Find the magnitude of: (a) the line current at the load, (b) the line voltage at the secondary side of the transformer, (c) the line current at the primary side of the transformer.


At the load,

V L = 240 V = V AB V AN = V L / 3 = 138.56 V

Since S =

3 V L I L then I L = 60,000/(240 3 ) = 144.34 A

1:n

0.05 

2640V

j0.1 

A 240V

j0.1 

0.05 

B 0.05 

(b)

j0.1 

C

Balanced Load 60kVA 0.85pf leading

Let V AN = |V AN |0 = 138.560 cos = pf = 0.85 or  = 31.79 I AA’ = I L  = 144.3431.79 V A’N’ = ZI AA’ + V AN = 138.560 + (0.05 + j0.1)(144.3431.79) = 138.036.69 V Ls = V A’N’

(c)

3 = 138.03

3 = 239.1 V

For Y- connections, n =

3 V Ls /V ps =

3 x238.7/2640 = 0.1569

f Lp = nI Ls / 3 = 0.1569x144.34/ 3 = 13.05 A


This is a single phase transformer.

V 1 = 13.2 kV, V 2 = 120 V

n = V 2 /V 1 = 120/13,200 = 1/110, therefore n = 1/110 or 110 turns on the primary to every turn on the secondary. (b)

P = VI or I = P/V = 100/120 = 0.8333 A I 1 = nI 2 = 0.8333/110 = 7.576 mA

Chapter 13, Solution 78. We convert the reactances to their inductive values.

X  L

  L

X



The schematic is as shown below.

AC = y es MAG = y es PHASE = y es

COUPLING = 0.5 L1_VALUE = 80H L2_VALUE = 60H

IPRINT

IPRINT R1

AC = y es MAG = y es PHASE = y es

TX1

20 100Vac ACPHASE = -30 V1 R2 50

0Vdc R3 40

0

FREQ


1.592E-01 1.347E+00 -8.489E+01 FREQ


1.592E-01 6.588E-01 -7.769E+01 Thus, I 1 = 1.347–84.89˚ amps and I 2 = 658.8–77.69˚ mA

Chapter 13, Solution 79. The schematic is shown below. k 1 = 15 / 5000 = 0.2121, k 2 = 10 / 8000 = 0.1118 In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the circuit is saved and simulated, the output includes

FREQ

IM(V_PRINT1)

IP(V_PRINT1)

1.592 E–01

4.068 E–01

–7.786 E+01

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

1.306 E+00

–6.801 E+01

FREQ

IM(V_PRINT3)

IP(V_PRINT3)

1.592 E–01

1.336 E+00

–5.492 E+01

Thus, I 1 = 1.306–68.01 A, I 2 = 406.8–77.86 mA, I 3 = 1.336–54.92 A

Chapter 13, Solution 80. The schematic is shown below. k 1 = 10 / 40x80 = 0.1768, k 2 = 20 / 40 x 60 = 0.4082 k 3 = 30 / 80 x 60 = 0.433 In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the simulation, we obtain the output file which includes

i.e.

FREQ

IM(V_PRINT1)

IP(V_PRINT1)

1.592 E–01

1.304 E+00

6.292 E+01

I o = 1.30462.92 A

Chapter 13, Solution 81. The schematic is shown below. k 1 = 2 / 4x8 = 0.3535, k 2 = 1 / 2 x8 = 0.25 In the AC Sweep box, we let Total Pts = 1, Start Freq = 100, and End Freq = 100. After simulation, the output file includes

FREQ 1.000 E+02

IM(V_PRINT1) 1.0448 E–01

IP(V_PRINT1) 1.396 E+01

FREQ 1.000 E+02

IM(V_PRINT2) 2.954 E–02

IP(V_PRINT2) –1.438 E+02

FREQ 1.000 E+02

IM(V_PRINT3) 2.088 E–01

IP(V_PRINT3) 2.440 E+01

i.e.

I 1 = 104.513.96 mA, I 2 = 29.54–143.8 mA, I 3 = 208.824.4 mA.

Chapter 13, Solution 82. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes FREQ 1.592 E–01

IM(V_PRINT1) 1.955 E+01

IP(V_PRINT1) 8.332 E+01

FREQ 1.592 E–01

IM(V_PRINT2) 6.847 E+01

IP(V_PRINT2) 4.640 E+01

FREQ 1.592 E–01

IM(V_PRINT3) 4.434 E–01

IP(V_PRINT3) –9.260 E+01

i.e.

V 1 = 19.5583.32 V, V 2 = 68.4746.4 V, I o = 443.4–92.6 mA.

These answers are incorrect, we need to adjust the magnitude of the inductances.

Chapter 13, Solution 83. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, the output file includes FREQ 1.592 E–01

IM(V_PRINT1) 1.080 E+00

IP(V_PRINT1) 3.391 E+01

FREQ 1.592 E–01

VM($N_0001) 1.514 E+01

VP($N_0001) –3.421 E+01

i.e.

i X = 1.0833.91 A, V x = 15.14–34.21 V.

This is most likely incorrect and needs to have the values of turns changed.

Checking with hand calculations. Loop 1.

–6 + 1I 1 + V 1 = 0 or I 1 + V 1 = 6

Loop 2.

–V 2 – j10I 2 + 8(I 2 –I 3 ) = 0 or (8–j10)I 2 – 8I 3 – V 2 = 0

Loop 3.

8(I 3 –I 2 ) + 6I 3 +2V x +V 3 = 0 or –8I 2 + 14I 3 + V 3 + 2V x = 0 but V x = 8(I 2 –I 3 ), therefore we get 8I 2 – 2I 3 + V 3 = 0

(3)

–V 4 + (4+j2)I 4 = 0 or (4+j2)I 4 – V 4 = 0

(4)

Loop 4.

(1) (2)

We also need the constraint equations, V 2 = 2V 1 , I 1 = 2I 2 , V 3 = 2V 4 , and I 4 = 2I 3 . Finally, I x = I 2 and V x = 8(I 2 – I 3 ).

We can eliminate the voltages from the equations (we only need I 2 and I 3 to obtain the required answers) by, (1)+0.5(2) = I 1 + (4–j5)I 2 – 4I 3 = 6 and 0.5(3) + (4) = 4I 2 – I 3 + (4+j2)I 4 = 0. Next we use I 1 = 2I 2 and I 4 = 2I 3 to end up with the following equations, (6–j5)I 2 – 4I 3 = 6 and 4I 2 + (7+j4)I 3 = 0 or I 2 = –[(7+j4)I 3 ]/4 = (–1.75–j)I 3 = (2.01556 –150.255°)I 3

This leads to (6–j5)(–1.75–j)I 3 – 4I 3 = (–10.5–5–4+j(8.75–6))I 3 = (–19.5+j2.75)I 3 = 6 or

I 3 = 6/(19.69296 171.973°) = 0.304677 –171.973° amps = –0.301692–j0.042545. I 2 = (–1.75–j)(0.304677 –171.973°)

= (2.01556 –150.255°)(0.304677 –171.973°)

= 614.096 37.772° mA = 0.48541+j0.37615

and I 2 – I 3 = 0.7871+j0.4187 = 0.89154 28.01°.

Therefore,

V x = 8(0.854876 22.97°) = 7.132 28.01° V

I x = I 2 = 614.1 37.77° mA.

Checking with MATLAB we get A and X from equations (1) – (4) and the four constraint equations.

>> A = [1 0 0 0 1 0 0 0;0 (8-10j) -8 0 0 -1 0 0;0 8 -2 0 0 0 1 0;0 0 0 (4+2j) 0 0 0 -1;0 0 0 0 -2 1 0 0;1 -2 0 0 0 0 0 0;0 0 0 0 0 0 1 -2;0 0 -2 1 0 0 0 0] A= 1.0000

0

0

0

1.0000

0

0

0 0

8.0000 -10.0000i -8.0000

0

8.0000

0

0

-1.0000

0

0 -2.0000

0

0

0

1.0000

0 0 1.0000 0 0 1.0000 0 0 2.0000 0 0

0

0

0

0

-2.0000

X= 6 0 0 0

0 0

0

0

0

-2.0000

>> X = [6;0;0;0;0;0;0;0]

4.0000 + 2.0000i

0

-2.0000 0

0 1.0000

0 1.0000

0

0

0

0 0

0 0

0

1.0000 0

-

0

0 0 0 0 >> Y = inv(A)*X Y=

0.9708 + 0.7523i = I 1 = 1.2817 37.773° amps

0.4854 + 0.3761i = I 2 = 614.056 37.769° mA = I x

-0.3017 - 0.0425i = I 3 = 0.30468 –171.982° amps -0.6034 - 0.0851i = I 4 5.0292 - 0.7523i = V 1 10.0583 - 1.5046i = V 2 -4.4867 - 3.0943i = V 3 -2.2434 - 1.5471i = V 4

I x = 614.1 37.77° mA

Finally, V x = 8(I 2 – I 3 ) = 8(0.7871+j0.4186) = 8(0.891489 28.01°)

= 7.132 28.01° volts

Chapter 13, Solution 84. The schematic is shown below. we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, the output file includes FREQ 1.592 E–01

IM(V_PRINT1) 4.028 E+00

IP(V_PRINT1) –5.238 E+01

FREQ 1.592 E–01

IM(V_PRINT2) 2.019 E+00

IP(V_PRINT2) –5.211 E+01

FREQ 1.592 E–01

IM(V_PRINT3) 1.338 E+00

IP(V_PRINT3) –5.220 E+01

i.e.

I 1 = 4.028–52.38 A, I 2 = 2.019–52.11 A, I 3 = 1.338–52.2 A.

Dot convention is wrong.


Z1

VS

+ 

Z L /n2

For maximum power transfer, Z 1 = Z L /n2 or n2 = Z L /Z 1 = 8/7200 = 1/900 n = 1/30 = N 2 /N 1 . Thus N 2 = N 1 /30 = 3000/30 = 100 turns.

Chapter 13, Solution 86. n = N 2 /N 1 = 48/2400 = 1/50 Z Th = Z L /n2 = 3/(1/50)2 = 7.5 k

Chapter 13, Solution 87. Z Th = Z L /n2 or n =

Z L / Z Th  75 / 300 = 0.5

Chapter 13, Solution 88. n = V 2 /V 1 = I 1 /I 2 or I 2 = I 1 /n = 2.5/0.1 = 25 A p = IV = 25x12.6 = 315 watts

Chapter 13, Solution 89. n = V 2 /V 1 = 120/240 = 0.5 S = I 1 V 1 or I 1 = S/V 1 = 10x103/240 = 41.67 A S = I 2 V 2 or I 2 = S/V 2 = 104/120 = 83.33 A


n = V 2 /V 1 = 240/2400 = 0.1

(b)

n = N 2 /N 1 or N 2 = nN 1 = 0.1(250) = 25 turns

(c)

S = I 1 V 1 or I 1 = S/V 1 = 4x103/2400 = 1.6667 A S = I 2 V 2 or I 2 = S/V 2 = 4x104/240 = 16.667 A


The kVA rating is S = VI = 25,000x75 = 1.875 MVA

(b)

Since S 1 = S 2 = V 2 I 2 and I 2 = 1875x103/240 = 7.812 kA


V 2 /V 1 = N 2 /N 1 = n, V 2 = (N 2 /N 1 )V 1 = (28/1200)4800 = 112 V

(b)

I 2 = V 2 /R = 112/10 = 11.2 A and I 1 = nI 2 , n = 28/1200 I 1 = (28/1200)11.2 = 261.3 mA

(c)

p = |I 2 |2R = (11.2)2(10) = 1254 watts.

Chapter 13, Solution 93. (a) For an input of 110 V, the primary winding must be connected in parallel, with series aiding on the secondary. The coils must be series opposing to give 14 V. Thus, the connections are shown below.

110 V 14 V

(b) To get 220 V on the primary side, the coils are connected in series, with series aiding on the secondary side. The coils must be connected series aiding to give 50 V. Thus, the connections are shown below.

220 V 50 V

Chapter 13, Solution 94. V 2 /V 1 = 110/440 = 1/4 = I 1 /I 2 There are four ways of hooking up the transformer as an auto-transformer. However it is clear that there are only two outcomes.

V1

V1

V1 V2

V2

(1)

V1

(2)

V2

(3)

V2

(4)

(1) and (2) produce the same results and (3) and (4) also produce the same results. Therefore, we will only consider Figure (1) and (3). (a)

For Figure (3), V 1 /V 2 = 550/V 2 = (440 – 110)/440 = 330/440 Thus,

(b)

V 2 = 550x440/330 = 733.4 V (not the desired result)

For Figure (1), V 1 /V 2 = 550/V 2 = (440 + 110)/440 = 550/440 Thus,

V 2 = 550x440/550 = 440 V (the desired result)


n = V s /V p = 120/7200 = 1/60

(b)

I s = 10x120/144 = 1200/144 S = VpIp = VsIs I p = V s I s /V p = (1/60)x1200/144 = 139 mA

*Chapter 13, Solution 96. Problem, Some modern power transmission systems now have major, high voltage DC transmission segments. There are a lot of good reasons for doing this but we will not go into them here. To go from the AC to DC, power electronics are used. We start with three-phase AC and then rectify it (using a full-wave rectifier). It was found that using a delta to wye and delta combination connected secondary would give us a much smaller ripple after the full-wave rectifier. How is this accomplished? Remember that these are real devices and are wound on common cores. Hint, using Figures 13.47 and 13.49, and the fact that each coil of the wye connected secondary and each coil of the delta connected secondary are wound around the same core of each coil of the delta connected primary so the voltage of each of the corresponding coils are in phase. When the output leads of both secondaries are connected through full-wave rectifiers with the same load, you will see that the ripple is now greatly reduced. Please consult the instructor for more help if necessary. Solution, This is a most interesting and very practical problem. The solution is actually quite easy, you are creating a second set of sine waves to send through the full-wave rectifier, 30˚ out of phase with the first set. We will look at this graphically in a minute. We begin by showing the transformer components. The key to making this work is to wind the secondary coils with each phase of the primary. Thus, a-b is wound around the same core as A 1 -N 1 and A 2 -B 2 . The next thing we need to do is to make sure the voltages come out equal. We need to work the number of turns of each secondary so that the peak of V A1 – V B1 is equal to V A2 –V B2 . Now, let us look at some of the equations involved.

If we let v ab (t) = 100sin(t) V, assume that we have an ideal transformer, and the turns ratios are such that we get v A1-N1 (t) = 57.74sin(t) V and V A2-B2 (t) = 100sin(t) V. Next, let us look at V bc (t) = 100sin(t+120˚) V. This leads to V B1-N1 (t) = 57.74sin(t+120˚) V. We now need to determine V A1-B1 (t). V A1-B1 (t) = 57.74sin(t) – 57.74sin(t+120˚) = 100sin(t–30˚) V. This then leads to the output per phase voltage being equal to v out (t) = [100sin(t) + 100sin(t–30˚)] V. We can do this for each phase and end up with the output being sent to the full-wave rectifier. This looks like v out (t) = [|100sin(t)| + |100sin(t–30˚)| + |sin(t+120˚)| |100sin(t+90˚)| + |100sin(t–120˚)| + |100sin(t–150˚)|] V. The end result will be more obvious if we look at plots of the rectified output.

A1

N1

B1

a C1 b A2 c C2

B2

In the plot below we see the normalized (1 corresponds to 100 volts) ripple with only one of the secondary sets of windings and then the plot with both. Clearly the ripple is greatly reduced!

Chapter 14, Solution 1. Vo R jRC   Vi R  1 jC 1  jRC j  0 1 H ()  , where  0  1  j  0 RC H () 

H  H () 

 0

  H () 

1  ( 0 ) 2

   tan -1   2  0 

This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that 0  1 RC . Thus, the sketches of H and  are shown below. H 1 0.7071

0

0 =





90 45 0

0 =



Chapter 14, Solution 2. Using Fig. 14.69, design a problem to help other students to better understand how to determine transfer functions. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Obtain the transfer function V o /V i of the circuit in Fig. 14.66. 10  + 2

+ _

Vo

1 F 8

Figure 14.66

Vo _

For Prob. 14.2.

Solution V H ( s)  o  Vi

2

1 s/8

10  20 

1 s/8



2  8/ s 1 s4  12  8 / s 6 s  0.6667

Chapter 14, Solution 3. 0.2 F

 

0.1F

 

1 1 5   jC s (0.2) s 1 10  s (0.1) s

The circuit becomes that shown below.

2

V1

5 s +

Vi

+ _

5

10 s

Vo _

10 5 10 1  s (5  ) 5( ) 10 5 s  s s  10( s  1) Let Z  //(5  )  s 15 5 s s s ( s  3) 5 (3  s ) s s Z V1  Vi Z 2 5 s s Z  Vi Vo  V1  V1  55/ s s 1 s 1 Z  2

10( s  1) s 10s 5s s ( s  3) V    2 H (s)  o  Vi s  1 10( s  1) 2 s ( s  3)  10( s  1) s  8s  5 2 s( s  3) H(s) = 5s/(s2+8s+5)


R ||

1 R  jC 1  jRC

R Vo R 1  jRC H ()    R Vi R  jL (1  jRC) jL  1  jRC

(b)

H () 

R -  RLC  R  jL

H () 

jC (R  jL) R  jL  R  jL  1 jC 1  jC (R  jL)

H () 

-  2 LC  jRC 1   2 LC  jRC

2

Chapter 14, Solution 5. (a) Let Z  R // sL 

Vo 

sRL R  sL

Z Vs Z  Rs

sRL Vo Z sRL H (s)    R  sL  Vs Z  Rs R  sRL RRs  s ( R  Rs ) L s R  sL

1 Rx 1 sC  R (b) Let Z  R //  sC R  1 1  sRC sC Z Vo  Vs Z  sL

V Z H(s)  o   Vi Z  sL

R R 1  sRC  R s 2 LRC  sL  R sL  1  sRC

Chapter 14, Solution 6. The 2 H inductors become jω2 or 2s. Let Z = 2s||2 = [(2s)(2)/(2s+2)] = 2s/(s+1) We convert the current source to a voltage source as shown below. 2

Is  2

2S

+

+ _

Vo

Z

_

V o = [(Z)/(Z+2s+2)](2I s ) = H(s) = I o /I s = [2s/(s2+3s+1)].

or


0.05  20 log10 H

2.5  10 -3  log10 H H  10 2.510  1.005773 -3

(b)

- 6.2  20 log10 H - 0.31  log10 H

H  10 -0.31  0.4898 (c)

104.7  20 log10 H 5.235  log10 H

H  10 5.235  1.718  105

Chapter 14, Solution 8. Design a problem to help other students to better calculate the magnitude in dB and phase in degrees of a variety of transfer functions at a single value of ω. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine the magnitude (in dB) and the phase (in degrees) of H() at  = 1 if H() equals (a) 0.05 (b) 125 10 j (c) 2  j 3 6 (d)  1  j 2  j Solution (a)

(b)

(c)

H  0.05 H dB  20 log10 0.05  - 26.02 ,

φ = 0

H  125 H dB  20 log10 125  41.94 ,

φ = 0

H dB (d)

j10  4.47263.43 2 j  20 log10 4.472  13.01 ,

H(1) 

φ = 63.43

3 6   3.9  j2.7  4.743 - 34.7 1 j 2  j H dB  20 log10 4.743  13.521, φ = –34.7˚

H(1) 


H ( ) 

10 10(1  j )(1  j 10)

H dB  20 log 10 1 - 20 log 10 1  j  20 log 10 1  j / 10

  - tan -1 ()  tan -1 ( / 10)

The magnitude and phase plots are shown below. 20 H dB 0.1

1

10



100 20 log 10

-20

1 1  j / 10 20 log10

-40

1 1  j

 0.1 -45

1

10



100 arg

1 1  j / 10

-90 arg

-135 -180

1 1  j

Chapter 14, Solution 10. Design a problem to help other students to better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of jω. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Sketch the Bode magnitude and phase plots of: H  j  

50 j  5  j 

Solution

H( j) 

50  j(5  j)

10 j   1 j1   5  

H dB 40 20 log1

20 10 0.1 -20

1

100

  1 20 log  j  1 5 

 1   20 log   j 

-40

 0.1 -45

     



1 10



100 arg

1 1  j / 5

-90 arg

-135 -180

1 j


H ( ) 

0.2 x10 (1  j 10) 2[ j (1  j 2)]

H dB  20 log10 1  20 log10 1  j 10  20 log10 j  20 log10 1  j 2

  -90  tan -1  10  tan -1  2 The magnitude and phase plots are shown below. H dB 40 20

0.1

1

10

100



1

10

100



-20 -40  90 45 0.1 -45 -90


T ( ) 

10(1  j ) j (1  j / 10)

To sketch this we need 20log 10 |T(ω)| = 20log 10 |10| + 20log 10 |1+jω| – 20log 10 |jω| – 20log 10 |1+jω/10| and the phase is equal to tan–1(ω) – 90° – tan–1(ω/10). The plots are shown below. |T|

(db)

20



0 0.1

1

10

100

-20 -40

arg T

90o



0 0.1 -90o

1

10

100


G ( ) 

0.1(1  j ) (1 100)(1  j )  2 ( j ) (10  j ) ( j ) 2 (1  j 10)

GdB  40  20 log10 1  j  40 log10 j  20 log10 1  j 10

  -180  tan -1  tan -1  10

The magnitude and phase plots are shown below. G dB 40 20 0.1 -20

1

10

100



1

10

100



-40

 90 0.1 -90 -180

Chapter 14, Solution 14. 250 25

H ( ) 

1  j 2  j10  j     j 1     25 5    

H dB  20 log10 10  20 log10 1  j  20 log10 j

 20 log10 1  j2 5  ( j 5) 2  10 25     -90  tan -1   tan -1  1  2 5 

The magnitude and phase plots are shown below. H dB 40 20 0.1 -20

1

10

100



1

10

100



-40  90 0.1 -90 -180


H ( ) 

2 (1  j ) 0.1(1  j )  (2  j )(10  j ) (1  j 2)(1  j 10)

H dB  20 log10 0.1  20 log10 1  j  20 log10 1  j 2  20 log10 1  j 10

  tan -1   tan -1  2  tan -1  10

The magnitude and phase plots are shown below. H dB 40 20 0.1 -20

1

10

100



1

10

100



-40  90 45 0.1 -45 -90


H(ω) =

H db = 20log 10 |0.1| – 20olg 10 |jω| – 20log 10 |1+jω+(jω/4)2| The magnitude and phase plots are shown below. H 20 20 log (j) 1

10

4

40

100



0.1

–20  j  20 log 1  j     4 

–40

–60 

 0.4 -90

1

4

10

40

90

100

-tan-1

 2

1

-180

16

2


G () 

(1 4) j (1  j)(1  j 2) 2

G dB  -20log10 4  20 log10 j  20 log10 1  j  40 log10 1  j 2   -90 - tan -1  2 tan -1  2

The magnitude and phase plots are shown below. G dB 20 0.1

1

10

100 

-12 -20 -40

 90 0.1 -90 -180

1

10

100



Chapter 14, Solution 18. The MATLAB code is shown below. >> w=logspace(-1,1,200); >> s=i*w; >> h=(7*s.^2+s+4)./(s.^3+8*s.^2+14*s+5); >> Phase=unwrap(angle(h))*57.23; >> semilogx(w,Phase) >> grid on

60

40

H (jw ) P h a s e

20

0

-2 0

-4 0

-6 0 -1 10

10 w

0

Now for the magnitude, we need to add the following to the above, >> H=abs(h); >> HdB=20*log10(H); >> semilogx(w,HdB); >> grid on

10

1

0

-5

HdB

-1 0

-1 5

-2 0

-2 5 -1 10

10 w

0

10

1

Chapter 14, Solution 19. H(ω) = 80jω/[(10+jω)(20+jω)(40+jω)] = [80/(10x20x40)](jω)/[(1+jω/10)(1+jω/20)(1+jω/40)] H db = 20log 10 |0.01| + 20log 10 |jω| – 20log 10 |1+jω/10| – 20log 10 |1+jω/20| – 20log 10 |1+jω/40| The magnitude and phase plots are shown below.

20 log j

20 db

0 db 1

0.1

100

10

ω

–20 log |1+jω/40| –20 db –20 log 1  20 log |1/80|

–40 db

j 10

–20 log |1+jω/20|

jω

90˚

0˚ 0.1

1

100

10

ω (1+jω/40)

–90˚ (1+jω/10) –180˚

(1+jω/20)

Chapter 14, Solution 20. Design a more complex problem than given in Prob. 14.10, to help other students to better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of jω. Include at least a second order repeated root. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Sketch the magnitude phase Bode plot for the transfer function

Solution

20log(1/100) = -40 For the plots, see the next page.

The magnitude and phase plots are shown below. 20 log j

40

1

20 log

j 10

1

20



0.1

1

5

10

50

100 20 log

-20 20 log

1 100

1 1  j

-40 20 log

-60

180˚

1 j   1   5  

2

jω (1+jω/10)

90˚ (1+jω) 0.1

 1

5

10

50

100

–90˚ (1+jω/5)2 –180˚

–270˚

Chapter 14, Solution 21. H(ω) = 10(jω)(20+jω)/[(1+jω)(400+60jω–ω2)] = [10x20/400](jω)(1+jω/20)/[(1+jω)(1+(3jω/20)+(jω/20)2)] H dB

j j 3  j   20 log(0.5)  20 log j  20 log 1   20 log 1  j  20 log 1    20 20  20 

2

The magnitude plot is as sketched below. 20log 10 |0.5| = –6 db

db 40 20log|jω| 20 log |1+jω/20|

20 1

20 log 0.5

10

20 

0.1

100

–20

–40

–20 log 1  j

–60 –20 log –80

Chapter 14, Solution 22. 20  20 log10 k

  k  10

A zero of slope  20 dB / dec at   2   1  j 2 1 A pole of slope - 20 dB / dec at   20   1  j 20 1  A pole of slope - 20 dB / dec at   100  1  j 100

Hence, H () 

10 (1  j 2) (1  j 20)(1  j 100)

10 4 ( 2  j) H ()  ( 20  j)(100  j)

Chapter 14, Solution 23. A zero of slope  20 dB / dec at the origin

 

j

1 1  j 1 1 A pole of slope - 40 dB / dec at   10   (1  j 10) 2

 A pole of slope - 20 dB / dec at   1 

Hence, H () 

j (1  j)(1  j 10) 2

H () 

100 j (1  j)(10  j) 2

(It should be noted that this function could also have a minus sign out in front and still be correct. The magnitude plot does not contain this information. It can only be obtained from the phase plot.)

Chapter 14, Solution 24. 40  20 log10 K   K  100 There is a pole at =50 giving 1/(1+j/50) There is a zero at =500 giving (1 + j/500). There is another pole at =2122 giving 1/(1 + j/2122). Thus, H(jω) = 100(1+jω)/[(1+jω/50)(1+jω/2122)] = [100(50x2122)/500](jω+500)/[(jω+50)(jω+2122)] or H(s) = 21220(s+500)/[(s+50)(s+2122)].

Chapter 14, Solution 25. 0 

1 LC



1 (40  10 -3 )(1  10 -6 )

 5 krad / s

Z(0 )  R  2 k  0 4   Z(0 4)  R  j  L  0 C   4

  5  10 3 4  Z(0 4)  2000  j   40  10 -3  3 -6 (5  10 )(1  10 )   4 Z(0 4)  2000  j (50  4000 5) Z(0 4)  2  j0.75 k  0 2   Z(0 2)  R  j  L  0 C   2

  (5  10 3 ) 2  Z(0 2)  2000  j  (40  10 -3 )  3 -6 (5  10 )(1  10 )  2  Z(ω 0 /2) = 200+j(100-2000/5) Z(0 2)  2  j0.3 k  1   Z(20 )  R  j  20 L  20 C  

  1  Z(20 )  2000  j  (2)(5  10 3 )(40  10 -3 )  3 -6 (2)(5  10 )(1  10 )   Z(20 )  2  j0.3 k  1   Z(40 )  R  j  40 L  40 C  

  1  Z(40 )  2000  j  (4)(5  10 3 )(40  10 -3 )  3 -6 (4)(5  10 )(1  10 )   Z(40 )  2  j0.75 k

Chapter 14, Solution 26. Design a problem to help other students to better understand ω o , Q, and B at resonance in series RLC circuits. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A coil with resistance 3  and inductance 100 mH is connected in series with a capacitor of 50 pF, a resistor of 6 , and a signal generator that gives 110V-rms at all frequencies. Calculate  o , Q, and B at resonance of the resultant series RLC circuit. Solution Consider the circuit as shown below. This is a series RLC resonant circuit.

6

50 pF

3

+ _

100 mH

R=6+3=9

o 

Q

1 1   447.21 krad/s 3 LC 100 x10 x50 x1012

o L R

o



447.21x103 x100 x103  4969 9

447.21x103 B   90 rad/s Q 4969


o 

1  40 LC

  LC 

1 402

R  10   R  10 L L If we select R =1 , then L = R/10 = 100 mH and B

C

1 1  2  6.25 mF 2 40 L 40 x0.1


R  10  . L

R 10   0.5 H B 20

C

1 1   2 F 2 0 L (1000) 2 (0.5)

Q

0 1000   50 B 20

Therefore, if R  10  then L  500 mH , C  2 F ,

Q  50

Chapter 14, Solution 29. We convert the voltage source to a current source as shown below.

12 k

is

is 

45 k

1 F

20 cos t , R = 12//45= 12x45/57 = 9.4737 k 12 1 1 o    4.082 krad/s = 4.082 krad/s 3 LC 60 x10 x1x106

B

1 1   105.55 rad/s = 105.55 rad/s RC 9.4737 x103 x106  4082 Q o   38.674 = 38.67 B 105.55

4.082 krads/s, 105.55 rad/s, 38.67

60 mH

Chapter 14, Solution 30. (a) f o = 15,000 Hz leads to ω o = 2πf o = 94.25 krad/s = 1/(LC)0.5 or LC = 1/8.883x109 or C = 1/(8.883x109x10–2) = 11.257x10–9 F = 11.257 pF. (b) since the capacitive reactance cancels out the inductive reactance at resonance, the current through the series circuit is given by I = 120/20 = 6 A. (c) Q = ω o L/R = 94.25x103(0.01)/20 = 47.12.


R  10  . R 10   0.05 H  50 mH 0 Q (10)(20) 1 1 C 2   0.2 F 0 L (100)(0.05) 1 1 B   0.5 rad/s RC (10)(0.2) L

Chapter 14, Solution 32. Design a problem to help other students to better understand the quality factor, the resonant frequency, and bandwidth of a parallel RLC circuit. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A parallel RLC circuit has the following values: R = 60 , L = 1 mH, and C = 50 F Find the quality factor, the resonant frequency, and the bandwidth of the RLC circuit. Solution

1 1   4.472 krad/s 3 LC 10 x50 x106 1 1 B   333.33 rad/s RC 60 x50 x106  4472 Q o   13.42 B 333.33

o 

Chapter 14, Solution 33. B = ω o /Q = 6x106/120 = 50 krad/s. ω 1 = ω o – B = 5.95x106 rad/s and ω 2 = ω o + B = 6.05x106 rad/s.


Q   o RC

Q

R o L

 

 

C

L

Q 80   56.84 pF 2f o R 2x5.6x10 6 x40x10 3

R 40 x10 3  = 14.21 µH 2f o Q 2x 5.6 x10 6 x80


1

o 

(b)

B

(c)

Q   o RC  1.443x10 3 x5x10 3 x60x10 6  432.9

LC



1

(a)

8x10  3 x60x10  6

 1.443 krad/s

1 1   3.33 rad/s 3 RC 5x10 x60x10  6

Chapter 14, Solution 36. At resonance,

1 1   40  Y 25  10 -3 Q 80 Q  0 RC   C    10 F 0 R (200  10 3 )(40) 1 1 1 0    L  2   2.5 H 10 0 C (4  10 )(10  10 -6 ) LC Y

1 R

  R 

0 200  10 3 B   2.5 krad / s Q 80 B 1  0   200  1.25  198.75 krad/s 2 B  2  0   200  1.25  201.25 krad/s 2

Chapter 14, Solution 37. 0 

1 LC

 5000 rad / s

Y(0 ) 

1 R

  Z(0 )  R  2 k

 1 4    0.5  j18.75 mS  j  0 C  R 0 L   4 1 Z(0 4)   (1.4212  j53.3)  0.0005  j0.01875 Y(0 4) 

 1 2    0.5  j7.5 mS  j  0 C  R 0 L   2 1 Z(0 2)   (8.85  j132.74)  0.0005  j0.0075 Y(0 2) 

 1 1    0.5  j7.5 mS  j  20 L  R 20 C   Z(20 )  (8.85  j132.74) 

Y ( 2 0 ) 

 1 1    0.5  j18.75 mS  j  40 L   R 4 C 0   Z(40 )  (1.4212  j53.3) 

Y ( 4 0 ) 


1  L  )   jLR  R  j(L  1 C C     Z  jL //( R  )  1 1 2 jC R R 2  (L  )  jL j C C jL(R 

1 ) jC

L 1   L   C C  0 1 2 2 R  ( L  ) C

LR 2  Im(Z) 

 

Thus, 

1 LC  R 2 C 2

 2 ( LC  R 2 C 2 )  1


Y

1 R  j L  jC  jC  R  jL R 2   2 L2

At resonance, Im(Y)  0 , i.e. 0 L 0 C  2 0 R  02 L2 L R 2  02 L2  C

0 

1 R2   LC L2

0  4.841 krad/s

 50     (40  10 -3 )(1  10 -6 )  40  10 -3  1

2


(a)

B   2  1  2(f 2  f1 )  2(90  86) x10 3  8krad / s

1 (1   2 )  2(88) x10 3  176X10 3 2 1 1 1 B   C    19.89nF RC BR 8x10 3 x 2x10 3

o 

1

(b)

o 

(c )

o  176  552.9krad / s

(d)

B  8  25.13krad / s

(e)

Q

LC

 

o 176   22 B 8

L

1 2 o C



1 (176X10 3 ) 2 x19.89 x10  9

= 164.45 µH

Chapter 14, Solution 41. Using Fig. 14.80, design a problem to help other students to better understand the quality factor, the resonant frequency, and bandwidth of an RLC circuit. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in Example 14.9. Problem For the circuits in Fig. 14.80, find the resonant frequency  0 , the quality factor Q, and the bandwidth B. Let C = 0.1 F, R 1 = 10 Ω, R 2 = 2 Ω, and L = 2 H. R2

C

R1

L

Figure 14.80 For Prob. 14.41. Solution To find ω o , we need to find the input impedance or input admittance and set imaginary component equal to zero. Finding the input admittance seems to be the easiest approach. Y = jω0.1 + 0.1 + 1/(2+jω2) = jω0.1 + 0.1 + [2/(4+4ω2)] – [jω2/(4+4ω2)] At resonance, 0.1ω = 2ω/(4+4ω2) or 4ω2 + 4 = 20 or ω2 = 4 or ω o = 2 rad/s and, Y = 0.1 + 2/(4+16) = 0.1 + 0.1 = 0.2 S The bandwidth is define as the two values of ω such that |Y| = 1.4142(0.2) = 0.28284 S. I do not know about you, but I sure would not want to solve this analytically. So how about using MATLAB or excel to solve for the two values of ω?

Using Excel, we get ω 1 = 1.414 rad/s and ω 2 = 3.741 rad/s or B = 2.327 rad/s We can now use the relationship between ω o and the bandwidth. Q = ω o /B = 2/2.327 = 0.8595


This is a series RLC circuit.

R  26  8,

L 1H,

1

 1.5811 rad / s

0 

(b)

LC



1 0.4

Q

 0 L 1.5811   0.1976 R 8

B

R  8 rad / s L

C  0.4 F

This is a parallel RLC circuit. (3)(6)  2 F 3 6 R  2 k , L  20 mH

3 F and 6 F   C  2 F , 0 

1 LC



1 (2  10 -6 )(20  10 -3 )

 5 krad / s

Q

R 2  10 3   20 0 L (5  10 3 )(20  10 -3 )

B

1 1   250 rad/s 3 RC (2  10 )(2  10 -6 )


Z in  (1 jC) || (R  jL)

Z in 

R  jL jC



R  jL 1  2 LC  jRC

1 jC (R  jL)(1  2 LC  jRC) Z in  (1  2 LC) 2  2 R 2 C 2 R  jL 

At resonance, Im(Z in )  0 , i.e. 0   0 L(1   02 LC)   0 R 2 C

02 L2 C  L  R 2 C 0 

(b)

L  R 2C L2 C



1 R2  LC L2

Z in  R || ( jL  1 jC)

R ( jL  1 jC) R (1  2 LC)  R  jL  1 jC (1   2 LC)  jRC R (1  2 LC)[(1  2 LC)  jRC] Z in  (1  2 LC) 2  2 R 2 C 2 Z in 

At resonance, Im(Z in )  0 , i.e. 0  R (1  2 LC) RC 1  2 LC  0 1 0  LC


1/jC

Z in

jL

R1

(a)

R2

Z in  (R 1 || jL) || (R 2  1 jC)

 R 1 jL   1   ||  R 2   Z in   jC   R 1  jL   jR 1 L  1    R 2  R 1  jL  jC  Z in  jR 1L 1 R2   jC R 1  jL jR 1 L (1  jR 2 C) Z in  (R 1  jL)(1  jR 2 C)  2 LCR 1

- 2 R 1 R 2 LC  jR 1 L Z in  R 1  2 LCR 1  2 LCR 2  j (L  R 1 R 2 C) (-2 R 1 R 2 LC  jR 1 L)[R 1  2 LCR 1  2 LCR 2  j (L  R 1 R 2 C)] Z in  (R 1  2 LCR 1  2 LCR 2 ) 2  2 (L  R 1 R 2 C) 2 At resonance, Im(Z in )  0 , i.e. 0  3 R 1 R 2 LC (L  R 1 R 2 C)  R 1 L (R 1  2 LCR 1  2 LCR 2 ) 0  3 R 12 R 22 LC 2  R 12 L  3 R 12 L2 C 0  2 R 22 C 2  1  2 LC 2 (LC  R 22 C 2 )  1

0  0 

1 LC  R 22 C 2 1

(0.02)(9  10 )  (0.1) 2 (9  10 -6 ) 2 0  2.357 krad / s -6

(b)

At   0  2.357 krad / s , jL  j(2.357  10 3 )(20  10 -3 )  j47.14 R 1 || jL 

R2 

j47.14  0.9996  j0.0212 1  j47.14

1 1  0.1   0.1  j47.14 j (2.357  10 3 )(9  10 -6 ) jC

Z in (0 )  (R 1 || jL) || (R 2  1 jC)

(0.9996  j0.0212)(0.1  j47.14) (0.9996  j0.0212)  (0.1  j47.14) Z in (0 )  1  Z in (0 ) 

Chapter 14, Solution 45. Convert the voltage source to a current source as shown below.

Is

30 k

50 F

R = 30//50 = 30x50/80 = 18.75 k This is a parallel resonant circuit. 1 1 o    447.21 rad/s LC 10 x103 x50 x106 1 1 B   1.067 rad/s RC 18.75 x103 x50 x106  447.21 Q o   419.13 B 1.067

447.2 rad/s, 1.067 rad/s, 419.1

10 mH

50 k


j , 1  j

1 || j 

1 ||

1 j 1 1   j  1  1 j 1  j

Transform the current source gives the circuit below.

j I 1  j

+ 

j 1  j

1

1 1  j

+ Vo

 1 j 1  j Vo  I  1 j 1  j   1 1  j 1  j

(b)

H () 

Vo j  I 2 (1  j) 2

H (1) 

1 2 (1  j) 2

H (1) 

1 2 ( 2)2

 0.25

Chapter 14, Solution 47. H () 

Vo R 1   Vi R  jL 1  jL R

H(0)  1 and H()  0 showing that this circuit is a lowpass filter. 1 At the corner frequency, H(c )  , i.e. 2 c L 1 1 R    1  or   c 2 L R 2 c L   1   R  Hence, c 

R  2f c L

fc 

1 R 1 10  10 3     796 kHz 2 L 2 2  10 -3

Chapter 14, Solution 48. R || H () 

1 jC

1 jC R jC R  1 jC H ()  R jC jL  R  1 jC R H ()  R  jL   2 RLC jL  R ||

H(0)  1 and H()  0 showing that this circuit is a lowpass filter.

Chapter 14, Solution 49. Design a problem to help other students to better understand lowpass filters described by transfer functions. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine the cutoff frequency of the lowpass filter described by H( ) 

4 2  j10

Find the gain in dB and phase of H() at  = 2 rad/s.

Solution At dc, H(0) 

Hence,

H() 

2 2



1 2

4  2. 2

H(0) 

2 2

4 4  100c2

4  100c2  8   c  0.2

H(2) 

4 2  2  j20 1  j10

H(2) 

2 101

 0.199

In dB, 20 log10 H(2)  - 14.023 arg H(2)  -tan -110  - 84.3 or ω c = 1.4713 rad/sec.


Vo jL  Vi R  jL

H(0)  0 and H()  1 showing that this circuit is a highpass filter. H (c ) 

or

fc 

1 2

c 



1  R   1   c L 

2

  1 

R  2f c L

1 R 1 200     318.3 Hz 2 L 2 0.1

R c L

Chapter 14, Solution 51. The lowpass RL filter is shown below. L

+

+ R

vs

vo -

-

H

c 

R  2 f c L

Vo R 1   Vs R  jL 1  jL / R  

R  2f c L  2x5x10 3 x40x10  3  1.256k

Chapter 14, Solution 52. Design a problem to help other students to better understand passive highpass filters. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem In a highpass RL filter with a cutoff frequency of 100 kHz, L = 40 mH. Find R. Solution

c 

R  2f c L

R  2f c L  (2)(10 5 )(40  10 -3 )  25.13 k

Chapter 14, Solution 53. 1  2f 1  20  10 3 2  2f 2  22  10 3 B  2  1  2  10 3 2  1 0   21  10 3 2 0 21 Q   10.5 B 2

0  L

1 LC

  L 

1 02 C

1  2.872 H (21  10 ) (80  10 -12 ) 3 2

R   R  BL L R  (2  10 3 )(2.872)  18.045 k

B

Chapter 14, Solution 54. We start with a series RLC circuit and the use the equations related to the circuit and the values for a bandstop filter. Q = ω o L/R = 1/(ω o CR) = 20; B = R/L = ω o /Q = 10/20 = 0.5; ω o = 1/(LC)0.5 = 10 (LC)0.5 = 0.1 or LC = 0.01. Pick L = 10 H then C = 1 mF. Q = 20 = ω o L/R =10x10/R or R = 100/20 = 5 Ω.


o 

Therefore,

1 LC



1 (25  10 )(0.4  10  6 ) 3

B

R 10   0.4 krad / s L 25  10 -3

Q

10  25 0.4

 10 krad / s

1  o  B 2  10  0.2  9.8 krad / s

or

2  o  B 2  10  0.2  10.2 krad / s

or

9.8  1.56 kHz 2 10.2 f2   1.62 kHz 2 f1 

1.56 kHz  f  1.62 kHz


From Eq 14.54, R s sRC L H (s)    1 1  sRC  s 2 LC R 1 R  sL  s2  s  sC L LC R

R 1 , and 0  L LC sB H (s)  2 s  sB   02

Since B 

(b)

From Eq. 14.56, H (s) 

sL 

1 sC

1 R  sL  sC

s 2   02 H (s)  2 s  sB   02



s2  s2  s

1 LC

R 1  L LC


Consider the circuit below.

R

I

I1

1/sC +

Vs

+ 

1/sC

R

Vo 

1  1 R   1  1 sC  sC  Z(s)  R  ||  R    R  2 sC  sC  R sC 1  sRC Z(s)  R  sC (2  sRC) Z(s) 

I

1  3sRC  s 2 R 2 C 2 sC (2  sRC)

Vs Z

I1 

Vs 1 sC I 2 sC  R Z (2  sRC)

Vo  I 1 R  H (s) 

R Vs sC (2  sRC)  2  sRC 1  3sRC  s 2 R 2 C 2

Vo sRC  Vs 1  3sRC  s 2 R 2 C 2

 3 s 1 RC H (s)   3 1 3 2 s  s 2 2  RC R C

1 or R C2 3  3 rad / s B RC

Thus, 02 

2

    

0 

1  1 rad / s RC

(b)

Similarly, Z(s)  sL  R || (R  sL)  sL  Z(s) 

I

R (R  sL) 2R  sL

R 2  3sRL  s 2 L2 2R  sL

Vs , Z

Vo  I 1  sL 

I1 

R Vs R I 2R  sL Z (2R  sL)

sLR Vs 2R  sL  2 2R  sL R  3sRL  s 2 L2

1  3R   s Vo sRL 3 L    H (s)  R2 3R Vs R 2  3sRL  s 2 L2 s 2 s2  L L

R  1 rad / s L 3R  3 rad / s B L

Thus, 0 


(b)

0 

1 LC



1 (0.1)(40  10 ) -12

 0.5  10 6 rad / s

R 2  10 3   2  10 4 L 0.1 0 0.5  10 6   25 Q B 2  10 4 B

As a high Q circuit, B 1  0   10 4 (50  1)  490 krad / s 2 B 2  0   10 4 (50  1)  510 krad / s 2

(c)

As seen in part (b),

Q  25

Chapter 14, Solution 59. Consider the circuit below. Ro + 1/sC Vi

+ 

R

Vo sL

 where L = 1 mH, C = 4 µF, R o = 6 Ω, and R = 4 Ω.  1  R (sL  1 sC) Z(s)  R || sL     sC  R  sL  1 sC R (1  s 2 LC) Z(s)  1  sRC  s 2 LC H

Vo R (1  s 2 LC) Z   Vi Z  R o R o  sRR o C  s 2 LCR o  R  s 2 LCR

Z in  R o  Z  R o 

Z in 

R (1  s 2 LC) 1  sRC  s 2 LC

R o  sRR o C  s 2 LCR o  R  s 2 LCR 1  sRC  s 2 LC

s  j R o  jRR o C  2 LCR o  R  2 LCR Z in  1  2 LC  jRC (R o  R  2 LCR o  2 LCR  jRR o C)(1   2 LC  jRC) Z in  (1  2 LC) 2  (RC) 2

Im(Z in )  0 implies that

- RC [R o  R  2 LCR o  2 LCR ]  RR o C (1  2 LC)  0

R o  R  2 LCR o  2 LCR  R o  2 LCR o  0

2 LCR  R 1

0 

H

LC

1



(1  10 )(4  10 -6 ) -3

 15.811 krad / s

R (1  2 LC) R o  jRR o C  R  2 LCR o  2 LCR

H max  H(0) 

or

H max

R Ro  R

 1  R  2  LC  R    H()  lim    R o  R RR o C j  LC (R  R o ) R  R o 2  

At 1 and 2 , H 

1 2

H mzx

R (1  2 LC)  R o  R   2 LC (R o  R )  jRR o C 2 (R o  R ) R

1 2 1 2 0





(R o  R )(1  2 LC) (RR o C) 2  (R o  R  2 LC(R o  R )) 2 10 (1  2  4  10 -9 ) (96  10 -6 ) 2  (10  2  4  10 -8 ) 2 10 (1  2  4  10 -9 ) (96  10 )  (10    4  10 ) -6

2

2

-8 2



1 2

(10  2  4  10 -8 )( 2 )  (96  10 -6 ) 2  (10  2  4  10 -8 ) 2  0 (2)(10  2  4  10 -8 ) 2  (96  10 -6 ) 2  (10  2  4  10 -8 ) 2 (96  10 -6 ) 2  (10  2  4  10 -8 ) 2  0 1.6  10 -15 4  8.092  10 -7 2  100  0

4  5.058  10 8  6.25  1016  0  2.9109  10 8 2    2.1471  10 8 Hence, 1  14.653 krad / s 2  17.061 krad / s B  2  1  17.061  14.653  2.408 krad / s


H () 

jRC j  1  jRC j  1 RC

(from Eq. 14.52)

This has a unity passband gain, i.e. H()  1 . 1  c  50 RC j10 H ^ ()  10 H ()  50  j j10 H ()  50  j


V 

1 jC V, R  1 jC i

V  Vo

Since V  V , 1 V  Vo 1  jRC i H () 

(b)

V 

Vo 1  Vi 1  jRC

R V, R  1 jC i

Since V  V , jRC V  Vo 1  jRC i H () 

Vo jRC  Vi 1  jRC

V  Vo

Chapter 14, Solution 62. This is a highpass filter.

jRC 1  1  jRC 1  j RC 1 1 H ()  , c   2 (1000) RC 1  j c  1 1 H ()   1  j f c f 1  j1000 f H () 

(a)

H (f  200 Hz) 

Vo 

(b)

1  j5

 23.53 mV

H (f  2 kHz) 

Vo 1  1  j0.5 Vi

120 mV

 107.3 mV

Vo 

(c)

120 mV

Vo 1  1  j5 Vi

1  j0.5

H (f  10 kHz) 

Vo 

120 mV 1  j0.1

Vo 1  1  j0.1 Vi

 119.4 mV

Chapter 14, Solution 63. For an active highpass filter, H(s)  

sC i R f 1  sC i R i

(1)

H(s)  

10s 1  s / 10

(2)

But

Comparing (1) and (2) leads to: C i R f  10

 

Rf 

10  10M Ci

C i R i  0.1

 

Ri 

0.1  100k Ci

Chapter 14, Solution 64. Z f  R f ||

Rf 1  jC f 1  jR f C f

Zi  R i 

1  jR i C i 1  jC i jC i

Hence, H () 

Vo - Z f - jR f C i   Vi Zi (1  jR f C f )(1  jR i C i )

This is a bandpass filter. H () is similar to the product of the transfer function of a lowpass filter and a highpass filter.


V 

R jRC Vi  V R  1 jC 1  jRC i

V 

Ri V Ri  Rf o

Since V  V , Ri jRC Vo  V Ri  Rf 1  jRC i H () 

Vo  R f   jRC     1  Vi  R i   1  jRC 

It is evident that as    , the gain is 1 

Rf 1 and that the corner frequency is . Ri RC


Proof

(b)

When R 1 R 4  R 2 R 3 , R4 s H (s)   R 3  R 4 s  1 R 2C

(c)

When R 3   , H (s) 

- 1 R 1C s  1 R 2C

Chapter 14, Solution 67. DC gain 

Rf 1  Ri 4

  R i  4R f

Corner frequency  c 

1  2 (500) rad / s R f Cf

If we select R f  20 k , then R i  80 k and 1  15.915 nF C (2)(500)(20  10 3 ) Therefore, if R f  20 k , then R i  80 k and C  15.915 nF

Chapter 14, Solution 68. Design a problem to help other students to better understand the design of active highpass filters when specifying a high-frequency gain and a corner frequency. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Design an active highpass filter with a high-frequency gain of 5 and a corner frequency of 200 Hz. Solution Rf   R f  5R i Ri 1  2 (200) rad / s Corner frequency  c  R i Ci High frequency gain  5 

If we select R i  20 k , then R f  100 k and 1  39.8 nF C (2)(200)(20  10 3 ) Therefore, if R i  20 k , then R f  100 k and C  39.8 nF

Chapter 14, Solution 69. This is a highpass filter with f c  2 kHz. 1 c  2f c  RC 1 1 RC   2f c 4  103

10 8 Hz may be regarded as high frequency. Hence the high-frequency gain is  R f  10  or R f  2 .5 R R 4 If we let R  10 k , then R f  25 k , and C 

1  7.96 nF . 4000  10 4


Vo (s) Y1 Y2  Vi (s) Y1 Y2  Y4 (Y1  Y2  Y3 ) 1 1 where Y1   G 1 , Y2   G 2 , Y3  sC1 , Y4  sC 2 . R1 R2 H (s) 

H (s) 

(b)

G 1G 2 G 1 G 2  sC 2 (G 1  G 2  sC1 )

G 1G 2  1, H()  0 G 1G 2 showing that this circuit is a lowpass filter. H ( 0) 


R  50  , L  40 mH , C  1 F L 

Km K L   1  m  (40  10 -3 ) Kf Kf

25K f  K m C 

C KmKf

10 6 K f 

(1)   1 

1 Km

Substituting (1) into (2), 1 10 6 K f  25K f K f  2x10–4 K m  25K f  5  10 -3

10 -6 KmKf

(2)

Chapter 14, Solution 72. Design a problem to help other students to better understand magnitude and frequency scaling. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem What values of K m and K f will scale a 4-mH inductor and a 20-F capacitor to 1 H and 2 F respectively? Solution

LC 

K f2 

LC K f2

  K f2 

LC L C

(4  10 -3 )(20  10 -6 )  4  10 -8 (1)(2)

K f  2  10 -4

L L 2  K C C m K 2m 

  K 2m 

L C  C L

(1)(20  10 -6 )  2.5  10 -3 (2)(4  10 -3 )

K m  5  10 -2

Chapter 14, Solution 73. R   K m R  (12)(800  10 3 )  9.6 M L 

Km 800 L (40  10 -6 )  32 F Kf 1000

C 300  10 -9 C    0.375 pF K m K f (800)(1000)

Chapter 14, Solution 74. R'1  K m R 1  3x100  300 R' 2  K m R 2  10x100  1 k

Km 10 2 L'  L  6 ( 2)  200 H Kf 10

1 C C'   108  1 nF K m K f 10


R'  K m R  20x10  200  L' 

Km 10 L  5 (4)  400 H Kf 10

C' 

C 1   1 F K m K f 10x10 5

1 Chapter 14, Solution 76.

R '  K m R  500 x5 x103  25 M K 500 L '  m L  5 (10mH )  50  H Kf 10 C'

C 20 x106   0.4 pF K m K f 500 x105

Chapter 14, Solution 77. L and C are needed before scaling.

B

R L

0 

  L 

1 LC

R 10  2H B 5

  C 

1 1   312.5 F 2 0 L (1600)(2)

(a)

L   K m L  (600)(2)  1.200 kH C 3.125  10 -4 C    0.5208 F Km 600

(b)

L 

L 2  3  2 mH K f 10

C 3.125  10 -4 C    312.5 nF Kf 10 3

(c)

L 

Km (400)(2) L  8 mH Kf 10 5

C 3.125  10 -4 C    7.81 pF KmKf (400)(10 5 )

Chapter 14, Solution 78. R   K m R  (1000)(1)  1 k Km 10 3 L  4 (1)  0.1 H L  10 Kf C 1 C    0.1 F 3 K m K f (10 )(10 4 )

The new circuit is shown below. 1 k + I

1 k

0.1 H

0.1 F

1 k

Vx 


Insert a 1-V source at the input terminals.

Ro Io

R

V1

V2 + 

+ 

1V

1/sC

+ sL

3V o

Vo 

There is a supernode. 1  V1 V2  R sL  1 sC But

V1  V2  3Vo

Also,

Vo 

  V2  V1  3Vo

sL V sL  1 sC 2

Combining (2) and (3)

V2  V1  3Vo 

(1)

 

Vo V2  sL sL  1 sC

Substituting (3) and (4) into (1) gives 1  V1 Vo sC V   R sL 1  4s 2 LC 1 sRC 1  4s 2 LC  sRC  1  V1  V V1 1  4s 2 LC 1 1  4s 2 LC 1  4s 2 LC V1  1  4s 2 LC  sRC 1  V1 sRC  R R (1  4s 2 LC  sRC)

Z in 

(3)

sL  1 sC Vo sL

s 2 LC Vo  V 1  4s 2 LC 1

Io 

(2)

1 1  sRC  4s 2 LC  sC Io

(4)

Z in  4sL  R 

1 sC

(5)

When R  5 , L  2 , C  0.1 , 10 Z in (s)  8s  5  s

At resonance, Im(Z in )  0  4L  or

(b)

0 

1 2 LC



1 C

1 2 (0.1)(2)

 1.118 rad / s

After scaling, R   K m R 4   40  5   50  L 

Km 10 L ( 2 )  0 .2 H Kf 100

C 

C 0.1   10 -4 K m K f (10)(100)

From (5), Z in (s)  0.8s  50 

0 

1 2 LC



10 4 s 1

2 (0.2)(10 -4 )

 111.8 rad / s


R   K m R  (200)(2)  400  K m L (200)(1)   20 mH Kf 10 4 C 0.5 C    0.25 F K m K f (200)(10 4 ) L 

The new circuit is shown below. 20 mH a Ix 0.25 F

400 

0.5 I x

b

(b)

Insert a 1-A source at the terminals a-b. a

sL

V1

V2

Ix 1A

1/(sC)

R

0.5 I x

b

At node 1, 1  sCV1 

V1  V2 sL

(1)

At node 2, V1  V2 V  0.5 I x  2 sL R But, I x  sC V1 . V1  V2 V2  0.5sC V1  sL R

(2)

Solving (1) and (2), sL  R V1  2 s LC  0.5sCR  1 Z Th 

V1 sL  R  2 1 s LC  0.5sCR  1

At   10 4 , Z Th 

( j10 4 )(20  10 -3 )  400 ( j10 4 ) 2 (20  10 -3 )(0.25  10 -6 )  0.5( j10 4 )(0.25  10 -6 )(400)  1

Z Th 

400  j200  600  j200 0.5  j0.5

Z Th  632.5 - 18.435 ohms


1 1 (G  jC)(R  jL)  1  G  jC   Z R  j L R  j L

which leads to

Z

jL  R 2

  LC  j(RC  LG)  GR  1

 R  C LC Z()  R G  GR  1   2  j    LC L C j

(1)

We compare this with the given impedance: Z() 

1000( j  1)

(2)

  2  2 j  1  2500

Comparing (1) and (2) shows that

1  1000 C R G  2 L C

   

C  1 mF,

R/L  1

 

RL

G  C  1 mS

GR  1 10 3 R  1 2501   LC 10 3 R



R  0.4  L

Thus, R = 0.4Ω, L = 0.4 H, C = 1 mF, G = 1 mS (b) By frequency-scaling, K f =1000. R’ = 0.4 Ω, G’ = 1 mS L' 

L 0.4  3  0.4mH , K f 10

C' 

C 10 3   3  1F K f 10


C 

C KmKf

Kf 

c 200   200  1

Km 

C 1 1 1   -6   5000 C K f 10 200

R   K m R  5 k,

thus,

R f  2R i  10 k

Chapter 14, Solution 83. 1 10 6 C  0.1 pF K mK f 100x10 5

1F

 

C' 

5F

 

C'  0.5 pF

10 k

 

R'  K m R  100x10 k  1 M

20 k

 

R '  2 M

Chapter 14, Solution 84. The schematic is shown below. A voltage marker is inserted to measure v o . In the AC sweep box, we select Total Points = 50, Start Frequency = 1, and End Frequency = 1000. After saving and simulation, we obtain the magnitude and phase plots in the probe menu as shown below.


We let I s  10 o A so that Vo / I s  Vo . The schematic is shown below. The circuit is simulated for 100 < f < 10 kHz.

Chapter 14, Solution 86. Using Fig. 14.103, design a problem to help other students to better understand how to use PSpice to obtain the frequency response (magnitude and phase of I) in electrical circuits. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Use PSpice to provide the frequency response (magnitude and phase of i) of the circuit in Fig. 14.103. Use linear frequency sweep from 1 to 10,000 Hz.

Figure 14.103 Solution The schematic is shown below. A current marker is inserted to measure I. We set Total Points = 101, start Frequency = 1, and End Frequency = 10 kHz in the AC sweep box. After simulation, the magnitude and phase plots are obtained in the Probe menu as shown below.

Chapter 14, Solution 87. The schematic is shown below. I n the AC Sweep box, we set Total Points = 50, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude response as shown below. It is evident from the response that the circuit represents a high-pass filter.

Chapter 14, Solution 88. The schematic is shown below. We insert a voltage marker to measure V o . In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude and phase plots of V o as shown below.

Chapter 14, Solution 89. The schematic is shown below. In the AC Sweep box, we type Total Points = 101, Start Frequency = 100, and End Frequency = 1 k. After simulation, the magnitude plot of the response V o is obtained as shown below.

Chapter 14, Solution 90. The schematic is shown below. In the AC Sweep box, we set Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After simulation, we obtain the magnitude plot of the response as shown below. The response shows that the circuit is a high-pass filter.

Chapter 14, Solution 91. The schematic is shown below. In the AC Sweep box, we select Total Points = 101, Start Frequency = 10, and End Frequency = 10 k. After simulation, the magnitude plot of the frequency response is obtained. From the plot, we obtain the resonant frequency f o is approximately equal to 800 Hz so that  o = 2f o = 5026 rad/s.

Chapter 14, Solution 92. The schematic is shown below. We type Total Points = 101, Start Frequency = 1, and End Frequency = 100 in the AC Sweep box. After simulating the circuit, the magnitude plot of the frequency response is shown below.


R Vs

+

+ _

V1



C

V1 

1 sC

Vs 

V 1  sRC

1 sC R sRC V2  Vs  Vs R  sC 1  sRC R

Vo  V1  V2  Hence, H (s) 

1  sRC Vs 1  sRC

Vo 1  sRC  V s 1  sRC

Vo

_



C V2 R


c 

1 RC

We make R and C as small as possible. To achieve this, we connect 1.8 k  and 3.3 k  in parallel so that

R

1.8x 3.3  1.164 k 1.8  3.3

We place the 10-pF and 30-pF capacitors in series so that

C = (10x30)/40 = 7.5 pF Hence, c 

1 1   114.55x10 6 rad/s 3 RC 1.164x10 x7.5x10 12


f0 

1 2 LC

When C  360 pF ,

f0 

1 2 (240  10 -6 )(360  10 -12 )

 0.541 MHz

When C  40 pF , f0 

1 2 (240  10 -6 )(40  10 -12 )

 1.624 MHz

Therefore, the frequency range is 0.541 MHz  f 0  1.624 MHz (b)

Q

2fL R

At f 0  0.541 MHz , Q

(2 )(0.541  10 6 )(240  10 -6 )  67.98 12

At f 0  1.624 MHz , Q

(2 )(1.624  10 6 )(240  10 -6 )  204.1 12


Ri

L

V1

Vo +

Vi

+ 

C1

C2

RL

Vo 

Z2 Z1  R L ||

Z1

RL 1  sC 2 1  sR 2 C 2

1 1  sL  R L  s 2 R L C 2 L   Z2  || (sL  Z1 )  ||  sC1 sC1  1  sR L C 2 

1 sL  R L  s 2 R L C 2 L  sC1 1  sR L C 2 Z2  sL  R L  s 2 R L C 2 L 1  sC1 1  sR L C 2 Z2 

sL  R L  s 2 R L LC 2 1  sR L C 2  s 2 LC1  sR L C1  s 3 R L LC1C 2

V1 

Z2 V Z2  R i i

Vo 

Z2 Z1 Z1 V V1   Z 2  R 2 Z1  sL i Z1  sL

Vo Z2 Z1   Vi Z 2  R 2 Z1  sL

where Z2  Z2  R 2

sL  R L  s 2 R L LC 2 sL  R L  s 2 R L LC 2  R i  sR i R L C 2  s 2 R i LC1  sR i R L C1  s 3 R i R L LC1C 2 Z1 RL  and Z1  sL R L  sL  s 2 R L LC 2

Therefore, Vo  Vi

R L (sL  R L  s 2 R L LC 2 ) (sL  R L  s 2 R L LC 2  R i  sR i R L C 2  s 2 R i LC 1  sR i R L C 1  s 3 R i R L LC 1 C 2 )( R L  sL  s 2 R L LC 2 )

where s  j .

Chapter 14, Solution 97. Ri

L

V1

Vo +

Vi

+ 

C1

C2

RL

Vo 

Z2

Z1

 1  sL (R L  1 sC 2 )  Z  sL ||  R L  , sC 2  R L  sL  1 sC 2 

s  j

V1 

Z V Z  R i  1 sC1 i

Vo 

RL RL Z V1   V R L  1 sC 2 R L  1 sC 2 Z  R i  1 sC1 i

H () 

Vo RL sL (R L  1 sC 2 )   Vi R L  1 sC 2 sL (R L  1 sC 2 )  (R i  1 sC1 )(R L  sL  1 sC 2 )

H () 

s 3 LR L C 1C 2 (sR i C 1  1)(s 2 LC 2  sR L C 2  1)  s 2 LC 1 (sR L C 2  1)

where s  j .

Chapter 14, Solution 98. B  2  1  2 (f 2  f 1 )  2 (454  432)  44

0  2f 0  QB  (20)(44 )

f0 

(20)(44)  (20)(22)  440 Hz 2


Xc  C

1 1  C 2f C

1 1 10 -9   2f X c (2 )(2  10 6 )(5  10 3 ) 20

X L  L  2f L

L

f0 

B

XL 300 3  10 -4   2f (2 )(2  10 6 ) 4 1 2 LC



1 3  10 -4 10 -9  2 4 20

 8.165 MHz

 4  R   4.188  10 6 rad / s  (100)   3  10 -4  L


c  2f c 

R

1 RC

1 1   15.91  2f c C (2)(20  10 3 )(0.5  10 -6 )


c  2f c 

R

1 RC

1 1   1.061 k 2f c C (2)(15)(10  10 -6 )


When R s  0 and R L   , we have a low-pass filter.

c  2f c  fc  (b)

1 RC

1 1   994.7 Hz 2RC (2)(4  10 3 )(40  10 -9 )

We obtain R Th across the capacitor. R Th  R L || (R  R s ) R Th  5 || (4  1)  2.5 k fc 

1 1  2R Th C (2 )(2.5  10 3 )(40  10 -9 )

f c  1.59 kHz


H (s) 

Vo R2 ,  Vi R 2  R 1 || 1 jC

s  j

R2 R 2 (R 1  1 sC)  R (1 sC) R 1R 2  (R 1  R 2 )(1 sC) R2  1 R 1  1 sC

H(s) 

R 2 (1  sCR 1 ) R 1  R 2  sCR 1 R 2

Chapter 14, Solution 104. The schematic is shown below. We click Analysis/Setup/AC Sweep and enter Total Points = 1001, Start Frequency = 100, and End Frequency = 100 k. After simulation, we obtain the magnitude plot of the response as shown.

Chapter 15, Solution 1. e at  e - at 2 1 1 1  s L  cosh(at )      2  2  s  a s  a  s  a2

(a)

cosh(at ) 

(b)

sinh(at ) 

e at  e - at 2 1 1 1  a L  sinh(at )      2  2 s  a s  a  s  a2


f ( t )  cos(t ) cos()  sin(t ) sin() F(s)  cos() L  cos(t )   sin() L  sin(t )  s cos()   sin() F(s)  s 2  2

(b)

f ( t )  sin(t ) cos()  cos(t ) sin() F(s)  sin() L  cos(t )   cos() L  sin(t )  s sin()   cos() F(s)  s 2  2


L  e -2t cos(3t ) u ( t )  

s2 (s  2 ) 2  9

(b)

L  e -2t sin(4 t ) u ( t )  

4 (s  2) 2  16

(c)

Since L  cosh(at )  

2

(d)

Since L  sinh(at )  

2

(e)

L  e - t sin( 2t )  

s s  a2 s3 L  e -3t cosh(2 t ) u ( t )   (s  3 ) 2  4

a s  a2 1 L  e -4t sinh( t ) u ( t )   (s  4) 2  1 2 (s  1) 2  4

f (t)   F(s) -d t f (t)   F(s) ds -d -1 Thus, L  t e - t sin(2 t )   2  (s  1) 2  4 ds 2   2 (s  1) ((s  1) 2  4) 2 4 (s  1) L  t e -t sin( 2t )   ((s  1) 2  4) 2

If





Chapter 15, Solution 4. Design a problem to help other students better understand how to find the Laplace transform of different time varying functions. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the Laplace transforms of the following: (a) g(t) = 6cos(4t – 1) (b) f(t) = 2tu(t) + 5e-3(t – 2)u(t – 2) Solution s 6se  s s e  s 2  42 s 2  16

(a)

G (s )  6

(b)

2 e 2 s F(s )  2  5 s3 s


s cos(30)  2 sin(30) s2  4 d 2  s cos(30)  1  L  t 2 cos(2t  30)   2  ds  s 2  4  L  cos(2t  30)  



 -1  d d  3 s  1 s 2  4   ds ds  2  



 3  -1 -2  d  3 2 s  1 s 2  4   s  4  2s  ds  2   2 

 3   3  3  3   2s   (8s 2 )     2 s 1 s 1 - 2s   2   2   2         2  2 2 2 3 2 2 2 2 s 4 s 4 s 4 s 4















 3  (8s 2 )  s  1 - 3s  3s  2  3s  2    2 3 2 2 s 4 s 4





L t



2





(-3 3 s  2)(s  4) 2

s

2

4



3

8  12 cos(2t  30)  



4!





4 3 s  8s 3

s

2

4

2



3

3 s  6s 2  3s 3

 s 2  4 3 72 (s  2 ) 5

(b)

L 3 t 4 e - 2t  3 

(c)

  2 d 2 L  2t u ( t )  4 ( t )   2  4(s  1  0)  2  4s   s s dt

(s  2)

5



(d)

2 e -(t-1) u ( t )  2 e -t u ( t ) 2e L  2 e -(t-1) u ( t )   s1

(e)

Using the scaling property, 1 1 1 5   5 2  L  5 u ( t 2)   5  2s s 1 2 s (1 2)

(f)

L  6 e -t 3 u ( t )  

18 6  s  1 3 3s  1



(g)

Let f ( t )  ( t ) . Then, F(s)  1 .  dn   dn  L  n ( t )   L  n f ( t )   s n F(s)  s n 1 f (0)  s n  2 f (0)    dt   dt 

 dn   dn  L  n ( t )   L  n f ( t )   s n  1  s n 1  0  s n  2  0    dt   dt  n  d  L  n ( t )   s n  dt 

Chapter 15, Solution 6. f(t) = 5t[u(t)–u(t–1)] – 5t[u(t–1)–u(t–2)] = 5[tu(t)–tu(t–1) – tu(t–1) + tu(t–2)] = 5[tu(t) – 2tu(t–1) + tu(t–2)] = 5[tu(t) – 2(t–1)u(t–1) – 2u(t–1) + (t–2)u(t–2) + 2u(t–2)] which leads to F(s) =5[(1/s2) – (1/s2)e–s – (2/s)e–s + (1/s2)e–2s + (2/s)e–2s]


F (s) 

(b) G ( s ) 

2 4  s2 s

4 3  s s2

(c ) H(s )  6

3 s 8s  18 8 2  2 s 9 s 9 s 9 2

(d) From Problem 15.1, s L{cosh at}  2 s  a2 s2 s2 X (s)   2 2 2 ( s  2)  4 s  4s  12 (a )

2 4 4 3 8s  18 s2 , (c ) 2 , (d ) 2  , (b )  2 s s s2 s s 9 s  4s  12

Chapter 15, Solution 8. (a) 2t=2(t-4) + 8 f(t) = 2tu(t-4) = 2(t-4)u(t-4) + 8u(t-4) 2 8  2 8 F ( s )  2 e 4 s  e 4 s   2   e 4 s s s s s 



(b) F ( s )   f (t )e dt   5cos t (t  2)e st dt 5cos te st 0

(c)

 st

0

t2

 5 5cos(2)e cos 2e2 s –2s

e  t  e  ( t  ) e  f (t )  e  e (t  )u (t   ) F ( s )  e  e  s

1 e  ( s 1)  s 1 s 1

(d) sin 2t  sin[2(t   )  2 ]  sin 2(t   ) cos 2  cos 2(t   )sin 2 f (t )  cos 2 sin 2(t   )u (t   )  sin 2 cos 2(t   )u (t   ) 2 s F ( s )  cos 2 e  s 2  sin 2 e  s 2 s 4 s 4


f ( t )  ( t  4) u ( t  2)  ( t  2) u ( t  2)  2 u ( t  2) e -2s 2 e -2s F(s)  2  2 s s

(b)

g( t )  2 e -4t u ( t  1)  2 e -4 e -4(t -1) u ( t  1) 2 e -s G (s)  4 e (s  4)

(c)

h ( t )  5 cos(2 t  1) u ( t ) cos(A  B)  cos(A) cos(B)  sin(A) sin(B) cos(2t  1)  cos(2t ) cos(1)  sin(2t ) sin(1)

h ( t )  5 cos(1) cos(2 t ) u ( t )  5 sin(1) sin(2t ) u ( t ) 2 s  5 sin(1)  2 s 4 s 4 2.702 s 8.415 H(s)  2  s  4 s2  4

H(s)  5 cos(1) 

(d)

2

p( t )  6u ( t  2)  6u ( t  4) P(s) 

6 - 2s 6 -4s e  e s s


By taking the derivative in the time domain, g( t )  (-t e -t  e -t ) cos( t )  t e -t sin( t ) g( t )  e -t cos( t )  t e -t cos( t )  t e -t sin( t ) G (s) 

 s 1 d  s 1  d  1      2 2 2 (s  1)  1 ds  (s  1)  1 ds  (s  1)  1

G (s) 

s 1 s 2  2s 2s  2   2 2 2  2 s  2s  2 (s  2s  2) (s  2s  2) 2 s 2 (s  2) (s 2  2s  2) 2

(b)

By applying the time differentiation property,

G (s)  sF(s)  f (0) where f ( t )  t e -t cos( t ) , f (0)  0 - d  s 1  (s)(s 2  2s)  G (s)  (s)    ds  (s  1) 2  1  (s 2  2s  2) 2

s 2 (s  2) (s 2  2s  2) 2


Since L  cosh(at )  

2

(b)

Since L  sinh(at )  

2

s s  a2 6 (s  1) 6 (s  1)  2 F(s)  2 (s  1)  4 s  2s  3

a s  a2 (3)(4) 12 L  3 e -2t sinh(4t )    2 2 (s  2)  16 s  4s  12

F(s)  L  t  3 e -2t sinh(4t )  

-d  12 (s 2  4s  12) -1  ds 24 (s  2) F(s)  (12)(2s  4)(s 2  4s  12) -2  2 (s  4s  12) 2

(c)

1  (e t  e - t ) 2 1 f ( t )  8 e -3t   (e t  e - t ) u ( t  2) 2 -2t  4 e u ( t  2)  4 e-4t u ( t  2)  4 e-4 e-2(t - 2) u ( t  2)  4 e-8 e-4(t - 2) u ( t  2) cosh( t ) 

L  4 e -4 e -2(t -2) u ( t  2)  4 e -4 e -2s  L  e -2 u ( t ) 4 e -(2s 4) L  4 e -4 e -2(t -2) u ( t  2)  s2 Similarly, L  4 e -8 e -4(t -2) u ( t  2) 

4 e -(2s8) s4

Therefore, 4 e -(2s 4) 4 e -(2s8) e -(2s 6)  (4 e 2  4 e -2 ) s  (16 e 2  8 e -2 ) F(s)    s2 s4 s 2  6s  8

Chapter 15, Solution 12. G(s) 

s2 s2  2 2 2 ( s  2)  4 s  4 s  20


 

(a) tf (t )



d F (s) ds

If f(t) = cost, then F(s)

s s2  1

and -

L (t cos t ) 

d (s 2  1)(1)  s(2s) F(s)  ds (s 2  1) 2 s2  1 (s 2  1) 2

(b) Let f(t) = e-t sin t.

F (s) 

1 1  2 2 ( s  1)  1 s  2s  2

dF ( s 2  2s  2)(0)  (1)(2s  2)  ds ( s 2  2s  2) 2 dF 2(s  1) L (e  t t sin t )    2 ds (s  2s  2) 2

(c )

f (t ) t



 

 F (s)ds s

Let f (t )  sin t , then F ( s )  

 s 2 2

 1 s  sin  t  L  2 ds   tan 1 2     t  s s 



s



s    tan 1  tan 1 2 s 

Chapter 15, Solution 14. Taking the derivative of f(t) twice, we obtain the figures below. f’(t) 5 0

t 2

4

6

-2.5

f’’(t) 5  (t)

0

2.5(t-6)

2

6

-7.5(t-2) f” = 5δ(t) – 7.5δ(t–2) + 2.5δ(t–6) Taking the Laplace transform of each term, s2F(s) = 5 – 7.5e–2s + 2.5e–6s or F(s ) 

5 e 2 s e 6 s  7.5 2  2.5 2 s s s

Please note that we can obtain the same answer by representing the function as, f(t) = 5tu(t) – 7.5u(t–2) + 2.5u(t–6).

Chapter 15, Solution 15. This is a periodic function with T=3. F ( s) F ( s )  1 3 s 1 e To get F 1 (s), we consider f(t) over one period.

f 1 (t)

f 1 ’(t)

5

f 1 ’’(t)

5 5(t)

0

1

t

0

1

t

–5(t-1)

0

1

t

–5(t-1) –5’(t-1)

f 1 ” = 5δ(t) –5δ(t–1) – 5δ’(t–1) Taking the Laplace transform of each term, s2F 1 (s) = 5 –5e–s – 5se–s or F 1 (s) = 5(1 – e–s – se–s)/s2 Hence, F(s) = 5

1  e  s  se  s s 2 (1  e  3s )

Alternatively, we can obtain the same answer by noting that f 1 (t) = 5tu(t) – 5tu(t–1) – 5u(t–1).


f ( t )  5 u ( t )  3 u ( t  1)  3 u ( t  3)  5 u ( t  4)

F(s) 

1  5  3 e -s  3 e - 3 s  5 e - 4 s  s

Chapter 15, Solution 17. Using Fig. 15.29, design a problem to help other students to better understand the Laplace transform of a simple, non-periodic waveshape. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the Laplace transform of f(t) shown in Fig. 15.29. f(t) 2 1 0

1

2

Figure 15.29

t(s) For Prob. 15.17.

Solution Taking the derivative of f(t) gives f’(t) as shown below. f’(t) 2(t)

t -(t-1) – (t-2) f’(t) = 2δ(t) – δ(t–1) – δ(t–2) Taking the Laplace transform of each term, sF(s) = 2 – e–s – e–2s which leads to F(s) = [2 – e–s – e–2s]/s We can also obtain the same answer noting that f(t) = 2u(t) – u(t–1) – u(t–2).


g ( t )  u ( t )  u ( t  1)  2  u ( t  1)  u ( t  2)  3  u ( t  2)  u ( t  3)  u ( t )  u ( t  1)  u ( t  2)  3 u ( t  3) 1 G (s)  (1  e -s  e - 2s  3 e - 3s ) s

(b)

h ( t )  2 t  u ( t )  u ( t  1)  2  u ( t  1)  u ( t  3)  (8  2 t )  u ( t  3)  u ( t  4)

 2t u ( t )  2 ( t  1) u ( t  1)  2 u ( t  1)  2 u ( t  1)  2 u ( t  3)  2 ( t  3) u ( t  3)  2 u ( t  3)  2 ( t  4) u ( t  4)

 2t u ( t )  2 ( t  1) u ( t  1)  2 ( t  3) u ( t  3)  2 ( t  4) u ( t  4) H(s) 

2 2 - 3s 2 - 4 s 2 -s  2 e  2 (1  e -s  e - 3s  e -4s ) 2 (1  e )  2 e s s s s

Chapter 15, Solution 19. Since L ( t )  1 and T  2 , F(s) 

1 1  e - 2s

Chapter 15, Solution 20. Using Fig. 15.32, design a problem to help other students to better understand the Laplace transform of a simple, periodic waveshape. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem The periodic function shown in Fig. 15.32 is defined over its period as sin  t , g (t )   0,

0  t 1 1 t  2

Find G(s).

Figure 15.32 Solution Let

g 1 ( t )  sin(t ), 0  t  1  sin( t )  u ( t )  u ( t  1) 0 < t < 2  sin(t ) u ( t )  sin(t ) u ( t  1)

Note that sin(( t  1))  sin(t  )  - sin(t ) . So, g1 ( t )  sin( t) u(t)  sin( ( t - 1)) u(t - 1) G 1 (s) 

 (1  e -s ) s  2 2

G 1 (s)  (1  e -s ) G (s)   1  e -2s (s 2   2 )(1  e - 2s )


T  2 Let

t   f1 ( t )  1    u ( t )  u ( t  2)  2  t 1 f1 ( t )  u ( t )  u(t)  ( t  2) u ( t  2) 2 2



1 1 e - 2s 2 s  - 1  e -2s F1 (s)     s 2s 2 2s 2 2s 2

F(s) 

F1 (s) 2s  1  e 2 s  1  e -Ts 2s 2 (1  e -2 s )




Let

g1 ( t )  2t, 0  t  1  2 t  u ( t )  u ( t  1)  2t u ( t )  2 ( t  1) u ( t  1)  2 u ( t  1)

2 2 e -s 2 -s G 1 (s)  2  2  e s s s

(b)

G (s) 

G 1 (s) , T 1 1  e -sT

G (s) 

2 (1  e -s  s e -s ) s 2 (1  e -s )

Let h  h 0  u ( t ) , where h 0 is the periodic triangular wave. Let h 1 be h 0 within its first period, i.e.  2t 0  t 1 h 1 (t)    4  2t 1  t  2 h 1 ( t )  2 t u ( t )  2 t u ( t  1)  4u ( t  1)  2 t u ( t  1)  2 ( t  2) u ( t  2) h 1 ( t )  2 t u ( t )  4 ( t  1) u ( t  1)  2 ( t  2) u ( t  2) 2 4 -s 2 e -2s 2 H 1 (s)  2  2 e  2  2 (1  e -s ) 2 s s s s

H 0 (s) 

H(s) 

2 (1  e -s ) 2 s 2 (1  e -2s )

1 2 (1  e -s ) 2  s s 2 (1  e - 2s )


Let

1 0  t 1 f1 ( t )   - 1 1  t  2

f 1 ( t )   u ( t )  u ( t  1)   u ( t  1)  u ( t  2) f 1 ( t )  u ( t )  2 u ( t  1)  u ( t  2)

1 1 F1 (s)  (1  2 e -s  e -2s )  (1  e -s ) 2 s s F1 (s) , T2 (1  e -sT ) (1  e -s ) 2 F(s)  s (1  e - 2s )

F(s) 

(b)

Let h 1 ( t )  t 2  u ( t )  u ( t  2)  t 2 u ( t )  t 2 u ( t  2) h 1 ( t )  t 2 u ( t )  ( t  2) 2 u ( t  2)  4 ( t  2) u ( t  2)  4 u ( t  2) H 1 (s) 

4 4 2 - 2s )  2 e -2s  e -2s 3 (1  e s s s

H 1 (s) , T2 (1  e -Ts ) 2 (1  e -2s )  4s e -2s (s  s 2 ) H(s)  s 3 (1  e - 2s )

H(s) 

Chapter 15, Solution 24. Design a problem to help other students to better understand how to find the initial and final values of a transfer function. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Given that F ( s) 

s 2  10s  6 s ( s  1)2 ( s  2)

Evaluate f(0) and f(  ) if they exist. Solution

0

f(0) = f ()  lim sF ( s )  lim s 0

s 0

s  10s  6 6   3= 3 2 ( s  1) ( s  2) (1)(2) 2

Chapter 15, Solution 25. 5s ( s  1) 5(1  1/ s )  lim 5 s  s  ( s  2)( s  3) s  (1  2 / s )(1  3 / s ) 5s ( s  1) f ()  lim sF ( s )  lim 0 s 0 s  0 ( s  2)( s  3)

(a)

f (0)  lim sF ( s )  lim

(b) F ( s ) 

5( s  1) A B   ( s  2)( s  3) s  2 s  3

5(1) 5(2)  5, B  10 1 1 5 10    f (t )  5e 2t  10e 3t F (s)  s2 s3 A

f(0) = -5 + 10 = 5 f(  )= -0 + 0 = 0


5s 3  3s f (0)  lim sF ( s )  lim 3 5 s  s  s  4 s 2  6 Two poles are not in the left-half plane. f () does not exist

(b)

s 3  2s 2  s s  s  4( s  2)( s 2  2 s  4) 2 1 1  2 s s  lim  0.25 s   2  2 4  1   1   2   s  s s 

f (0)  lim sF ( s )  lim

One pole is not in the left-half plane. f () does not exist


f ( t )  u(t )  2 e -t u(t )

(b)

G (s) 

3 (s  4)  11 11  3 s4 s4

g( t )  3 (t )  11 e -4t u(t ) (c)

4 A B   (s  1)(s  3) s  1 s  3 A  2, B  -2 2 2 H(s)   s 1 s  3

H(s) 

h ( t )  [2 e -t  2 e -3t ] u(t) (d)

12 A B C   2 2  (s  2) (s  4) s  2 (s  2) s4 12 12 B  6, C 3 2 (-2) 2 12  A (s  2) (s  4)  B (s  4)  C (s  2) 2 J (s) 

Equating coefficients : s2 : 0 AC   A  -C  -3

s1 : s0 :

0  6A  B  4C  2A  B   B  -2A  6 12  8A  4B  4C  -24  24  12  12

J (s) 

-3 6 3  2  s  2 (s  2) s4

j( t )  [3 e -4t  3 e -2t  6 t e -2t ] u(t)

Chapter 15, Solution 28. Design a problem to help other students to better understand how to find the inverse Laplace transform. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the inverse Laplace transform of the following functions: 20( s  2) (a) F ( s )  s ( s 2  6s  25) (b) P ( s ) 

6 s 2  36 s  20 ( s  1)( s  2)( s  3)

Solution (a) F ( s ) 

20( s  2) A Bs  C   2 2 s ( s  6s  25) s s  6s  25

20( s  2)  A( s 2  6s  25s )  Bs 2  Cs Equating components, s2 : 0 = A + B or B= - A s: 20 = 6A + C constant: 40 – 25 A or A = 8/5, B = -8/5, C= 20 – 6A= 52/5 8 52 8 24 52  s  ( s  3)   8 8 5 5 F (s)   5 2 5 2   5 2 2 5s ( s  3)  4 5s ( s  3)  4

8 8 19 f (t )  u (t )  e 3t cos 4t  e 3t sin 4t 5 5 5 6 s 2  36 s  20 A B C    ( s  1)( s  2)( s  3) s  1 s  2 s  3 6  36  20 A  5 (1  2)(1  3) 24  72  20 B  28 (1)(1) 54  108  20 C  17 (2)(1)

(b) P ( s ) 

P( s) 

5 28 17   s 1 s  2 s  3

p (t )  (5e  t  28e 2t  17e 3t )u (t )

Chapter 15, Solution 29. V(s) 

2 As  B  ; 2s 2  8s  26  As 2  Bs  2s  26  A  2 and B  6 2 2 s (s  2)  3

V(s) 

2 2(s  2) 2 3   2 2 s (s  2)  3 3 (s  2) 2  3 2

2 v(t) = ( 2  2e  2t cos 3t  e 2 t sin 3t )u(t ), t  0 3


A Bs  C 6 s 2  8s  3   2 F1 ( s )  2 s ( s  2 s  5) s s  2s  5

6 s 2  8s  3  A( s 2  2 s  5)  Bs 2  Cs We equate coefficients. 6=A+B s2 : s: 8= 2A + C constant: 3=5A or A=3/5 B=6-A = 27/5, C=8-2A = 34/5 F1 ( s ) 

3 / 5 27 s / 5  34 / 5 3 / 5 27( s  1) / 5  7 / 5  2   s s  2s  5 s ( s  1) 2  22

7  3 27  f1 (t )    e  t cos 2t  e t sin 2t  u (t ) 10 5 5  s 2  5s  6 A B C    2 2 ( s  1) ( s  4) s  1 ( s  1) s4 2 s  5s  6  A( s  1)( s  4)  B ( s  4)  C ( s  1) 2 Equating coefficients,

(b) F2 ( s ) 

1=A+C s2 : s: 5=5A+B+2C constant: 6=4A+4B+C Solving these gives A=7/9, B= 2/3, C=2/9 F2 ( s ) 

7/9 2/3 2/9   2 s  1 ( s  1) s  4

2 2 7  f 2 (t )   e  t  te  t  e 4t  u (t ) 3 9 9  10 A Bs  C   2 2 ( s  1)( s  4s  8) s  1 s  4 s  8 2 10  A( s  4 s  8)  B ( s 2  s )  C ( s  1) s2 : 0 = A + B or B = -A s: 0=4A+ B + C constant: 10=8A+C Solving these yields (c ) F 3 ( s ) 

A=2, B= -2, C= -6 2 2 s  6 2 2( s  1) 4 F 3 (s)   2    2 2 s  1 s  4s  8 s  1 ( s  1)  2 ( s  1) 2  22 f 3 (t) = (2e–t – 2e–tcos(2t) – 2e–tsin(2t))u(t).


F(s) 

10s A B C    (s  1)(s  2)(s  3) s  1 s  2 s  3

- 10  -5 2 - 20 B  F(s) (s  2) s -2   20 -1 - 30 C  F(s) (s  3) s -3   -15 2 A  F(s) (s  1) s -1 

F(s) 

-5 20 15   s 1 s  2 s  3

f ( t )  (-5 e -t  20 e -2t  15 e -3t )u(t ) (b)

F(s) 

2s 2  4s  1 A B C D   3  2  (s  1)(s  2) s  1 s  2 (s  2) (s  2) 3

A  F(s) (s  1) s -1  -1 D  F(s) (s  2) 3

s  -2

 -1

2s 2  4s  1  A(s  2)(s 2  4s  4)  B(s  1)(s 2  4s  4)  C(s  1)(s  2)  D(s  1) Equating coefficients : s3 : 0 AB   B  -A  1

s2 : s1 : s0 : F(s) 

2  6A  5B  C  A  C   C  2  A  3 4  12A  8B  3C  D  4A  3C  D 4  6A D   D  -2  A  -1 1  8A  4B  2C  D  4A  2C  D  -4  6  1  1 -1 1 3 1   2  s  1 s  2 (s  2) (s  2) 3

t 2 -2t e 2  t2  f ( t )  (-e -t   1  3 t   e - 2t )u(t ) 2 

f(t)  -e - t  e -2t  3 t e -2t 

(c)

F(s) 

s 1 A Bs  C   2 2 (s  2)(s  2s  5) s  2 s  2s  5

A  F(s) (s  2) s -2 

-1 5

s  1  A (s 2  2s  5)  B (s 2  2s)  C (s  2) Equating coefficients : s2 : s1 : s0 : F(s) 

1 5 1  2A  2B  C  0  C   C  1 1  5A  2C  -1  2  1 0 AB   B  -A 

-1 5 1 5 s 1 -1 5 1 5 (s  1) 45   2 2  2 2  s  2 (s  1)  2 s  2 (s  1)  2 (s  1) 2  2 2

f ( t )  (-0.2 e -2t  0.2 e -t cos( 2t )  0.4 e -t sin( 2t ))u(t )


F(s) 

8 (s  1)(s  3) A B C    s (s  2)(s  4) s s  2 s  4

(8)(3) 3 (2)(4) (8)(-1) 2 B  F(s) (s  2) s-2  (-4) (8)(-1)(-3) 3 C  F(s) (s  4) s-4  (-4)(-2) A  F(s) s s 0 

F(s) 

3 2 3   s s2 s4

f ( t )  3 u(t )  2 e -2t  3 e -4t (b)

F(s) 

s 2  2s  4 A B C   2  (s  1)(s  2) s  1 s  2 (s  2) 2

s 2  2s  4  A (s 2  4s  4)  B (s 2  3s  2)  C (s  1) Equating coefficients : 1 A B   B  1  A s2 : 1 - 2  4A  3B  C  3  A  C s : 0 s : 4  4A  2B  C  -B  2   B  -6 A  1 B  7

F(s) 

C  -5 - A  -12

7 6 12   s  1 s  2 (s  2) 2

f ( t )  7 e -t  6 (1  2 t ) e -2t (c)

s2 1 A Bs  C F(s)    2 2 (s  3)(s  4s  5) s  3 s  4s  5 s 2  1  A (s 2  4s  5)  B (s 2  3s)  C (s  3) Equating coefficients : 1 A B   B  1  A s2 :

s1 :

0  4A  3B  C  3  A  C   A  C  -3

s0 :

1  5A  3C  -9  2A   A  5

B  1  A  -4 F(s) 

C  -A  3  -8

4 (s  2) 5 4s  8 5    2 s  3 (s  2)  1 s  3 (s  2) 2  1

f ( t )  5 e -3t  4 e -2t cos(t )


F(s) 

6 (s  1) 6 As  B C    s4 1 (s 2  1)(s  1) s 2  1 s  1

6  A (s 2  s)  B (s  1)  C (s 2  1) Equating coefficients : s2 : 0 AC   A  -C

s1 :

0 AB   B  -A  C

0

6  B  C  2B   B  3

s :

A  -3 ,

F(s) 

B  3,

C3

3 - 3s  3 3 - 3s 3  2   2  2 s 1 s 1 s 1 s 1 s 1

f ( t )  ( 3 e -t  3 sin( t )  3 cos(t ))u(t ) (b)

F(s) 

s e - s s2 1

f ( t )  cos(t   ) u(t   ) (c)

8 A B C D   3  2  s (s  1) s s  1 (s  1) (s  1) 3 A  8, D  -8 F(s) 

8  A (s 3  3s 2  3s  1)  B (s 3  2s 2  s)  C (s 2  s)  D s Equating coefficients : s3 : 0 AB   B  -A s2 :

0  3A  2B  C  A  C   C  -A  B

s1 : s0 :

0  3A  B  C  D  A  D   D  -A A  8, B  8, C  8, D  8

F(s) 

8 8 8 8   2  s s  1 (s  1) (s  1) 3

f ( t )  8  1  e -t  t e -t  0.5 t 2 e -t  u(t )





(a) (3 e-t  3 sin( t )  3 cos(t ))u ( t ) , (b) cos( t  ) u ( t  ) , (c) 8 1  e -t  t e-t  0.5 t 2 e-t u ( t )


s2  4  3 3 F(s)  10  2  11  2 s 4 s 4

f ( t )  11 (t )  1.5 sin( 2t ) (b)

e -s  4 e -2s G (s)  (s  2)(s  4) 1 A B   (s  2)(s  4) s  2 s  4

Let A 1 2

G (s) 

B 1 2

 1 1  e -s  1 1     2 e -2s     s  2 s  4 2 s  2 s  4

g( t )  0.5  e -2(t -1)  e -4(t -1)  u(t  1)  2  e -2(t - 2)  e -4(t - 2)  u(t  2) (c)

Let

s 1 A B C    s (s  3)(s  4) s s  3 s  4

A  1 12 ,

B  2 3,

C  -3 4

1 1 23 3 4  -2s e H(s)      12 s s  3 s  4  1 2  3 h ( t )    e - 3(t - 2)  e -4(t - 2)  u(t  2)  12 3  4


G (s) 

Let

A  2, G (s) 

s3 A B   (s  1)(s  2) s  1 s  2

B  -1 2 1  s 1 s  2

  g( t )  2 e - t  e -2t

F(s)  e -6s G (s)   f ( t )  g( t  6) u ( t  6)

f ( t )   2 e -(t -6)  e -2(t -6)  u(t  6) (b)

Let

G (s) 

A  1 3,

1 A B   (s  1)(s  4) s  1 s  4 B  -1 3

G (s) 

1 1  3 (s  1) 3 (s  4)

g( t ) 

1 -t  e  e -4t  3

F(s)  4 G (s)  e -2t G (s) f ( t )  4 g( t ) u ( t )  g ( t  2) u ( t  2) 4 1 f ( t )   e -t  e -4t  u(t )   e -(t - 2)  e -4(t - 2)  u(t  2) 3 3 (c)

Let

G (s) 

s A Bs  C   2 2 (s  3)(s  4) s  3 s  4

A  - 3 13

s  A (s 2  4)  B (s 2  3s)  C (s  3) Equating coefficients : 0 AB   B  -A s2 : 1 s : 1  3B  C

s0 :

0  4A  3C

A  - 3 13 ,

13 G (s) 

B  3 13 ,

C  4 13

- 3 3s  4  s  3 s2  4

13 g( t )  -3 e -3t  3 cos(2t )  2 sin(2t ) F(s)  e -s G (s)

f ( t )  g( t  1) u ( t  1) 1  - 3 e -3(t-1)  3 cos( 2 (t  1))  2 sin( 2 (t  1)) u(t  1) f (t)  13


X ( s)  3 B  1 6,

C D  1 A B  3  2    s ( s  2)( s  3) s  2 s  3 s s 2

C 1 4,

D  -1 9

1  A (s 3  5s 2  6s)  B (s 2  5s  6)  C (s 3  3s 2 )  D (s 3  2s 2 ) Equating coefficients : 0  ACD s3 : 2 s : 0  5A  B  3C  2D  3A  B  C 1 0  6 A  5B s : 0 s : 1  6B   B  1 6 A  - 5 6 B  - 5 36

19   - 5 36 1 6 1 4  2   X ( s )  3  s s  2 s  3  s 1 3 1 5  x(t)   u(t )  t  e- 2t  e- 3t u(t ) 4 3 2  12 

(b)

A B C  1   2   2 2  s ( s  1)  s s  1 ( s  1)  A  1, C  -1

Y (s)  2

1  A (s 2  2s  1)  B (s 2  s)  C s Equating coefficients : s2 : 0 AB   B  -A s1 : s0 :

0  2A  B  C  A  C   C  -A 1  A, B  -1, C  -1

1 1 1    Y ( s )  2  2   s s  1 ( s  1) 





y( t )  2  2e-t  2t e-t u(t )

(c)

B Cs  D  A Z ( s )  5   2   s s  1 s  6s  10  A  1 10 ,

B  -1 5

1  A (s 3  7s 2  16s  10)  B (s 3  6s 2  10s)  C (s 3  s 2 )  D (s 2  s) Equating coefficients : 0  A BC s3 : 2 0  7 A  6 B  C  D  6 A  5B  D s : 1 s : 0  16A  10B  D  10A  5B   B  -2A

s0 :

1  10A   A  1 10

A  1 10 ,

B  -2A  - 1 5 ,

D  4A 

C  A  1 10 ,

s4 10 1 2 Z ( s)    2 s s  1 s  6s  10 5 s3 1 2 1  2 Z ( s)    2 s s  1 ( s  3)  1 ( s  3) 2  1





z( t )  0.5 1  2 e-t  e-3t cos(t )  e-3t sin(t ) u(t )

4 10

Chapter 15, Solution 37. (a) H ( s ) 

s4 A B   s ( s  2) s s  2

s+4 =A(s+2) + Bs Equating coefficients, s: 1=A+B constant: 4= 2 A

A =2, B=1-A = -1

2 1  s s2 h(t )  2u (t )  e 2t u (t )  (2  e 2t )u (t ) H (s) 

(b)

G ( s) 

A Bs  C  2 s  3 s  2s  2

s 2  4s  5  ( Bs  C )( s  3)  A( s 2  2 s  2) Equating coefficients, s2: 1= B + A (1) s: 4 = 3B + C + 2A (2) Constant: 5 =3C + 2A (3) Solving (1) to (3) gives 2 3 7 A , B , C 5 5 5 0.4 0.6s  1.4 0.4 0.6( s  1)  0.8 G ( s)   2   s  3 s  2s  2 s  3 ( s  1) 2  1 g (t )  0.4e 3t  0.6e t cos t  0.8e  t sin t u(t)

(c) f (t )  e 2(t  4)u (t  4) (d) D( s ) 

10s As  B Cs  D  2  2 2 ( s  1)( s  4) s  1 s 4 2

10 s  ( s 2  4)( As  B)  ( s 2  1)(Cs  D) Equating coefficients, 0=A+C s3 : s2 : 0=B+D s: 10 = 4A + C constant: 0 = 4B+D Solving these leads to A = -10/3, B = 0, C = -10/3, D = 0

10s / 3 10s / 3  s2  1 s2  4 10 10 d (t )  cos t  cos 2t u(t) 3 3 D( s ) 


s 2  4s s 2  10s  26  6s  26 F(s)  2  s  10s  26 s 2  10s  26 6s  26 F(s)  1  2 s  10s  26 6 (s  5) 4 F(s)  1  2 2  (s  5)  1 (s  5) 2  12

f ( t )  (t )  6 e -t cos(5t )  4 e -t sin( 5t ) (b)

F(s) 

5s 2  7s  29 A Bs  C   2 2 s (s  4s  29) s s  4s  29

5s 2  7s  29  A (s 2  4s  29)  B s 2  C s Equating coefficients : 29  29A   A  1 s0 :

s1 : 2

s : A  1,

7  4A  C   C  7  4A  3 5 AB   B  5  A  4

B  4,

C3

4 (s  2) 1 4s  3 1 5 F(s)   2   2 2  s s  4s  29 s (s  2)  5 (s  2) 2  5 2

f ( t )  u(t )  4 e -2t cos(5t )  e -2t sin( 5t )


2s 3  4s 2  1 As  B Cs  D F(s)  2   (s  2s  17)(s 2  4s  20) s 2  2s  17 s 2  4s  20 s 3  4s 2  1  A(s 3  4s 2  20s)  B(s 2  4s  20)  C(s 3  2s 2  17s)  D(s 2  2s  17) Equating coefficients : s3 : 2 AC 2 4  4 A  B  2C  D s : 1 s : 0  20A  4B  17C  2D 0 1  20B  17 D s :

Solving these equations (Matlab works well with 4 unknowns), A  -1.6 , B  -17.8 , C  3.6 , D  21 - 1.6s  17.8 3.6s  21  2 2 s  2s  17 s  4s  20 (-1.6)(s  1) (-4.05)(4) (3.6)(s  2) (3.45)(4) F(s)  2 2  2 2  2 2  (s  1)  4 (s  1)  4 (s  2)  4 (s  2) 2  4 2

F(s) 

f (t)  [ - 1.6 e -t cos(4t )  4.05 e -t sin( 4t )  3.6 e -2t cos(4t )  3.45 e -2t sin( 4t ) ]u(t) (b)

s2  4 As  B Cs  D  2 F(s)  2  2 2 (s  9)(s  6s  3) s  9 s  6s  3 s 2  4  A (s 3  6s 2  3s)  B (s 2  6s  3)  C (s 3  9s)  D (s 2  9) Equating coefficients : s3 : 0 AC   C  -A s2 : 1  6A  B  D 1 0  3A  6B  9C  6B  6C   B  -C  A s : 0 4  3B  9D s : Solving these equations, A  1 12 , B  1 12 ,

12 F(s) 

s 1 -s 5  2 2 s  9 s  6s  3

C  - 1 12 ,

D  5 12

s 2  6s  3  0  

Let

G (s) 

-s5 s  5.449 -s5 F s  0.551

E

G (s) 

- 6  36 - 12  -0.551, - 5.449 2

-s5 E F   s  6s  3 s  0.551 s  5.449 2

s  -0.551

 1.133

s  -5.449

 - 2.133

1.133 2.133  s  0.551 s  5.449

12 F(s) 

s 1 3 1.133 2.133  2  2  2  s 3 3 s 3 s  0.551 s  5.449 2

f (t)  [ 0.08333 cos( 3t )  0.02778 sin( 3t )  0.0944 e -0.551t  0.1778 e -5.449t ]u(t)

Chapter 15, Solution 40.  4s 2  7s  13  A Bs  C  Let H(s)    2 2  (s  2)(s  2s  5)  s  2 s  2s  5 4s 2  7s  13  A(s 2  2s  5)  B(s 2  2s)  C(s  2) Equating coefficients gives: s2 :

4AB

s:

7  2A  2B  C

 

C  1

13  5A  2C

 

5A  15 or A  3, B  1

constant : H(s) 

3 s 1 3 (s  1)  2    s  2 s 2  2s  5 s  2 (s  1) 2  2 2

Hence, h ( t )  3e 2 t  e  t cos 2t  e  t sin 2t  3e 2 t  e  t (A cos  cos 2t  A sin  sin 2t ) where A cos   1,

A sin   1

 

  45 o

A  2,

Thus, h( t ) 





2e  t cos( 2t  45 o )  3e 2t u(t )

Chapter 15, Solution 41. We fold x(t) and slide on y(t). For t<0, no overlapping as shown below. x(t) =0. y(  ) 4

2

4

6

8



t 2

4

6

8



0 t -4

For 0 < t < 2, there is overlapping, as shown below. y(  ) 4

0

-4 t

z (t )   (2)(4)dt  8t 0

For 2 < t < 6, the two functions overlap, as shown below.

y(  ) 4

0

t4

6

8



2

4

6

t 8



2

 8

2

t -4

2

t

0

0

z (t )   (2)(4)d    (2)(4)d   16  8t For 6
0

-4

2

z (t ) 



t 6

6

t

2

6

(2)(4)d    (2)(4)d    (2)(4)d   8

For 8< t <12, they overlap as shown below.

t 6

6 2

 8

t 6

 16

y(  ) 4

0

2

4

6

8

10 t 12 

4

6

8

10

-4

6



z (t ) 

t 6

8

(2)(4)d    (2)(4)d   8 6

6 8  8  8t  80 t 6 6

For 12 < t < 14, they overlap as shown below. y(  ) 4

0

2

-4 8

z (t ) 

8

 (2)(4)d   8 t  6  112  8t

t 6

Hence, z(t) = 8t, 16–8t, –16, 8t–80, 112–8t, 0,

0
12 t 

1 Chapter 15, Solution 42. Design a problem to help other students to better understand how to convolve two functions together. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Suppose that f(t) = u(t) – u(t-2). Determine f(t)*f(t). Solution For 0
1

t-2

0

t



2

t

y (t )  f (t ) * f (t )   (1)(1)d   t 0

For 2 < t< 4, they overlap as shown below.

1

0

2

y (t ) 

t 2

Thus,

2

 (1)(1)d   t t  2  4  t

t-2

2

t



2  t, 0t 2  y (t )  4  t , 2t 4  0, otherwise 


For 0  t  1 , x ( t  ) and h () overlap as shown in Fig. (a). t 2 t t 2  y( t )  x ( t )  h ( t )  0 (1)() d  2 0 2

x(t - )

1

1 h()

t-1

0 t

1



t-1 1

0

(a)

t



(b)

For 1  t  2 , x ( t  ) and h () overlap as shown in Fig. (b). 1 t 2 1 -1 2 t y( t )  t 1 (1)() d  1 (1)(1) d  t  2t  1 t 1   1  2 2 For t  2 , there is a complete overlap so that

y( t )  t 1 (1)(1) d   tt 1  t  ( t  1)  1 t

Therefore,  t 2 2, 0t1  2 - (t 2)  2t  1, 1  t  2 y( t )   1, t2   0, otherwise

(b)

For t  0 , the two functions overlap as shown in Fig. (c). y( t )  x ( t )  h ( t )  0 (1) 2 e - d  -2 e - t

x(t-)

2

t 0

h() = 2e-

1

0

t (c)

Therefore, y( t )  2 (1  e -t ), t  0



(c)

For - 1  t  0 , x ( t  ) and h () overlap as shown in Fig. (d). t 1 2 t 1 1 y( t )  x ( t )  h ( t )  0 (1)() d   ( t  1) 2 2 0 2 x(t - )

1

t-1 -1

h()

t 0

t+1 1

2



(d)

For 0  t  1 , x ( t  ) and h () overlap as shown in Fig. (e). y( t )  0 (1)() d  1 (1)(2  ) d t 1

1

y( t ) 

2 2

1 0

 2  -1 1   2   1t 1  t 2  t  2 2 2 

1

-1 t-1

0 t

1 t+1 2



(e)

For 1  t  2 , x ( t  ) and h () overlap as shown in Fig. (f). y( t )  t 1 (1)() d  1 (1)(2  ) d 1

y( t ) 

2 2

2

1 t 1

 2  -1 1   2   12  t 2  t  2 2 2 

1

0

t-1 1

t (f)

2 t+1



For 2  t  3 , x ( t  ) and h () overlap as shown in Fig. (g).  2 2  2 9 1  y( t )  t 1 (1)(2  ) d  2   t 1   3t  t 2 2 2 2 

1

0

1 t-1 2

t

(g)

Therefore,  (t 2 2)  t  1 2, - 1  t  0  2 - ( t 2 )  t  1 2 , 0  t  2 y( t )   2 (t 2)  3t  9 2, 2  t  3  0, otherwise

t+1




For 0  t  1 , x ( t  ) and h () overlap as shown in Fig. (a). y( t )  x ( t )  h ( t )  0 (1)(1) d  t t

x(t - )

h()

1

0 t

t-1

1

2



-1 (a)

For 1  t  2 , x ( t  ) and h () overlap as shown in Fig. (b). y( t )  t 1 (1)(1) d  1 (-1)(1) d   1t 1   1t  3  2 t 1

t

For 2  t  3 , x ( t  ) and h () overlap as shown in Fig. (c). y( t )  t 1 (1)(-1) d  - 2

2 t 1

1

 t3

1

0

t-1 1

t

2



0

1 t-1

-1

-1 (b)

Therefore, 0t 1  t,  3  2t , 1  t  2  y( t )   2t3  t  3,  0, otherwise

(c)

2 t



(b)

For t  2 , there is no overlap. For 2  t  3 , f1 ( t  ) and f 2 () overlap, as shown in Fig. (d). y( t )  f 1 ( t )  f 2 ( t )  2 (1)( t  ) d t

 2   t  2 

 t t2  2   2 t  2 2 

f 1 (t - )

f 2 ( )

1

1 t-1 2

0

3

t

4

5



4

5



(d)

1

0

2 t-1 3

1

t

(e)

For 3  t  5 , f1 ( t  ) and f 2 () overlap as shown in Fig. (e).  t 2  1 y( t )  t 1 (1)( t  ) d   t   tt 1  2 2  For 5  t  6 , the functions overlap as shown in Fig. (f).  5 2  -1 y( t )  t 1 (1)( t  ) d   t   5t 1  t 2  5t  12 2 2 

1

0

1

2

3 (f)

4 t-1 5

t



Therefore,  (t 2 2)  2t  2, 2t3  1 2, 3t5  y( t )   2 - (t 2)  5t  12, 5  t  6  0, otherwise


y (t )  h(t ) * x(t )   4e 2t u (t )  *  (t )  2e2t u (t )  t

 4e2t u (t ) *  (t )  4e2t u (t ) * 2e2t u (t )  4e2t u (t )  8e2t  eo d  0

2 t

2 t

 4e u (t )  8te u (t )

Chapter 15, Solution 46. (a) x(t ) * y (t )  2 (t ) * 4u (t )  8u (t ) (b) x(t ) * z (t )  2 (t ) * e 2t u (t )  2e 2t u (t ) t

(c ) y (t ) * z (t )  4u (t ) * e 2t u (t )  4  e 2  d   0

4e 2  t  2(1  e2t ) 2 0

(d) y (t ) *[ y (t )  z (t )]  4u (t ) *[4u (t )  e2t u (t )]  4  [4u ( )  e2  u ( )]d  t

 4  [4  e2  ]d   4[4t  0

e 2  t ]  16t  2e2t  2 2 0

Chapter 15, Solution 47. s A B   ( s  1)( s  2) s  1 s  2 s=A(s+2) + B(s+1) We equate the coefficients.

(a) H ( s ) 

s: 1= A+B constant: 0 =2A +B Solving these, A = -1, B= 2.

1 2  s 1 s  2 h(t )  (e  t  2e 2t )u (t ) H (s) 

Y ( s) s 1   Y (s)  H (s) X (s)  ( s  1)( s  2) s X ( s) 1 C D Y (s)    ( s  1)( s  2) s  1 s  2

(b) H ( s ) 

C=1 and D=-1 so that 1 1 Y (s)   s 1 s  2 y (t )  (e  t  e 2t )u (t )


Let G (s) 

2 2  s  2s  5 (s  1) 2  2 2 2

g( t )  e -t sin(2 t )

F(s)  G (s) G (s) f ( t )  L -1  G (s) G (s)  0 g () g ( t  ) d t

f ( t )  0 e - sin(2) e -( t ) sin( 2( t  )) d t

sin(A) sin(B) 

1  cos(A  B)  cos(A  B) 2

1 - t t - e  e  cos(2t )  cos(2( t  2)) d 2 0 t e -t e -t t f (t)  cos(2 t ) 0 e -2 d  0 e -2 cos(2 t  4) d 2 2 -t -t -2  e e t e t -2  f (t)  cos(2 t )    e  cos(2t ) cos(4)  sin(2t ) sin(4) d 2 -2 0 2 0 t 1 e -t f ( t )  e -t cos(2 t ) (-e -2 t  1)  cos(2 t ) 0 e -2  cos(4) d 4 2 -t t e  sin( 2 t ) 0 e -2 sin( 4) d 2 f (t) 

f (t) 

1 -t e cos(2t ) (1  e -2 t ) 4  e -2  e -t - 2cos(4)  4 sin(4) 0t  cos(2t )  2  4  16  

 e -2  e -t - 2sin(4)  4 cos(4) 0t sin(2t )  2  4  16 

e -t e -3t e -t e -3t cos( 2t )  cos( 2t )  cos( 2t )  cos( 2t ) cos(4t ) f (t)  2 4 20 20 e -3t e -t cos( 2t ) sin( 4t )   sin( 2t ) 10 10 e -t e -t  sin( 2t ) sin( 4t )  sin( 2t ) cos(4t ) 20 10

(b)

Let

X(s) 

2 , s 1

Y(s) 

s s4

y( t )  cos(2t ) u ( t )

x ( t )  2 e -t u ( t ) , F(s)  X(s) Y(s)

f ( t )  L -1  X(s) Y (s)  0 y() x ( t  ) d 

f ( t )  0 cos(2)  2 e -(t ) d t

f ( t )  2 e -t 

e cos(2)  2 sin(2) 0t 1 4

2 -t t e  e  cos(2t )  2 sin(2t )  1  5 2 4 2 f ( t )  cos( 2t )  sin( 2t )  e -t 5 5 5 f (t) 

Chapter 15, Solution 49. (a) t*eαtu(t) = e a e ( t ) d t     0 a t

t

a

0

t

e a t 1 e at  2 (a  1)  (e at  1)  2  2 (at  1) u(t) a a a a 0

t

t

0

0

(b) cos t *cos tu (t )   cos  cos(t   )d    {cos t cos  cos   sin t sin  cos }d  t t   1 1 sin 2 t cos  ]  sin t  cos t  [1  cos 2 ]d   sin t  cos  d ( cos  )    cos t[  0 2 2 2 0 0    2

=[ 0.5cos(t)(t+0.5sin(2t)) – 0.5sin(t)(cos(t) – 1)]u(t).

t  0 

Chapter 15, Solution 50. Take the Laplace transform of each term.

s

2

V(s)  s v(0)  v (0)  2  s V(s)  v(0)  10 V(s) 

3s s 4 2

3s s 4 3 3s s  7s (s 2  2s  10) V (s)  s  2  2 s 4 s 4 3 s  7s As  B Cs  D V(s)  2  2  2 2 (s  4)(s  2s  10) s  4 s  2s  10

s 2 V(s)  s  2  2s V(s)  2  10 V(s) 

2

s 3  7s  A (s 3  2s 2  10s)  B (s 2  2s  10)  C (s 3  4s)  D (s 2  4) Equating coefficients : 1 AC   C  1  A s3 : 2 s : 0  2A  B  D 1 7  10A  2B  4C  6A  2B  4 s : 0 0  10B  4D   D  -2.5 B s :

Solving these equations yields 9 12 A , B , 26 26

C

17 , 26

D

- 30 26

1  9s  12 17s  30   2 2  26  s  4 s  2s  10   1  9s 2 s 1 47  6 2  17   V(s)   2 26  s  4 s 4 (s  1) 2  3 2 (s  1) 2  3 2  V(s) 

v( t ) 

9 6 17 47 cos( 2t )  sin( 2t )  e -t cos( 3t )  e -t sin( 3t ) 26 26 26 78

Chapter 15, Solution 51. Taking the Laplace transform of the differential equation yields

s V(s)  sv(0)  v' (0) 5sV(s)  v(0)]  6V(s)  s10 1 s  5s  6V(s)  2s  4  10  s10 1  V(s)  (s 2s1)(s 16 2s)(s24 3) 2

or

2

2

Let V(s) 

A B C   , s 1 s  2 s  3

A  5,

B  0,

C  3

Hence,





v(t) = 5e  t  3e 3 t u(t ) .


s

2

I(s)  s i(0)  i (0)  3  s I(s)  i(0)  2 I(s)  1  0

(s 2  3s  2) I(s)  s  3  3  1  0 I(s) 

s5 A B   (s  1)(s  2) s  1 s  2

A  4, I(s) 

B  -3

4 3  s 1 s  2

i( t )  (4 e -t  3 e -2t ) u(t )

Chapter 15, Solution 53. Transform each term. We begin by noting that the integral term can be rewritten as, t

0 x ()e

 (t  )

d which is convolution and can be written as e–t*x(t).

Now, transforming each term produces, X(s) =

X(s) 

s 2

s 1

s 1 2

s 1





1 s  s  1 1 X(s)   X(s)  2 s 1  s 1  s 1

s 2

s 1



1 2

s 1 x(t) = cos(t) + sin(t).

If partial fraction expansion is used we obtain, x(t) = 1.4142cos(t–45˚). This is the same answer and can be proven by using trigonometric identities.

Chapter 15, Solution 54. Design a problem to help other students to better understand solving second order differential equations with a time varying input. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Using Laplace transform, solve the following differential equation for t >0 d 2i di  4  5i  2e 2t 2 dt dt subject to i(0)=0, i’(0)=2. Solution Taking the Laplace transform of each term gives

2  s 2 I ( s )  si (0)  i '(0)   4  sI ( s )  i (0)  5I ( s )  s2 2  s 2 I ( s)  0  2   4  sI ( s )  0  5I ( s )  s2

2 2s  6 2 s2 s2 2s  6 A Bs  C I ( s)    2 2 ( s  2)( s  4s  5) s  2 s 4s  5 2s  6  A( s 2  4s  5)  B( s 2  2 s) C ( s  2) I ( s)( s 2  4 s  5) 

We equate the coefficients. s2 : 0 = A+ B s: 2= 4A + 2B + C constant: 6 = 5A + 2C Solving these gives A = 2, B= -2, C = -2 I ( s) 

2 2s  2 2 2( s  2) 2  2    2 s  2 s 4 s  5 s  2 ( s  2) 1 ( s  2) 2 1

Taking the inverse Laplace transform leads to: i (t )   2e 2t  2e 2t cos t  2e 2t sin t  u (t )  2e 2t (1  cos t  sin t )u (t )


s

3

Y(s)  s 2 y(0)  s y(0)  y(0)  6  s 2 Y(s)  s y(0)  y(0) s 1  8  s Y(s)  y(0)  (s  1) 2  2 2

Setting the initial conditions to zero gives

(s 3  6 s 2  8s) Y(s) 

Y(s) 

A

s 1 s 2  2s  5

(s  1) A B C Ds  E     2 2 s (s  2)(s  4)(s  2s  5) s s  2 s  4 s  2s  5

1 , 40

B

1 , 20

C

-3 , 104

D

-3 , 65

E

-7 65

Y(s) 

3s  7 1 1 1 1 3 1 1        40 s 20 s  2 104 s  4 65 (s  1) 2  2 2

Y(s) 

3 (s  1) 1 1 1 1 3 1 1 1 4         2 2  40 s 20 s  2 104 s  4 65 (s  1)  2 65 (s  1) 2  2 2

1 3 -4t 3 -t 2  1  y( t )    e - 2t  e  e cos( 2t )  e -t sin( 2t )  u(t ) 104 65 65  40 20 

Chapter 15, Solution 56. Taking the Laplace transform of each term we get:

4  s V(s)  v(0) 

12 V(s)  0 s

 12   4 s  s  V(s)  8 V(s) 

8s 2s  2 4s  12 s  3 2

v( t )  2 cos



3t




Although there is no correct way to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Solve the following integrodifferential equation using the Laplace transform method: t dy (t )  9 y (t )dt  cos 2t , 0 dt

y (0)  1

Solution

Take the Laplace transform of each term.

 s Y(s)  y(0)  9 Y(s)  s

s s 4 2

s2  9  s s2  s  4   Y(s)  1  2  2 s 4 s 4  s 

Y(s) 

s 3  s 2  4s As  B Cs  D   2 2 2 (s  4)(s  9) s  4 s 2  9

s 3  s 2  4s  A (s 3  9s)  B (s 2  9)  C (s 3  4s)  D (s 2  4) Equating coefficients : s0 : 0  9B  4D 1 4  9 A  4C s : 2 s : 1 B D 3 s : 1 AC

Solving these equations gives A  0,

Y(s) 

B  - 4 5,

C  1,

D9 5

-4 5 s9 5 -4 5 95 s  2  2  2  2 2 s 4 s 9 s 4 s 9 s 9 y( t )  [ - 0.4 sin( 2t )  cos( 3t )  0.6 sin( 3t ) ]u(t)

Chapter 15, Solution 58. We take the Laplace transform of each term.

5 4 [ sV ( s )  v(0)]  2V ( s)  V ( s )  s s 5 4 4s [ sV ( s)  1]  2V ( s)  V ( s )    V (s)  2 s s s  2s  5 V ( s) 

( s  1)  5 ( s  1) 2  5 2 2 2 2 2 ( s  1) 2  22 ( s  1)  2 ( s  1)  2 v(t )  (e  t cos 2t  2.5e  t sin 2t )u (t )

Chapter 15, Solution 59. Take the Laplace transform of each term of the integrodifferential equation.

 s Y(s)  y(0)  4 Y(s)  3 Y(s)  s

6 s2

 6  (s 2  4s  3) Y(s)  s   1 s  2  Y(s) 

s ( 4  s) ( 4  s) s  2 (s  2)(s  4s  3) (s  1)(s  2)(s  3)

Y(s) 

A B C   s 1 s  2 s  3

A = –2.5,

Y(s) 

C  -10.5

B = 12,

 2.5 12 10.5   s 1 s  2 s  3

y( t )  [–2.5e–t + 12e–2t – 10.5e–3t]u(t)

Chapter 15, Solution 60. Take the Laplace transform of each term of the integrodifferential equation.

4 4 3 2  s X(s)  x (0)  5 X(s)  X(s)   2 s s  16 s 4s 2s 3  4s 2  36s  64 (2s  5s  3) X(s)  2s  4  2  s  16 s 2  16 2

X(s) 

2s 3  4s 2  36s  64 s 3  2s 2  18s  32  (2s 2  5s  3)(s 2  16) (s  1)(s  1.5)(s 2  16)

X(s) 

A B Cs  D   2 s  1 s  1.5 s  16

A  (s  1) X(s) s -1  -6.235 B  (s  1.5) X(s) s  -1.5  7.329 When s  0 , - 32 B D  A  (1.5)(16) 1.5 16

  D  0.2579

s3  2s 2  18s  32  A (s3  1.5s 2  16s  24)  B (s3  s 2  16s  16)  C (s 3  2.5s 2  1.5s)  D (s 2  2.5s  1.5) Equating coefficients of the s3 terms,

1 A BC   C  -0.0935 X(s) 

- 6.235 7.329 - 0.0935s  0.2579   s 1 s  1.5 s 2  16

x ( t )  - 6.235 e -t  7.329 e -1.5t  0.0935 cos(4t )  0.0645 sin( 4t )

Chapter 16, Solution 61. Solve the following differential equations subject to the specified initial conditions (a) d2v/dt2 + 4v = 12, v(0) = 0, dv(0)/dt = 2 (b) d2i/dt2 + 5di/dt + 4i = 8, i(0) = -1, di(0)/dt = 0 (c) d2v/dt2 + 2dv/dt + v = 3, v(0) = 5, dv(0)/dt = 1 (d) d2i/dt2 + 2di/dt + 5i = 10, i(0) = 4, di(0)/dt = -2 8.29 Solution (a)

Converting into the s-domain we get s2V(s)–sv(0–)–v’(0–)+4V(s) = 12/s = s2V(s)–s0–2+4V(s) or (s2+4)V(s) = 2+12/s = 2(s+6)/s or V(s) = 2(s+6)/[s(s+j2)(s–j2)] = [A/s] + [B/(s+j2)] + [C/(s–j2)] where A = 12/4 = 3, B = 2(–j2+6)/[–j2(–j4)] = 2(6.325–18.43˚)/(8180˚) = 1.5812161.57˚ and C = s(j2+6)/[j2(j4)] = 2(6.32518.43˚)/(8180˚) = 1.5812–161.57˚ v(t) = [3+1.5812e161.57˚e–j2t+1. 5812e–161.57˚ej2t]u(t) volts = [3+3.162cos(2t–161.12˚)]u(t) volts.

(b)

Converting into the s-domain we get s2I(s)–si(0–)–i’(0–)+5sI(s)–5i(0–)+4I(s) = 8/s = s2I(s)–s(–1)–0+5sI(s)–5(–1)+4I(s) or (s2+5s+4)I(s) = (–s–5) + 8/s = –(s2+5s–8)/s I(s) = –(s2+5s–8)/[s(s+1)(s+4)] = [A/s] + [B/(s+1)] + [C/(s+4)] where A = 8/[(1)(4)] = 2; B = –[(–1)2+5(–1)–8]/[(–1)(–1+4)] = 12/(–3) = –4; C = –[(–4)2+5(–4)–8]/[(–4)(–4+1)] = 12/(12) = 1 therefore i(t) = [2–4e–t+e–4t]u(t) amps.

(c)

s2V(s)–sv(0–)–v’(0–)+2sV(s)–2v(0–)+V(s) = 3/s = s2V(s)–s5–1+2sV(s)–2x5+V(s) = (s2+2s+1)V(s) – (5s+11) or (s2+2s+1)V(s) = (5s+11) + 3/s = (5s2+11s+3)/s or V(s) = (5s2+11s+3)/[s(s+1)2] = [A/s] + [B/(s+1)] + [C/(s+1)2] where A = 3; C = [5(–1)2+11(–1)+3]/(–1) = (–3)/(–1) = 3; going back to the original and eliminating the denominators we get 5s2+11s+3 = 3(s2+2s+1)+Bs2+Bs+3s or B = 2, thus, v(t) = [3+2e–t+3te–t]u(t) volts.

(d)

s2I(s)–si(0–)–i’(0–)+2sI(s)–2i(0–)+5I(s) = 10/s = s2I(s)–s(4)–(–2)+2sI(s)–2(4)+5I(s) or (s2+2s+5)I(s) – (4s–2+8) = 10/s or (s2+2s+5)I(s) = (4s2+6s+10)/s or I(s) = (4s2+6s+10)/[s(s+1+j2)(s+1–j2)] = [A/s] + [B/(s+1+j2)] + [C/(s+1–j2)] where A = 10/5 = 2; B = [4(–1–j2)2+6(–1–j2)+10]/[(–1–j2)(–j4)] = [4(1+j4–4)–6–j12+10]/[–8+j4] = [–12+j16–6–j12+10]/(8.944153.43˚) = [–8+j4]/(8.944153.43˚) = 1; C = [4(–1+j2)2+6(–1+j2)+10]/[(–1+j2)(j4)] = [4(1–j4–4)–6+j12+10]/[–8–j4] = [–12–j16–6+j12+10]/(8.944–153.43˚) = [–8–j4]/(8.944153.43˚) = 1 thus i(t) = [2+e–te–j2t+ e–tej2t]u(t) amps = [2+2e–tcos(2t)]u(t) amps.

(a) [3+3.162cos(2t–161.12˚)]u(t) volts, (b) [2–4e–t+e–4t]u(t) amps, (c) [3+2e–t+3te–t]u(t) volts, (d) [2+2e–tcos(2t)]u(t) amps

Chapter16, Solution 1. The current in an RLC circuit is described by d 2i di  10  25i  0 2 dt dt If i(0) = 2 and di(0)/dt = 0, find i(t) for t > 0. Solution Step 1.

Transform the equation into the s-domain and solve for I(s). s2I(s) – (di(0–)/dt) – si(0–) + 10sI(s) – 10i(0–) +25 I(s) = 0 (s2+10s+25)I(s) + [–(di(0–)/dt)–si(0–)–10i(0–)] = 0 (s2+10s+25)I(s) + [–2s–20] = 0 or (s2+10s+25)I(s) = 2(s+10) or I(s) = 2(s+10)/ (s2+10s+25)

Step 2.

Perform a partial fraction expansion and then solve for i(t) in the time domain.

 10  10  10 = –5, repeated roots. 2 I(s) = 2(s+10)/(s+5)2 = A/(s+5) + B/(s+5)2 = (As+A5+B)/(s+5)2 or s2 + 10s + 25 = 0, thus s 1,2 =

A = 2 and 5A+B = 20 or B = 20 – 10 = 10 or I(s) = 2/(s+5) + 10/(s+5)2 or i(t) = [(2+10t)e–5t]u(t) A

Chapter 16, Solution 2. The differential equation that describes the voltage in an RLC network is d 2v dv  5  4v  0 2 dt dt Given that v(0) = 0, dv(0)/dt = 5, obtain v(t).

Solution Step 1.

Transform the equation into the s-domain and solve for V(s). s2V(s) – (dv(0–)/dt) – sv(0–) + 5sV(s) – 5v(0–) +4V(s) = 0 or (s2 + 5s +4)V(s) – 5 = 0 or V(s) = 5/(s2+5s+4)

Step 2.

Perform a partial fraction expansion of V(s) and then solve for v(t). s2 + 5s + 4 = 0, thus s 1,2 =

 5  25  16 = –4, –1. 2

Thus, V(s) = 5/[(s+1)(s+4)] = A/(s+1) + B/(s+4) where A = 5/3 and B = –5/3 Thus, v(t) = [(5/3)e–t – (5/3)e–4t]u(t) V

Chapter 16, Solution 3. The natural response of an RLC circuit is described by the differential equation d 2v dv 2 v 0 2 dt dt for which the initial conditions are v(0) = 20 V and dv(0)/dt = 0. Solve for v(t). Solution

Step 1.

Transform the equation into the s-domain and solve for v(t). s2V(s) – (dv(0–)/dt) – sv(0–) + 2sV(s) – 2v(0–) + V(s) = 0 or (s2+2s+1)V(s) – 20s – 40 = 0 or V(s) = 20(s+2)/(s2+2s+1)

Step 2. Perform a partial fraction expansion and solve for V(s). Inverse transform into the time-domain and solve for v(t).

2 44 = -1, repeated roots. 2 V(s) = 20(s+2)/(s2+2s+1) = 20(s+2)/(s+1)2 = A/(s+1) + B/(s+1)2 s2 + 2s + 1 = 0, thus s 1,2 =

As+A+B = 20s+40 or A = 20 and A+B = 40 = 20 + B or B = 40 – 20 = 20 Thus, v(t) = [(20 + 20t)e-t]u(t) V

Chapter 16, Solution 4. If R = 20 , L = 0.6 H, what value of C will make an RLC series circuit: (a) overdamped, (b) critically damped, (c) underdamped? Solution Step 1.

Since we are working with a series RLC circuit, we can express our values in terms of I(s) and the s equation that multiplies it in the s-domain. From here we can easily find the values that produce over damped, critically damped, and underdamped conditions. Equating the mesh equation we get, RI(s) + LsI(s) + (1/C)I(s)/s – V(s) = 0 or (0.6s + 20 + 1/(Cs))I(s) = V(s) or [s2+(20/0.6)+(1/(0.6Cs)]I(s) = V(s)/0.6 or [ss+(20/0.6)s+1/(0.6C)]I(s) = sV(s)/0.6 The roots for the denominator are s 1,2 =

Step 2.

 (20 / 0.6)  (400 / 0.36)  4 /(0.6C) . 2

To find the values of our roots that produces overdamped, critically damped, and underdamped conditions, we note that when s 1 and s 2 values that produces these values, overdamped is when s 1 and s 2 are real with no complex values critically damped is when s 1 = s 2 underdamped is when both s 1 and s 2 have complex roots and s 1 = s 2 * Now all we need to do is to solve for these conditions. (a)

Overdamped is when [4/(0.6C)] is less than 400/0.36 or C > 4x0.36/(400x0.6) = 6x10-3, or C > 6 mF

(b)

Critically damped is when [4/(0.6C)] is equal to 400/0.36 or C = 4x0.36/(400x0.6) = 6x10-3 = 6 mF

(c)

Underdamped is when [4/(0.6C)] is greater than 400/0.36 or C < 4x0.36/(400x0.6) = 6x10-3 or C < 6mF

Chapter 16, Solution 5. The responses of a series RLC circuit are v C (t) = [30 – 10e–20t + 30e–10t]u(t) V i L (t) = [40e–20t – 60e–10t]u(t) mA where v C (t) and i L (t) are the capacitor voltage and inductor current, respectively. Determine the values of R, L, and C. Solution Step 1.

We can start with the generalized mesh equation for a series RLC network. We can lump the initial conditions (v C (0) = 30–10+30 = 50 volts and i L (0) = 40–60 = 20 amps) with the source in the loop since all we are currently after are the values of R, L, and C. RI(s) + Ls(I(s)+20/s) + (1/C)I(s)/s – 50/s – V(s) = 0 or [s2 + (R/L)s + 1/(LC)]I(s) = (V(s)/L) – 20 + 50/(Ls)

Step 2.

The values of R, L, and C will come from the roots of the denominator s equation. We already know that they are equal to –10 and –20. We note however, that this will give us only two equations. Obviously we need a third, and that will come from knowing the current through the capacitor and the voltage across it.

 (R / L)  (R / L) 2  4 /(LC) = –10, –20 s 1,2 = 2 We can simplify our effort by noting that s 1 + s 2 = –R/L[(1/2)+(1/2)] = –30 or R = 30L. 2 (R / L) 2  4 /(LC) = 10 or (R/L)2 – 4/(LC) = 100. Since (R/L) 2 = 30, we then get 900 – 100 = 4/(LC) or LC = 4/800 = 1/200. Next, s 1 – s 2 =

Now we work with i C (t) = Cdv C (t)/dt or 40e–20t – 60e–10t mA = C[200e–20t – 300e–10t] V or C = 0.2x10–3 = 200 µF. Since LC = 1/200 then L = 1/(200x200x10–6) = 1/0.04 = 25 H. Finally R = 30L = 30x25 = 750 Ω. 750 Ω, 25 H, 200 µF

Chapter 16, Solution 6. Design a parallel RLC circuit that has the characteristic equation .s2+100s+106 = 0. Solution Step 1. Develop a general equation for a parallel RLC circuit with initial conditions lumped into a parallel current source i(t). [Cs + (1/R) + (1/(Ls))]V(s) – I(s) = 0 or [s2+(1/(RC))s+1/(LC)]V(s) = sI(s)/C Step 2. The next step is to equate the unknowns to the parameters in the characteristic equation. This does become a design problem in that we have two equations with three unknowns. We need to pick one of the values so that the other values are realistic. 1/(RC) = 100 and 1/(LC) = 106 or RC = 0.01 and LC = 10–6. We can start with some values of R and see what happens to the values of L and C. R

L

1Ω 10 Ω 100 Ω 1kΩ 10 k Ω 100 k Ω

100 µH 1 mH 10 mH 100 mH 1H 10 H

C 10 mF 1 mF 100 µF 10 µF 1 µF 0.1 µF

We now need to pick reasonable values, R = 10 kΩ, L = 1 H, and C = 1 µF represents an acceptable set since their values are relatively common and inexpensive.

Chapter 16, Solution 7. The step response of an RLC circuit is given by d 2i di  2  5i  10 2 dt dt Given that i(0) = 6 and di(0)/dt = 12, solve for i(t). Solution Step 1.

We start by transforming the equation into the s-domain. We then solve for I(s). s2I(s) – (di(0–)/dt) – si(0–) + 2sI(s) – 2i(0–) + 5 I(s) = 10/s or s2I(s) – (12) – 6s + 2sI(s) – 2x6 + 5I(s) = 10/s = (s2+2s+5)I(s) – 6(s+4) or (s2+2s+5)I(s) = [6(s2+4s)+10]/s or I(s) = [6(s2+4s)+10]/[s(s2+2s+5)]

Step 2. We need to find the roots of (s2+2s+5) and then perform a partial fraction expansion and then transform back into the time domain and realize i(t).  2  4  20 s2 + 2s + 5, has the roots s 1,2 = = –1j2 2 I(s) = [6(s2+4s)+10]/[s(s+1+j2)(s+1–j2)] = [A/s] + [B/(s+1+j2)] + [C/(s+1–j2)]

A = 10/5 = 2; B = [6(1+j4–4–4–j8)+10]/[(–1–j2)(–j4)] = [6(–7–j4)+10]/[–8+j4] = (–32–j24)/[4(–2+j)] = 4(–8–j6)/[4(–2+j)] = 2(–4–j3)/(–2+j) = [2(–4–j3)(–2–j)]/[(–2+j)(–2–j)] = 2(8–3+j6+j4)/5 = 2(1+j2); C = [6(1–j4–4–4+j8)+10]/[(–1+j2)(j4)] = [6(–7+j4)+10]/[–8–j4] = (–32+j24)/[4(–2–j)] = 4(–8+j6)/[4(–2–j)] = 2(–4+j3)/(–2–j) = [2(–4+j3)(–2+j)]/[(–2–j)(–2+j)] = 2(8–3–j6–j4)/5 = 2(1–j2). I(s) = [2/s] + [(2+j4)/(s+1+j2)] + [(2–j4)/(s+1–j2)] or i(t) = [2+4e–t(cos(2t)+2sin(2t))]u(t) A

Chapter 16, Solution 8. A branch voltage in an RLC circuit is described by d2v dv  4  8v  48 2 dt dt If the initial conditions are v(0) = 0 = dv(0)/dt, find v(t). Solution

Step 1.

First we transform the equation into the s-domain. Then we solve for V(s). s2V(s) – (dv(0–)/dt) – sv(0–) + 4sV(s) – 4v(0–) + 8V(s) = 48/s or s2V(s) + 4sV(s) + 8V(s) = 48/s = (s2+5s+8)V(s) or V(s) = 48/[s(s2+4s+8)]

Step 2.

Now we need to solve for the roots of the denominator and perform a partial fraction expansion. Then we can inverse transform the answer back into the time domain.  4  16  32 s2 + 4s + 8 has the roots s 1,2 =  2  j2 thus, 2 V(s) = 48/[s(s+2+j2)(s+2–j2)] = [A/s] + [B/(s+2+j2)] + [C/(s+2–j2)]

where A = 48/4 = 6; B = 48/[(–2–j2)(–j4)] = 48/(–8+j8) = 48(–1–j)/[8(–1+j)( –1–j)] = 6(–1–j)/2 = 3(–1–j); and C = 48/[(–2+j2)(j4)] = 48/(–8–j8) = 48(–1+j)/[8(–1–j)(–1+j)] = 6(–1+j)/2 = 3(–1+j). Therefore, V(s) = [8/s] + [3(–1–j)/(s+2+j2)] + [3(–1+j)/(s+2–j2)] v(t) = [6 – 6e-2t(cos2t + sin2t)]u(t) volts

Chapter 16, Solution 9. A series RLC circuit is described by

Find the response when L = 0.5 H, R = 4 , and C = 0.2 F. Let i(0–) = 1 A and [di(0–)/dt] = 0. Solution Step 1.

First transform the equation into the s-domain. Then solve for I(s). 0.5s2I(s) – 0.5(di(0–)/dt) – 0.5si(0–) + 4sI(s) – 4i(0–) + 5I(s) = 2/s or s2I(s) – s + 8sI(s) – 8 + 10I(s) = 4/s or (s2+8s+10)I(s) = s+8+4/s = (s2+8s+4)/s or I(s) = (s2+8s+4)/[s(s2+8s+10)]

Step 2. Next we need to find the roots of (s2+8s+10) and then perform a partial fraction expansion and then inverse transform back into the time domain. s 1,2 = –1.5505 and –6.45 (s2+8s+2)/[s(s2+8s+10)] = [A/s] + [B/(s+1.5505)] + [C/(s+6.45)] A = 0.4; B = 0.7898; and C = –0.1898 thus, I(s) = [0.4/s] + [0.7898/(s+1.5505)] + [–0.1898/(s+6.45)] and i(t) = [400+789.8e–1.5505t–189.8e–6.45t] mA.

Chapter 16, Solution 10. The step responses of a series RLC circuit are

vc  40  10e2000 t  10e4000t V, t>0 iL(t)  3e2000t  6e4000t mA, t>0 (a) Find C. (b) Determine what type of damping exhibited by the circuit. Solution (a)

iL(t)  iC(t)  C

dvo dt

(1)

dv  2000 x10e2000t  4000 x10e4000t  2 x104(e2000 t  2e4000 t ) dt

(2)

But iL(t)  3[e2000t  2e4000 t ]x10-3 Substituting (2) and (3) into (1), we get

(3)

2 x104 xC  3 x10 3

  C  1.5 x10 7  150 nF

(b) Since s 1 = - 2000 and s 2 = - 4000 are real and negative, it is an overdamped case.

Chapter 16, Solution 11. The step response of a parallel RLC circuit is v =10 + 20e-300t (cos 400t – 2 sin 400t) V, t  0 when the inductor is 50 mH. Find R and C. Solution Step 1.

There are different ways to approach this problem so, we will convert everything into the s-domain and then solve for the unknowns. We should also note that the steady-state voltage is 10 volts, then the circuit is a step input voltage across a parallel combination of a capacitor and an inductor all in series with an output resistor. The nodal equation for this circuit is given by, [(V–10/s)/R] + [(V–0)/(0.05s)] + [(V–0)/(1/sC)] + = 0 or [(1/R)+(1/(0.05s))+sC]V = 10/(Rs) = [(20R+RCs2+s)/(Rs)]V or V = [10/(Rs)][Rs/(RCs2+s+20R)] = 10/[(RCs)(s2+(1/(RC))s+(20/C))]

Step 2.

From the value of v(t) we can determine the value of the roots of the polynomial (s2+(1/(RC))s+(20/C)) = (s+300+j400)(s+300–j400) thus, 20/C = 3002+4002 = 90,000 + 160,000 = 250,000 or C = 20/250,000 = 80 µF and 1/(RC) = 600 or R = 1/(600x80x10–6) = 20.83 Ω.

Chapter 16, Solution 12. Consider the s-domain form of the circuit which is shown below. 1

1/s

I(s)

+ 

1/s

s I(s) 

i( t ) 

1s 1 1  2  1  s  1 s s  s  1 (s  1 2) 2  ( 3 2) 2

 3  t  u(t) A e - t 2 sin  2 3  

2

i( t )  1.155 e -0.5t sin (0.866t )u(t ) A

Chapter 16, Solution 13. Using Fig. 16.36, design a problem to help other students to better understand circuit analysis using Laplace transforms. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find v x in the circuit shown in Fig. 16.36 given v s = 4u(t) V.

Figure 16.36 For Prob. 16.13. Solution s

8/s +

4 s

+ 

Vx 

2

4

4 s  Vx  0  Vx  0  V (4s  8)  (16s  32)  (2s 2  4s)V  s 2 V  0 x x x 8 s s 2 4 s

Vx 

Vx (3s 2  8s  8) 

16s  32 s

  s2 0.25  0.125  0.125 Vx  16  16   2  s s(3s  8s  8) 4 8 4 8 s  j s  j  3 3 3 3  v x  (4  2e (1.3333 j0.9428) t  2e (1.3333 j0.9428) t )u ( t ) V

 2 2  v x = 4  4e  4 t / 3 cos t  3   

  u( t )V 

     

Chapter 16, Solution 14. In the s-domain, the circuit becomes that shown below. 1 I 4 2 s

2

0.2s

We transform the current source to a voltage source and obtain the circuit shown below. 2 1 I 8 4 s

+ _

0.2s

8 4 20 s  40 A B    I s 3  0.2 s s ( s  15) s s  15

40 8 15 x 20  40 52  , B  15 3 15 3 8 / 3 52 / 3 I  s s  15 A

i(t) = [(2.667+17.333e–15t]u(t) A

Chapter 16, Solution 15. For the circuit in Fig. 16.38, calculate the value of R needed to have a critically damped response.

Figure 16.38 For Prob. 16.15. 8.13 Solution Step 1.

Let R||60 = R o . Next, convert the circuit into the s-domain and solve for T(s) = R o + [1/(0.01s)] + 4s = R o + (100/s) + 4s = [(4s2+R o s+100)/s]. Now to solve for the roots that represent a critically damped system. s 1,2 = {-R o ±[(R o )2-4(400)]0.5}/2. The system is critically damped when [(R o )2-4(400)] = 0.

Step 2.

(R o )2 = 1600 or R o = 40. Since R o = [Rx60/(R+60)] = 40 or 60R = 40R+2400 or 20R = 2400 or R = 120 Ω.

Chapter 16, Solution 16. The circuit in the s-domain is shown below. I

2

4I

+ 5

+ _

Vo

1/s

–

I  4I 

But I 

Vo 1/ s

  5 I  sVo

5  Vo 2

 5  Vo  5   sVo  2 

  Vo 

12.5 s 5/ 2

v o (t) = 12.5e–2.5tu(t) V

1


Io

1 s2

2

s

2 s

(-1+-sqrt(1-8))/2 = (-1+-(j2.646))/2 = -0.5+-j(1.3229)    1  2s  2s 1  1   V  2  s  2  1  1  s  s  2  s  s  2  (s  2)(s  0.5  j1.3229)(s  0.5  j1.3229)   s 2 2 Vs s2 Io   2 (s  2)(s  0.5  j1.3229)(s  0.5  j1.3229) (0.5  j1.3229) 2 (0.5  j1.3229) 2 1 (1.5  j1.3229)( j2.646) (1.5  j1.3229)( j2.646)    s  0.5  j1.3229 s  0.5  j1.3229 s2





i o ( t )  e  2 t  0.3779e 90e  t / 2e  j1.3229 t  0.3779e90e  t / 2e j1.3229 t u ( t ) A or





 e  2 t  0.7559e  0.5 t sin 1.3229 t u ( t ) A

  7  2  0.5 t or i o (t) =  e  2t  e sin t  u(t )A   2 7   

Chapter 16, Solution 18. For t<0, v(0) = v s = 20 V For t>0, the circuit in the s-domain is as shown below.

I +

10 s 20 s

100mF  0.1F

+ _

10 

v _

 

1 10  sC s

20

s  2 s 1 10  10 s 20 V  10 I  s 1 I

v(t )  20e  t u (t )

Chapter 16, Solution 19. The switch in Fig. 16.42 moves from position A to position B at t=0 (please note that the switch must connect to point B before it breaks the connection at A, a make before break switch). Find v(t) for t >0. t=0 30  A 4H B + + 10  v(t) 0.25 F 20V  –


First find all the initial conditions and then transform into the s-domain. Since the capacitor is not connected to a circuit, we do not know its initial condition so we can assume it is zero (v(0) = 0). We can find i L (0) by letting the inductor be a short and i L (0) = 20/40 = 0.5 amp. 0.5/s 30 

t=0 V2

A

V1

4s

B 20V

+ 

4/s

10 

[(V 1 –0)/(4/s)] + [(V 1 –V 2 )/(4s)] + (0.5/s) = 0 and [(V 2 –V 1 )/(4s)] + (–0.5/s) + [(V 2 –0)/10] = 0 where V = V 1 . Next, add these together, [sV 1 /4] + [V 2 /10] = 0 or V 2 = –2.5sV 1 . Now we can solve for V 1 and V. Step 2.

[(s/4)+(1/(4s))+(2.5s/(4s))]V 1 = –0.5/s = [(s +2.5s+1)/(4s)]V 1 or V 1 = –0.5(4)/(s2+2.5s+1) = –2/[(s+0.5)(s+2)] = [–1.3333/(s+0.5)]+[1.3333/(s+2)] or v(t) = [–1.3333e–t/2+1.3333e–2t]u(t) volts. 2

Chapter 16, Solution 20. Find i(t) for t > 0 in the circuit of Fig. 8.43.

20V

Figure 16.43 For Prob. 16.20. 8.16 Step 1.

Convert the circuit into the s-domain and write one loop equation noting that i(0) = 0 and v c (0) = 16 volts. 60  I 103/s 40  16/s

+ 

2.5s

[(1000/s)]I +[16/s] + [2.5s]I + [40+60]I = 0 or [(2.5s2+100s+1000)/s]I = –16/s or I = –16/[2.5(s2+40s+400)] = –.64/(s+20)2 Step 2.

I = [A/(s+20)] + [B/(s+20)2] where B = –64 so A = 0. Thus, i(t) = 6.4te–20tu(t) A.

Chapter 16, Solution 21. In the circuit of Fig. 16.44, the switch moves (make before break switch) from position A to B at t = 0. Find v(t) for all t  0.


First we need to find our initial conditions, clearly i(0) = 0 and v(0) = 4x15 = 60 volts. Next we convert the circuit into the s-domain. We can then write a mesh equation and solve for v(t). s/4 I 10  60/s

25/s

+ 

[10+(s/4)+(25/s)]I + 60/s = 0 and V = (25/s)I + 60/s Step 2.

I = –(60/s)/[ 10+(s/4)+(25/s)] = –(60/s){4s/[s2+40s+100] = –240/[(s+2.679)(s+37.32)] = [A/(s+2.679)]+[B/(s+37.32)] where A = –240/(–2.679+37.32) = –6.928 and B = –240/(–37.32+2.679) = 6.928. This now leads to V = (25/s)I+60/s = {(25)(–6.928)/[s(s+2.679)]}+{(25)(6.928)/[s(s+37.32)]}+60/s = {–173.2/[s(s+2.679)]}+{173.2/[s(s+37.32)]}+60/s = [a/s]+[b/(s+2.679)]+[c/(s+37.32)] where a = [–173.2/2.679]+[173.2/37.32]+60 = –64.65+4.641+60 = –0.009 (In practice and theoretically, this term must be equal to be zero since there will be no energy in the circuit at t = ∞!); b = [–173.2/(–2.679)] = 64.65; and c = 173.2/(–37.32) = –4.641 or –4.65 if we correct the rounding errors. Thus, v(t) = [64.65e–2.679t–4.65e–37.32t]u(t) volts.

Chapter 16, Solution 22. Find the voltage across the capacitor as a function of time for t > 0 for the circuit in Fig. 16.45. Assume steady-state conditions exist at t = 0-.


Solution Step 1.

First we need to calculate the initial conditions, v C (0) = 0 and i L (0) = 20/5 = 4 amps. Next we need to convert the circuit into the s-domain and solve for the node voltage V = V C . Convert this back into the time domain and obtain v C (t). V

1

4/s

s/4

1/s

[(V–0)/1]+[4/s]+[(V–0)/(s/4)]+[(V–0)/(1/s)] = 0 then solve for V, next complete a partial fraction expansion, and then convert back into the time domain. Step 2.

[1+(4/s)+s]V = [(s2+s+4)/s]V = –4/s or V = –4/[(s+0.5+j1.9365)(s+0.5–j1.9365)] = [A/(s+0.5+j1.9365)] + [B/(s+0.5–j1.9365)] where A = –4/(–j3.873) = 1.0328–90˚ and B = –4/(j3.873) = 1.032890˚. Thus, v C (t) = 1.0328e–t/2[e–(j1.9365t+90˚)+e(j1.9365t+90˚)]u(t) = 2.066e–t/2cos(1.9365t+90˚)u(t) volts.

Chapter 16, Solution 23. Obtain v(t) for t > 0 in the circuit of Fig. 16.46.

90 V


First we need to calculate the initial conditions. Clearly since the inductor looks like a short, v(0) = 0 and i L (0) = 90/10 = 9 amps. Next we convert the circuit into the s-domain and solve for V and then obtain the partial fraction expansion and convert back into the time domain. + 1/s

V 9/s 

4s

[(V–0)/(1/s)]+(–9/s)+[(V–0)/4s] = 0 Step 2. [s+(1/(4s))]V = 9/s = [(s2+0.25)/(4s)]V or V = 36/[(s+j0.5)(s–j0.5)] = [A/(s+j0.5)] + [B/(s–j0.5)] where A = 36/(–j) = 3690˚ and B = 36/(j) = 36–90˚. Thus, v(t) = 36[e–(j0.5t–90˚)+e(j0.5t–90˚)]u(t) = 18cos(0.5t–90˚)u(t) volts.

Chapter 8, Solution 24. The switch in the circuit of Fig. 16.47 has been closed for a long time but is opened at t = 0. Determine i(t) for t > 0.

Figure 16.47 For Prob. 16.24. 8.20

Solution Step 1.

First we solve for the initial conditions and then convert the circuit into the s-domain and then solve for I, perform a partial fraction expansion, then convert back into the time domain. We recognize that the capacitor becomes an open circuit and the inductor becomes a short circuit at t = 0–. Therefore, v(0) = 12 volts and i(0) = 12/2 = 6 amps. 0.5s 2

I

6/s 4/s + 

12/s We can use mesh analysis, –(12/s) + (4/s)I + (0.5s)(I–6/s) + 2I = 0. Step 2.

[(4/s)+0.5s+2]I = (12/s)+(3) = (3s+12)/s = [(s2+4s+8)/(2s)]I or I = (6s+24)/[(s+2+j2)(s+2–j2)] = [A/(s+2+j2)] + [B/(s+2–j2)] where A = (–12–j12+24)/(–j4) = 16.97–45˚/4–90˚= 4.24345˚and B = (–12+j12+24)/(j4) = 4.243–45˚. Thus, i(t) = 4.243e–2t[e–(j2t–45˚)+e(j2t–45˚)]u(t) = 8.486e–2tcos(2t–45˚) amps.

Chapter 16, Problem 25. Calculate v(t) for t > 0 in the circuit of Fig. 16.48.


Step 1.

First solve for the initial conditions. Then simplify the circuit and then convert it into the s-domain and then solve for v(t). Since the capacitor becomes an open circuit, i L (0) = 0 and v(0) = (24)24/36 = 16 volts. 12 

t=0

6

I

3s 24V

+ 

+

24  V



27/s + 

16/s

We can now use mesh analysis to solve for V(s). (30+3s+27/s)I + 16/s = 0 or [3(s2+10s+9)/s]I = –16/s or I = –(16/3)/[(s+1)(s+9)] and v = (27/s)I + 16/s. Step 2.

V = –(16/3)(27)/[s(s+1)(s+9)] + 16/s = –144/[s(s+1)(s+9)] + 16/s. Thus, V = [A/s]+[B/(s+1)]+[C/(s+9)] where A = –(144/9)+16 = 0 (as expected) and B = –144/[(–1)(–1+9)] = 144/8 = 18 and C = –144/[(–9)(–9+1)] = –144/72 = –2. v(t) = [18e–t–2e–9t]u(t) volts.

Chapter 16, Problem 26. The switch in Fig. 16.49 moves from position A to position B at t=0 (please note that the switch must connect to point B before it breaks the connection at A, a make before break switch). Determine i(t) for t >0. Also assume that the initial voltage on the capacitor is zero. A

t=0 i(t)

B 12 A

20  10 mF

10  0.25H


Determine the initial conditions and then convert the circuit into the sdomain. Then solve for V and then find I. Convert it into the time domain. It is clear from the circuit that i L (0) = 12 A. I V

100/s

10

0.25s

12/s

Applying nodal analysis we get, [(V–0)/(100/s)]+[(V–0)/10]+[(V–0)/(0.25s)]+12/s = 0 where I = [(V–0)/(0.25s)]+12/s. Step 2.

[(s2+10s+400)/(100s)]V = –12/s or V = –1200/[(s+5+j19.365)(s+5–j19.365)] or I = –{4800/[s(s+5+j19.365)(s+5–j19.365)]}+12/s = [A/s]+[B/(s+5+j19.365)]+[C/(s+5–j19.365)] where A = –12+12 = 0, as to be expected; B = –4800/[(–5–j19.365)(–j38.73)] = 4800180˚/[(20–104.48)(38.73–90˚)] = 6.19714.48˚; and C = –4800/[(–5+j19.365)(j38.73)] = 4800180˚/[(20104.48)(38.7390˚)] = 6.197–14.48˚. i(t) = [12.394e–5tcos(19.365t+14.48˚)]u(t) amps.

Chapter 16, Problem 27. Find v(t) for t > 0 in the circuit in Fig. 16.50.

Figure 16.50 For Problem 16.27.

Solution Step 1.

First we need to determine the initial conditions. We note that the source on the right is equal to zero until the switch opens. So, all initial conditions come from the 3-amp source on the left. Since the capacitor looks like an open and the inductor looks like a short we get, v(0) = 3[5x10/(5+10)] = 10 volts and i L (0) = 10/5 = 2 amps. Next we convert the circuit (t > 0) into the s-domain with initial conditions. Then we can solve for V, perform a partial fraction expansion and solve for v(t). 2/s 5 + 1/(4s) V 

+ 

s I

20/s

+ 

10/s

–[10/s]+[1/(4s)]I+[s(I–(2/s))]+5I+[20/s] = 0 and V = [1/(4s)](–I) + [10/s] Step 2.

{[1/(4s)]+s+5}I = {[s2+5s+0.25]/(s)}I = 2–10/s = 2(s–5)/s or I = 2(s–5)/[(s+0.05051)(s+4.949)] and V = {–2(s–5)/[(4s)(s+0.05051)(s+4.949)]}+10/s = {–0.5(s–5)/[s(s+0.05051)(s+4.949)]}+10/s V = [A/s] + [B/(s+0.05051)] + [C/(s+4.949)] where

A = [2.5/[(0.05051)(4.949)]]+10 = 20; B = –0.5(–0.05051–5)/[(–0.05051)(–0.05051+4.949)] = 2.52525/(–0.24742) = –10.206; and C = –0.5(–4.949–5)/[(–4.949)(–4.949+0.05051)] = 4.9745/24.243 = 0.2052. v(t) = [20–10.206e–0.05051t+0.2052e–4.949t]u(t) volts. dv/dt = –10.206(–0.05051)+0.2052(–4.949) = 0.5155–1.0155 = –0.5 or Cdv/dt = –4x0.5 = –2 amps, the answer checks!

Chapter 16, Problem 28. For the circuit in Fig. 16.51, find v(t) for t > 0. 4u(–t) A

100u(t) V


Determine the initial conditions (at t = 0, the 4 amp current source turns off and the 100 volt voltage source becomes active). Since the capacitor becomes an open circuit, i L (0) = 0 and v(0) = –4x6 = –24 volts. Now convert the circuit into the s-domain and solve for V and then convert it into the time domain to obtain v(t), 25/s 24/s s  +

+ 4

V



I

2

+ 

100/s Now for the mesh equation, [4+s+(25/s)+2]I–(24/s)–(100/s) = 0. V = (25/s)I–24/s. Step 2.

[(s2+6s+25)/s]I = 124/s or I = 124/(s2+6s+25) = 124/[(s+3+j4)(s+3–j4)] thus, V = {3100/[s(s+3+j4)(s+3–j4)]}–24/s = [A/s]+[B/(s+3+j4)]+[C/(s+3–j4)] where A = (3100/25)–24 = 124–24 = 100; B = 3100/[(–3–j4)(–j8)] = 3100/[(5–126.87˚)(8–90˚)] = 77.5–143.13˚; and C = 3100/[(j8)(–3+j4)] = 3100/[(5126.87˚)(890˚)] = 77.5143.13˚. v(t) = [100+155e–3tcos(4t+143.13˚)]u(t) volts.

Chapter 16, Problem 29. Calculate i(t) for t > 0 in the circuit in Fig. 16.52.

20u(–t) V

5


Calculate the initial conditions and then convert the above circuit into the s-domain. Then solve for I, perform a partial fraction expansion, and convert into the time domain. v(0) = 20 volts and i(0) = 0. 16/s + 

I

20/s 0.25s

[16/s]I + [20/s] + 0.25sI = 0. Step 2.

{[16/s]+0.25s}I = –20/s = {[s2+64]/(4s)}I or I = –80/[(s+j8)(s–j8)] or I = [A/(s+j8)] + [B/(s–j8)] where A = –80/(–j16) = 5–90˚ and B = –80/(j16) = 590˚ thus, i(t) = [5e–j8t–90˚+5ej8t+90˚]u(t) = 10cos(8t+90˚)u(t) amps.

Chapter 16, Solution 30. The circuit in the s-domain is shown below. Please note, i L (0) = 0 and v o (0) = o because both sources were equal to zero for all t<0. 1

1 Vo

V1 2/s

+ _

s

2/s 1/s

At node 1 [(V 1 –2/s)/1] + [(V 1 –0)/s] + [(V 1 –V 2 )/1] = 0 or [1+(1/s)+1]V 1 –V 2 = 2/s or [(2s+1)/s]V 1 –V 2 = 2/s At node o [(V o –V 1 )/1] + [(V o –0)/(2/s)] – (1/s) = 0 or –V 1 +[(s+2)/2]V o = 1/s In matrix form we get,

or

s2+1.5s+1 = (s+0.75+j0.6614)(s+0.75–j0.6614) V o = s[(2/s)+(2s+1)/s2]/[(s+0.75+j0.6614)(s+0.75–j0.6614)] = (4s+1)/[s(s+0.75+j0.6614)(s+0.75–j0.6614)] = [A/s] + [B/(s+0.75+j0.6614)] + [C/(s+0.75–j0.6614] where A = 1; B = [4(–0.75–j0.6614)+1]/[(–0.75–j0.6614)(–j1.3228)] = [–3–j2.6456+1]/[(1–138.59˚)(1.3228–90˚)] = (3.3165–127.09˚)/[(1–138.59˚)(1.3228–90˚)] = 2.507101.5˚ C = [4(–0.75+j0.6614)+1]/[(–0.75+j0.6614)(j1.3228)] = [–3+j2.6456+1]/[(1138.59˚)(1.322890˚)] = (3.3165127.09˚)/[(1138.59˚)(1.322890˚)] = 2.507–101.5˚

Therefore, v o (t) = [1+2.507e–0.75te–(j0.6614t–101.5˚)+2.507e–0.75te(j0.6614t–101.5˚)]u(t) volts or = [1+5.014e–0.75tcos(0.6614t–101.5˚)]u(t) volts. An alternate solution is, Vo 

(4 s  1) A Bs  C   2 s ( s  1.5s  1) s s  1.5s 1 2

4 s  1  A( s 2  1.5s  1)  Bs 2  Cs

We equate coefficients. s2 : 0 = A+ B or B = –A s: 4=1.5A + C constant: 1 = A, B=–1, C = 4–1.5A = 2.5 3.25 7 x 4 7  s  2.5 s  3/ 4 1 1 4 where Vo   2    2 2 s s  1.5s  1 s  7  7 2 2 ( s  3 / 4)    ( s  3 / 4)     4   4  This now leads to, v o (t) = [1–e–3t/4cos(0.6614t)+4.914e–3t/4sin(0.6614t)]u(t) volts.

= 0.6614.

Chapter 16, Solution 31. Obtain v(t) and i(t) for t > 0 in the circuit in Fig. 16.54.


First, find the initial conditions and then transform the above circuit into the s-domain after converting the current source in parallel with the 5-ohm resistor into a 15 volts voltage source in series with a 5-ohm resistor. Then solve for V and I, perform a partial fraction expansion on each and then convert back into the time domain. The steady state the values are i(0) = 0 and v(0) = 20 volts. 5 I

1

5s

+

5/s 15/s

+ 

V 20/s

2

20/s

+ 



+ 

–[15/s]+5I+(5s)I+1I+[5/s]I+[20/s]–[20/s]+2I = 0 and V = {[5/s]I+[20/s]. Step 2.

{5+[5s]+1+[5/s]+2}I = [15/s]–[20/s]+[20/s] = 15/s or {5[s2+1.6s+1]/s}I = 15/s or I = 3/[(s+0.8+j0.6)(s+0.8–j0.6)] and V = {[5/s]I+[20/s] = 15/[s(s+0.8+j0.6)(s+0.8–j0.6)] + 20/s = [A/s] + [B/(s+0.8+j0.6)] + [C/(s+0.8–j0.6)] where A = [15/(0.64+0.36)]+20 = 35; B = 15/[(–0.8–j0.6)(–j1.2)] = 12.590˚/1–143.13˚ = 12.5–126.87˚; and C = 15/[(–0.8+j0.6)(j1.2)] = 12.5–90˚/(1143.13˚ = 12.5126.87˚.

v(t) = [35+12.5e–0.8t–j0.6t–126.87˚+12. 5e–0.8t+j0.6t+126.87˚]u(t) volts = [35+25e–0.8tcos(0.6t+126.87˚)]u(t) volts. I = 3/[(s+0.8+j0.6)(s+0.8–j0.6)] = [A/(s+0.8+j0.6)] + [B/(s+0.8–j0.6)] where A = 3/(–j1.2) = 2.590˚ and B = 3/(j1.2) = 2.5–90˚. Thus, i(t) = 2.5e–0.8t[e–j0.6t+90˚+ej0.6t–90˚]u(t) = 5e–0.8t[cos(0.6t–90˚)]u(t) amps.

Chapter 16, Solution 32. For the network in Fig. 16.55, solve for i(t) for t > 0.


First we need to find all the initial conditions. Then we need to transform the circuit into the s-domain and solve for I. We then perform a partial fraction expansion and convert the results into the time domain. The inductor becomes a short and the capacitor becomes an open circuit. Thus, i(0) = [20/6] + [20/12] = 5 amps and v C (0) = 10 + 10 = 20 volts.

4 I 8/s 0.5s

5/s

10/s

20/s

+ 

+ 

Loop equation, –[10/s]–0.5s(I–5/s)–4I–[8/s]I+[20/s] = 0. Step 2.

[0.5s+4+8/s]I = [(s2+8s+16)/(2s)]I = –[10/s]+2.5+20/s = (s+4)/(0.4s) or I = 5(s+4)/[(s+4)2] = [A/(s+4)]+[B/(s+4)2] where A(s+4)+B = 5(s+4) or A = 5 and B = 0. Therefore, i(t) = [5e–4t]u(t) amps.

Chapter 16, Solution 33. Using Fig. 16.56, design a problem to help other students to better understand how to use Thevenin’s theorem (in the s-domain) to aid in circuit analysis. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Use Thevenin’s theorem to determine v o (t), t>0 in the circuit of Fig. 16.56. 1

1H +

10e-2t u(t) V

0.25 F

+ _

vo 2 –

Figure 16.56

For Prob. 16.33.

Solution 1H   1s and i L (0) = 0 (the sources is zero for all t<0). 1 1 4    and v C (0) = 0 (again, there are no source F 4 sC s contributions for all t<0).

To find Z Th , consider the circuit below. 1

s

Z Th 2

s2 s3 To find V Th , consider the circuit below. ZTh  1//( s  2) 

1

s

+ 10 s2

+ _

V Th

2

-

s  2 10 10   s3 s2 s3 The Thevenin equivalent circuit is shown below VTh 

Z Th +

V Th

+ _

4/s

Vo –

40 4 3 10 40 3 s . Vo  VTh     4 4 s  2 s  3 s 2  6s  12 ( s  3)3  ( 3) 2  ZTh  s s s3 4 s

vo (t )  23.094e 3t sin 3t V

Chapter 16, Solution 34. In the s-domain, the circuit is as shown below.

1

10 s

+ _

I1

4

1 H 4

I2

10 s 1  (1  ) I1  sI 2 s 4 4 1 5  sI1  I 2 (4  s)  0 4 4 In matrix form,

1s

(1)

(2)

1   s  s  10  1   I1  1 2 9 4  s  4     s  s4    1 I 5 4 4  0   s 4  s   2   4 4  10 1 s 10 1  s 5 40 50 4 s s 4 1    2   1 5 2 s 4  s 0 0 4 s 4 4 40 25  1 50s  160 s 2 I1    2  0.25s  2.25s  4 s(s 2  9s  16) I2 

2 2.5 10   2 2  0.25s  2.25s  4 s  9s  16

Chapter 16, Solution 35. Find v o (t) in the circuit in Fig. 16.58.


First we note that the initial condition on the capacitor and inductor must be equal to zero since the circuit is unexcited until t = 0. Next we transform the circuit into the s-domain. s

10/(s + 1)

+ 

Vo

1/(2s)

4

3/s

We then can solve for V o using nodal analysis.

Finally we solve for V o , perform a partial fraction expansion and then convert into the time-domain.

Step 2. Next

where A = B=

C=

=

=3.9306 –117.553°

Thus, v o (t) =

or .

Chapter 16, Solution 36. Refer to the circuit in Fig. 16.59. Calculate i(t) for t > 0. 6(1–u(t)) A


First we need to determine the initial conditions and then transform the circuit into the s-domain. 6A + i 10  v

i1 5



10 

Clearly i = 6 A. The current then travels through the parallel combination of the 10 ohm resistor and the combined 15 ohm resistance. i 1 = 6[(15)(10)/(15+10)]/(15) = 2.4 A. Therefore, v(0) = 5x2.4 = 12 V and i(0) = 6 A. We also note that the two 10 ohm resistors are in series and the combination is in parallel with the 5 ohm resistor resulting in a 100/25 = 4 ohm resistor. The circuit in the s-domain is shown below. –[12/s] + [3/s]I + [0.75s](I–6/s) + 4I = 0.

I 3/s 0.75s 7/s

+ 

6/s

I 4

Step 2.

[(s2+5.333s+4)/(4s/3)]I = 4.5+12/s = 4.5(s+2.667)/s or I = 6(s+2.667)/[(s+0.903)(s+4.43)] = [A/(s+0.903)]+[B/(s+4.43)] where A = 6(–0.903+2.667)/(–0.903+4.43) = 6x1.764/3.527 = 3.001 and B = 6(–4.43+2.667)/(–4.43+0.903) = 6x(–1.763)/(–3.527) = 2.999 or i(t) = [3.001e–0.903t+2.999e–4.43t]u(t) amps.

Chapter 16, Solution 37. Determine v(t) for t > 0 in the circuit in Fig. 16.60. 30 

+ v  20 

+ 

60u(t)V

0.25H

0.5F

+ 

30u(t)V


First we need to determine the initial conditions for this circuit. Since both sources were zero (shorts) until t = 0, the initial conditions for this circuit are equal to zero (v(0) = 0 and i L (0) = 0). Next we transform the circuit into the sdomain. Then we can write node equations and then solve for V. Then perform a partial fraction expansion and convert back into the time domain. 30 

60/s

s/4

2/s + V  20 

+ 

+ 

30/s

From this circuit there are different ways of solving for v(t). Perhaps the easiest is to replace the circuit seen by the capacitor and inductor with a Thevenin equivalent circuit. V Thev = [(60/s)/(30+20)]20 – 30/s = (24/s)–(30/s) = –6/s and R eq = 20x30/(20+30) = 12 Ω. Thus we now have the following circuit where we can now find I. Once we have I we can find V and then perform a partial fraction expansion and then convert into the time domain to solve for v(t). 12 

–6/s

+ 

2/s + V 

s/4

I

–[–6/s] + 12I + [2/s]I + [s/4]I = 0 and V = [2/s]I. Step 2.

[(s/4)+12+(2/s)]I = –6/s = [(s2+48s+8)/(4s)]I or

I = (–6/s)(4s)/ [(s2+48s+8)] = –24/[(s+0.165)(s+47.84)] and V = –48/[s(s+0.1672)(s+47.84)] = [A/s]+[B/(s+0.1672)]+[C/(s+47.84)] where A = –48/[0.1672x47.84] = 6; B = –48/[–0.1672(–0.1672+47.84)] = 6.022; and C = –48/[–47.84(–47.84+0.1672)] = –0.021 Therefore, v(t) = [–6+6.022e–0.1672t–0.021e–47.84t]u(t) volts.

Chapter 16, Solution 38. The switch in the circuit of Fig. 16.61 is moved from position a to b (a make before break switch) at t = 0. Determine i(t) for t > 0.


We first determine the initial conditions. We assume that v C (0) = 0 since we are not given otherwise. i(0) = [4(2x6)/(2+6)]/2 = 3 amps. Next we need to convert the circuit for t > 0 into the s-domain converting the current source in parallel with the 6Ω into a voltage source in series with 6Ω. 3/s 3/s 50/s 50/s 14  I I 2s 6

24/s

12/s

+ 

+ 

I

2s 20 

12/s + 

Using the simplified circuit on the right, 2s(I–3/s) + [50/s]I –(12/s) +20I = 0. Now we solve for I, perform a partial fraction expansion, and then convert into the time domain. Step 2.

[2s+(50/s)+20]I = 6+12/s = [(s2+10s+25)/(0.5s)]I = 6(s+2)/s or I = [3(s+2)/(s+5)2] = [A/(s+5)] + [B/(s+5)2] where As+5A+B = 3s+6 or A = 3 and B = –5A+6 = –15+6 = –9. Thus, i(t) = [(3–9t)e–5t]u(t) amps.

Chapter 16, Solution 39. For the network in Fig. 16.62, find i(t) for t > 0.

50 V


First determine the initial conditions at t = 0. Clearly i(0) = 0 and v(0) = [50/(20+5+5)]5 = 25/3 volts. Next simplify and convert the circuit for t > 0 into the s-domain. s 4 I 10/s

+ 

25/s

I + 

25/(3s)

–[10/s] + [4+s+(25/s)]I + [25/(3s)] = 0 Now we need to solve for I, perform a partial fraction expansion, and then convert into the time domain. Step 2.

[(s2+4s+25)/s]I = [10/s] – 25/(3s) = [5/(3s)] or I = 1.6667/[(s+2+j4.583)(s+2–j4.583)] = [A/(s+2+j4.583)]+[B/(s+2–j4.583)] where A = 1.6667/(–j9.166) = 0.1818290˚ and B = 0.18182–90˚. Therefore, i(t) = [0.18182e–2t(e–j4.583t+90˚+ej4.583t–90˚)]u(t) amps or = [363.6e–2tcos(4.583t–90˚)]u(t) mA.

Chapter 16, Solution 40. Given the network in Fig. 16.63, find v(t) for t > 0.


First we determine initial conditions and then simplify the circuit and then transform it into the s-domain. Just before the switch closes, the capacitor is an open circuit (i L (0) = 0) with v(0) = 4 –12 = –8 volts. 6

 +

s

+

25/s 12/s 8/s

V

 +



We can write a node equation at V and then solve for V. Then we perform a partial fraction expansion and then solve for v(t). [(V–(–12/s))/(s+6)]+[(V–(–8/s))/(25/s)] = 0. Step 2.

[(1/(s+6))+s/25]V = [(s2+6s+25)/(25(s+6))]V = –[12/(s(s+6))]–[8/25] = –[(12+0.32s2+1.92s)/(s(s+6))] = –0.32[(s2+6s+37.5)/(s(s+6))] or V = –8[(s2+6s+37.5)/(s(s+3+j4)(s+3–j4))] = [A/s]+[B/(s+3+j4)]+[C/(s+3–j4)] where A = –8[37.5/25] = –12; B = –8[((–3–j4)2+6(–3–j4)+37.5)/((–3–j4)(–j8)] = –8[(–7+j24–18–j24+37.5)/(–32+j24)] = –8[(12.5)/(40143.13˚)] = 2.536.87˚; and

C = –8[((–3+j4)2+6(–3+j4)+37.5)/((–3+j4)(j8)] = –8[(–7–j24–18+j24+37.5)/(–32–j24)] = –8[(12.5)/(40–143.13˚)] = 2.5–36.87˚ V = [–12/s] + [2.536.87˚/(s+3+j4)] + [2.5–36.87˚/(s+3–j4)] or v(t) = [–12+2.5e–3t(e–j4t+36.87˚+ej4–36.87˚)]u(t) amps = [–12+5e–3t(cos(4t–36.87˚)]u(t) volts.

Chapter 16, Solution 41. Find the output voltage v o (t) in the circuit of Fig. 16.64.


First we need to determine the initial conditions. We see that v o (0) = 0 since the inductor becomes a short. We also note that the initial current through the inductor is the same as the current through the 10 Ω resistor or i L (0) = [3(5x10)/(5+10)]/10 = 1 amp. Then we simplify the circuit and convert it into the s-domain and solve for V o . We then perform a partial fraction expansion and convert into the time domain. Vo

3/s

5

100/s

s

–[3/s] + [(V o –0)/5] + [(V o –0)/(100/s)] + [(V o –0)/s] + [1/s] = 0. Step 2.

[0.2+(s/100)+(1/s)]V o = 2/s = [(s2+20s+100)/(100s)]V o or V o = 200/[(s+10)2] and v o (t) = [200te–10t]u(t) volts.

1/s

Chapter 16, Solution 42. Given the circuit in Fig. 16.65, find i(t) and v(t) for t > 0.

12 V


First we need to find the initial conditions. Since the inductor becomes a short and the capacitor becomes an open circuit, all the current flows through the 1 Ω and 2 Ω resistors or i(0) = –12/3 = –4 amps and v(0) = 4x1 = 4 volts. Next we need to convert the circuit into the s-domain and solve for V and I. Once we have done that, we can perform partial fraction expansions and convert back into the time domain. V

4/s s

1 4/s

4/s

+ 

[(V–0)/1] + [(V–4/s)/(4/s)] + [(V–0)/s] – [4/s] = 0 and I = [(V–0)/s] – [4/s]. Step 2.

[(1+(s/4)+(1/s)]V = [(s2+4s+4)/(4s)]V = 1+4/s = (s+4)/s or V = (4s+16)/[(s+2)2] = [A/(s+2)]+[B/(s+2)2] where As+2A + B = 4s+16 and A = 4 and B = 16–2A =8. I = [4/(s(s+2))]+[8/(s(s+2)2)]–4/s

The partial fraction expansion is straight forward for the first and third terms, but the second term takes a little work. 8/(s(s+2)2 = [a/s]+[b/(s+2)]+[c/(s+2)2] or as2+a4s+a4+bs2+b2s+cs = 8 or a = 2, b = –2, and c = –4. Thus, I = [2/s]+[–2/(s+2)]+[2/s]+[–2/(s+2)]+[–4/(s+2)2]–4/s = –[4/(s+2)] – [4/(s+2)2] and we finally get, v(t) = [4e–2t + 8te–2t]u(t) volts and i(t) = [–4e–2t – 4te–2t]u(t) amps.

Chapter 16, Solution 43. Determine i(t) for t > 0 in the circuit of Fig. 16.66.


First we need to determine the initial conditions. Then we need to transform the circuit into the s-domain. Once in the s-domain we can calculate V and I. We then perform a partial fraction expansion on I and convert back into the time domain. Since the inductor looks like a short just before the switch opens, v C (0) = 0 and i(0) = (12/4)+3 = 6 amps. V I

5s

6/s

20/s

5

3/s

[(V–0)/(5s)] + [6/s] + [(V–0)/(20/s)] + [(V–0)/5] – [3/s] = 0 and I = [(V–0)/(5s)] + [6/s]. Step 2.

[(1/(5s))+(s/20)+(1/5)]V = [(s2+4s+4)/(20s)]V = –3/s or V = –60/(s+2)2 and I = –[12/(s(s+2)2)] + 6/s = [A/s] + [B/(s+2)] + [C/(s+2)2] + (6/s) where A = –3 and A(s2+4s+4) + B(s2+2s) + Cs = –12 = –3s2 – 12s –12 + Bs2 + B(2s) + Cs or –3 B = 0 or B = 3 and –12 + 6 + C = 0 or C = 6. i(t) = [3+3e–2t+6te–2t]u(t) amps.

Chapter 16, Solution 44. For the circuit in Fig. 16.67, find i(t) for t > 0.


First we identify the initial conditions. Then we simplify the circuit (for t > 0) and then transform it into the s-domain. We then solve for the node voltage, V, and then find I. Finally we perform a partial fraction expansion and convert the answer into the time domain. For t < 0, the inductor looks like a short circuit producing v C (0) = 0 and i(0) = 30/10 = 3 amps. V I 9/s

100/s

8

4s

3/s

–[9/s] + [(V–0)/(100/s)] + [(V–0)/8] + [(V–0)/(4s)] + [3/s] = 0 and I = [(V–0)/(4s)] + [3/s]. Step 2.

[(s/100)+(1/8)+1/(4s)]V = [(s2+12.5s+25)/(100s)]V = 6/s or V = 600/[(s+2.5)(s+10)] and I = 150/[s(s+2.5)(s+10)]+[3/s] = [A/s] + [B/(s+2.5)] + [C/(s+10)] where A = 6+3 = 9; B = 150/[–2.5(–2.5+10)] = –8; and C = 150/[–10(–10+2.5)] = 2. i(t) = [9–8e–2.5t+2e–10t]u(t) amps.

Chapter 16, Solution 45. Find v(t) for t > 0 in the circuit in Fig. 16.68.


First, determine the initial conditions. Next convert the circuit into the sdomain and solve for V. Perform a partial fraction expansion and convert back into the time domain. For t < 0, the inductor looks like a short circuit so that v(0) = 0 and i L (0) = i o . V

1/(Cs)

Ls

i o /s

[(V–0)/(1/(Cs))] + [(V–0)/(Ls)] + [i o /s] = 0. Step 2.

[Cs+(1/(Ls))]V = [C{s2+(1/(LC))}/s]V = –i o /s or V = –(i o /C)/[(s2+1/(LC))]. If we let ω2 = 1/(LC) then we get, V = –(i o /C)/[(s2+ω2)] = –(i o /C)/[(s+jω)(s–jω)] = [A/(s+jω)]+[B/(s–jω)] where A = –(i o /C)/(–j2ω) = [i o /(2ωC)]–90˚ and B = –(i o /C)/(j2ω) = [i o /(2ωC)]–90˚. Thus, v(t) = [i o /(2ωC)][e–jωt–90˚+ejωt+90˚] = [i o /(ωC)]cos(ωt+90˚)u(t) volts.

Chapter 16, Solution 46. Consider the following circuit. 1/s

2s

Vo

Io 1/(s + 2)

2

1

Applying KCL at node o, Vo Vo 1 s 1 V    s  2 2s  1 2  1 s 2s  1 o 2s  1 Vo  (s  1)(s  2)

Io 

Vo 1 A B    2s  1 (s  1)(s  2) s  1 s  2

A  1, Io 

B  -1

1 1  s 1 s  2 i o ( t )   e -t  e -2t  u(t ) A

Chapter 16, Solution 47. We first find the initial conditions from the circuit in Fig. (a). 1

4 +

5V

+ 

v c (0)

io

 (a)

i o (0  )  5 A , v c (0  )  0 V We now incorporate these conditions in the s-domain circuit as shown in Fig.(b). 1

4

Vo Io

15/s

+ 

2s

(b)

At node o, Vo  15 s Vo 5 Vo  0    0 1 2s s 4  4 s 15 5  1 s  V   1   s s  2s 4 (s  1)  o 5s 2  6s  2 10 4s 2  4s  2s  2  s 2 Vo Vo   4s (s  1) s 4s (s  1) 40 (s  1) Vo  2 5s  6s  2 Vo 5 4 (s  1) 5    2 2s s s (s  1.2s  0.4) s 5 A Bs  C Io    2 s s s  1.2s  0.4 Io 

5/s

4/s

4 (s  1)  A (s 2  1.2s  0.4)  B s s  C s Equating coefficients :

s0 :

4  0.4A   A  10

1

4  1.2A  C   C  -1.2A  4  -8

2

0 AB   B  -A  -10

s :

s :

5 10 10s  8   2 s s s  1.2s  0.4 10 (s  0.6) 10 (0.2) 15 Io   2 2  s (s  0.6)  0.2 (s  0.6) 2  0.2 2 Io 

i o ( t )   15  10 e -0.6t  cos(0.2 t )  sin( 0.2 t )  u(t ) A

Chapter 16, Solution 48. First we need to transform the circuit into the s-domain.

s/4

10

Vo +

3V x

+ 

Vx

5/s

+ 

5 s2

5 Vo  Vo  3Vx Vo  0 s2 0   s/4 5/s 10 5s 5s  0  (2s 2  s  40)Vo  120Vx  40Vo  120Vx  2s 2 Vo  sVo  s2 s2 But, Vx  Vo 

5 5  Vo  Vx  s2 s2

We can now solve for V x . 5  5s  ( 2s 2  s  40) Vx  0   120Vx  s 2 s2  (s 2  20) 2(s 2  0.5s  40)Vx  10 s2 Vx   5

(s 2  20) (s  2)(s 2  0.5s  40)

Chapter 16, Solution 49. We first need to find the initial conditions. For t  0 , the circuit is shown in Fig. (a).

2

+



1

1F

V o /2

Vo

+ 

+ 

1H

3V

io

(a) To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit. Hence, -3  -1 A , v o  -1 V i L (0)  i o  3  - 1 v c (0)  -(2)(-1)     2.5 V 2

We now incorporate the initial conditions for t  0 as shown in Fig. (b). 2

+

Vo



1

1/s

s 5/(s + 2)

+ 

2.5/s I1 V o /2

+ 

I2  +

+ 

Io (b)

For mesh 1, 2.5 Vo 1 - 5  1  0   2   I1  I 2  s 2 s s s2 

-1 V

But,

Vo  I o  I 2  1 1  1 5 2.5  2   I1     I 2   2 s  s2 s s

For mesh 2, V  2.5 1 1 1  s   I 2  I1  1  o  0  2 s s s 1 2.5 1 1 1 - I1    s   I 2  2 s s s

(1)

(2)

Put (1) and (2) in matrix form.  1 2  s   -1  s

 5 1 1  2.5     I1    2 s   s2 s     1 1    2.5   s   I 2   1   s  2 s

3   2s  2  , s Io  I2 

 2  -2 

4 5  s s (s  2)

2 - 2s 2  13 A Bs  C    2 2  (s  2)(2s  2s  3) s  2 2s  2s  3

- 2s 2  13  A (2s 2  2s  3)  B (s 2  2s)  C (s  2) Equating coefficients : s2 : s1 : s0 :

- 2  2A  B 0  2A  2 B  C 13  3A  2C

Solving these equations leads to A  0.7143 , B  -3.429 , C  5.429 0.7143 3.429 s  5.429 0.7143 1.7145 s  2.714    s2 2s 2  2s  3 s2 s 2  s  1.5 0.7143 1.7145 (s  0.5) (3.194)( 1.25 ) Io    s2 (s  0.5) 2  1.25 (s  0.5) 2  1.25 Io 





i o ( t )  0.7143 e -2t  1.7145 e -0.5t cos(1.25t )  3.194 e -0.5t sin(1.25t ) u(t ) A

Chapter 16, Solution 50. For the circuit in Fig. 16.73, find v(t) for t > 0. Assume that v(0+) = 4 V and i(0+) = 2 A.


Determine the initial condition of the second capacitor and then convert the circuit into the s-domain. Finally, solve for V, perform a partial fraction expansion and convert the answer back into the time domain. Since v(0) = 4 volts and i(0) = 2 amps then –4–2(2)+v 2 (0) = or v 2 (0) = 8. 2 +

I

10/s

2/s V

4/s

+  

I/4 8/s

+ 

–

[(V–4/s)/(10/s)]–[I/4]+[(V–8/s)/(2+2/s)] = 0 and I = [((8/s)–V)/(2+2/s)] = [4/(s+1)]–0.5sV/(s+1) Step 2.

[(V–4/s)/(10/s)]+[0.125sV/(s+1)]–[1/(s+1)]+[(V–8/s)/(2+2/s)] [(s/10)+(0.125s/(s+1))+(0.5s/(s+1))]V = [0.4+4/(s+1)]+[1/(s+1)] = (0.4s+5.4)/(s+1) = [(s2+s+6.25s)/(10(s+1))]V = [s(s+7.25)/(10(s+1))]V or V = 4(s+13.5)/[s(s+7.25)] = [A/s]+[B/(s+7.25)] where A = 4(13.5)/7.25 = 7.748 and B = 4(–7.25+13.5)/(–7.25) = –3.448 or v(t) = [7.748–3.448e–7.25t]u(t) volts.

Chapter 16, Solution 51. In the circuit of Fig. 16.74, find i(t) for t > 0.

50 V


First we note that the initial conditions for the capacitor and inductor have to be equal to zero. Next we simplify the circuit and then convert the circuit into the s-domain and solve for V. Then we can solve for I and then perform a partial fraction expansion and convert I back into the time domain. 0.25s 6 V 50/s

+ 

4

25/s I

[(V–50/s)/(0.25(s+24))]+[s(V–0)/25]+[(V–0)/4] = 0 and I = [(0–V)/4] = –V/4. Step 2.

[(4/(s+24))+(s/25)+0.25]V = [(s2+24s+6.25s+100+150)/(25(s+24))]V = [(s +30.25s+250)/(25(s+24))]V = [{(s+15.125+j4.608)(s+15.125–j4.608)}/(25(s+24))]V = [200/(s(s+24))] or V = 5,000/[s(s+15.125+j4.608)(s+15.125–j4.608)] and I = –1250/[s(s+15.125+j4.608)(s+15.125–j4.608)] = [A/s]+[B/(s+15.125+j4.608)] +[C/(s+15.125–j4.608)] where A = –1250/250 = –5; B = –1250/[(–15.125–j4.608)(–j9.216)] = 1250180˚/[(15.811–163.06˚)(9.216–90˚)] = 8.57873.06˚; and C = 1250180˚/[(15.811163.06˚)(9.21690˚)] = 8.578–73.06˚. Thus, i(t) = [–5+8.578e–15.125t(e–j4.608t+73.06˚+ej4.608–73.06˚)]u(t) amps 2

i(t) = [–5+17.156e–15.125tcos(4.608t–73.06˚)]u(t) amps.

Chapter 16, Solution 52. If the switch in Fig. 16.75 has been closed for a long time before t = 0 but is opened at t = 0, determine i x and v R for t > 0.


Fist we need to determine the initial conditions. Just before the switch opens, v C (0) = 16 volts and i L (0) = 2 amps. Next we convert the circuit into the s-domain.

12 

+ VR –

Ix

8

36/s s 16/s

2/s

+ 

We can now write a mesh equation (this time going in the counter-clockwise direction). [s(I x +2/s)]+[8I x ]+[12I x ]+[(36/s)I x ]+(16/s) = 0 and V R = –8I x . Step 2.

[s+8+12+(36/s)]I x = [(s2+20s+36)/s]I x = –2–16/s = –[2(s+8)/s] or I x = –2(s+8)/[(s+2)(s+18)] = [A/(s+2)]+[B/(s+18)] where A = –2(–2+8)/(–2+18) = –2x6/16 = –0.75 and B = –2(–18+8)/(–18+2) = –1.25 thus, i x (t) = [–0.75e–2t–1.25e–18t]u(t) amps and v R (t) = –8i x (t) = [6e–2t+10e–18t]u(t) volts.

Chapter 16, Solution 53. In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to position 2 at t = 0. Find: (a) v(0+), dv(0+)/dt (b) v(t) for t  0.


Clearly i L (0) = 0 and v(0) = 4 volts. When the switch moves to 2, i C (0+) = Cdv(0)/dt = –4/0.5 = –8 volts/second = 1dv(0)/dt. Next we convert the circuit into the s-domain and solve for V. Then we perform a partial fraction expansion and then convert back into the time domain. + 1/s 0.25s

V

0.5  4/s

+ 



[(V–0)/(0.25s)]+[(V–0)/0.5]+[(V–4/s)s/1] = 0. Step 2.

[(4/s)+2+s]V = [(s2+2s+4)/s]V = 4 or V = 4s/[(s+1+j1.7321)(s+1– j1.7321)] = [A/(s+1+j1.7321)] + [B/(s+1–j1.7321)] where A = 4(–1–j1.7321)/(3.464–90˚) = 4(2–120˚)/(3.464–90˚) = 2.309–30˚ and B = 4(2120˚)/(3.46490˚) = 2.30930˚ or v(t) = 2.309e–t[e–j1.7321t–30˚+ej1.7321t+30˚]u(t) volts or v(t) = [4.618e–tcos(1.7321t+30˚)]u(t) volts.

Chapter 16, Solution 54. The switch in Fig. 16.77 has been in position 1 for t < 0. At t =0, it is moved from position 1 to the top of the capacitor at t = 0. Please note that the switch is a make before break switch, it stays in contact with position 1 until it makes contact with the top of the capacitor and then breaks the contact at position 1. Determine v(t).


First determine the initial conditions and then transform the circuit into the s-domain and solve for V. Then perform a partial fraction expansion and then find v(t). We will assume that the value of v(0) = 0. i L (0) = 40/20 = 2 amps. 4s

+ V

16/s 

2/s I

16 

[16/s]I + [4s](I–2/s) + 16I = 0 and V = [16/s](–I). Step 2.

[(16/s)+4s+16]I = [4(s2+4s+4)/s]I = 8 or I = 8s/[4(s+2)2] = 2s/[(s+2)2] and V = –32/[(s+2)2] v(t) = [–32te–2t]u(t) volts.

Chapter 16, Solution 55. Obtain i 1 and i 2 for t > 0 in the circuit of Fig. 16.78.


The first thing we do is to determine the initial conditions. Since there is no excitation of the circuit before t = 0, all initial conditions must be zero. Next we convert the circuit into the s-domain. Then use nodal analysis and eventually solve for I 1 and I 2 , then perform a partial fraction expansion and convert back into the time domain. 3 V1 I2 I1 4/s

2

s

s

–[4/s]+[(V 1 –0)/2]+[(V 1 –0)/s]+[(V 1 –0)/(s+3)] = 0 and I 1 = [(V 1 –0)/s] and I 2 = [(V 1 –0)/(s+3)]. Step 2.

{[1/2]+[1/s]+[1/(s+3)]}V 1 = 4/s = {[s2+3s+2s+6+2s]/[2s(s+3)]}V 1 or V 1 = 8(s+3)/[s2+7s+6] = 8(s+3)/[(s+1)(s+6)] and I 1 = 8(s+3)/[s(s+1)(s+6)] = [A/s]+[B/(s+1)]+[C/(s+6)] where A = 8x3/6 = 4; B = 8(–1+3)/[(–1)(–1+6)] = –16/5 = –3.2; C = 8(–6+3)/[(–6)(–6+1)] = –24/30 = –0.8. Thus, i 1 (t) = [4–3.2e–t–0.8e–6t]u(t) amps. I 2 = [(V 1 –0)/(s+3)] = 8/[(s+1)(s+6)] = [A/(s+1)] + [B/(s+6)] where A = 8/5 = 1.6 and B = 8/(–6+1) = –1.6. Thus, i 2 (t) = [1.6e–t–1.6e–6t]u(t) amps.

[4–3.2e–t–0.8e–6t]u(t) amps, [1.6e–t–1.6e–6t]u(t) amps

Chapter 16, Solution 56. We apply mesh analysis to the s-domain form of the circuit as shown below. 2/(s+1) + 

I3 1/s

1

s

I1

I2

1

4/s For mesh 3,  1 1 2   s   I 3  I1  s I 2  0 s s 1  s

For the supermesh,  1 1  1   I1  (1  s) I 2    s  I 3  0  s s 

(1)

(2)

Adding (1) and (2) we get, I 1 + I 2 = –2/(s+1)

(3)

But

(4)

–I 1 + I 2 = 4/s

Adding (3) and (4) we get, I 2 = (2/s) – 1/(s+1)

(5)

Substituting (5) into (4) yields, I 1 = –(2/s) – (1/(s+1))

(6)

Substituting (5) and (6) into (1) we get, 2 s2



 s2  1 1 s I 3   2 2  s(s  1) s  1  s  s 1

2 1.5  0.5 j 1.5  0.5 j I3     s s j s j

Substituting (3) into (1) and (2) leads to 2(s 2  2s  2)  1  1 -  s   I 2   s   I3   s  s s 2 (s  1)

(4)

1 4(s  1)   1  2  s   I 2   s   I3   s s   s2

(5)

We can now solve for I o . I o = I 2 – I 3 = (4/s) – (1/(s+1)) + ((–1.5+0.5j)/(s+j)) + ((–1.5–0.5)/(s–j)) or i o (t) = [4 – e–t + 1.5811e–jt+161.57˚ + 1.5811ejt–161.57˚]u(t)A This is a challenging problem. I did check it with using a Thevenin equivalent circuit and got the same exact answer.

Chapter 16, Solution 57. 3 e s 3  (1  e  s ) v s (t) = 3u(t) – 3u(t–1) or V s =  s s s 1 + Vs

+ 

1/s

2

Vo 

Vo  Vs V  sVo  o  0  (s  1.5)Vo  Vs 1 2 Vo 

3 2  2 s (1  e  s )    (1  e  ) s(s  1.5) s s  1 . 5  

v o (t )  [( 2  2e 1.5 t )u(t )  ( 2  2e 1.5( t 1) )u(t  1)] V

(a) (3/s)[1–e–s], (b) [(2–2e–1.5t)u(t) – (2–2e–1.5(t–1))u(t–1)] V

Chapter 16, Solution 58. Using Fig. 16.81, design a problem to help other students to better understand circuit analysis in the s-domain with circuits that have dependent sources. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem In the circuit of Fig. 16.81, let i(0) = 1 A, v o (0) = 2 V, and v s = 4 e-2t u(t) V. Find v o (t) for t > 0.

Figure 16.81 For Prob. 16.58. Solution We incorporate the initial conditions in the s-domain circuit as shown below. 2I 2 Vo V1  + I 4/(s + 2)

+ 

1/s 1/s

2

s

At the supernode, (4 (s  2))  V1 V 1  2  1   sVo 2 s s  1 1 1 2  2     V1   s Vo 2 s s s2

(1)

But

Vo  V1  2 I and Vo  V1 

2 (V1  1) s

I

V1  1 s

  V1 

Vo  2 s s Vo  2  (s  2) s s2

(2)

Substituting (2) into (1) 2 1  s  2   s  2  2   s Vo    Vo  s  2  s2 s  2s   s  2  2 1 1  1    2       s  Vo s s  2   s2 2s  4  2 2s  6   (s  1 / 2)Vo (s  2) s2 2s  6 A B   Vo  (s  2)(s  1 / 2) s  1 / 2 s  2

A  (1  6) /(0.5  2)  3.333 , B  (4  6) /(2  1 / 2)  1.3333 3.333 1.3333 Vo   s  1/ 2 s  2

Therefore, v o ( t )  (3.333e-t/2 – 1.3333e-2t)u(t) V

Chapter 16, Solution 59. We incorporate the initial conditions and transform the current source to a voltage source as shown. 2/s

1

1/s

Vo

+ 

1/(s + 1)

+ 

1

At the main non-reference node, KCL gives 1 (s  1)  2 s  Vo Vo Vo 1    11 s 1 s s s s 1  2  s Vo  (s  1)(1  1 s) Vo  s 1 s s s 1   2  (2s  2  1 s) Vo s 1 s - 2s 2  4s  1 Vo  (s  1)( 2s 2  2s  1) - s  2s  0.5 A Bs  C   2 Vo  2 (s  1)(s  s  0.5) s  1 s  s  0.5 A  (s  1) Vo

s  -1

1

- s 2  2s  0.5  A (s 2  s  0.5)  B (s 2  s)  C (s  1) Equating coefficients : s2 : -1  A  B   B  -2 s1 : s0 :

Vo 

-2  ABC   C  -1 - 0.5  0.5A  C  0.5  1  -0.5

2 (s  0.5) 1 2s  1 1  2   s  1 s  s  0.5 s  1 (s  0.5) 2  (0.5) 2 v o ( t )   e -t  2 e -t 2 cos(t 2) u(t ) V

s

1/s

Chapter 16, Solution 60. Find the response v R (t) for t > 0 in the circuit in Fig. 16.83. Let R = 3 , L = 2 H, and C = 1/18 F.


First convert the circuit into the s-domain. Then use nodal analysis and eventually solve for V R , then perform a partial fraction expansion and convert back into the time domain. 3 V1 +  VR 10/s

+ 

18/s

2s

[(V 1 -10/s)/3]+[(V 1 -0)/(18/s)]+[(V 1 -0)/(2s)] = 0 and V R = (10/s)-V 1 . Step2.

[(1/3)+(s/18)+1/(2s)]V 1 = 3.333/s = [(s2+6s+9)/(18s)]V 1 or V 1 = 60/[(s+3)2] and V R = (10/s)-60/[(s+3)2]. Thus, v R (t) = [10-60te-3t]u(t) volts.

Chapter 16, Solution 61. The s-domain version of the circuit is shown below. 1

s V1

+

10/s

2/s

Vo 2

1/s

-

At node 1,

10 s  V1  Vo  s (V  0)  0 1 1 s 2

V1 

 

  s2   s  1V1  (  1)Vo  10  2

(1)

At node 2, Vo  V1 Vo  0   s(Vo  0)  0 s 2

 

V1  (s 2  0.5s  1)Vo

Substituting (2) into (1) gives 10  [0.5(s 2  2s  2)(s 2  0.5s  1)Vo  Vo  0.5(s 4  2.5s 3  4s 2  3s  2  2)Vo

Vo 

20 s(s  2.5s 2  4s  3) 3

Use MATLAB to find the roots. >> p=[1 2.5 4 3] p= 1.0000 2.5000 4.0000 3.0000 >> r=roots(p) r= -0.6347 + 1.4265i -0.6347 - 1.4265i -1.2306

(2)

Thus, 20 s(s  1.2306)(s  0.6347  j1.4265)(s  0.6347  j1.4265) A B C D     s (s  1.2306) (s  0.6347  j1.4265) (s  0.6347  j1.4265) Vo 

Where A = 20/3 = 6.667; B = 20 (1.2306)(1.2306  0.6347  j1.4265)(1.2306  0.6347  j1.4265)  16.252   6.8 (0.3551  2.035) 20 C (0.6347  j1.4265)(0.6347  j1.4265  1.2306)( j2.853) 20 20    2.904  88.68 (1.5613  113.99)(1.546  67.33)(2.853  90) 6.88688.68 20 (0.6347  j1.4265)(0.6347  j1.4265  1.2306)( j2.853) 20 20    2.90488.68 (1.5613113.99)(1.54667.33)(2.85390) 6.886  88.68

D

Vo 

 6.8 6.667 2.904  88.68 2.90488.68    or s (s  1.2306) (s  0.6347  j1.4265) (s  0.6347  j1.4265)

v o (t) = [6.667–6.8e–1.2306t+2.904e–0.6347t(e–(1.4265t+88.68˚)+e(1.4265t+88.68˚))]u(t) volts or = [6.667–6.8e–1.2306t+5.808e–0.6347tcos(1.4265t+88.68˚)]u(t) V. Answer does check for initial values and final values.

Chapter 16, Solution 62. Using Fig. 16.85, design a problem to help other students better understand solving for node voltages by working in the s-domain. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the node voltages v 1 and v 2 in the circuit of Fig. 16.85 using Laplace transform technique. Assume that i s = 12e-t u(t) A and that all initial conditions are zero.

Figure 16.85 For Prob. 16.62. Solution The s-domain version of the circuit is shown below. 4s V1

12 s 1

1

V2

2

3/s

At node 1, V V  V2 12  1 1 s 1 1 4s At node 2,



12 1 V   V1 1    2 s 1  4s  4s

(1)

V1  V2 V2 s   V2 4s 2 3



 4 V1  V2  s 2  2s  1  3

(2)

Substituting (2) into (1),  4 12 1  1  4 7 3   V2  s 2  2s  11       s 2  s  V2 2 s 1 3  4s  4s   3  3

V2 

9 7 9 (s  1)(s 2  s  ) 4 8

9  A(s 2 



A Bs  C  (s  1) (s 2  7 s  9 ) 4 8

7 9 s  )  B(s 2  s)  C(s  1) 4 8

Equating coefficients: s2 :

0AB 7 3 0  ABC  AC 4 4 9 3 9  A  C A   8 8

s: constant :

 

3 C A 4

A  24, B  -24, C  -18

3 24s  18 24 24(s  7 / 8)    7 23 7 9 7 23 (s  1) (s  ) 2  (s 2  s  ) (s  ) 2  8 64 4 8 8 64 Taking the inverse of this produces: V2 

24  (s  1)





v 2 (t )  24e  t  24e 0.875 t cos(0.5995t )  5.004e 0.875 t sin( 0.5995t ) u(t )V

Similarly, 4  9 s 2  2s  1 Es  F 3   D  V1  7 9 7 9 (s  1) (s  1)(s 2  s  ) (s 2  s  ) 4 8 4 8

7 9 4  9 s 2  2s  1  D(s 2  s  )  E(s 2  s)  F(s  1) 4 8 3  Equating coefficients:

s2 :

12  D  E 7 3 3 s: 18  D  E  F or 6  D  F   F  6 D 4 4 4 9 3 constant : 9  D  F or 3  D   D  8, E  4, F  0 8 8 8 4s 8 4(s  7 / 8) 7/2 V1      7 9 7 23 7 23 (s  1) (s  1) (s 2  s  ) (s  ) 2  (s  ) 2  4 8 8 64 8 64 Thus, v 1 (t )  8e  t  4e 0.875 t cos(0.5995t )  5.838e 0.875 t sin( 0.5995t ) u(t )V





Chapter 16, Solution 63. The s-domain form of the circuit with the initial conditions is shown below. V I 4/s

sL

R

-2/s

1/sC

5C

At the non-reference node, V V 4 2   5C    sCV R sL s s 6  5 sC CV  2 s 1  s     s s  RC LC  5s  6 C V 2 s  (s RC)  (1 LC)

But

1 1   8, RC 10 80

V

1 1   20 LC 4 80

5s  480 5 (s  4 ) ( 230)( 2)  2 2  s  8s  20 (s  4)  2 (s  4 ) 2  2 2 2

v( t )  [5 e -4t cos( 2t )  230 e -4t sin( 2t )]u(t ) V I

V 5s  480  sL 4s (s 2  8s  20)

I

1.25s  120 A Bs  C   2 2 s (s  8s  20) s s  8s  20

A  6, I

B  -6 , C  -46.75

6 6s  46.75 6 6 (s  4) (11.375)(2)  2   2 2  s s  8s  20 s (s  4)  2 (s  4) 2  2 2

i( t )  [6  6 e -4t cos( 2t )  11.375 e -4t sin( 2t )]u( t ) A Checking, Ldi/dt = 4{24 e–4tcos(2t) + 12 e–4tsin(2t) + 45.5 e–4tsin(2t) – 22.75 e–4tcos(2t)}u(t) = [5 e–4tcos(2t) + 230 e–4tsin(2t)]u(t). Answer checks.

Chapter 16, Solution 64. When the switch is position 1, v(0)=12, and i L (0) = 0. When the switch is in position 2, we have the circuit as shown below. s/4

+ 100/s v

–

1 100  sC s 12 / s 48 ,  I s / 4  100 / s s 2  400

10mF  0.01F

+ _

12/s

 

V  sLI 

s 12s I 2 4 s  400

v(t) = [12cos(20t)]u(t) V

Chapter 16, Solution 65. For t  0 , the circuit in the s-domain is shown below. 6

s

I + 9/s

(2s)/(s2 + 16)

+ 

V



+ 

2/s

Applying KVL,  2s  9 2  6  s   I   0 2 s s s  16   32 I (s 2  6s  9)(s 2  16) V

9 2 2  288 I   s s s s (s  3) 2 (s 2  16) 2 A B C Ds  E      2 2 s s s  3 (s  3) s  16

- 288  A (s 4  6s 3  25s 2  96s  144)  B (s 4  3s 3  16s 2  48s)  C (s 3  16s)  D (s 4  6s 3  9s 2 )  E (s 3  6s 2  9s) Equating coefficients :  288  144A s0 : 1 0  96A  48B  16C  9E s : 2 0  25A  16B  9D  6E s : 3 0  6A  3B  C  6D  E s : 4 0  A  B D s : Solving equations (1), (2), (3), (4) and (5) gives A  2 , B  2.202 , C  3.84 , D  -0.202 , V(s) 

(1) (2) (3) (4) (5)

E  2.766

0.202 s (0.6915)(4) 2.202 3.84    2 s  3 (s  3) s 2  16 s 2  16 v( t )  {2.202e-3t + 3.84te-3t – 0.202cos(4t) + 0.6915sin(4t)}u(t) V

Chapter 16, Solution 66. Consider the op-amp circuit below where R 1 = 20 kΩ, R 2 = 10 kΩ, C = 50 µF, and v s (t) = [3e-5t]u(t) V. R2

1/sC R1

Vs

0

+ 

 +

+ Vo 

At node 0,

Vs  0 0  Vo   (0  Vo ) sC R1 R2  1   sC   - Vo  Vs  R 1  R2  Vo -1  Vs sR 1C  R 1 R 2

But

So,

R 1 20   2, R 2 10 Vo -1  Vs s  2 v s (t )  3 e -5t

R 1C  (20  103 )(50  10-6 )  1

  Vs  3 ( s  5)

Vo 

-3 A B  where A = –1 and B = 1. = (s  2)(s  5) s  2 s  5

Vo 

1 1  s5 s2 v o ( t )   e -5t  e -2t  u(t ) V.

Chapter 16, Solution 67. Given the op amp circuit in Fig. 16.90. If v 1 (0+) = 2 V and v 2 (0+) = 0 V, find v o for t > 0. Let R = 100 k and C = 1 F. R C C +

v1

 + R

 +

v2



 +

+ vo 


Convert the circuit into the s-domain and insert initial conditions. Next, solve for V o (s), then obtain the partial fraction expansion and convert back into the time domain. 105 106/s Va

2/s + 

+

V1  + Vb

106/s 

Vd 105

+

V2



 +

Vc Ve

+ Vo 

[(V a –(V c +2/s))/(106/s)]+[(V a –V o )/105]+0 = 0; V a = V b = 0 and [(V d –V c )/105]+[(V d –V o )/(106/s)]+0 = 0; V d = V e = 0. Step 2.

sV c + 10V o = –2 and 10V c + sV o = 0 or V c = –0.1sV o thus, (–0.1s2+10)V o = –2 or V o = 20/(s2–100) = [A/(s–10)]+[B/(s+10)] where A = 20/(10+10) = 1 and B = 20/(–10–10) = –1. This now leads to v o (t) = [e10t–e–10t]u(t) volts. It should be noted that this is an unstable circuit!

Chapter 16, Solution 68. Obtain V o /V s in the op amp circuit in Fig. 8.91.

10 pF 60 k

v S (t)

 +

60 k

+ 

+

20 pF v o (t) 


Convert the circuit into the s-domain and then solve for V o (s) in terms of V s (s). Then solve for V o /V s = T(s). 1011/s 60 k

60 k d

VS

+ 

a b

 +

c + Vo

5x1010/s



At a, V a = V b = V c = V o . At b, [(V b –V d )/60k]+[(V b –0)/(5x1010/s)]+0 = 0 or [(V o –V d )/60k]+[(V o –0)/(5x1010/s)] = 0 or [(1/60k)V d = [(1/60k)+(s/(5x1010))]V o or V d = [(1.2x10–6)s+1]V o . At d, [(V d –V s )/60k]+[(V d –V c )/(1011/s)]+(V d –V b )/60k] = 0 or [(2/60k)+(s/1011)]V d –(s/1011)V o –(1/60k)V o = (1/60k)V s or [(2/60k)+(s/1011)] [(1.2x10–6)s+1]V o –(s/1011)V o –(1/60k)V o = (1/60k)V s or [2+(6x10–7)s] [(1.2x10–6)s+1]V o –(6x10–7)sV o –V o = V s or [7.2x10–13s2+(2.4x10–6+0.6x10–6–0.6x10–6)s+(2–1)]V o = V s or T(s) = V o /V s = 1/[(7.2x10–13)s2+(2.4x10–6)s+1].

Chapter 16, Solution 69. Find I 1 (s) and I 2 (s) in the following circuit. 2H

(10e–3t)u(t) V

+ 

H

i1

2H

i2

1Ω

1Ω

Solution Step 1.

We note that the initial conditions in this case are equal to zero. Next, we need to convert the circuit into the s-domain and use the model for mutually coupled circuits. Then we can write the mesh equations and solve for I 1 and I 2 . sI 2

2s

sI 1

2s

+  10/(s+3)

Step 2.

+ 

I1

+  1Ω

I2

1Ω

–[10/(s+3)] + 2sI 1 + sI 2 + 1(I 1 –I 2 ) = 0 and 1(I 2 –I 1 ) + 2sI 2 + sI 1 + 1I 2 = 0. Simplifying we get, (2s+1)I 1 + (s–1)I 2 = 10/(s+3) and (s–1)I 1 + (2s+1)I 2 = 0.

We can solve this directly using substitution or use matrices. Let us use matrices.  10  2s  1 s  1   I1     s  1 2s  1 I    (s  3)  The matrix inverse   2   0   2s  1  s  1  2s  1  s  1 1  s  1 2s  1   s  1 2s  1  2s  1 s  1         s  1 2s  1 2 2 3 s ( s  2 ) 4s  4s  1  s  2s  1   Therefore, I 1 = 6.667(s+0.5)/[s(s+2)(s+3)] and I 2 = –3.333(s–1)/[s(s+2)(s+3)] 6.667(s+0.5)/[s(s+2)(s+3)], –3.333(s–1)/[s(s+2)(s+3)]

Chapter 16, Solution 70. Using Fig. 16.93, design a problem to help other students better understand how to do circuit analysis with circuits that have mutually coupled elements by working in the s-domain. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the circuit in Fig. 16.93, find v o (t) for t > 0.

Figure 16.93 For Prob. 16.70. Solution Consider the circuit shown below. s 1 + 6/s

+ 

I1

2s

s

I2

Vo

2

 For mesh 1, 6  (1  2s) I1  s I 2 s

For mesh 2, 0  s I1  (2  s) I 2  2 I1  - 1   I 2  s

(1)

(2)

Substituting (2) into (1) gives  2 - (s 2  5s  2) 6  -(1  2s)1   I 2  s I 2  I2  s s s -6 or I2  2 s  5s  2

Vo  2 I 2 

- 12 - 12  s  5s  2 (s  0.438)(s  4.561) 2

Since the roots of s 2  5s  2  0 are -0.438 and -4.561, A B Vo   s  0.438 s  4.561 A

- 12  -2.91 , 4.123

Vo (s) 

B

- 12  2.91 - 4.123

- 2.91 2.91  s  0.438 s  4.561 v o ( t )  2.91  e -4.561t  e 0.438 t  u(t ) V

Chapter 16, Solution 71. Consider the following circuit. 1

1:2

+ 

10/(s + 1)

Let

Io

ZL  8 ||

4/s

4 (8)(4 s) 8   s 8  4 s 2s  1

When this is reflected to the primary side, Z Zin  1  L2 , n  2 n 2 2s  3 Zin  1   2s  1 2s  1 10 1 10 2s  1    s  1 Zin s  1 2s  3 10s  5 A B Io    (s  1)(s  1.5) s  1 s  1.5 Io 

A  -10 , I o (s) 

B  20

- 10 20  s  1 s  1 .5





i o ( t )  10 2 e -1.5t  e  t u(t ) A

8

Chapter 16, Solution 72. Y (s)  H (s) X (s) ,

X(s) 

4 12  s  1 3 3s  1

12 s 2 4 8s  4 3  2  (3s  1) 3 (3s  1) 2 4 8 s 4 1 Y (s)      3 9 (s  1 3) 2 27 (s  1 3) 2

Y (s) 

Let G (s) 

-8 s  9 (s  1 3) 2

Using the time differentiation property,  -8 d - 8  -1 g( t )   ( t e -t 3 )   t e -t 3  e -t 3   9 dt 93 8 -t 3 8 -t 3 g( t )  te  e 27 9 Hence,  4 8 -t 3 8 -t 3 4 -t 3  y( t )    te  e  t e  u(t) 9 27   3 27

4 -t 3  4 8 t e  u( t ) y( t )    e -t 3  27 3 9 

Chapter 16, Solution 73. x(t)  u(t)

 

y( t )  10 cos( 2 t )

H(s) 

X (s) 

 

Y(s) 10s 2  X(s) s 2  4

1 s

Y(s) 

10s s2  4

Chapter 16, Solution 74. Design a problem to help other students to better understand how to find outputs when given a transfer function and an input. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A circuit is known to have its transfer function as

H (s) 

s3 s  4s  5 2

Find its output when: (a) the input is a unit step function (b) the input is 6te-2t u(t). Solution

(a)

Y(s)  H(s) X(s)

s3 1  s  4s  5 s s3 A Bs  C   2  2 s (s  4s  5) s s  4s  5 

2

s  3  A (s 2  4s  5)  Bs 2  Cs Equating coefficients : 3  5A   A  3 5 s0 :

s1 : 2

s :

1  4A  C   C  1  4A  - 7 5 0 AB   B  -A  - 3 5

35 1 3s  7   2 s 5 s  4s  5 0.6 1 3 (s  2)  1   Y(s)  s 5 (s  2) 2  1

Y(s) 

y( t )   0.6  0.6 e -2t cos(t )  0.2 e -2t sin( t ) u(t )

(b)

x ( t )  6 t e -2t

  X(s) 

6 (s  2) 2

s3 6  2 s  4s  5 (s  2) 2 6 (s  3) A B Cs  D   Y(s)  2 2 2  2 (s  2) (s  4s  5) s  2 (s  2) s  4s  5 Y(s)  H(s) X(s) 

Equating coefficients : s3 : 0 AC   C  -A 2 0  6 A  B  4C  D  2 A  B  D s : 1 s : 6  13A  4B  4C  4D  9A  4B  4D 0 18  10A  5B  4D  2A  B s : Solving (1), (2), (3), and (4) gives A6, B  6, C  -6 ,

D  -18

6 6 6s  18  2  s  2 (s  2) (s  2) 2  1 6 6 6 (s  2) 6  Y(s)   2  2 s  2 (s  2) (s  2)  1 (s  2) 2  1

Y(s) 

y( t )   6 e -2t  6 t e -2t  6 e -2t cos(t )  6 e -2t sin( t ) u(t )

(1) (2) (3) (4)


1 s

H(s) 

Y(s) , X(s)

Y(s) 

4 1 2s (3)(4)    2 s 2 (s  3) (s  2)  16 (s  2) 2  16

H(s)  s Y(s)  4 

X(s) 

2 s(s  2) 12 s s   2 2 2 (s  3) s  4s  20 s  4s  20

Chapter 16, Solution 76. Consider the following circuit.

2

s

Vo +

kV s

+ 

4

10/s

V o (s) 

Using nodal analysis, kVs  Vo Vo Vo   s2 4 10 s 1 s  1  1  1     Vo  (1 / k )1  (s  2)  (s 2  2s)  Vo Vs  (1 / k )(s  2)  10  s  2 4 10   4  1 2s 2  9s  30 Vo Vs  20k





Vo  10k/(s2+4.5s+15) Vs

Chapter 16, Solution 77. Consider the following circuit. 2/s

I

s

V1

+ Vs

+ 

2I

Vo 

At node 1, 2I  I 

3

V1 , s3

where I 

Vs  V1 V  1 2s s3

V1 3s 3s  Vs  V1 s3 2 2  1 3s  3s    V1  Vs s  3 2  2 3s (s  3) V1  2 V 3s  9s  2 s Vo 

3 9s V1  2 V s3 3s  9s  2 s

H (s) 

9s Vo  2 Vs 3s  9s  2

Vs  V1 2s

3

Chapter 16, Solution 78. Taking the inverse Laplace transform of each term gives h(t )   5e t  3e 2t  6e 4t  u (t )


(a)

Consider the circuit shown below. 3

2s +

Vs

+ 

I1

2/s

Vx

I2

+ 

4V x

 For loop 1,  2 2 Vs  3   I1  I 2  s s For loop 2,  2 2 4Vx   2s   I 2  I1  0  s s

But,

So,

(1)

2 Vx  ( I1  I 2 )   s  8 2 2 (I1  I 2 )   2s   I 2  I1  0  s s s 6  -6 0 I1    2s  I 2 s  s

In matrix form, (1) and (2) become  Vs  3  2 s - 2 s  I1   0    - 6 s 6 s  2s  I      2     6  2  2  6   3    2s         s  s  s  s 18    6s  4 s 6   1    2s  Vs , s 

I1 

2 

1 (6 s  2s)  V  18 s  4  6s s

6 V s s

(2)

I1 3 ss s2  3   2 Vs 9 s  2  3 3s  2s  9

(b)

I2 

2 

2  2  2  ( I1  I 2 )   1 s   s 2 s Vs (6 s  2s  6 s) - 4Vs  Vx    Vx 

6 s Vs - 3 I2   Vx - 4Vs 2s


Consider the following circuit. Is

1

V1

s

Vo

Io +

Vs

+ 

1/s

1/s

1

Vo 

At node 1, Vs  V1 V1  Vo  s V1  1 s  1 1 Vs  1  s   V1  Vo  s s At node o, V1  Vo  s Vo  Vo  (s  1) Vo s V1  (s 2  s  1) Vo Substituting (2) into (1) Vs  (s  1  1 s)(s 2  s  1) Vo  1 s Vo Vs  (s 3  2s 2  3s  2) Vo

H 1 (s) 

(b)

Vo 1  3 2 Vs s  2s  3s  2

I s  Vs  V1  (s 3  2s 2  3s  2) Vo  (s 2  s  1) Vo I s  (s 3  s 2  2s  1)Vo

H 2 (s) 

(c)

Io 

Vo 1  3 2 Is s  s  2s  1

Vo 1

H 3 (s) 

I o Vo 1   H 2 (s)  3 2 Is Is s  s  2s  1

(1)

(2)

(d)

H 4 (s) 

I o Vo 1   H1 (s)  3 2 Vs Vs s  2s  3s  2

Chapter 16, Solution 81. For the op-amp circuit in Fig. 16.99, find the transfer function, T(s) = I o (s)/V s (s). Assume all initial conditions are zero. C R  +

i o (t

+ 

V S (t)

L


Solution Step 1.

Convert the circuit into the s-domain. The write the node equations at the input to the op amp and solve for T(s). 1/(Cs) R

VS

+ 

Va

 +

Vc

Io

Vb Ls

[(V a –V s )/R]+[(V a –V c )/(1/(Cs))]+0 = 0; V a = V b = 0 and I o = (V c –0)/(Ls). Step 2.

CsV c = –V s /R or V c = –V s /(RCs) and I o = –V s /(RLCs2) or T(s) = –1/(RLCs2).

Chapter 16, Solution 82. Consider the circuit below. Va Vb Vs

+ 

 +

+

R

Vo 1/sC

Io 

Since no current enters the op amp, I o flows through both R and C.  1  Vo  -I o  R    sC 

Va  Vb  Vs 

H(s) 

- Io sC

Vo R  1 sC   sRC  1 Vs 1 sC


(b)

H (s) 

Vo R R L   Vs R  sL s  R L

h(t) 

R - Rt L e u( t ) L

v s (t )  u(t )   Vs (s)  1 s Vo 

R L R L A B Vs    sR L s (s  R L ) s s  R L

A  1,

B  -1

1 1 Vo   s sR L v o ( t )  u ( t )  e -Rt L u ( t )  (1  e -Rt L ) u(t )

Chapter 16, Solution 84. Consider the circuit as shown below. Vo I2 4

Is

Vo Vo 2   4 s s 2 But I S  s 1 2 1 1 2  Vo     s 1 4 s s

s

Is 

Vo 

s4 2 2 4s  2 )   4s s  1 s s ( s  1)

8(2 s  1) ( s  1)( s  4)

IL 

A

  Vo (

Vo 8(2s  1) A B C     s s( s  1)( s  4) s s  1 s  4

8(1) 8(2  1)  2, B  8 / 3, (1)(4) (1)(2) V 2 8 / 3 14 / 3 IL  o    s s s 1 s  4 8 14   iL (t )   2  e t  e 4t  u (t ) 3 3  

C

8(8  1)  14 / 3 (4)(3)

2 s

Chapter 16, Solution 85. s4 A B C    2 ( s  1)( s  2) s  1 s  2 ( s  2) 2 s  4  A( s  2) 2  B( s  1)( s  2)  C ( s  1)  A( s 2  2s  4)  B( s 2  3s  2)  C ( s  1) We equate coefficients. 0=A+B or B=-A s2 : s: 1=4A+3B+C=B+C constant: 4=4A+2B+C =2A+C Solving these gives A=3, B=-3, C=-2 H (s) 

H (s) 

3 3 2   s  1 s  2 ( s  2) 2

h(t )  (3e t  3e2t  2te 2t )u (t )


1Ω

u(t)V

+ 

i(t)

1F

1H First select the inductor current i L and the capacitor voltage v C to be the state variables. Applying KVL we get:

 u ( t )  i  v C  i'  0; i  v 'C Thus,

vC  i i  vC  i  u (t ) Finally we get,

1   v C   0  vC   v C   0  i     1  1  i   1 u(t ) ; i(t )  0 1 i   0u(t )         

Chapter 16, Solution 87. Develop the state equations for the problem you designed in Prob. 16.13. Although there is no correct way to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Develop the state equations for Problem 16.13. Chapter 16, Problem 13.

Find v x in the circuit shown in Fig. 16.36 given v s = 4u(t) V.

Figure 16.36 Solution 1/8 F

1H +

4u ( t )

+ 

vx

2Ω

4Ω



First select the inductor current i L and the capacitor voltage v C to be the state variables. Applying KCL we get:

vx vC   0; or vC  8iL  4vx 2 8 iL  4u (t )  vx v v vx  vC  4 C  vC  C  vC  4iL  2vx ; or vx  0.3333vC  1.3333iL 8 2  iL 

vC  8iL  1.3333vC  5.333iL  1.3333vC  2.666iL iL  4u (t )  0.3333vC  1.3333iL

Now we can write the state equations. 2.666   v C  0  vC   v C    1.3333  i     0.3333  1.3333  i   4 u(t ); v x  0.3333 1.3333 i   L     L  L 


First select the inductor current i L (current flowing left to right) and the capacitor voltage v C (voltage positive on the left and negative on the right) to be the state variables. Applying KCL we get: vC vo   iL  0 or vC  4iL  2vo 4 2 iL  vo  v2 

v o   v C  v1

vC  4iL  2vC  2v1 iL  vC  v1  v2  i L  0  1  i L  1  1  v1 (t )   iL   v1 ( t )   ; v o (t )  0  1   1 0           v 2 ( t )  vC   v C  4  2  v C   2 0   v 2 (t )


First select the inductor current i L (left to right) and the capacitor voltage v C to be the state variables. Letting v o = v C and applying KCL we get:

vC  is  0 or vC  0.25vC  iL  is 4 iL  vC  vs  iL  vC 

Thus,

 v C   0.25 1  vC  0 1  v s   ; v o ( t )   i       0  i L  1 0  i s   L  1

1   v C   0 0  v s   0  i    0 0  i   s    L  


First select the inductor current i L (left to right) and the capacitor voltage v C (+ on the left) to be the state variables. Letting i 1 =

vC and i 2 = i L and applying KVL we get: 4

Loop 1:  v   v1  vC  2 C  iL   0 or vC  4iL  2vC  2v1  4 

Loop 2: v   2 iL  C   iL  v2  0 or 4  4i  2vC  2v1 iL  2iL  L  v2  vC  v1  v2 2

i1 

4i L  2v C  2v1  i L  0.5v C  0.5v1 4

 i L  0  1  i L  1  1  v1 (t )   i1 (t )  1  0.5  i L  0.5 0  v1 (t )       v    0 0   v ( t )   v    2 0   v (t ) ; i (t )  1  4 2  0 v  2    C      C  2 2        C

Chapter 16, Solution 91. Let x 1 = y(t). Thus, x1  y  x2 and x2  y  3x1  4 x2  z (t ) This gives our state equations.

 x 1   x    2

1   x1  0  x1   0   3  4  x   1 z(t ); y(t )  1 0 x   0z(t )  2      2

Chapter 16, Solution 92. Let x1  y (t ) and x2  x1  z  y  z or y  x2  z

Thus, x2  y  z  9 x1  7( x2  z )  z  2 z  z  9 x1  7 x2  5 z

This now leads to our state equations,

1   x1   1   x1   x 1   0  x     9  7  x     5 z(t ); y(t )  1 0  x   0 z(t )  2     2  2 

Chapter 16, Solution 93. Let x 1 = y(t), x 2 = x1 , and x3  x2 . Thus, x3  6 x1  11x2  6 x3  z (t )

We can now write our state equations. 1 0   x1   0   x 1   0  x1          0 1   x 2  0 z (t ); y ( t )  1 0 0 x 2   0 z( t ) x2   0  x 3    6  11  6  x 3  1   x 3 

Chapter 16, Solution 94. We transform the state equations into the s-domain and solve using Laplace transforms. 1 sX (s)  x (0)  AX(s)  B  s

Assume the initial conditions are zero. 1 (sI  A ) X (s)  B  s s  4  4 X (s)   s   2

1

4  0  0  1  s 1  2 s   2  2 s  4  ( 2 / s )     s  4s  8   

s4 1   2 s(s  4s  8) s s  4s  8 1 1 s4  (s  2) 2      s (s  2) 2  2 2 s (s  2) 2  2 2 (s  2) 2  2 2

Y(s)  X1 (s) 

8

2





y(t) = 1  e 2t cos 2t  sin 2t  u(t )

Chapter 16, Solution 95. Assume that the initial conditions are zero. Using Laplace transforms we get,

1  s  2 X(s)      2 s  4

X1 



1

3s  8 2

2

s((s  3)  1 )



s  4  1  3 / s  1 1  1 / s  1   4 0  2 / s   2 s  2 4 / s     s  6s  10  2 

0.8  0.8s  1.8  s (s  3) 2  12

s3 1 0.8  0.8  .6 s (s  3) 2  12 (s  3) 2  12

x1 ( t )  (0.8  0.8e 3t cos t  0.6e 3t sin t )u ( t )

X2 



4s  14 s((s  3) 2  12



1.4  1.4s  4.4  s (s  3) 2  12

1.4 s3 1  1.4  0.2 s (s  3) 2  12 (s  3) 2  12

x 2 ( t )  (1.4  1.4e 3t cos t  0.2e 3t sin t )u ( t ) y 1 (t )  2x 1 (t )  2x 2 (t )  2u(t )  ( 2.4  4.4e  3 t cos t  0.8e  3 t sin t )u(t ) y 2 ( t )  x 1 ( t )  2u( t )  ( 1.2  0.8e 3 t cos t  0.6e 3 t sin t )u( t )

[–2.4 + 4.4e–3tcos(t) – 0.8e–3tsin(t)]u(t), [–1.2 – 0.8e–3tcos(t) + 0.6e–3tsin(t)]u(t)

Chapter 16, Solution 96. If Vo is the voltage across R, applying KCL at the non-reference node gives Is 

Vo V 1 1  sC Vo  o    sC   Vo R sL  R sL  Is sRL Is  1 1 sL  R  s 2 RLC  sC  R sL

Vo 

Io 

Vo sL Is  2 R s RLC  sL  R

H (s ) 

Io sL s RC  2  2 Is s RLC  sL  R s  s RC  1 LC

The roots -1 1 1  2  2RC (2RC) LC both lie in the left half plane since R, L, and C are positive quantities. s1, 2 

Thus, the circuit is stable.


(a)

H1 (s) 

3 , s 1

H 2 (s) 

H (s )  H 1 (s ) H 2 (s ) 

1 s4

3 (s  1)(s  4)

 A B   h ( t )  L-1  H(s)  L-1   s  1 s  4 

A  1,

B  -1

h ( t )  (e -t  e -4t ) u(t )

(b)

Since the poles of H(s) all lie in the left half s-plane, the system is stable.


Vo1 be the voltage at the output of the first op amp. Vo1  1 sC 1 ,   Vs R sRC H (s) 

Vo 1  2 2 2 Vs s R C

h(t) 

t R C2

Vo 1  Vo1 sRC

2

lim h ( t )   , i.e. the output is unbounded. t 

Hence, the circuit is unstable.


1 sL  1 sC  sL sL ||  1 1  s 2 LC sC sL  sC sL 2 V2 sL  1  s LC  2 sL V1 s RLC  sL  R R 2 1  s LC 1 s  V2 RC  1 1 V1 s2  s   RC LC Comparing this with the given transfer function, 1 1 , 2 6 RC LC If R  1 k ,

1  500 F 2R 1 L  333.3 H 6C

C

Chapter 16, Solution 100. The circuit is transformed in the s-domain as shown below. 1/sC 2

R2

1/sC 1 Vi – R1

Vo

+

1 R1 1 sC1 Let Z1  R 1 //   1  sR1C1 sC1 R  1 1 sC1 R1

1 R2 1 sC2 Z 2  R 2 //   sC2 R  1 1  sR2C2 2 sC2 This is an inverting amplifier. R2

R2 1  1    s s    Z R RC C 1  sR2C2 R1C1 R1C1  V  1  H (s)  o   2   2 1 1  Vi R1 Z1 R1 R2C2  s  1  C2  s  1    R2C2  R2C2  1  sR1C1 Comparing this with ( s  1000) H (s)   2( s  4000) we obtain: C1  1/ 2   C2  2C1  20 F C2 1 1 1  1000   R1   3  100 R1C1 1000C1 10 x10 x106 1 1 1 12.5Ω  4000   R2    12.8 3 4000C2 4 x10 x 20 x106 R2C2 

Chapter 16, Solution 101. We apply KCL at the noninverting terminal at the op amp. (Vs  0) Y3  (0  Vo )(Y1  Y2 ) Y3 Vs  - (Y1  Y2 )Vo Vo - Y3  Vs Y1  Y2 Let

Y1  sC1 , Y2  1 R 1 , Y3  sC 2 Vo - sC 2 - sC 2 C1   Vs sC1  1 R 1 s  1 R 1C1

Comparing this with the given transfer function, C2 1  1,  10 C1 R 1 C1 If R 1  1 k , C1  C 2 

1  100 F 10 4

Chapter 16, Solution 102. Consider the circuit shown below. We notice that V3  Vo and V2  V3  Vo .

Y4

Y1

V in

+ 

Y2

V2

V1

 +

Vo

Y3

At node 1, (Vin  V1 ) Y1  (V1  Vo ) Y2  (V1  Vo ) Y4 Vin Y1  V1 (Y1  Y2  Y4 )  Vo (Y2  Y4 ) At node 2, (V1  Vo ) Y2  (Vo  0) Y3 V1 Y2  (Y2  Y3 ) Vo Y2  Y3 V1  Vo Y2 Substituting (2) into (1), Y2  Y3 Vin Y1   (Y1  Y2  Y4 ) Vo  Vo (Y2  Y4 ) Y2

(1)

(2)

Vin Y1 Y2  Vo ( Y1 Y2  Y22  Y2 Y4  Y1 Y3  Y2 Y3  Y3 Y4  Y22  Y2 Y4 )

Vo Y1 Y2  Vin Y1 Y2  Y1 Y3  Y2 Y3  Y3 Y4

Y1 and Y2 must be resistive, while Y3 and Y4 must be capacitive. 1 1 Y4  sC 2 , , Y3  sC1 , Let Y1  Y2  R1 R2 1 Vo R 1R 2  sC1 sC1 1 Vin    s 2 C1 C 2 R 1R 2 R 1 R 2

1 Vo R 1 R 2 C1 C 2   R1  R 2  Vin 1  s2  s   R 1 R 2 C 2  R 1 R 2 C1 C 2

Choose R 1  1 k , then 1  10 6 R 1 R 2 C1 C 2

and

R1  R 2  100 R 1R 2 C 2

We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is

R 2  1 k ,

C1  50 nF ,

C 2  20 F

Chapter 16, Solution 103. Using the result of Practice Problem 16.14, Vo - Y1 Y2  Vi Y2 Y3  Y4 (Y1  Y2  Y3 ) When Y1  sC1 , 1 , Y2  R1 Y3  Y2 , Vo Vi Vo Vi Vo Vi Vo Vi

C1  0.5 F R 1  10 k Y4  sC 2 ,

C 2  1 F

- sC1 R 1 - sC1 R 1  1 R  sC 2 (sC1  2 R 1 ) 1  sC 2 R 1 (2  sC1 R 1 ) - sC1 R 1  2 s C1C 2 R 12  s  2C 2 R 1  1 - s (0.5  10 -6 )(10  10 3 )  2 s (0.5  10 -6 )(1  10 -6 )(10  10 3 ) 2  s (2)(1  10 -6 )(10  10 3 )  1 



2 1

- 100 s s  400 s  2  10 4 2

Therefore,

a  - 100 ,

b  400 ,

c  2  10 4


(a)

Y(s) 

Let

K (s  1) s3

K (s  1) K (1  1 s)  lim K s  s  s3 1 3 s 0.25  K .

Y ()  lim

i.e.

Hence, Y (s)  (b)

s1 4 (s  3)

Consider the circuit shown below. t=0

Vs = 8 V

I

+ 

Vs  8 u ( t )   Vs  8 s Vs 8 s  1 2 (s  1)   Y(s) Vs (s)   4s s  3 s (s  3) Z A B I  s s3 I

A  2 3, i( t ) 



B = 2(–3+1)/(–3) = 4/3



1 2  4 e - 3t u(t ) A 3

YS

Chapter 16, Solution 105. The gyrator is equivalent to two cascaded inverting amplifiers. Let V1 be the voltage at the output of the first op amp. -R V1  V  -Vi R i Vo 

- 1 sC 1 V1  V R sCR i

Io 

Vo Vo  R sR 2 C

Vo  sR 2 C Io Vo  sL, when L  R 2 C , so if you let L = R2C then V o /I o = sL. Io


This is periodic with  =  which leads to T = 2/ = 2.

(b)

y(t) is not periodic although sin t and 4 cos 2t are independently periodic.

(c) Since sin A cos B = 0.5[sin(A + B) + sin(A – B)], g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t which is harmonic or periodic with the fundamental frequency  = 1 or T = 2/ = 2. (d)

h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and a constant is also periodic, h(t) is periodic.  = 2 or T = 2/ = .

(e)

The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic.  = 0.2 or T = 2/ = 10.

(f)

p(t) = 10 is not periodic.

(g)

g(t) is not periodic.

1 Chapter 17, Solution 2. The function f(t) has a DC offset and is even. We use the following MATLAB code to plot f(t). The plot is shown below. If more terms are taken, the curve is clearly indicating a triangular wave shape which is easily represented with just the DC component and three, cosinusoidal terms of the expansion. for n=1:100 tn(n)=n/10; t=n/10; y1=cos(t); y2=(1/9)*cos(3*t); y3=(1/25)*cos(5*t); factor=4/(pi*pi); y(n)=0.5- factor*(y1+y2+y3); end plot(tn,y)

Chapter 17, Solution 3. T = 4,  o = 2/T = /2 g(t) = 5, 10, 0,

0
T

1

0

0

2

a o = (1/T)  g( t )dt = 0.25[  5dt +  10dt ] = 3.75 a n = (2/T)

T

 g( t ) cos(n t )dt o

0

1

= (2/4)[

0

1

n t )dt ] 2

2

an =

b n = (2/T)

2

t )dt +  10 cos(

2 n 2 n sin t + 10 sin t ] = (–1/(n))5 sin(n/2) n 2 0 n 2 1 1

= 0.5[ 5

n

1

 5 cos( 2

(5/(n))(–1)(n+1)/2, 0,

T

 g( t ) sin(no t )dt = (2/4)[ 0

1

 5 sin( 0

n = odd n = even

2 n n t )dt +  10 sin( t )dt ] 1 2 2

 2x5 2x10 n n cos cos = 0.5[ t – t ] = (5/(n))[3 – 2 cos n + cos(n/2)] n n 2 0 2 1 1

n 1 2 3 4 5 6 7 8

an –1.59 0 0.53 0 –0.32 0 0.23 0

2

bn 7.95 0 2.65 0.80 1.59 0 1.15 0.40

An 8.11 0 2.70 0.80 1.62 0 1.17 0.40

phase –101.31 0 –78.69 –90 –101.31 0 –78.69 –90

8

An

–78.69˚ 90˚ –101.31˚

φ

Figure D. 35 For Prob. 17.3.

0

π 2π 3π 4π 5π 6π 7π 8π

ω

0

π 2π 3π 4π 5π 6π 7π 8π

ω

Chapter 17, Solution 4. f(t) = 10 – 5t, 0 < t < 2, T = 2,  o = 2/T =  a o = (1/T)

a n = (2/T) =

T

2

2

0

0

0

2  f ( t )dt = (1/2)  (10  5t )dt = 0.5[10t  (5t / 2)] = 5

T

 f ( t ) cos(n t )dt o

0

2

2

0

0

= (2/2)

0

 (10) cos(nt )dt –  (5t ) cos(nt )dt 5t 5 cos nt + sin nt = [–5/(n22)](cos 2n – 1) = 0 2 2 n  n 0 0 2

=

b n = (2/2) =

2

 (10  5t ) cos(nt )dt

2

2

 (10  5t ) sin( nt )dt 0

2

2

0

0

 (10) sin(nt )dt –  (5t ) sin( nt )dt

5 5t = 2 2 sin nt + cos nt = 0 + [10/(n)](cos 2n) = 10/(n) n n  0 0 2

2

Hence f(t) = 5 

10  1  sin(nt ) .  n 1 n

Chapter 17, Solution 5. T  2,

  2 / T  1

T

ao 

1 1 z (t )dt  [2 x  4 x ]  1  T0 2 T



2

2 1 1 2 an   z (t ) cos(no )dt   2 cos(nt )dt   4 cos(nt )dt  sin( nt ) T0 n  0   T

bn 



2

2 1 1 2 z (t ) sin( no )dt   2 sin(nt )dt   4 sin( nt )dt   cos(nt )  T 0 n 0  

 12  , n  odd   n  0, n  even

Thus, z(t )   1 





n 1 n  odd

12 sin(nt ) n

 0

 0



4 sin( nt ) n



4 cos(nt ) n

2

 2



0

Chapter 17, Solution 6. T=2,  o =2/T = 1 T  2  1 1  1 ao   f(t)dt  (5  10 )  7.5   5dt   10dt   T0 2  0   2

T  2  2 2    f t n  tdt ntdt ( )cos 5cos 10cos ntdt   0  o   T0 2  0   T  2  1 1  1 2  2 2  bn   f(t)sin notdt     5 sin ntdt   10 s inntdt     cos nt  cos nt T0 2  0 0 n       n  10 , n  odd 5   cos  n  1   n n  0, n  even

an 

Thus, f(t)  7.5 

10 sin nt n odd n 



Chapter 17, Solution 7. T  3,

o  2 / T  2 / 3 T 2 3  1 1 1 ao   f(t)dt    2dt   (1)dt   (4  1)  1 T0 3 0 2  3

an 



T 2 3 2 2n t 2 2n t 2n t    f ( t )cos dt 2cos dt (1)cos dt     T0 3 3 0 3 3 2 

2 3 2nt 3 2nt 2 sin sin 1 3  2n 3 0 2n 3  2

3

3 4 n  sin 3 2  n

T 2 3 2 2n t 2 2n t 2n t    f ( t )sin dt 2 sin dt (1)s in dt     T0 3 3 0 3 3 2  2 3 2n t 2 3 2n t 3  3 4n   2 x  cos cos (1 2cos )  3  2n 3 0 2n 3 2  n 3 1  4n  3  4 n  =  2  3 cos  1   1  cos n  3 3   n 

bn 

Hence,   3 4n 2nt 3  4n  2nt  cos  f(t) = 1    sin  sin  1  cos 3 3 n  3  3  n  0  n

We can now use MATLAB to check our answer,

>> t=0:.01:3; >> f=1*ones(size(t)); >> for n=1:1:99, f=f+(3/(n*pi))*sin(4*n*pi/3)*cos(2*n*pi*t/3)+(3/(n*pi))*(1cos(4*n*pi/3))*sin(2*n*pi*t/3); end >> plot(t,f)

2 .5 2 1 .5 1 0 .5 0 -0 . 5 -1 -1 . 5

0

0 .5

1

1 .5

2

Clearly we have nearly the same figure we started with!!

2 .5

3

Chapter 17, Solution 8. Using Fig. 17.51, design a problem to help other students to better understand how to determine the exponential Fourier Series from a periodic wave shape. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Obtain the exponential Fourier series of the function in Fig. 17.51.

f(t)

5

t 0

1

2

Figure 17.51

3

4

5

For Prob. 17.8.

Solution

o  2 / T  

T  2,

5(1 t), 0  t  1 f(t)    0, 1  t  2 1

1 T 1 cn   f(t)e jnot dt   5(1 t)e jn t dt 0 20 T 

1 1 5 1  jn t 5 5 e jn t 1 5 e jn t  jn t ( jn t  1) e dt te dt    2   2 0 20 2  jn 0 2 ( jn ) 0

 jn  1 5 e jn 5 e 5 (1)   ( jn  1)  2 2 2  jn 2 n  2 n2 2

But e jn  cos  n  j sin n  cos n  0  (1)n cn 

2.5[1 (1)n ] 2.5(1)n[1 jn ] 2.5   2 2 jn n2 2 n


f(t) is an even function, b n =0. T  8,   2 / T   / 4

ao 

1 T

T

 0

2  10 4 2 f (t )dt    10 cos t / 4dt  0  ( ) sin t / 4 8 0  4 

T /2

2

2 0



10



 3.183

2

4 40 f (t ) cos n o dt  [  10 cos t / 4 cos nt / 4dt 0]  5 cos t (n  1) / 4  cos t (n  1) / 4dt  T 0 8 0 0 For n = 1, an 

2

 2 a1  5 [cos t / 2  1]dt  5 sin t / 2dt  t   10 0  0 For n>1, 2

20 20 (n  1) t (n  1) t an  sin sin  4 4 (n  1) (n  1)

2

 0

20 20 (n  1) (n  1) sin sin  2 2 (n  1) (n  1)

20 20 sin 1.5  sin  / 2  2,122066 sin( 270)  6.3662 sin(90) 3  20 10  2.122066  6.3662  4.244, a 3  sin 2  sin   0 4 

a2 

Thus, a 0  3.183,

a 1  10,

a 2  4.244,

a 3  0,

b1  0  b 2  b 3

Chapter 17, Solution 10. T  2 ,

o  2 / T  1

T  Vo Vo e jnt  1  jnot  jnt cn   f(t)e dt  e dt (1 )  T0 2 0 2  jn 0



Vo jV  je jn  j   o (cos n  1) 2n 2n

f(t) 



jVo

 2n (cos n  1)e

n

jnt


 o  2 / T   / 2 T

cn 

1 1 1 0 y (t )e  jnot dt    10(t  1)e  jnt / 2 dt   (10)e  jnt / 2 dt    0 T0 4  1

cn 

10  e  jnt / 2 2  jnt / 2 e  2 2 ( jnt / 2  1)  4  n  / 4 jn



0 1



2  jnt / 2 e jn

1 0

  

10  4 2 4 2 jn / 2 2  jn / 2 2    2 2 e jn / 2 ( jn / 2  1)  e  e   2 2 4 n  jn n  jn jn jn 

But e jn / 2  cos(n / 2)  j sin( n / 2)  j sin( n / 2), e  jn / 2  cos(n / 2)  j sin( n / 2)   j sin( n / 2) 10 cn  2 2 1  j ( jn / 2  1) sin(n / 2)  n sin(n / 2) n

y(t ) 





n  

10 1  j([ jn / 2]  1) sin(n / 2)  n sin(n / 2) e jnt / 2 2 2 n

Chapter 17, Solution 12. A voltage source has a periodic waveform defined over its period as v(t) = 10t(2 – t) V, for all 0 < t < 2 Find the Fourier series for this voltage. v(t) = 10(2 t – t2), 0 < t < 2, T = 2,  o = 2/T = 1 ao = T

(1/T)  f (t )dt  0



1 2

10 (t 2  t 3 / 3) 2



2 0

2

0



10(2t  t 2 )dt 40 3 20 2 (1  2 / 3)  2 3 2

2 T 10  2 2t  a n =  10(2t  t 2 ) cos(nt )dt   2 cos(nt )  sin(nt ) T 0 n  n 0

bn =









10 2nt cos(nt )  2 sin(nt )  n 2t 2 sin(nt ) n 3



20 10  40 (1  1)  3 4n cos(2n)  2 2 n n n



2 0

20 T 10 T 2 ( 2 nt  t ) sin( nt ) dt  (2nt  t 2 ) sin( nt )dt   0 0 T  2 2n 10 10  2 2 (sin( nt )  nt cos( nt ))  ( 2 nt sin( nt )  2 cos( nt )  1 n t cos( nt )) 0 0  n2 n 3

 40 40  0 n n

Hence,

f(t) =

20 2  40   2 cos(nt ) 3 n 1 n

Chapter 17, Solution 13. Design a problem to help other students to better understand obtaining the Fourier series from a periodic function. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A periodic function is defined over its period as 10sin t , h(t )    20sin(t   ),

0t    t  2

Find the Fourier series of h(t). Solution T = 2,  o = 1 T

a o = (1/T)  h ( t )dt  0



a n = (2/T)

2 1  [  10 sin t dt +  20 sin( t  ) dt ]  2 0





1 30  2  10 cos t 0  20 cos( t  )   2  T

 h( t ) cos(n t )dt o

0

= [2/(2)]   10 sin t cos( nt )dt   0 



2



20 sin( t  ) cos( nt )dt  

Since sin A cos B = 0.5[sin(A + B) + sin(A – B)] sin t cos nt = 0.5[sin((n + 1)t) + sin((1 – n))t] sin(t – ) = sin t cos  – cost sin  = –sin t sin(t – )cos(nt) = –sin(t)cos(nt) an =

2 1   10 [sin([1  n ]t )  sin([1  n ]t )]dt  20 [sin([1  n ]t )  sin([1  n ]t )]dt    2  0

5 = 

 cos([1  n ]t ) cos([1  n ]t )    2 cos([1  n ]t ) 2 cos([1  n ]t )  2           1 n 1 n 1 n 1 n 0     

5 3 3 3 cos([1  n ]) 3 cos([1  n ])        1  n 1  n 1 n 1 n

an = But,

[1/(1+n)] + [1/(1-n)] = 2/(1–n2) cos([n–1]) = cos([n+1]) = cos  cos n – sin  sin n = –cos n

a n = (5/)[(6/(1–n2)) + (6 cos(n)/(1–n2))] = [30/((1–n2))](1 + cos n) = [–60/((n2–1))], n = even = 0, n = odd T

b n = (2/T)  h( t ) sin no t dt 0



2

0



= [2/(2)][  10 sin t sin nt dt +  20(  sin t ) sin nt dt

This is an interesting function which will have a value for b 1 but not for any of the other b n terms (they will be zero). 1  cos(2 t ) dt  5 2 2 2 +  20( sin t ) sin t dt  20  (sin t ) 2 dt  10   5   

b 1 = [2/(2)][   10 sin t sin t dt  10 0  0 



Now we can calculate the rest of the b n for values of n = 2 and greater than 2. We note that, sin A sin B = 0.5[cos(A–B) – cos(A+B)] sin t sin nt = 0.5[cos([1–n]t) – cos([1+n]t)] 

b n = (5/){[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ (1  n )] 0 2

+ [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ (1  n )]  }

=

5 

 sin([1  n ]) sin([1  n ])      = 0 1 n 1 n

{Note, that if we substitute 1 for n, the first term is undefined!}

Thus,

h(t) =

30 60  cos( 2kt )  5 sin( t )     k 1 (4k 2  1)

This does make a very good approximation!

Chapter 17, Solution 14. Since cos(A + B) = cos A cos B – sin A sin B.  25  25  cos(n / 4) cos( 2nt )  3 sin(n / 4) sin( 2nt )  f(t) = 5    3 n 1  n 1  n  1

Chapter 17, Solution 15. Dcos t + Esin t = A cos(t - )

(a) where

f(t) = 10 

A =

D 2  E 2 ,  = tan-1(E/D)

A =

16 1  6 ,  = tan-1((n2+1)/(4n3)) 2 ( n  1) n





n 1

2

2  16 1 1 n  1      cos 10 nt tan  ( n 2  1) 2 n 6 4n 3  

Dcos t + Esin t = A sin(t + )

(b) where

f(t) = 10 

A =





n 1

D 2  E 2 ,  = tan-1(D/E)

 16 1 4n 3  1     sin 10 nt tan  ( n 2  1) 2 n 6 n 2  1  

Chapter 17, Solution 16. If v 2 (t) is shifted by 1 along the vertical axis, we obtain v 2 *(t) shown below, i.e. v 2 *(t) = v 2 (t) + 1. v 2 *(t) 2 1 -2 -1

0

1

2

3

4

5

t

Comparing v 2 *(t) with v 1 (t) shows that v 2 *(t) = 2v 1 ((t + t o )/2) where (t + t o )/2 = 0 at t = -1 or t o = 1 Hence

v 2 *(t) = 2v 1 ((t + 1)/2)

But

v 2 *(t) = v 2 (t) + 1 v 2 (t) + 1 = 2v 1 ((t+1)/2) v 2 (t) = -1 + 2v 1 ((t+1)/2)

= -1 + 1 

v 2 (t) = 

8 2

8 2

1    t  1 1  t  1  t  1 cos  2   9 cos 3 2   25 cos 5 2           

1   t   1   3t 3   5t 5  cos 2  2   9 cos 2  2   25 cos 2  2           

v 2 (t) = 

8 2

  t  1  1  3 t   5 t  sin 2   9 sin 2   25 sin 2           

Chapter 17, Solution 17. We replace t by –t in each case and see if the function remains unchanged. (a)

1 – t,

neither odd nor even.

(b)

t2 – 1,

even

(c)

cos n(-t) sin n(-t) = - cos nt sin nt,

odd

(d)

sin2 n(-t) = (-sin t)2 = sin2 t,

even

(e)

e t,

neither odd nor even.


T = 2 leads to  o = 2/T =  f 1 (-t) = -f 1 (t), showing that f 1 (t) is odd and half-wave symmetric.

(b)

T = 3 leads to  o = 2/3 f 2 (t) = f 2 (-t), showing that f 2 (t) is even.

(c)

T = 4 leads to  o = /2 f 3 (t) is even and half-wave symmetric.


o  2 / T   / 2

T  4,

 10t, 0  t  1 f(t)   10(2  t), 1  t  2 a0 

T 1 2 1 1 1 1 2 1 10 t2 2 ( ) 10 10(2 ) 5 (2 )  2.5 f t dt  tdt   t dt  t  t  4 0 4 1 4 2 1 T 0 0 4 T

1

2

2 2 2 an   f(t)cos notdt   10t cos notdt   10(2  t)cos notdt 40 41 T0

1 10 2 2 20 t 5 5t cos not  sin not  sin not  2 2 cos not  sin not no no no 0 no 1 n o 1 20 1 10 5 (cos n / 2  1)  sin n / 2  (sin n  sin n / 2)  2 2 cos n  no no no n  /4 



5 n  /4 2

2

cos n / 2 

T

bn 





10 5 sin n  sin n / 2 no n / 2 1

2

2 2 2 f(t)s innotdt   10t s innotdt   10(2  t)sin notdt  40 41 T0

1 10 1 2 2 5 5 t sin not  cos not  2 2 sin not  cos not no 0 no 0 n o 1 no 1 5 n 2

2 o

sin n / 2  

10 5 (cos  n  cos n / 2)  2 2 (sin  n  sin n / 2) no n o

2 cos  n / 2 cos n  no no

Chapter 17, Solution 20. This is an even function. b n = 0, T = 6,  = 2/6 = /3

2 T

ao =



T/2

f ( t )dt 

0

3 2 2 ( 4 t  4)dt  4 dt    2 6  1

2 1 2 ( 2 t  4 t )  4(3  2) = 2 1  3 

=

4 T

an =



T/4

f ( t ) cos( nt / 3)dt

0

2

= (4/6)[  ( 4 t  4) cos( nt / 3)dt + 1



3

2

4 cos( nt / 3)dt ] 2

3

16  3 3 16  9  nt   nt   nt   nt  3t = sin sin sin cos      2 2   6  n  3  2 6 n   3  n   3  n  3   1 = [24/(n22)][cos(2n/3)  cos(n/3)] f(t) = 2 

Thus

24  1  2 n 1 n2

  2n   nt   n   cos 3   cos 3   cos 3        

At t = 2, f(2) = 2 + (24/2)[(cos(2/3)  cos(/3))cos(2/3) + (1/4)(cos(4/3)  cos(2/3))cos(4/3) + (1/9)(cos(2)  cos())cos(2) + -----] = 2 + 2.432(0.5 + 0 + 0.2222 + -----) f(2) = 3.756

Chapter 17, Solution 21. This is an even function. b n = 0, T = 4,  o = 2/T = /2. f(t) = 2  2t, = 0,

0
 t2  2 1 ao = 2(1  t )dt   t   = 0.5 2 0 4 0  an =

4 T



T/2

0

f ( t ) cos( no t )dt 

4 1  nt  2(1  t ) cos dt  4 0  2 

= [8/(2n2)][1  cos(n/2)] f(t) =

1  2



8

n  n1

2

2

  n   nt  1  cos 2  cos 2      

Chapter 17, Solution 22. Calculate the Fourier coefficients for the function in Fig. 16.54.

f(t) 4

-5 -4 -3 -2 -1

0

1

Figure 16.54

2

3

4

5

t

For Prob. 16.15

This is an even function, therefore b n = 0. In addition, T=4 and  o = /2. ao =

an =

2 T



4 T

T2

0



f ( t )dt 

T2

0

1 2 1 4 tdt  t 2  1  0 4 0

f ( t ) cos(o nt )dt 

4 1 4 t cos( nt / 2)dt 4 0 1

2t   4 sin( nt / 2)  4  2 2 cos( nt / 2)  n 0 n  an =

16 8 (cos( n / 2)  1)  sin( n / 2) 2 2 n n 

Chapter 17, Solution 23. Using Fig. 17.61, design a problem to help other students to better understand finding the Fourier series of a periodic wave shape. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the Fourier series of the function shown in Fig. 17.61.

Figure 17.61 Solution f(t) is an odd function. f(t) = t, 1< t < 1 a o = 0 = a n , T = 2,  o = 2/T =  bn =

=

4 T



T/2

0

f ( t ) sin( no t )dt 

4 1 t sin( nt )dt 2 0

2 sin(nt )  nt cos(nt ) 10 2 n  2

= [2/(n)]cos(n) = 2(1)n+1/(n) f(t) =

2 

( 1) n  1 sin( nt )  n n 1 


This is an odd function. a o = 0 = a n , T = 2,  o = 2/T = 1 bn =

4 T



T/2

0

f ( t ) sin(o nt )dt f(t) = 1 + t/,

bn =

0
4  (1  t / ) sin( nt )dt 2 0 

=

2 1 1 t   cos( nt )  2 sin( nt )  cos( nt )   n n  n 0

= [2/(n)][1  2cos(n)] = [2/(n)][1 + 2(1)n+1] a 2 = 0, b 2 = [2/(2)][1 + 2(1)] = 1/ = 0.3183 (b)

n = n o = 10 or n = 10 a 10 = 0, b 10 = [2/(10)][1  cos(10)] = 1/(5) Thus the magnitude is A 10 = and the phase is

 10 = tan1(b n /a n ) = 90 

(c)

f(t) =

2 a 210  b10 = 1/(5) = 0.06366

2

 n [1  2 cos(n)] sin(nt )  n 1



f(/2) =

2

 n [1  2 cos(n)] sin(n / 2)  n 1

For n = 1,

f 1 = (2/)(1 + 2) = 6/

For n = 2,

f2 = 0

For n = 3,

f 3 = [2/(3)][1  2cos(3)]sin(3/2) = 6/(3)

For n = 4,

f4 = 0

For n = 5,

f 5 = 6/(5), ----

f(/2) = 6/  6/(3) + 6/(5)  6/(7) ---------

Thus,

= (6/)[1  1/3 + 1/5  1/7 + --------] f(/2)  1.3824 which is within 8% of the exact value of 1.5. (d)

From part (c) f(/2) = 1.5 = (6/)[1  1/3 + 1/5  1/7 + - - -] (3/2)(/6) = [1  1/3 + 1/5  1/7 + - - -] or /4 = 1  1/3 + 1/5  1/7 + - - -

Chapter 17, Solution 25. This is a half-wave (odd) function since f(tT/2) = f(t). a o = 0, a n = b n = 0 for n = even, T = 3,  o = 2/3. For n = odd,

4 1.5 4 1 f ( t ) cos n0 tdt   t cos n0 tdt  3 0 3 0

an =

1

4 9  2nt   2nt  3t sin  =  cos    3  4 2 n 2  3  2n  3  0

 3 =  2 2  n

bn =

  2n   2  2n  sin   cos   1     3   n  3 

4 1.5 4 1 f ( t ) sin( no t )dt   t sin(2nt / 3)dt  0 3 3 0 1

4 9 3t  2nt   2 nt  = sin cos   2 2  3  4 n  3  2n  3  0  3  2n  2  2n  sin  =  cos   2 2  3  n  3   n

  3   2n   2  2n    2nt       cos 1 sin      cos    2 2      n   3   n  3    3   f(t) =    n 1   3 2 n  2 2n    2nt     n  odd     2 n 2 sin 3   n cos 3   sin 3      

Chapter 17, Solution 26. T = 4,  o = 2/T = /2 ao =

1 T 1 1 f ( t )dt    1 dt   T 0 4  0

an =

2 T f ( t ) cos(no t )dt T 0

an =

2 2 1 cos( nt / 2)dt  4  1



3

1



3

2

2 dt   1 dt  = 1  3 4

2 cos( nt / 2)dt   1 cos( nt / 2)dt   3 4

2 3 4 2 nt 4 nt 2 nt    = 2  sin sin sin  2 1 n 2 2 n 2 3   n

=

4 n

n   3n sin 2  sin 2 

bn =

2 T f ( t ) sin( no t )dt T 0

=

2 2 nt 1 sin dt   4  1 2



3

2

2 sin

nt dt  2



4

3

1 sin

nt  dt  2 

2 3 4  2 nt 4 nt 2 nt    = 2 cos cos cos  2 1 n 2 2 n 2 3   n

=

4 cos(n)  1 n

Hence f(t) =

1



 n (sin( 3n / 2)  sin(n / 2)) cos(nt / 2)  (cos( n)  1) sin(nt / 2) n 1

4

Chapter 17, Solution 27. (a) (b)

odd symmetry. a o = 0 = a n , T = 4,  o = 2/T = /2 f(t)

= t,

0
= 0,

1
nt  nt 2 t nt 4 1  4 bn = cos dt   2 2 sin t sin   2  0 2 n 2 4 0 n  =

4 n 2 n sin cos 0  2 n  2 n 2 2

= 4(1)(n1)/2/(n22),

n = odd

2(1)n/2/(n),

n = even

a 3 = 0, b 3 = 4(1)/(92) = –0.04503 (c)

b 1 = 4/2, b 2 = 1/, b 3 = 4/(92), b 4 = 1/(2), b 5 = 4/(252) F rms =

a 2o 

1  a 2n  b 2n  2

F rms 2 = 0.5b n 2 = [1/(22)][(16/2) + 1 + (16/(812)) + (1/4) + (16/(6252))] = (1/19.729)(2.6211 + 0.27 + 0.00259) F rms =

0.14667 = 0.383

Compare this with the exact value of F rms = (0.383/0.4082)x100 = 93.83%, close.

2 T

1

 t dt 0

2

 1 / 6 = 0.4082 or

Chapter 17, Solution 28. This is half-wave symmetric since f(t  T/2) = f(t). a o = 0, T = 2,  o = 2/2 =  an =

4 T



T/2

0

f ( t ) cos( no t )dt 

4 1 ( 2  2 t ) cos( nt )dt 2 0 1

1 t 1  = 4  sin( nt )  2 2 cos( nt )  sin( nt ) n  n  n 0 = [4/(n22)][1  cos(n)] =

8/(n22), 0,

n = odd n = even

1

b n = 4  (1  t ) sin( nt )dt 0

1

1 t  1  = 4  cos( nt )  2 2 sin( nt )  cos( nt ) n  n  n 0 = 4/(n), n = odd 

f(t) =



  n k 1

8 4  cos( nt )  sin(nt )  , n = 2k  1 2 n  

2

Chapter 17, Solution 29. This function is half-wave symmetric. T = 2,  o = 2/T = 1, f(t) = t, 0 < t <  For odd n,

an =

2 T





bn =

2 





0

0

(  t ) cos( nt )dt = 

(  t ) sin( nt )dt  

2 cos(nt )  nt sin(nt ) 0 = 4/(n2) 2 n 

2 sin(nt )  nt cos(nt ) 0 = 2/n 2 n 

Thus, 

1   2 f(t) = 2  2 cos( nt )  sin(nt ) , n  k 1  n 

n = 2k  1


1 cn  T

(a)

T/2



T / 2

f ( t )e  jno t dt 

T/2 1  T/2   f ( t ) cos n tdt j f ( t ) sin no tdt  o    T / 2 T / 2   T 

The second term on the right hand side vanishes if f(t) is even. Hence

cn  (b)

(1)

2 T

T/2

 f (t ) cos n tdt o

0

The first term on the right hand side of (1) vanishes if f(t) is odd. Hence,

j2 cn   T

T/ 2

 f (t ) sin n tdt o

0


If h(t )  f (t ),

T'  T / 

 

T'

o ' 

2 2    o T' T /  T'

2 2 an '  h(t ) cos n o ' tdt   f (t ) cos n o ' tdt  T' 0 T' 0 Let t   , ,

d t  d /  ,

2 f ( ) cos n o  d /   a n T 0 T

an '  Similarly,

T'  T

bn ' bn

Chapter 17, Solution 32. When i s = 1 (DC component) i = 1/(1 + 2) = 1/3 For n  1,

 n = 3n, I s = 1/n20 I = [1/(1 + 2 + j n 2)]I s = I s /(3 + j6n) 1 0 2 1 n    tan(2n ) = 3 1  4n 2  tan 1 (6n / 3) 3n 2 1  4n 2

Thus, 1 i(t) =  3



 n 1

1 3n

2

1  4n

2

cos( 3n  tan 1 ( 2n ))


For the DC case, the inductor acts like a short, V o = 0. For the AC case, we obtain the following:

Vo  Vs V jnVo  o  0 10 j2n 4  5   1  j 2.5n   Vo  Vs n     Vo 

Vs 5   1  j 2.5n   n  

A n  n 

4 n

An 

1 5   1  j 2.5n   n  



4 n  j(2.5n 2  2  5)

 2.5n 2  2  5   ;  n   tan  2 2 2 2 2  n n   ( 2.5n   5)   4

1



v o (t )   A n sin(nt   n ) V n 1

Chapter 17, Solution 34. Using Fig. 17.70, design a problem to help other students to better understand circuit responses to a Fourier series. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Obtain v o (t) in the network of Fig. 17.70 if

10 n  cos  nt  2 4  n 1 n 

v(t )  

 V 


For any n, V = [10/n2](n/4),  = n. 1 H becomes j n L = jn and 0.5 F becomes 1/(j n C) = j2/n 2

jn +

V

+ 

j2/n

Vo 

V o = {j(2/n)/[2 + jn  j(2/n)]}V = {j2/[2n + j(n2  2)]}[(10/n2)(n/4)]

20((n / 4)   / 2)

 n

2

4n  (n 2  2) 2  tan 1 ((n 2  2) / 2n ) 2

20

 n

2

n 4 4

[(n / 4)  ( / 2)  tan 1 ((n 2  2) / 2n )] 

v o (t) =

 n 1

 n  n2  2     tan 1 cos nt  4 2 2n  n4  4 

20 n2

Chapter 17, Solution 35. If v s in the circuit of Fig. 17.72 is the same as function f 2 (t) in Fig. 17.57(b), determine the dc component and the first three nonzero harmonics of v o (t). 1

1H +

vS

+ 

1

1F

vo

 Figure 16.64

For Prob. 16.25

f 2 (t) 2 1 -2 -1

0

1

2

Figure 16.50(b)

3

4

5

t

For Prob. 16.25

The signal is even, hence, b n = 0. In addition, T = 3,  o = 2/3. v s (t)

ao =

an =

=

= 1 for all 0 < t < 1 = 2 for all 1 < t < 1.5

2 1 1dt  3  0

4 2dt    3

1.5



1

4 1 cos(2nt / 3)dt  3  0

1.5



1

2 cos(2nt / 3)dt  

4 3 6 2 1 1.5  sin(2nt / 3) 0  sin(2nt / 3) 1    sin(2n / 3)  3  2n 2n n  4 2  1 v s (t) =   sin(2n / 3) cos(2nt / 3) 3  n 1 n

Now consider this circuit, j2n/3

1

+ vS

+ 

-j3/(2n)

1

vo



Let Z = [-j3/(2n)](1)/(1 – j3/(2n)) = -j3/(2n - j3) Therefore, v o = Zv s /(Z + 1 + j2n/3). Simplifying, we get vo =

 j9 v s 12n  j( 4 n 2  2  18)

For the dc case, n = 0 and v s = ¾ V and v o = v s /2 = 3/8 V. We can now solve for v o (t) 3    2nt v o (t) =    A n cos   n   volts   3  8 n 1

where A n 

6 sin( 2n / 3) n

3   n  and  n  90 o  tan 1   2  3 2 n 2 2    4n    6  16n 2  2    3 

where we can further simplify A n to this, A n 

9 sin( 2n / 3) n 4n 4  4  81

Chapter 17, Solution 36. We first find the Fourier series expansion of v s . T  1,

 o  2 / T  2

T 1 1 1 t2 0 a0   f(t)dt   10(1 t)tdt  10(t  )  5 20 2 1 T0 1

T

2 an   f (t ) cos notdt  2 10(1  t ) cos 2n tdt T 0 0 1 t  1 1 sin2n t  2 2 cos 2n t  sin2n t   0  20  2n 4n   2 n o 1

T

bn 

2 2 f(t)sin notdt   10(1 t)t sin notdt  20 T0

1 1  1  1 10  20   cos 2n t  2 2 sin 2n t  cos 2n t   4n  2n  2n  0 n 

10 sin 2n t n 1 n 1H   jn L  jn 1 1  j100 10mF     jn C jn 0.01 n vs (t )  5  

Io 

Vs 5  jn 

j100

n

For dc component, ω 0 = 0 which leads to I 0 = 0. For the nth harmonic, 10  0 10 n  In    A n  n j100 5n  j( 2n 2  2  50) 5  j2n  2n

where

An 

10 25n 2 2  (2n 2 2  50) 2

n   tan 1

,



2n 2 2  50 5n

io (t )   An sin(2n t  n ) n 1

Chapter 17, Solution 37. We first need to express i s in Fourier series. T  2,

ao 

o  2 / T  

T 1 2  1 1 1 f t dt dt ( ) 3 1dt   (3  1)  2      T0 2 0 1  2

T 1 2  3 1 1 2 2 2 an   f(t)cos notdt    3 cos n tdt   cos n tdt   sin n t  sin n t  0 T0 2 0 0 n 1 1  n

bn 

T 1 2  3 1 1 2 2 2 2   f t inn  tdt inn  tdt co s n t  ( )s 3 s s inn tdt   cos n t  (1 cos n  o   T0 2 0 0 n 1 n 1  n

2 (1 cos n )sin n t n1 n 

is(t)  2   By current division, Io 

Is 1 Is  1 2  jnL 3  j 3n

jn 3Is jnIs  3  j3n 1 jn For dc component (n=0), V o = 0. For the nth harmonic, Vo  jnLIo 

Vo 

2(1  cos n) jn 2 (1  cos n)  90  (90  tan 1 n  90) 1  jn n 1  n 22 

2(1 cos  n)

n1

1 n 

vo(t)  

2

2

cos(n t  tan1 n )


1 2  1 v s ( t )    sin nt , 2  k 1n Vo 

For dc,  n  0,

j n Vs , 1  j n

Vs  0.5,

For nth harmonic, Vs 

Vo 

n  2k  1

 n  n

Vo  0

2   90 o n

2 2  tan 1 n 90 o  1  n 2  2  tan 1 n n 1  n 22 n90 o



v o (t )  

k 1



2 1 n  2

2

cos(nt  tan 1 n ),

n  2k  1

Chapter 17, Solution 39. Comparing v s (t) with f(t) in Figure 15.1, v s is shifted by 2.5 and the magnitude is 5 times that of f(t). Hence 10  1 v s (t) = 5  n = 2k  1  sin(nt ),  k 1 n

T = 2,  o = 2//T = ,  n = n o = n For the DC component,

i o = 5/(20 + 40) = 1/12

For the kth harmonic,

V s = (10/(n))0

100 mH becomes j n L = jnx0.1 = j0.1n 50 mF becomes 1/(j n C) = j20/(n) I 20 

VS

40 

Io

j20/(n)

+ 

j0.1n

Z

j20 ( 40  j0.1n) n  Let Z = j20/(n)||(40 + j0.1n) = j20   40  j0.1n n 

= Z in

 j20( 40  j0.1n 2 n  j800  2 2 40n  j(0.1n 2  2  20)  j20  40n  j0.1n 

802n  j( 2n 2  2  1200) = 20 + Z = 40n  j(0.1n 2  2  20)

Vs 400n  j( n 2  2  200)  Z in n[802n  j( 2n 2  2  1200)] j20 I   j20I n  Io = 2 2 j20   ( 40  j0.1n) 40n  j(0.1n   20) n

I =

=

=

 j200 n[802n  j( 2n 2  2  1200)]

200  90  tan 1{(2n 2  2  1200) /(802n)} n (802) 2  ( 2 n 2  2  1200) 2

Thus i o (t) =

where

1 200    I n sin(n t  n ) , 20  k 1

2n 2  2  1200 n  90  tan 802n 1

In =

1 n (804n )  ( 2n 2  2  1200) 2

n = 2k  1

Chapter 17, Solution 40. T = 2,  o = 2/T =  1

1 ao = T an =

2 T

 1 1 t2  v ( t ) dt  ( 2  2 t ) dt  t   1/ 2  0 2 0 2  0  T

T

1

0

0

 v( t ) cos(nt )dt   2(1  t ) cos(nt )dt 1

1 t 1  = 2  sin( nt )  2 2 cos( nt )  sin( nt ) n  n  n 0 2 = 2 2 (1  cos n)  n  bn =

2 T

n  even

0,

4 4 , n  odd  2 2 n   ( 2n  1) 2 2

T

1

0

0

 v( t ) sin(nt )dt  2 (1  t ) sin(nt )dt 1

2 t 1   1 = 2 cos( nt )  cos( nt )  2 2 sin( nt )  n n   0 n  n v s (t) =

1  2

A

where n  tan 1

n

cos( nt   n )

( 2n  1) 2 , An  2n

4 16  4 2 n   ( 2n  1) 4 2

For the DC component, v s = 1/2. As shown in Figure (a), the capacitor acts like an open circuit. 1

Vx

2V x

 +

+ 0.5V

i

+ 

Vo +

Vx

3



Vo



(a) 1

Vx

 +

2V x

Vo +

VS

+ 

3

(1/4)F

Vo

 (b)

Applying KVL to the circuit in Figure (a) gives

But Adding (1) and (2),

–0.5 – 2V x + 4i = 0

(1)

–0.5 + i + V x = 0 or –1 + 2V x + 2i = 0

(2)

–1.5 + 6i = 0 or i = 0.25 V o = 3i = 0.75

For the nth harmonic, we consider the circuit in Figure (b).  n = n, V s = A n –, 1/(j n C) = –j4/(n) At the supernode, (V s – V x )/1 = –[n/(j4)]V x + V o /3 V s = [1 + jn/4]V x + V o /3 But

(3)

–V x – 2V x + V o = 0 or V o = 3V x

Substituting this into (3), V s = [1 + jn/4]V x + V x = [2 + jn/4]V x = (1/3)[2 + jn/4]V o = (1/12)[8 + jn]V o V o = 12V s /(8 + jn) =

Vo =

12 64  n  2

2

12A n    64  n 2  2  tan 1 (n / 8)

4 16  4 [tan 1 (n / 8)  tan 1 ((2n  1) /(2n ))] 2 4 n   (2n  1) 2

Thus v o (t) =

3  4



V n 1

n

cos( nt   n ) volts

where Vn =

12 64  n 2  2

4 16 and  4 2 n   ( 2n  1) 4 2

 n = tan–1(n/8) – tan–1((2n – 1)/(2n))

Chapter 17, Solution 41. For the full wave rectifier, T = ,  o = 2/T = 2,  n = n o = 2n Hence 2 4  1  v in (t) =    2 cos 2nt  volts     n1 4n  1 For the DC component, V in = 2/ The inductor acts like a short-circuit, while the capacitor acts like an open circuit. V o = V in = 2/ For the nth harmonic, V in = [–4/((4n2 – 1))]0 2 H becomes j n L = j4n 0.1 F becomes 1/(j n C) = –j5/n Z = 10||(–j5/n) = –j10/(2n – j) V o = [Z/(Z + j4n)]V in = –j10V in /(4 + j(8n – 10))  j10 40     =  4  j(8n  10)  (4n 2  1)  =

40{90  tan 1 (2n  2.5)} (4n 2  1) 16  (8n  10) 2

Hence 2   v o (t) =    A n cos( 2nt  n ) volts   n 1  where An =

20 ( 4n 2  1) 16n 2  40n  29

 n = 90 – tan–1(2n – 2.5)


20  1 vs  5   sin nt, n  2k - 1  k 1n Vs  0  j n C(0  Vo ) R

For n = 0 (dc component),



Vo 

j Vs ,  n  no  n  n RC

V o =0.

For the nth harmonic, Vo 

190 o 20 20 10 5   90 o   nRC n n 2  2 x10 4 x 40 x10 9 2n 2  2

Hence,

v o (t ) 

10 5 2 2



1

n k 1

2

cos nt , n  2k - 1

Alternatively, we notice that this is an integrator so that

v o (t )  

10 5 1  v dt s RC  2 2



1

n k 1

2

cos nt , n  2k - 1


V rms =

(b)

I rms =

(c)

a 02 

1  2 1 (a n  b 2n )  30 2  (20 2  10 2 ) = 33.91 V  2 n 1 2

1 6 2  (4 2  2 2 ) = 6.782 A 2 1 P = V dc I dc +  Vn I n cos( n   n ) 2 = 30x6 + 0.5[20x4cos(45o-10o) – 10x2cos(-45o+60o)]

= 180 + 32.76 – 9.659 = 203.1 W

Chapter 17, Solution 44. Design a problem to help other students to better understand how to find the rms voltage across and the rms current through an electrical element given a Fourier series for both the current and the voltage. In addition, have them calculate the average power delivered to the element and the power spectrum. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem The voltage and current through an element are respectively v(t) = [30 cos (t + 35) + 10cos(2t – 55˚) + 4 cos(3t - 10)] V i(t) = [2 cos (t–10˚) + cos(2t – 10)] A (a) Find the average power delivered to the element (b) Plot the power spectrum. Solution





1 60 cos(25 o )  10 cos 45 o  0  27.19  3.536 = 30.73 W. 2

(a)

p  vi 

(b)

The power spectrum is shown below. p

27.19

3.536

0

1

2

3



Chapter 17, Solution 45.  n = 1000n j n L = j1000nx2x10–3 = j2n 1/(j n C) = –j/(1000nx40x10–6) = –j25/n Z = R + j n L + 1/(j n C) = 10 + j2n – j25/n I = V/Z For n = 1, V 1 = 100, Z = 10 + j2 – j25 = 10 – j23 I 1 = 100/(10 – j23) = 3.98773.89 For n = 2, V 2 = 50, Z = 10 + j4 – j12.5 = 10 – j8.5 I 2 = 50/(10 – j8.5) = 3.8140.36 For n = 3, V 3 = 25, Z = 10 + j6 – j25/3 = 10 – j2.333 I 3 = 25/(10 – j2.333) = 2.43513.13 I rms =

0.5(3.987 2  3.812  2.435 2 ) = 4.263 A

p = R(I rms )2 = 181.7W

Chapter 17, Solution 46. (a)The MATLAB commands are: t=0:0.01:5; y=5*cos(3*t) – 2*cos(3*t-pi/3); plot(t,y)

5 4 3 2 1 0 -1 -2 -3 -4 -5 0

0.5

1

1.5

2

(b) The MATLAB commands are: t=0:0.01:5; » x=8*sin(pi*t+pi/4)+10*cos(pi*t-pi/8); » plot(t,x) » plot(t,x)

2.5

3

3.5

4

4.5

5

20 15 10 5 0 -5 -10 -15 -20 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5


o  2 / T  

T 1 2  1 1 1 ao   f(t)dt    4dt   (2)dt   (4  2)  1 T0 2 0 1  2 T 1 2  R 2 R 2 2 P  Rirms   f (t)dt    4 dt   (2)2 dt   10R 2 0 T0 1 

The average power dissipation caused by the dc component is

P 0  Rao2  R  10% of P

Chapter 17, Solution 48. (a) For the DC component, i(t) = 20 mA. The capacitor acts like an open circuit so that v = Ri(t) = 2x103x20x10–3 = 40 For the AC component,  n = 10n, n = 1,2 1/(j n C) = –j/(10nx100x10–6) = (–j/n) k Z = 2||(–j/n) = 2(–j/n)/(2 – j/n) = –j2/(2n – j) V = ZI = [–j2/(2n – j)]I For n = 1,

V 1 = [–j2/(2 – j)]1645 = 14.311–18.43 mV

For n = 2,

V 2 = [–j2/(4 – j)]12–60 = 5.821–135.96 mV

(b)

v(t) = 40 + 0.014311cos(10t – 18.43) + 0.005821cos(20t – 135.96) V 1  p = V DC I DC +  Vn I n cos( n   n ) 2 n 1

= 20x40 + 0.5x10x0.014311cos(45 + 18.43) +0.5x12x0.005821cos(–60 + 135.96) = 800.1 mW


(a)

Z

2

rms

T  2  1 1 2 1  (20 )  10   z (t )dt    4dt   16dt   T0 2  0   2

Z rms = 3.162 (b) Z 2 rms  a02 

1 1  144 72  1 1  2  (an  bn2 )  1   2 2  1  2 1  0   0   ...   9.396    25 2 n1 n  9 2 n1  n odd

Z rms = 3.065 (c )

 3.065  %error  1   x100  3.068%  3.162 

Chapter 17, Solution 50. cn =

=

1 T



T

0

f ( t )e  jo nt dt,

o 

2n  1

1 1  jnt te dt 2 

Using integration by parts, u = t and du = dt dv = e–jntdt which leads to v = –[1/(2jn)]e–jnt t cn =  e  jnt 2 jn



1

 1

1 1  jnt e dt 2 jn 1



1 j  jn  e jnt  e  jnt e = 2 2 2 2n  (  j) n

1

1

= [j/(n)]cos(n) + [1/(2n22)](e–jn – ejn) cn =

j( 1) n j( 1) n 2j    sin( n ) n n 2n 2  2

Thus 

f(t) =

 c n e jnot =

n 



 (1)

n 

n

j jnt e n

Chapter 17, Solution 51. Design a problem to help other students to better understand how to find the exponential Fourier series of a given periodic function. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Given the periodic function f(t) = t2,

0
obtain the exponential Fourier series for the special case T = 2. Solution T  2,

o  2 / T  





1 1 1 e  jnt 2 c n   f ( t )e  jno t dt   t 2 e  jnt dt   n 2  2 t 2  2 jnt  2 0 3 T 2 2 ( jn) 0 0 T

cn 

2

1 3 3

j2n 

(4n 2  2  j4n) 

f (t ) 



n 

2

n

n  

2 2 2

2

2

(1  jn)

(1  jn )e jnt

Chapter 17, Solution 52. cn =

=

1 T



T

0

f ( t )e  jo nt dt,

o 

2n  1

1 1  jnt te dt 2 

Using integration by parts, u = t and du = dt dv = e–jntdt which leads to v = –[1/(2jn)]e–jnt t cn =  e  jnt 2 jn

1

 1



1 1  jnt e dt 2 jn 1



1 j  jn  e jnt  e  jnt e = 2 2 2 2n  (  j) n

1

1

= [j/(n)]cos(n) + [1/(2n22)](e–jn – ejn) cn =

j( 1) n j( 1) n 2j    sin( n ) n n 2n 2  2

Thus 

f(t) =

 c n e jnot =

n  



 ( 1)

n  

n

j jnt e n

Chapter 17, Solution 53.  o = 2/T = 2 cn =



T

0

1

e  t e  jno t dt   e (1 jno ) t dt 0

1 = e  (1  j2 n ) t 1  j2 n

1

 0





1 e  (1  j 2 n  )  1 1  j2n

= [1/(j2n)][1 – e–1(cos(2n) – jsin(2n))] = (1 – e–1)/(1 + j2n) = 0.6321/(1 + j2n 0.6321e j2 nt  n   1  j2n 

f(t) =

Chapter 17, Solution 54. T = 4,  o = 2/T = /2 cn =

1 T



T

0

f ( t )e  jo nt dt

=

1  1  jnt / 2 2e dt  4  0

=

j 2e  jn / 2  2  e  jn  e  jn / 2  e  j2 n  e  jn 2n



2

1

1e  jnt / 2 dt 



4

2

1e  jnt / 2 dt  



=



j 3e  jn / 2  3  2e  jn 2n 

f(t) =

c e

n  

n

jn o t





Chapter 17, Solution 55. T = 2,  o = 2/T = 1 cn =

But

i(t) =

cn =

1 T



T

0

i( t )e  jno t dt

sin( t ), 0,

0t   t  2

1  1  1 jt sin( t )e  jnt dt  (e  e  jt )e  jnt dt  2 0 2 0 2 j

1  e jt (1 n ) e  jt (1 n )      4j  j(1  n ) j(1  n ) 







0

1  e j(1 n )  1 e  j( n 1)  1    4  1  n 1 n 



1 e j (1  n )  1  ne j (1  n )  n  e  j (1  n )  1  ne  j (1 n )  n 4( n 2  1)



But ej = cos() + jsin() = –1 = e–j cn =

1 1  e  jn  jn  jn  jn  jn  e  e  ne  ne  2  4( n 2  1) 2 (1  n 2 )





Thus 

i(t) =



n  

1  e  jn jnt e 2(1  n 2 )

Chapter 17, Solution 56. c o = a o = 10,  o =  c o = (a n – jb n )/2 = (1 – jn)/[2(n2 + 1)] f(t) = 10 



(1  jn ) jnt e 2  1)

 2(n

n   n0

Chapter 17, Solution 57. a o = (6/–2) = –3 = c o c n = 0.5(a n –jb n ) = a n /2 = 3/(n3 – 2) f(t) =  3 



n

n   n0

3

3 e j50nt 2

Chapter 17, Solution 58. c n = (a n – jb n )/2, (–1)n = cos(n),  o = 2/T = 1 c n = [(cos(n) – 1)/(2n2)] – j cos(n)/(2n) Thus f(t) =

 cos(n )  jnt  cos(n)  1   j e 2 4 2n   2 n

Chapter 17, Solution 59. For f(t), T = 2,  o = 2/T = 1. a o = DC component = (1x + 0)/2 = 0.5 For

h(t), T = 2,  o = 2/T = . a o = (2x1 – 2x1)/2 = 0

Thus by replacing  o = 1 with  o =  and multiplying the magnitude by four, we obtain h(t) = 

j4e  j( 2n1) t  n   ( 2n  1) 

n0

Chapter 17, Solution 60. From Problem 17.24, a o = 0 = a n , b n = [2/(n)][1 – 2 cos(n)], c o = 0 c n = (a n – jb n )/2 = [j/(n)][2 cos(n) – 1], n  0.

Chapter 17, Solution 61.  o = 1.

(a) f(t) = a o +

A

n

cos(n o t   n )

= 6 + 4cos(t + 50) + 2cos(2t + 35) + cos(3t + 25) + 0.5cos(4t + 20) = 6 + 4cos(t)cos(50) – 4sin(t)sin(50) + 2cos(2t)cos(35) – 2sin(2t)sin(35) + cos(3t)cos(25) – sin(3t)sin(25) + 0.5cos(4t)cos(20) – 0.5sin(4t)sin(20) = 6 + 2.571cos(t) – 3.73sin(t) + 1.635cos(2t) – 1.147sin(2t) + 0.906cos(3t) – 0.423sin(3t) + 0.47cos(4t) – 0.171sin(4t) (b)

f rms =

a o2 

1  2  An 2 n 1

f rms 2 = 62 + 0.5[42 + 22 + 12 + (0.5)2] = 46.625 f rms = 6.828

Chapter 17, Solution 62. (a) f(t)  12  10cos(2ot  90o )  8 cos(4ot  90o )  5cos(6ot  90o )  3cos(8ot  90 o ) (b) f(t) is an even function of t.

Chapter 17, Solution 63. This is an even function. T = 3,  o = 2/3, b n = 0. f(t) =

ao =

an =

1, 0  t  1 2, 1  t  1.5 1.5 2 T/2 2 1 f ( t ) dt 1 dt   1 2 dt  = (2/3)[1 + 1] = 4/3 T 0 3  0

1.5 4 T/2 4 1 f ( t ) cos(no t )dt    1cos(2nt / 3)dt   2 cos(2nt / 3)dt     0 0 1 3 T

4 3 6  2nt   2nt  =  sin  sin     3  2n  3  0 2n  3  1 1

1.5

  

= [–2/(n)]sin(2n/3) f 2 (t) =

4 2  1  3n   2nt    sin   cos  3  n 1 n  3   3 

a o = 4/3 = 1.3333,  o = 2/3, a n = –[2/(n)]sin(2nt/3) An =

a 2n  b 2n 

2  2n  sin   n  3 

A 1 = 0.5513, A 2 = 0.2757, A 3 = 0, A 4 = 0.1375, A 5 = 0.1103

The amplitude spectra are shown below. 1.333

An

0.551

0.275 0.1378 0.1103 0

0

1

2

3

4

5

n

Chapter 17, Solution 64. Design a problem to help other students to better understand the amplitude and phase spectra of a given Fourier series. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Given that v(t) = (10/π)[1 + (1/2)cos(2πt) + (2/3)cos(4πt) – (2/15)cos(6πt)] V draw the amplitude and phase spectra for v(t). Solution The amplitude and phase spectra are shown below. An 3.183 2.122 1.591 0.4244 0

2

4

2

4

6



n

0

6

-180o



Chapter 17, Solution 65. a n = 20/(n22), b n = –3/(n),  n = 2n A n = a 2n  b 2n 

=

400 9  2 2 4 4 n  n 

3 44.44 1  2 2 , n = 1, 3, 5, 7, 9, etc. n n 

n 1 3 5 7 9

An 2.24 0.39 0.208 0.143 0.109

 n = tan–1(b n /a n ) = tan–1{[–3/(n)][n22/20]} = tan–1(–nx0.4712)

n 1 3 5 7 9 

n –25.23 –54.73 –67 –73.14 –76.74 –90

2.24

0

2

6

10

14

18 n

–30

–25.23

An 0.39 –60 0.208

0

2

6

10

–54.73

0.0.143 0.109

14

18

n

–90

n

–67 –73.14

–76.74

Chapter 17, Solution 66. The schematic is shown below. The waveform is inputted using the attributes of VPULSE. In the Transient dialog box, we enter Print Step = 0.05, Final Time = 12, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, the output plot is shown below. The output file includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 5.099510E+00 HARMONIC NO 1 2 3 4 5 6 7 8 9

FREQUENCY FOURIER NORMALIZED (HZ) COMPONENT COMPONENT 5.000E-01 1.000E+00 1.500E+00 2.000E+00 2.500E+00 3.000E+00 3.500E+00 4.000E+00 4.500E+00

3.184E+00 1.593E+00 1.063E+00 7.978E-01 6.392E-01 5.336E-01 4.583E-01 4.020E-01 3.583E-01

1.000E+00 5.002E-01 3.338E-01 2.506E-01 2.008E-01 1.676E-01 1.440E-01 1.263E-01 1.126E-01

PHASE (DEG) 1.782E+00 3.564E+00 5.347E+00 7.129E+00 8.911E+00 1.069E+01 1.248E+01 1.426E+01 1.604E+01

NORMALIZED PHASE (DEG) 0.000E+00 1.782E+00 3.564E+00 5.347E+00 7.129E+00 8.911E+00 1.069E+01 1.248E+01 1.426E+01

TOTAL HARMONIC DISTORTION = 7.363360E+01 PERCENT

From Prob. 17.4, we know the phase angle should be zero. Why do we have a phase angle equal to n(1.782)? The answer is actually quite straight forward. The angle comes from the approximation of the leading edge of the pulse. The graph shows an instantaneous rise whereas PSpice needs a finite rise time, thus artificially creating a phase shift.

Chapter 17, Solution 67. The Schematic is shown below. In the Transient dialog box, we type “Print step = 0.01s, Final time = 36s, Center frequency = 0.1667, Output vars = v(1),” and click Enable Fourier. After simulation, the output file includes the following Fourier components,

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 2.000396E+00 HARMONIC FREQUENCY FOURIER NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9

1.667E-01 3.334E-01 5.001E-01 6.668E-01 8.335E-01 1.000E+00 1.167E+00 1.334E+00 1.500E+00

2.432E+00 6.576E-04 5.403E-01 3.343E-04 9.716E-02 7.481E-06 4.968E-02 1.613E-04 6.002E-02

1.000E+00 -8.996E+01 2.705E-04 -8.932E+01 2.222E-01 9.011E+01 1.375E-04 9.134E+01 3.996E-02 -8.982E+01 3.076E-06 -9.000E+01 2.043E-02 -8.975E+01 6.634E-05 -8.722E+01 2.468E-02 9.032E+01

PHASE NORMALIZED PHASE (DEG)

0.000E+00 6.467E-01 1.801E+02 1.813E+02 1.433E-01 -3.581E-02 2.173E-01 2.748E+00 1.803E+02


Chapter 17, Solution 68. Since T=3, f =1/3 = 0.333 Hz. We use the schematic below.

We use VPWL to enter in the signal as shown. In the transient dialog box, we enable Fourier, select 15 for Final Time, 0.01s for Print Step, and 10ms for the Step Ceiling. When the file is saved and run, we obtain the Fourier coefficients as part of the output file as shown below.

Why is this problem wrong? Clearly the source is not periodic. The DC value must be +1!!!!!!!!!! FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = -1.000000E+00 HARMONIC FREQUENCY FOURIER NORMALIZED NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9

3.330E-01 6.660E-01 9.990E-01 1.332E+00 1.665E+00 1.998E+00 2.331E+00 2.664E+00 2.997E+00

1.615E-16 5.133E-17 6.243E-16 1.869E-16 6.806E-17 1.949E-16 1.465E-16 3.015E-16 1.329E-16

1.000E+00 3.179E-01 3.867E+00 1.158E+00 4.215E-01 1.207E+00 9.070E-01 1.867E+00 8.233E-01

PHASE PHASE (DEG)

1.762E+02 0.000E+00 2.999E+01 -3.224E+02 6.687E+01 -4.617E+02 7.806E+01 -6.267E+02 1.404E+02 -7.406E+02 -1.222E+02 -1.179E+03 -4.333E+01 -1.277E+03 -1.749E+02 -1.584E+03 -9.565E+01 -1.681E+03

Chapter 17, Solution 69. The schematic is shown below. In the Transient dialog box, set Print Step = 0.05 s, Final Time = 120, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, we obtain V(1) as shown below. We also obtain an output file which includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)

DC COMPONENT = 5.048510E-01 HARMONIC FREQUENCY FOURIER NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9


5.000E-01 4.056E-01 1.000E+00 -9.090E+01 0.000E+00 1.000E+00 2.977E-04 7.341E-04 -8.707E+01 3.833E+00 1.500E+00 4.531E-02 1.117E-01 -9.266E+01 -1.761E+00 2.000E+00 2.969E-04 7.320E-04 -8.414E+01 6.757E+00 2.500E+00 1.648E-02 4.064E-02 -9.432E+01 -3.417E+00 3.000E+00 2.955E-04 7.285E-04 -8.124E+01 9.659E+00 3.500E+00 8.535E-03 2.104E-02 -9.581E+01 -4.911E+00 4.000E+00 2.935E-04 7.238E-04 -7.836E+01 1.254E+01 4.500E+00 5.258E-03 1.296E-02 -9.710E+01 -6.197E+00


Chapter 17, Solution 70. Design a problem to help other students to better understand how to use PSpice to solve circuit problems with periodic inputs. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Rework Prob. 17.40 using PSpice. Chapter 17, Problem 40. The signal in Fig. 17.77(a) is applied to the circuit in Fig. 17.77(b). Find v o (t).

Figure 17.77 Solution The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s, Final Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable Fourier.

After simulation, we compare the output and output waveforms as shown. The output includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)

DC COMPONENT = 7.658051E-01 HARMONIC FREQUENCY FOURIER NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9

5.000E-01 1.070E+00 1.000E+00 3.758E-01 1.500E+00 2.111E-01 2.000E+00 1.247E-01 2.500E+00 8.538E-02 3.000E+00 6.139E-02 3.500E+00 4.743E-02 4.000E+00 3.711E-02 4.500E+00 2.997E-02

1.000E+00 3.512E-01 1.973E-01 1.166E-01 7.980E-02 5.738E-02 4.433E-02 3.469E-02 2.802E-02

1.004E+01 -3.924E+01 -3.985E+01 -5.870E+01 -5.680E+01 -6.563E+01 -6.520E+01 -7.222E+01 -7.088E+01


0.000E+00 -4.928E+01 -4.990E+01 -6.874E+01 -6.685E+01 -7.567E+01 -7.524E+01 -8.226E+01 -8.092E+01


Chapter 17, Solution 71. The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s, Final Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable Fourier. After simulation, we compare the output and output waveforms as shown. The output includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 7.658051E-01 HARMONIC FREQUENCY FOURIER NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9

5.000E-01 1.070E+00 1.000E+00 3.758E-01 1.500E+00 2.111E-01 2.000E+00 1.247E-01 2.500E+00 8.538E-02 3.000E+00 6.139E-02 3.500E+00 4.743E-02 4.000E+00 3.711E-02 4.500E+00 2.997E-02

1.000E+00 3.512E-01 1.973E-01 1.166E-01 7.980E-02 5.738E-02 4.433E-02 3.469E-02 2.802E-02

1.004E+01 -3.924E+01 -3.985E+01 -5.870E+01 -5.680E+01 -6.563E+01 -6.520E+01 -7.222E+01 -7.088E+01


0.000E+00 -4.928E+01 -4.990E+01 -6.874E+01 -6.685E+01 -7.567E+01 -7.524E+01 -8.226E+01 -8.092E+01


Chapter 17, Solution 72. T = 5,  o = 2/T = 2/5 f(t) is an odd function. a o = 0 = a n bn =

4 T/2 4 10 f ( t ) sin(no t )dt   10 sin(0.4nt )dt  T 0 5 0 1

= 

8x 5 20 cos(0.4nt ) = [1  cos(0.4n)] n 2 n 0

f(t) =

20  1  [1  cos(0.4n)]sin(0.4nt )  n 1 n

Chapter 17, Solution 73. 2 VDC 1 Vn2   p = R 2 R

= 0 + 0.5[(22 + 12 + 12)/10] = 300 mW


An =

a 2n  b 2n ,

 = tan–1(b n /a n )

A1 =

6 2  8 2 = 10,

 1 = tan–1(6/8) = 36.87

A2 =

3 2  4 2 = 5,

 2 = tan–1(3/4) = 36.87

i(t) = {4 + 10cos(100t – 36.87) – 5cos(200t – 36.87)} A

(b)

p = I 2DC R  0.5 I 2n R = 2[42 +0.5(102 + 52)] = 157 W


The lowpass filter is shown below. R

+

+ C

vs

vo

-

vs 

-

n A 2A  1  cos no t sin  T T T n 1n

1 j n C 1 Vs , Vs  Vo  1 1  j n RC R j n C For n=0, (dc component), Vo  Vs 

 n  no  2n / T

A T

(1)

For the nth harmonic, Vo 

When n=1, | Vo |

2A n sin   90 o T 1   2 n R 2 C 2  tan 1  n RC nT 1

2A n sin  T T



1

(2)

4 2 2 2 1 R C T

From (1) and (2),  A 2A  50 x sin T T 10

1 1

4 2 2 2 R C T

 

1

4 2 2 2 30.9  3.09 x10 4 R C   T

1

4 2 2 2 R C  10 10 T

 

C

T 10 2 x 3.09x10 4 10 5   24.59 mF 2R 4x10 3

Chapter 17, Solution 76. v s (t) is the same as f(t) in Figure 16.1 except that the magnitude is multiplied by 10. Hence v o (t) = 5 

20  1  sin(nt ) , n = 2k – 1  k 1 n

T = 2,  o = 2/T = 2,  n = n o = 2n j n L = j2n; Z = R||10 = 10R/(10 + R) V o = ZV s /(Z + j2n) = [10R/(10R + j2n(10 + R))]V s Vo =

10R  tan 1{(n / 5R )(10  R )} 100R 2  4n 2  2 (10  R ) 2

Vs

V s = [20/(n)]0 The source current I s is

Is =

Vs Vs  10R Z  j2n  j2n 10  R

20 n  10R  j2n(10  R ) (10  R )

20   tan 1{( n / 3)(10  R )} n 100R 2  4n 2  2 (10  R ) 2

(10  R )

=

p s = V DC I DC +

1  Vsn I sn cos( n   n ) 2

For the DC case, L acts like a short-circuit. Is =

5(10  R ) 5 , Vs = 5 = Vo  10R 10R 10  R

  1    tan  (10  R )   2 (10  R ) cos   25(10  R ) 1  20   5  ps     2 2 2  10R 2   100R  4 (10  R )      2  (10  R ) 2 cos tan 1  (10  R )     5   10       2 2 2     100R  16 (10  R )   2

ps =

=

VDC 1  Von   R 2 n 1 R  25 1  100R 100R      2 2 2 2 2 2 R 2 100R  4 (10  R ) 100R  10 (10  R ) 

We want p o = (70/100)p s = 0.7p s . Due to the complexity of the terms, we consider only the DC component as an approximation. In fact the DC component has the latgest share of the power for both input and output signals.

25 7 25(10  R )  x R 10 10R 100 = 70 + 7R which leads to R = 30/7 = 4.286 

Chapter 17, Solution 77. (a) For the first two AC terms, the frequency ratio is 6/4 = 1.5 so that the highest common factor is 2. Hence  o = 2. T = 2/ o = 2/2 =  (b) The average value is the DC component = –2 (c)

V rms =

ao 

1  2 (a n  b 2n )  2 n 1

1 2 Vrms  (2) 2  (10 2  8 2  6 2  3 2  12 ) = 121.5 2 V rms = 11.02 V

Chapter 17, Solution 78. 2

(a)

V V2 V2 V2 1 p = DC   n  DC   n ,rms R 2 R R R

= 0 + (402/5) + (202/5) + (102/5) = 420 W (b)

5% increase = (5/100)420 = 21 p DC = 21 W = V DC = 10.25 V

2 VDC 2 which leads to VDC  21R  105 R

Chapter 17, Solution 79. From Table 17.3, it is evident that a n = 0, b n = 4A/[(2n – 1)], A = 10. A Fortran program to calculate b n is shown below. The result is also shown. C

10

FOR PROBLEM 17.79 DIMENSION B(20) A = 10 PIE = 3.142 C = 4.*A/PIE DO 10 N = 1, 10 B(N) = C/(2.*FLOAT(N) – 1.) PRINT *, N, B(N) CONTINUE STOP END

n 1 2 3 4 5 6 7 8 9 10

bn 12.731 4.243 2.546 1.8187 1.414 1.1573 0.9793 0.8487 0.7498 0.6700

Chapter 17, Solution 80. From Problem 17.55, c n = [1 + e–jn]/[2(1 – n2)] This is calculated using the Fortran program shown below. The results are also shown. C

10

FOR PROBLEM 17.80 COMPLEX X, C(0:20) PIE = 3.1415927 A = 2.0*PIE DO 10 N = 0, 10 IF(N.EQ.1) GO TO 10 X = CMPLX(0, PIE*FLOAT(N)) C(N) = (1.0 + CEXP(–X))/(A*(1 – FLOAT(N*N))) PRINT *, N, C(N) CONTINUE STOP END

n 0 1 2 3 4 5 6 7 8 9 10

cn 0.3188 + j0 0 –0.1061 + j0 0 –0.2121x10–1 + j0 0 –0.9095x10–2 + j0 0 –0.5052x10–2 + j0 0 –0.3215x10–2 + j0

Chapter 17, Solution 81. (a) A

0

2T

T f(t) =

2A 4A  1 cos(no t )   2   n 1 4n  1

The total average power is

p avg = F rms 2R = F rms 2 since R = 1 ohm.

P avg = F rms 2 =

(b)

3T

1 T 2 f ( t )dt = 0.5A2  0 T

From the Fourier series above |c o | = 2A/π, |c n | = |a n |/2 = 2A/[(4n2 – 1)] n 0 1 2 3 4

o 0 2 o 4 o 6 o 8 o

|c n | 2A/ 2A/(3) 2A/(15) 2A/(35) 2A/(63)

(c)

81.1%

(d)

0.72%

|c o |2 or 2|c n |2 4A2/(2) 8A2/(92) 8A2/(2252) 8A2/(12252) 8A2/(39692)

% power 81.1% 18.01% 0.72% 0.13% 0.04%

Chapter 17, Solution 82. 2 VDC 1  Vn2   P = R 2 n 1 R

Assuming V is an amplitude-phase form of Fourier series. But |A n | = 2|C n |, c o = a o |A n |2 = 4|C n |2 Hence,  c o2 c 2n  2 P = R n 1 R

Alternatively, 2 Vrms P = R

where 2 Vrms  a o2 

  1  2 2 2 A  c  2 c  c 2n    n o n 2 n 1 n 1 n  

= 102 + 2(8.52 + 4.22 + 2.12 + 0.52 + 0.22) = 100 + 2x94.57 = 289.14 P = 289.14/4 = 72.3 W


f ' ( t )  ( t  2)  ( t  1)  ( t  1)  ( t  2) jF()  e j2   e j  e  j  e  j2  2 cos 2  2 cos  F() =

2[cos 2  cos ] j

Chapter 18, Solution 2. Using Fig. 18.27, design a problem to help other students to better understand the Fouier transform given a wave shape. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem What is the Fourier transform of the triangular pulse in Fig. 18.27?

Figure 18.27 Solution t, f (t)   0,

0  t 1 otherwise f ”(t)

f ‘(t) 1

(t)

0

t

1

–’(t-1) -(t-1)

f"(t) = (t) - (t - 1) - '(t - 1) Taking the Fourier transform gives -2F() = 1 - e-j - je-j F() =

t

(1  j)e  j  1 2

-(t-1)

1

or F()   t e  jt dt 0

eax But  x e dx  2 (ax  1)  c a 1 e  j ( jt  1) 10  2 1  je  j  1 F()  2   j ax






f (t) 

1 t ,  2  t  2, 2

1 f ' (t)  ,  2  t  2 2

1 j t e  j t ( jt  1) 2 2 t e dt  2 2 2( j) 2 1   2 e  j2 ( j2  1)  e j2 ( j2  1) 2 1   j2 e  j2  e j2  e j2  e  j2 2 2 1   2  j4 cos 2  j2 sin 2 2

F()  

2







F() =





j ( 2 cos 2  sin 2) 2



Chapter 18, Solution 4. We can solve the problem by following the approach demonstrated in Example 18.5. 2δ(t+1) g’ 2 –1 0

1

t

–2 –2δ(t–1)

4δ(t) 2δ’(t+1) g” –1 0 –2δ(t+1)

1

t

–2 –2δ(t–1) –2δ’(t–1)

g   2( t  1)  2( t  1)  4( t )  2( t  1)  2( t  1) ( j) 2 G ()  2e j  2 je j  4  2e  j  2 je  j  4 cos   4 sin   4

G ( ) 

4 (cos    sin   1) 2

Chapter 18, Solution 5. h’(t) 1

0

–1

t

1

–2δ(t)

h”(t) 1

δ(t+1)

1 t

0

–1 –2δ’(t)

–δ(t–1)

h ( t )  ( t  1)  ( t  1)  2( t ) ( j) 2 H()  e j  e  j  2 j  2 j sin   2 j

H(ω) =

2j 2j  sin   2

Chapter 18, Solution 6. (a) The derivative of f(t) is shown below. f’(t) 5(t) 0

5(t-1) 1

2

t -10(t-2)

f '(t )  5 (t )  5 (t  1)  10 (t  2) Taking the Fourier transform of each term, j F ( )  5  5e  j  10e  j 2 F ( ) 

5  5e  j  10e  j 2 j

(b) The derivative of g(t) is shown below. g’(t) 10(t) 0

1

2 -5

-5(t-1)

The second derivative of g(t) is shown below. g’’(t) 10’(t) 0

5(t-2)

1

2

t

-5’(t-1) -5(t-1) g”(t) = 10δ’(t) – 5δ’(t–1) – 5δ(t–1) + 5δ(t–2) Take the Fourier transform of each term. (jω)2G(jω) = 10jω – 5jωe–jω – 5e–jω + 5e–j2ω which leads to G(jω) = (–10jω + 5jωe–jω + 5e–jω – 5e–j2ω)/ω2

Chapter 18, Solution 7. (a) Take the derivative of f 1 (t) and obtain f 1 ’(t) as shown below. 2(t)

0

1

2

t

-(t-1) -(t-2) f1' (t )  2 (t )   '(t  1)   (t  2) Take the Fourier transform of each term, j F1 ( )  2  e  j  e  j 2 F1 ( ) 

2  e  j  e  j 2 j

(b) f 2 (t) = 5t

F2 ( ) 







f 2 (t )e

 j

2

dt   5te  j dt  0

F2 ( ) 

2 5  j t e  j   ( 1) 0 ( j ) 2

5e  j 2



2

(1  j 2) 

5

2


1

F()   2e

 jt

dt   (4  2 t )e  jt dt

0

(a) 

1

2  jt 1 4  jt 2 2  jt 2 e  e  e ( jt  1) 1 0 1 2  j  j 

F(  ) 

(b)

2

2 4  j2 2 2  j 2   e  2 (1  j2)e  j2  e  2 j j j  

g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ]

G ( ) 

4 sin 2 2 sin    


y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ]

Y( ) 

1

(b) Z()   ( 2t )e  jt dt  0

2 4 sin 2  sin   

 2e  jt (  jt  1)  2

1 0



2 2e  j  (1  j) 2 2

Chapter 18, Solution 10. x(t) = e2tu(t)

(a)

X() = 1/(–2 + j) (b)

e  t , t  0 e ( t )   t e , t  0 1

0

1

1

1

0

Y()   y( t )e jt dt   e t e jt dt   e  t e  jt dt



e (1 j) t 1  j



0 1



e  (1 j) t  (1  j)

1 0

 cos   jsin  cos   jsin   2   e 1   2 1  j 1  j 1   

Y() =



2 1  e 1 (cos    sin ) 2 1 



Chapter 18, Solution 11. f(t) = sin  t [u(t) - u(t - 2)] 2

F()   sin t e  jt dt  0







1 2 j t e  e  j t e  jt dt  0 2j

1  2  j(     ) t (e  e  j(  ) t )dt     0 2j 

e  j(    ) t 2  1  1 e  j(   ) t 02   0  j(  )  2 j   j(  ) 1  1  e  j2  1  e  j2       2     







1 2  2e  j2  2 2(   ) 2

 F() =





 e  j 2  1 2   2

Chapter 18, Solution 12. (a) F1 ( ) 

10 (3  j ) 2  100

(b) F2 ( ) 

4  j (4  j ) 2  100


(a) We know that F[cos at ]  [(  a )  (  a )] .

Using the time shifting property, F[cos a(t   / 3a )]  e  j / 3a [(  a )  (  a )]  e  j / 3 (  a )  e j / 3 (  a )

(b) sin ( t  1)  sin t cos   cos t sin    sin t g(t) = -u(t+1) sin (t+1) Let x(t) = u(t)sin t, then X() 

1 2

( j)  1



1 1  2

Using the time shifting property, G ()  

e j 1 j e  1  2 2  1

(c ) Let y(t) = 1 + Asin at, then Y()  2()  jA[(  a )  (  a )] h(t) = y(t) cos bt Using the modulation property,

1 H()  [Y(  b)  Y(  b)] 2 H()  (  b )  (  b ) 

4

(d) I( )   (1  t )e 0

 jt

jA (  a  b)  (  a  b)  (  a  b)  (  a  b) 2

e  jt e  jt dt   (  jt  1)  j   2

4 0

1 e  j4  e  j4   2  2 ( j4  1) j  

Chapter 18, Solution 14. Design a problem to help other students to better understand finding the Fourier transform of a variety of time varying functions (do at least three). Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the Fourier transforms of these functions: (a) (b) (c) (d) (e)

f(t) = e-t cos (3t + ) u(t) g(t) = sin  t [ u(t + 1) - u(t-1)] h(t) = e-2t cos  t u(t-1) p(t) = e-2t sin 4t u(-t) q(t) = 4 sgn (t - 2) + 3(t) - 2 u(t - 2)

Solution (a)

cos(3t  )  cos 3t cos   sin 3t sin   cos 3t (1)  sin 3t (0)   cos(3t ) f ( t )  e  t cos 3t u ( t )  1  j F() = 1  j2  9

(b) g(t) 1 -1 1

t

-1

g’(t)

 -1

1 t

-

g ' ( t )   cos tu ( t  1)  u ( t  1)

g" ( t )   2 g( t )  ( t  1)  ( t  1)   2 G ()    2 G ()  e j  e  j  2  2 G()  (e j  e  j )  2 j sin  2 j sin  G() = 2   2 Alternatively, we compare this with Prob. 17.7 f(t) = g(t - 1) F() = G()e-j

 e  j  e j 2    j2 sin   2   2 2 j sin  G() =  2  2

G   F()e j 

2





cos ( t  1)  cos t cos   sin t sin   cos t (1)  sin t (0)   cos t Let x ( t )  e 2( t 1) cos ( t  1)u ( t  1)  e 2 h ( t ) y( t )  e 2 t cos(t )u ( t ) 2  j Y()  (2  j) 2   2 y( t )  x ( t  1) Y()  X()e  j

(c) and

X() 

2  je j 2  j2   2

X()  e 2 H() H()  e 2 X() =

 2  je j 2

2  j2   2

Let x ( t )  e 2 t sin( 4t )u ( t )  y( t ) p( t )   x ( t ) where y( t )  e 2 t sin 4t u ( t ) 2  j Y ()  2  j2  4 2 2  j X()  Y()  2  j2  16

(d)

p()  X() 

(e)

j  2

j  2 2  16

 1  8  j 2  3  2 ()  e  j2 e j  j  6 j 2 Q() = e  3  2()e  j 2 j Q() 


F()  e j3  e  j3  2 j sin 3

(b)

Let g ( t )  2( t  1), G ()  2e  j t F()  F   g ( t ) dt     G ()   F(0)() j



2e  j  2(1)() j

2e  j = j (c)

F (2 t ) 

1 1 2 1 1 1 j F()   1  j   3 2 3 2

Chapter 18, Solution 16. (a) Using duality properly

t 

or

2 2

2  2  t2 4   4  t2 4 F() = F  2    4  t 

(b)

e

at

2a a  2 2

2a a  t2

2 e

a 

8 a  t2

4 e

2 

2

2

 8  2   4 e G() = F  2  4t 


1 F  0   F  0  2 1 where F() = F u t     , 0  2 j

(a) Since H() = F cos 0 t f ( t )  

H 

1 1 1     2      2    2 j   2  j   2 



   2    2  j    2    2  2 2    2  2  H() =

(b)

   2    2   2j 2  4

G() = F sin 0 t f ( t ) 

j F  0   F  0  2 1 where F() = F u t      j  j 1 1 G     10     10  2 j  10 j  10   

j   10     10  j  j  j  2 2    10   10  =

j   10    10   2 10 2   100


(a) F [ f (t  to )] 



f (t  to )e jt dt



Let t  to  

  t    to , 



F [ f (t  to )] 

dt  d 

f ( )e  j e jto d  e jto F ( )



(b) Given that

f (t )  F 1[ F ( )] 

1 2







F ( )e jt d



j f '(t )  F ( )e jt dt  j F 1[ F ( )]  2  or F [ f '(t )]  j F ( ) (c ) This is a special case of the time scaling property when a = –1. Hence, F [ f (t )] 

1 F ( )  F ( ) | 1|

(d) F ( )  





f (t )e  jt dt

Differentiating both sides respect to  and multiplying by t yields   dF ( ) j  j  ( jt ) f (t )e  jt dt   tf (t )e  jt dt d   Hence, dF ( ) j  F [tf (t )] d


F   f ( t )e jt dt  

F 







1 1 j2 t e  e  j2 t e  jt dt 2 0



1 1  j   2   t e  e  j 2  t dt  0 2 1

 1 1 1 e  j   2   t  e  j   2   t    2   j   2   j  2  0 1  e  j  2    1 e  j  2    1      2  j   2 j  2  

But

e j2   cos 2  j sin 2  1  e  j2  1  e  j  1  1 1   F      2 j    2   2 

=





j e  j  1 2   4 2


F (c n ) = c n ()





F c n e jno t  c n   no 

(b) cn 



n  

T  2

o 



 c   n  n

o

2 1 T

1 T 1    jnt f t  e  jno t dt   1 e dt  0    T 0 2  0 1  1 jnt  e 2  jn

But e  jn

cn 

   F   c n e jno t    n   

 j   e  jn  1 2  n   cos n  j sin n  cos n  (1) n



 0



 j  1n  1   0,j , 2n  n



n  even n  odd , n  0

for n = 0 cn 

1  1 1 dt   0 2 2

Hence f (t) 

 1 j jnt e   2 n   n n 0 n  odd

F() =

 1 j      n  2 n   n n0 n  odd




Using Parseval’s theorem, 

 

f 2 ( t )dt 

1  | F() | 2 d    2

If f(t) = u(t+a) – u(t+a), then 



a

f 2 ( t )dt   (1) 2 dt  2a  a

or 2

4a   sin a    a  d  4a 2  a as required. 

2

1   sin a  4a 2   d    2  a 

Chapter 18, Solution 22. F f ( t ) sin o t   







e f (t)

j o t



 e  jo t  jt e dt 2j

 1  f ( t )e  j o t dt   e  j o t dt     2 j   

=

1 F   o   F  o  2j

Chapter 18, Solution 23. 1 10 30   3 2  j / 35  j / 3 6  j15  j 30 F f  3t   6  j15  j (a) f(3t) leads to

(b) f(2t)

1 10 20   2 2  j / 215  j / 2 4  j10  j

20e  j / 2 4  j10  j 1 1 (c) f(t) cos 2t F  2  F  2  2 2 5 5 =  2  j  25  j  2 2  j  2 5  j  2

f(2t-1) = f [2(t-1/2)]

(d) F f ' t   j F  

(e)

j10 2  j5  j

F  F0  j 10 x10    j2  j5  j 2x5 10 =   j2  j5  j

 f t  dt t



Chapter 18, Solution 24. (a) X  F + F [3] j = 6  e  j  1 





(b) yt   f t  2  Y  e  j2 F 

je  j2  j e 1 



(c) If h(t) = f '(t)

H  jF  j







j  j e  1  1  e  j 

3 3  3 3  5  2  (d) g t   4f  t   10f  t , G ()  4 x F    10x F   2 2  5 5  3  3   6

=

j 3  2



e

 j3 / 2







1 

6 j  j3 / 5 e 1 3  5







j4  j3 / 2 j10  j3 / 5 e 1  e 1  

Chapter 18, Solution 25. (a) g(t) = 5e2tu(t) (b) h(t) = 6e–2t A B (c ) X ( )   , s 1 s  2

s  j

10 10  10, B  10 1 2 2 1 10 10  X ( )  j  1 j  2 x(t) = (–10etu(t)+10e2t)u(t)

A

(a) 5e2tu(t), (b) 6e–2t , (c) (–10etu(t)+10e2t)u(t)

Chapter 18, Solution 26. (a) f (t )  e  ( t  2 ) u(t ) (b) h(t )  te 4 t u(t ) (c) If x(t )  u(t  1)  u(t  1)

 

X()  2

sin  

By using duality property,

G ()  2u(  1)  2u(  1)

 

g( t ) 

2 sin t t

Chapter 18, Solution 27. (a) Let Fs  

100 A B   , s  j s s  10 s s  10 100 100 A  10, B   10  10 10 10 10 F   j j  10 f(t) = (5 sgnt   10e 10 t )ut 

(b) G s  

10s A B   , s  j 2  s 3  s  2  s s  3 20  30 A  4, B   6 5 5 4 6 G      j  2 j   3 g(t) = 4e 2 t u t   6e 3 t ut  60 60 (c) H   2  j  j40  1300  j  202  900 h(t) = 2e 20 t sin( 30t ) ut 

(d) yt  

1  e jt d 1 1 1      2  2  j j  1 2 2 4

Chapter 18, Solution 28. () e jt 1  1  j t F ( ) e d    d 2  2  (5  j)(2  j) 1 1 1    0.05 2 (5)(2) 20

(a)

f (t) 

(b)

1  10(  2) jt 10 e  j2 t f (t)  e d  2  j( j  1) 2 ( j2)( j2  1)



(c)

f (t) 



(d)

j5 e  j2 t ( 2  j)e  j2 t  2 2 1  j2 20 e jt 1  20(  1)e jt d   2  (2  j)(3  5) 2 (2  j)(3  j)

(1  j)e jt 20e jt   2(5  5 j)

Let

5() 5   F1 ()  F2 () (5  j) j(5  j) 1  5() jt 5 1 f1 ( t )  e d    0.5  2    5  j 2 5 F() 

5 A B   , A  1, B  1 s(5  s) s s  5 1 1 F2 ()   j j  5 F2 (s) 

f 2 (t) 

1 1 sgn( t )  e 5 t    u ( t )  e 5 t 2 2 f ( t )  f 1 ( t )  f 2 ( t )  u( t )  e 5 t


(a)

(b)

f(t) = F -1 [()]  F -1 4(  3)  4(  3) 1 1 4 cos 3t 1  8 cos 3t    = 2  2

If h ( t )  u ( t  2)  u ( t  2)

H() 

2 sin 2 

G ()  4H()

g(t) = (c)

g( t ) 

4 sin 2t t

Since cos(at)

(  a )  (  a )

Using the reversal property, 2 cos 2  ( t  2)  ( t  2) or F -1 6 cos 2  3(t  2)  3(t  2)

1 8 sin 2 t  2 t


2 , j 2a Y() 2(a  j) H()   2  j X() j

y( t )  sgn( t )

(b) X() 

1 , 1  j

H( ) 

 

Y() 

Y() 

X() 

 

1 a  j

h(t )  2(t )  a[u(t )  u(  t )]

1 2  j

1  j 1  1 2  j 2  j

(c ) In this case, by definition, h(t ) 

 

h(t )  (t )  e  2 t u(t )

y (t )  e  at sin bt u(t )


Y() 

1 (a  j) 2

X() 

,

H() 

Y(  ) 1  H() a  j

1 a  j

 

x(t )  e at u(t )

(b)

By definition, x(t )  y(t )  u(t  1)  u(t  1)

(c )

Y() 

X() 

1 (a  j)

,

H() 

Y(  ) j 1 a    H() 2(a  j) 2 2(a  j)

2 j

 

x(t ) 

1 a (t )  e at u(t ) 2 2


(a)

e  j j  1 and F 

e   t 1 u ( t  1)

Since

f(-t)

j

F1  

e  j  1 f 1 (t) = e t 1 u t  1

(b)

f 1 t   e   t1 u  t  1

From Section 17.3,

2  2e t 1  If F2   2e , then 2 f 2 (t) = 2  t 1 2



(b)



By partial fractions

F3  

1

 j  1  j  12 2

1 1 1 1 4 4   4   4  j  12  j  1  j  12 j  1

1 t te  e t  te t  e t ut  4 1 1 = t  1e  t ut   t  1e t ut  4 4

Hence f 3 t  

(d)

f 4 t  

1 1  e jt 1  jt      = F e d 1       2 1  j2 2 2


(a)

Let x t   2 sin tu t  1  u t  1 From Problem 17.9(b),

4 j sin   2  2 Applying duality property, X 

1 2 j sin  t  X t   2 2 2  t 2 j sin t f(t) = 2 t  2 f t  

(b)

F 

j cos 2  j sin 2  j cos   j sin     j j e e j2   e j2   e  j     j j 1 1 f t   sgn t  1  sgn t  2 2 2 But sgn( t )  2u ( t )  1 1 1 f t   u t  1   u t  2   2 2 = ut  1  ut  2 


First, we find G() for g(t) shown below. g t   10u t  2   u t  2   10u t  1  u t  1 g ' t   10t  2   t  2   10t  1  t  1

The Fourier transform of each term gives g(t) 20 10

–2

–1

0

1

t

2

g ‘(t) 10(t+2)

–2

10(t+1)

–1

0

1

–10(t-1)









2

–10(t-2)

jG   10 e j2  e  j2  10 e j  e  j  20 j sin 2  20 j sin  20 sin 2 20 sin  G     40 sinc(2) + 20 sinc()   Note that G() = G(-). F  2G   1 f t   G t  2 = (20/)sinc(2t) + (10/)sinc(t)

t


(a)

x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties, 1 1 e  j / 3 e  j / 3  3 2  j / 3 (6  j)

X() 

(b)

Using the modulation property,

Y(  ) 

 1 1 1 1  [F(  5)  F(  5)]   2  2  j(  5) 2  j(  5)  2

j 2  j

(c )

Z()  jF() 

(d)

H()  F()F() 

(e)

I( )  j

1 ( 2  j) 2

d (0  j) 1 F( )  j  2 d ( 2  j) ( 2  j) 2

Chapter 18, Solution 36. H ( ) 

Y ( ) X ( )

 

x(t )  vs (t )  e 4t u (t ) Y ( ) 

Y ( )  H ( ) X ( )

 

X ( ) 

2 2  , ( j  2)(4  j ) ( s  2)( s  4)

1 4  j

s  j

A B  s2 s4 2 2 A  1, B  1 2  4 4  2 1 1 Y (s)   s2 s4 y (t )   e2t  e4t  u (t )

Y (s) 

Please note, the units are not known since the transfer function does not give them. If the transfer function was a voltage gain then the units on y(t) would be volts.


2 j 

j2 2  j

By current division, j2 I  j2 2  j H  o    j 2 I s  j2  8  j4 4 2  j H() =

j 4  j3

Chapter 18, Solution 38. Using Fig. 18.40, design a problem to help other students to better understand using Fourier transforms to do circuit analysis. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Suppose v s (t) = u(t) for t>0. Determine i(t) in the circuit of Fig. 18.40 using Fourier transform. 1 i

vs

+ _

1H

Figure 18.40

For Prob. 18.38.

Solution Vs   ( ) 

1 j

Vs 1  1     ( )   1  j 1  j  j   ( ) 1 Let I ( )  I1 ( )  I 2 ( )   1  j j (1  j ) 1 A B I 2 ( )    , s  j j (1  j ) s s  1 I ( ) 

1 where A   1, 1 1 1  j j  1  ( ) I1 ( )  1  j I 2 ( ) 

B

1  1 1

1   i2 (t )  sgn(t )  e t 2

i1 (t ) 

1 2

 ( ) jt 1 e jt 1   e d   1  j 2 1  j   0 2 

Hence, i (t )  i1 (t )  i2 (t ) 

1 1  sgn(t )  e t 2 2


Vs () 

 (1  t )e

 jt



I( ) 

Vs () 10 3  jx10  3



dt 

1 1 1  j   e 2 j  2

 10 3  1 1 1   2  2 e  j  6  10  j  j  


v( t )  ( t )  2( t  1)  ( t  2)   2 V()  1  2e  j  e j2

Now

1  2e  j  e  j2 V   2 1 1  j2 Z  2   j j

V 2e j  e j2  1 j   2 Z 1  j2  1  0.5  0.5e  j2 e  j j0.5  j 1 A B But   A = 2, B = -2 s(s  0.5) s s  0.5 2 2 I  0.5  0.5e j2  e  j  0.5  0.5e  j2  e  j j 0.5  j 1 1 i(t) = sgn( t )  sgn( t  2)  sgn( t  1)  e  0.5t u(t )  e  0.5( t  2 ) u(t  2)  2e  0.5( t 1) u(t  1) 2 2 I














2

1 2  j

+ 

V

+ V

1/s

2

0.5s



1 2V 2  j  jV  20 2 j

j  22  4V  j4  2 jj   42 jj9 2

V(ω) =

2 j(4.5  j2) ( 2  j)(4  2 2  j)

Chapter 18, Solution 42. By current division, I o  (a)

2  I 2  j

For i(t) = 5 sgn (t), 10 I  j 2 10 20 Io    2  j j j(2  j) 20 A B Let I o    , A  10, B  10 s(s  2) s s  2 10 10 I o    j  2  j i o (t) = 5 sgn( t )  10e 2 t u(t )A

i(t)

(b)

i’(t)

4

4(t)

1 1

t

t –4(t–1)

i' ( t )  4( t )  4( t  1) j I  4  4e  j I 



4 1  e  j j







 1 8 1  e  j 1   1  e  j Io   4  j(2  j)  j  2  j  4 4 4e  j 4e  j     j 2  j j 2  j i o (t) = 2 sgn( t )  2 sgn( t  1)  4e 2 t u(t )  4e 2( t 1) ut  1A






20 mF

 

Vo 

Vo 

1 1 50   ,  3 jC j20x10  j

50 250 40 , Is   50 j (s  1)(s  1.25) 40  j

i s  5e  t

s  j

A B 1   1   1000  s  1 s  1.25  s  1 s  1.25  v o (t )  1(e 1t  e 1.25 t )u(t ) kV

 

Is 

5 1  j

Chapter 18, Solution 44. 1H j We transform the voltage source to a current source as shown in Fig. (a) and then combine the two parallel 2 resistors, as shown in Fig. (b).

Io +

V s /2 2

2 Vo

Io +

V s /2

1 Vo

j

 (a)

(b)

V 1  s 1  j 2 j Vs Vo  j I o  2(1  j) v s ( t )  10t   10( t  2) 2 2  1, I o 

j Vs   10  10e  j2 

Vs  



10 1  e  j2 j







5 1  e  j2  5 5   e  j2  1  j 1  j 1  j t v o ( t )  5e u ( t )  5e  ( t  2) u ( t  2)

Hence Vo 



v o (1)  5e 1  0  1.839 V

j

Chapter 18, Solution 45. We may convert the voltage source to a current source as shown below.

1H v s /2

2

2

Combining the two 2- resistors gives 1 . The circuit now becomes that shown below. I

1H v s /2

1

1 Vs 1 5 5     , s  j 1  j 2 1  j 2  j ( s  1)( s  2) A B   s 1 s  2 B  5 /  1  5 where A  5 /1  5, 5 5 I  s 1 s  2 i (t )  5(e  t  e 2t )u (t ) A I

Chapter 18, Solution 46. 1 F 4

1 j j2 3

2H 3( t )

1 4



 j4 

1 1  j

e  t u(t)

The circuit in the frequency domain is shown below: 2

Vo I o () –j4/

1/(1+j)

+  + 

3

j2

At node V o , KCL gives 1  Vo 3  Vo V 1  j   o  j4 2 j2 

2 j2Vo  2Vo  j3  jVo   1  j  2  j3 1  j Vo  j2 2  j   2  j3  3 2 V 1  j I o   o  j2  j2  j2 2  j     I o () =

2  j 2  3 2 4  6 2  j(8  2 3 )

Chapter 18, Solution 47. 1 F 2

 

Io 

1 2  jC j

1

I 2 s 1 j

2 2 8 2 j Is  Io  Vo  2 2  j 1  j j 1 j 

16 , s  j (s  1)(s  2)



A B  s 1 s  2

where A = 16/1 = 16, B = 16/(–1) = –16 Thus, v o (t) = 16(e–t – e–2t)u(t) V.

Chapter 18, Solution 48. 0.2F

1 j5  jC 

As an integrator, RC  20 x 10 3 x 20 x 10 6  0.4

1 t v i dt RC o  1  Vi Vo    Vi (0)  RC  j  vo  

 Io 

 1  2     0.4  j 2  j 

  Vo 2 mA  0.125      20  j 2  j 

0.125 0.125   0.125  j 2  j 0.125 i o ( t )  0.125 sgn( t )  0.125e  2 t u t    e jt dt  2 0.125  0.125  0.25u ( t )  0.125e 2 t u ( t )  2 

i o (t) = [0.625  0.25u(t )  0.125e 2 t u(t )] mA

Chapter 18, Solution 49. Consider the circuit shown below: j j2

j +

VS

+ 

i1

i2 2

1  vo



Vs     1     2  For mesh 1,  Vs  2  j2I1  2I 2  jI 2  0 Vs  2 1  j I1  2  jI 2 For mesh 2, 0  3  jI 2  2I1  jI1 3  I 2 I1  2  

(2)

Substituting (2) into (1) gives 2 1  j3  jI 2  2  jI 2 2  j Vs 2    2 3  j4   2  4  j4   2 I 2 Vs  2



 I 2 2  j4   2  s  2Vs I2  2 , s  j s  4s  2

Vo  I 2 

 



 j  2     1     1  j2  j4  2

1  v o e jt d    2 1  j  2 e jt    1d 1  j  2e jt   1d  2  2 2   j  j4  2  j2  j4  2

v o (t) 

(1)

1  j  2e jt 1  j  2e jt  2  2  1  j4  2  1  j4  2 1 1 2  j1  j4 2  j1  j4e jt jt 2 2 v o (t)  e  17 17 1 1 6  j7 e jt  6  j7 e jt  34 34  j  t 13.64   0.271e  0.271e j  t 13.64 

v o (t) = 542 cos(t  13.64 ) mV

Chapter 18, Solution 50. Consider the circuit shown below:

j0.5 1 + VS

For loop 1, For loop 2,

+ 

i2

i1

j

1

vo

j



 2  1  jI1  j0.5I 2  0

(1)

1  jI 2  j0.5I1  0

(2)

From (2),

I1 

1  jI 2  j0.5

 2

1  jI 2 j

Substituting this into (1),  21  jI 2 j 2  I2 j 2 3   2 j   4  j4   2 I 2 2   2 j I2  4  j4  1.5 2  2 j Vo  I 2  2 4  j4  1.5 j 4 j 3 Vo  8 8 2 j   j  3 3 4   4  j  3    2 4   8    j    3   3 

2

16 3 2 4   8    j    3   3 

2

 8   8  Vo ( t )   4e  4t / 3 cos t u(t ) V t  u(t )  5.657e  4t / 3 sin 3 3    

Chapter 18, Solution 51. In the frequency domain, the voltage across the 2- resistor is 2 2 10 20 V ( )  Vs    , s  j 2  j 2  j 1  j ( s  1)( s  2) A B V ( s)   s 1 s  2 A  20 /1  20, B  20 /  1  20

V ( ) 

20 20  j  1 j  2

v(t )   20e  t  20e 2t  u (t )

W





1  2 v ( t )dt  0.5 400 e  2 t  e  4 t  3e  3t dt  0 2

 e  2 t e  4 t 2e  3 t  200    2 4 3 



  = 16.667 J.  0


J = 2  f 2 ( t ) dt  0

1  2 F() d   0 

1  1 1 1  d  tan 1 ( / 3)  = = (1/6) 2 2  3 3 2  0 9  0

Chapter 18, Solution 53. If f(t) = e-2|t|, find J  





J =



  F()

2

F ( ) d  . 2

d  2





f(t) =

t0

e 2t , e

2 t

f 2 ( t ) dt

,

t0

  4t 0  4 t  0 4t e e 4 t     = 2[(1/4) + (1/4)] =  J = 2   e dt   e dt   2 0  4 4      0  

Chapter 18, Solution 54. Design a problem to help other students better understand finding the total energy in a given signal. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Given the signal f(t) = 4 e-t u(t), what is the total energy in f(t)? Solution

W 1 =











0

0

f 2 ( t ) dt  16  e  2 t dt   8e  2 t

= 8J

Chapter 18, Solution 55. f(t) = 5e2e–tu(t) F() = 5e2/(1 + j), |F()|2 = 25e4/(1 + 2) W 1

1  25e 4 2 =    F ( ) d  0 





0

= 12.5e4 = 682.5 J or

W 1 =









f 2 ( t ) dt  25e 4  e  2 t dt  12.5e4 = 682.5 J 0



25e 4 1 d   tan 1 () 2  1  0


 2 e 4t 2 (a) W   V (t )dt   t e dt  (16 t 8 t 2)     0.0313 J 0 64 (4)3  0 



2

2 4 t

(b) In the frequency domain, 1 V ( )  (2  j ) 2 1 (4  j ) 2 2 2 1 2 1 2 Wo  | V ( ) | d   d   2 2 2 0 (4   2 ) 2 | V ( ) |2  V ( )V * ( ) 

 1 1      0.5 tan 1 0.5  2x 4  2  4  Fraction =

2

 0

1 1   0.0256 32 64

Wo 0.0256   81.79% W 0.0313

Chapter 18, Solution 57. W 1 =







0

0





i 2 ( t ) dt   4e 2 t dt  2e 2 t

I() = 2/(1 – j),

= 2 J or

|I()|2 = 4/(1 + 2) 

W 1

1  4  1 4 4 2 = = 2J d  tan 1 ()  I ( ) d    2       2 2  2 (1   ) 0

In the frequency range, –5 <  < 5,

5

4 4 4 tan 1   tan 1 (5)  (1.373) = 1.7487 W =    0

W/ W 1 = 1.7487/2 = 0.8743 or 87.43%

Chapter 18, Solution 58.  m = 200 = 2f m

which leads to f m = 100 Hz

(a)

 c = x104 = 2f c which leads to f c = 104/2 = 5 kHz

(b)

lsb = f c – f m = 5,000 – 100 = 4,900 Hz

(c)

usb = f c + f m = 5,000 + 100 = 5,100 Hz

Chapter 18, Solution 59. 10 6  V () 2  j 4  j 5 3 H()  o    Vi () 2 2  j 4  j

 5 3  4  Vo ()  H()Vi ()     2  j 4  j  1  j 20 12   , s  j (s  1)(s  2) (s  1)(s  4) Using partial fraction, Vo () 

A B C D 16 20 4       s  1 s  2 s  1 s  4 1  j 2  j 4  j

Thus,





v o (t )  16e  t  20e 2 t  4e 4 t u(t ) V


2 + 1/jω

Is

jω

V 

V  jI s

1 j 1  2  j j



jI s 1   2  j2

Since the voltage appears across the inductor, there is no DC component. V1 

2908 1  4 2  j4

V2 



50.2790  1.2418  71.92  38.48  j12.566

4905 1  16 2  j8



62.8390  0.3954  80.9  156.91  j25.13

v(t )  1.2418 cos( 2t  41.92 )  0.3954 cos(4t  129.1 ) mV

Chapter 18, Solution 61. y (t )  (2  cos o t ) x(t ) We apply the Fourier Transform Y(ω) = 2X(ω) + 0.5X(ω+ω o ) + 0.5X(ω–ω o ).

Chapter 18, Solution 62. For the lower sideband, the frequencies range from 10,000,000 – 3,500 Hz = 9,996,500 Hz to 10,000,000 – 400 Hz = 9,999,600 Hz For the upper sideband, the frequencies range from 10,000,000 + 400 Hz = 10,000,400 Hz to 10,000,000 + 3,500 Hz = 10,003,500 Hz

Chapter 18, Solution 63. Since f n = 5 kHz, 2f n = 10 kHz i.e. the stations must be spaced 10 kHz apart to avoid interference. f = 1600 – 540 = 1060 kHz The number of stations = f /10 kHz = 106 stations

Chapter 18, Solution 64. f = 108 – 88 MHz = 20 MHz The number of stations = 20 MHz/0.2 MHz = 100 stations

Chapter 18, Solution 65.  = 3.4 kHz f s = 2 = 6.8 kHz

Chapter 18, Solution 66.  = 4.5 MHz f c = 2 = 9 MHz T s = 1/f c = 1/(9x106) = 1.11x10–7 = 111 ns

Chapter 18, Solution 67. We first find the Fourier transform of g(t). We use the results of Example 17.2 in conjunction with the duality property. Let Arect(t) be a rectangular pulse of height A and width T as shown below. Arect(t) transforms to Atsinc(2/2) f(t)

F()

A 

t –T/2

T/2 G()

 – m /2 According to the duality property, Asinc(t/2)

becomes 2Arect()

g(t) = sinc(200t) becomes 2Arect() where A = 1 and /2 = 200 or T = 400 i.e. the upper frequency  u = 400 = 2f u or f u = 200 Hz The Nyquist rate = f s = 200 Hz The Nyquist interval = 1/f s = 1/200 = 5 ms

m/

Chapter 18, Solution 68. The total energy is WT =







v 2 ( t ) dt

Since v(t) is an even function, WT =





0

2500e

4 t

e 4 t dt  5000 4



= 1250 J 0

V() = 50x4/(4 + 2) W =

But

 (a

2

1 5 (200) 2 1 5 2    d | V ( ) | d 2 1 (4   2 ) 2 2 1 1 1  x 1  dx  2  2  tan 1 ( x / a )  C 2 2 2 a x ) 2a  x  a  5

2x10 4 1   1   tan 1 ( / 2) W = 2   8 4   2 1

= (2500/)[(5/29) + 0.5tan-1(5/2) – (1/5) – 0.5tan–1(1/2) = 267.19 W/W T = 267.19/1250 = 0.2137 or 21.37%

Chapter 18, Solution 69. The total energy is WT =

=

W =

1  1  400 2 d F() d   2   2  4 2   2



400 (1 / 4) tan 1 ( / 4) 



 0



100  = 50  2



1 2 400 2 F() d  (1 / 4) tan 1 ( / 4)  0 2 2



2 0

= [100/(2)]tan–1(2) = (50/)(1.107) = 17.6187 W/W T = 17.6187/50 = 0.3524 or 35.24%

Chapter 19, Solution 1. To get z 11 and z 21 , consider the circuit in Fig. (a).

2

8

I2 = 0

+ I1

12 

V1

Io

+

4

V2



 (a)

z11 

V1  2  12 || (8  4)  8  I1

1 I , 2 1 V z 21  2  2  I1

Io 

V2  4 I o  I1

To get z 22 and z 12 , consider the circuit in Fig. (b). I1 = 0 2 

8 Io'

+

+

12 

V1

4



V2 

(b)

z 22 

V2  4 || (8  12)  3.333  I2

4 1 I2  I2 , 4  20 6 V z12  1  2  I2

Io  '

V1  12I o '  2I 2

Hence,

2  8 [z ]     2 3.333 

I2

Chapter 19, Solution 2. Consider the circuit in Fig. (a) to get z 11 and z 21 .

1

Io'

1

1

1

+ I1

1

V1

1

I2 = 0

Io

+

1

V2





1

1

1

1

(a) z 11 

V1  2  1 || [ 2  1 || (2  1) ] I1

(1)(11 4)  11 3  2   2.733 z 11  2  1 ||  2    2   15 1  11 4 4

1 1 Io'  Io' 1 3 4 1 4 Io'  I1  I1 1  11 4 15 1 4 1 I o   I1  I1 4 15 15

Io 

V2  I o  z 21 

1 I 15 1

V2 1   z 12  0.06667 I 1 15

To get z 22 , consider the circuit in Fig. (b). I1 = 0 1 

1

1

1

+

+ 1

V1

1

1



V2 

1

1

1 (b)

1

I2

z 22 

V2  2  1 || (2  1 || 3)  z 11  2.733 I2

Thus,  2.733 0.06667  [z ]     0.06667 2.733 


We can use Figure 19.5 to determine the z-parameters. z 12 = j12 = z 21 z 11 – z 12 = 8 or z 11 = (8+j12) Ω z 22 – z 12 = –j20 or z 22 = (–j8) Ω

Chapter 19, Solution 4. Using Fig. 19.68, design a problem to help other students to better understand how to determine z parameters from an electrical circuit. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Calculate the z parameters for the circuit in Fig.19.68.

Figure 19.68 Solution Transform the  network to a T network.

Z1

Z3

Z2

(12)( j10) j120  12  j10  j5 12  j5 - j60 Z2  12  j5 50 Z3  12  j5 Z1 

The z parameters are

z 12  z 21  Z 2 

(-j60)(12 - j5)  -1.775 - j4.26 144  25

z 11  Z1  z 12 

( j120)(12  j5)  z 12  1.775  j4.26 169

z 22  Z 3  z 21 

(50)(12  j5)  z 21  1.7758  j5.739 169

Thus,  1.775  j4.26 - 1.775  j4.26  [z ]     - 1.775  j4.26 1.775  j5.739 

Chapter 19, Solution 5. Consider the circuit in Fig. (a).

s

1

I2 = 0 Io

+ I1

1/s

1

V1

1/s

+ V2





(a)  1  1  1  s    1  1 1   s  1  s || 1  s    z 11  1 || || 1  s    1  1 s  s s  1  1   1 s  s  s  1 s 2 s  s 1 z 11  3 s  2s 2  3s  1 1 s

Io 

1 ||

1 s

1 s 1

s s 1

I  I  I1 1 1 1 1 1 1 s 2 1 s   s  s 1 1 ||  1  s  s s s 1 s s 1 s Io  3 I1 2 s  2s  3s  1 I1 1 V2  I o  3 s s  2s 2  3s  1

z 21 

V2 1  3 2 I 1 s  2s  3s  1

Consider the circuit in Fig. (b). s

1

I1 = 0 + V1

+ 1

1/s

1/s



V2 

(b)

I2

z 22 

V2 1  1 1  1    || 1  s  1 ||   || 1  s  s s  s  1 I2 s 

z 22

 1  1  1  1  s   1  s   s  s  1 s 1   1 1 s 1 s  1 s  s2  s s 1 s 1

z 22

s 2  2s  2  3 s  2s 2  3s  1

z 12  z 21

Hence,

  s2  s  1 1  3  s  2s 2  3s  1 s 3  2s 2  3s  1  [z ]     1 s 2  2s  2  3  2 3 2  s  2s  3s  1 s  2s  3s  1 

Chapter 19, Solution 6. To find z 11 and z 21 , consider the circuit below.

I1

5

10

4I 1

I 2 =0

Vo –

+ +

V1

20 

+ _

V2 –

z11 

Vo 

V1 (20  5) I1   25  I1 I1

20 V1  20 I1 25

V o 4 I 2  V2  0

 

V2  Vo  4 I1  20 I1  4 I1  24 I1

V2  24  I1 To find z 12 and z 22 , consider the circuit below. z21 

I 1 =0 5

10

4I 1

I2

– + + V1 – V2  (10  20) I 2  30 I 2

V2  30  I1 V1  20 I 2 z22 

20 

+ _

V2

z12 

V1  20  I2

Thus,  25 20  [ z]     24 30 

Chapter 19, Solution 7. To get z 11 and z 21 , we consider the circuit below. I1

20 

I 2 =0

100 

+ + vx

50 

60 

+

V1 -

V2

-

12v x

-

+

V1  Vx Vx Vx  12Vx   20 50 160

V  Vx 81 V1 I1  1 ( )  20 121 20

 



Vx 

z11 

40 V1 121

V1  29.88 I1

13Vx 57 57 40 57 40 20x121 )  12Vx   Vx   ( )V1   ( ) I1 160 8 8 121 8 121 81 V  70.37 I1  z 21  2  70.37 I1

V2  60(

To get z 12 and z 22 , we consider the circuit below.

I 1 =0

20 

I2

100 

+ +

50 

vx

60 

+

V1 -

V2

-

12v x

-

+ Vx 

50 1 V2  V2 , 100  50 3

z 22 

V2  1 / 0.09  11.11 I2

V1  Vx 

I2 

V2 V2  12Vx   0.09V2 150 60

1 11.11 V2  I 2  3.704I 2 3 3

 

Thus,  29.88 3.704 [z ]      70.37 11.11

V z12  1  3.704 I2

Chapter 19, Solution 8. To get z 11 and z 21 , consider the circuit below. I1

j4 

-j2 



5

I 2 =0



j6 

+

j8 

+ V2 V1 10 

-

-

V1  (10  j2  j6)I1

V2  10I1  j4I1

V z11  1  10  j4 I1

 

 

z 21 

V2  (10  j4) I1

To get z 22 and z 12 , consider the circuit below. j4 

I 1 =0 -j2 



j6 

5

I2



j8 

+

+ V2 V1 10  -

-

V2  (5  10  j8)I 2

V1  (10  j4)I 2

 

 

z 22 

V2  15  j8 I2

V z12  1  (10  j4) I2

Thus,  (10  j4)  (10  j4) [z ]      (10  j4) (15  j8) 


 y  y11 y22  y12 y21  0.5 x0.4  0.2 x0.2  0.16 z11 

y22 0.4   2.5 Ω  y 0.16

z12 

 y12 0.2   1.25 Ω = z 21 0.16 y

z 22 

y11 0.5   3.125 Ω  y 0.16

Thus,

 2.5 1.25  Z  z     1.25 3.125


(a)

This is a non-reciprocal circuit so that the two-port looks like the one shown in Figs. (a) and (b).

z 11

I1

z 22

I2

+

+ z 12 I 2

V1

+ 

+ 

z 21 I 1

V2





(a)

25 

I1

10 

I2

+

+ 20 I 2

V1

+ 

+ 

5 I1

V2





(b) (b)

This is a reciprocal network and the two-port look like the one shown in Figs. (c) and (d).

I1

z 11 – z 12

z 22 – z 12

+

+ z 12

V1 

z 22  z 12  2s 1 z 12  s

V2 

(c) z 11  z 12  1 

I2

1 2  1 0.5 s s

I1

1

0.5 F

2H

+

I2 +

1F

V1 

V2 

(d)


This is a reciprocal network, as shown below. 1+j5 3+j

5-j2

1

j5 

3

5 -j2 

j1 

Chapter 19, Solution 12. V1  10 I1  6 I 2 V2  4 I 2  12 I 2 V2  10 I 2

(1) (2) (3)

If we convert the current source to a voltage source, that portion of the circuit becomes what is shown below. 4 2 I1 +

12 V

V1

+ _

–

12  6 I1  V1  0

  V1  12  6 I1

(4)

Substituting (3) and (4) into (1) and (2), we get 12  6 I 1  10 I1  6 I 2

  12  16 I1  6 I 2

10 I 2  4 I1  12 I 2

  0  4 I1  22 I 2

(5)

  I1  5.5 I 2

From (5) and (6),

12  88I 2  6 I 2  82 I 2   I 2  0.1463 A I1  5.5 I 2  0.8049 A V2  10 I 2  1.463 V V1  12  6 I1  7.1706 V

(6)

Chapter 19, Solution 13. Consider the circuit as shown below. 10  I 1

I2

+ 500˚ V

+ _

+ V1

[z] V2

ZL

_ – V1  40I1  60I2 (1) V2  80I1  100I2 (2) V2  I2 ZL  I2(5  j 4) (3) 50  V1  10I1   V1  50  10I1 (4) Substituting (4) in (1) 50  10I1  40I1  60I2   5  5I1  6I2 (5) Substituting (3) into (2), I2(5  j 4)  80I1  100I2   0  80I1  (105  j4)I2 Solving (5) and (6) gives I 2 = –7.423 + j3.299 A

We can check the answer using MATLAB. First we need to rewrite equations 1-4 as follows, 1 0  0  1

0  40  60   V1  0   0  1  80  100  V2   A*X     U 0 1 0 5  j4  I1      0 10 0  I2  50

>> A=[1,0,-40,-60;0,1,-80,-100;0,1,0,(5+4i);1,0,10,0] A= 1.0e+002 * 0.0100 0 -0.4000 -0.6000 0 0.0100 -0.8000 -1.0000 0 0.0100 0 0.0500 + 0.0400i 0.0100 0 0.1000 0 >> U=[0;0;0;50] U=

(6)

0 0 0 50 >> X=inv(A)*U X= -49.0722 +39.5876i 50.3093 +13.1959i 9.9072 - 3.9588i -7.4227 + 3.2990i P = |I 2 |25 = 329.9 W.

Chapter 19, Solution 14. To find Z Th , consider the circuit in Fig. (a). I1

I2

+ ZS

+ 

V1

Vo = 1

 (a) V1  z 11 I 1  z 12 I 2 V2  z 21 I 1  z 22 I 2

(1) (2)

But V2  1 ,

V1  - Z s I 1

0  (z 11  Z s ) I 1  z 12 I 2

Hence,

  I 1 

- z 12 I z 11  Z s 2

 - z 21 z 12  1  z 22  I 2  z 11  Z s  Z Th 

V2 z z 1   z 22  21 12 z 11  Z s I2 I2

To find VTh , consider the circuit in Fig. (b). ZS

VS

+ 

I1

I2 = 0

+

+

V1

V 2 = V Th



 (b)

I2  0 ,

Substituting these into (1) and (2),

V1  Vs  I 1 Z s

Vs  I 1 Z s  z 11 I 1 V2  z 21 I 1 

  I 1 

Vs z 11  Z s

z 21 Vs z 11  Z s VTh  V2 

z 21 Vs z 11  Z s


(a) From Prob. 18.12,

ZTh  z 22 

z12z 21 80x 60  120   24 z11  Zs 40  10 ZL  ZTh  24

(b) VTh 

z 21 80 Vs  (120)  192 z11  Zs 40  10  V Pmax   Th  2R Th

2

  R Th  4 2 x 24  384W 

Chapter 19, Solution 16. As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).

5

10 – j6 

4 – j6 

+ 

150 V

j6 

j4 

b

(a) At terminals a-b, Z Th  (4  j6)  j6 || (5  10  j6) j6 (15  j6) Z Th  4  j6   4  j6  2.4  j6 15 Z Th  6.4 

VTh 

j6 (150)  j6  690 V j6  5  10  j6

The Thevenin equivalent circuit is shown in Fig. (b).

6.4  + 690 V

+ 

Vo 

(b) From this,

Vo 

a

j4 ( j6)  3.18148 6.4  j4

v o ( t )  3.18 cos( 2t  148) V

j4 

Chapter 19, Solution 17. To obtain z 11 and z 21 , consider the circuit in Fig. (a).

8 Io'

+ I1

V1

Io

I2 = 0 +

4

V2

16 



 12  (a)

In this case, the 8- and 16- resistors are in series, since the same current, I o , passes through them. Similarly, the 4- and 12- resistors are in series, since the same current, I o ' , passes through them.

z11 

Io  But

V1 (24)(16)  (8  16) || (4  12)  24 || 16   9.6 Ω 40 I1

16 2 I1  I1 16  24 5

Io' 

3 I 5 1

- V2  8 I o  4 I o  0 - 16 12 -4 ' V2  -8 I o  4 I o  I1  I 1  I1 5 5 5 V -4 z 21  2   –0.8 Ω I1 5 '

To get z 22 and z 12 , consider the circuit in Fig. (b). 8

I1 = 0 + V1

+

4

V2

16 

I2



 12  (b)

z 22 

(12)(28) V2  (8  4) || (16  12)  12 || 28   8.4 Ω 40 I2

z12  z 21  –0.8 Ω

Thus,  9.6 - 0.8  [z ]     - 0.8 8.4 

We may take advantage of Table 18.1 to get [y] from [z].  z  (9.6)(8.4)  (0.8) 2  80 z 8.4 -z 0.8  0.105 S  0.01 S y 11  22  y12  12  80  z 80 z z -z 0.8 9.6  0.01 S  0.12 S y 21  21  y 22  11  80 z  z 80 Thus,  0.105 0.01  [y ]   S  0.01 0.12 


To get y 11 and y 21 , consider the circuit in Fig.(a).

6

I1

3

I2 +

+ 

V1

6

3

V2 = 0 

(a) V1  (6  6 || 3) I 1  8 I 1 I1 1  y 11  V1 8

-6 - 2 V1 - V1 I1   63 3 8 12 I 2 -1   V1 12

I2  y 21

To get y 22 and y 12 , consider the circuit in Fig.(b).

I1

6

Io

3

I2

+ V1 = 0

6

3



(b) y 22 

I2 1 1 1    V2 3 || (3  6 || 6) 3 || 6 2

- Io 3 1 , Io  I2  I2 2 3 6 3 - I 2  - 1  1  - V2 I1     V2    6  2  12 6 I1 

+ 

V2

y 12 

I1 -1   y 21 V2 12

Thus,  1 -1   8 12  [y ]   S 1 1    12 2 

Chapter 19, Solution 19. Using Fig. 19.80, design a problem to help other students to better understand how to find y parameters in the s-domain. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the y parameters of the two-port in Fig.19.80 in terms of s.

Figure 19.80 Solution Consider the circuit in Fig.(a) for calculating y 11 and y 21 . 1

I1

I2 +

V1

+ 

1/s

s 1 (a)

2s 1  2 V1   || 2  I 1  I1  I s  2  (1 s) 2s  1 1 I 2s  1 y 11  1   s  0.5 V1 2 I2 

- I1 - V1 (- 1 s ) I1   (1 s)  2 2s  1 2

V2 = 0 

y 21 

I2  -0.5 V1

To get y 22 and y 12 , refer to the circuit in Fig.(b).

1

I1

I2

+ V1 = 0

1/s

s



1 (b)

V2  (s || 2) I 2  y 22 

2s I s2 2

I2 s  2 1   0.5  s V2 2s

- V2 -s -s s 2 I2   V2  s2 s  2 2s 2 I1   -0.5 V2

I1  y 12

Thus,  s  0.5 - 0.5  [y ]   S  - 0.5 0.5  1 s 

+ 

V2

Chapter 19, Solution 20. To get y 11 and y 21 , consider the circuit below. 3i x

2

I1

I2

+

ix 4

V1

I1

+

6

V 2 =0

-

-

Since 6-ohm resistor is short-circuited, i x = 0

V1  I1(4 // 2) 

I2  

8 I1 6



I y11  1  0.75 V1

4 2 6 1 I1   ( V1)   V1 42 3 8 2



I y 21  2  0.5 V1

To get y 22 and y 12 , consider the circuit below. 3i x

2

I1 +

ix 4

V 1 =0

+

6  V2

-

I2

ix 

V2 , 6

V V I 2  i x  3i x  2  2 2 6

V I1  3i x  2  0 2

 



I y12  1  0 V2

Thus, 0   0.75 [y ]   S   0.5 0.1667

I 1 y 22  2   0.1667 V2 6

Chapter 19, Solution 21. To get y 11 and y 21 , refer to Fig. (a).

0.2 V 1

I1

I2

V1

+ + 

V1

5

10 

V2 = 0 

(a) At node 1, I1 

V1  0.2 V1  0.4 V1 5

I 2  -0.2 V1

  y 11 

  y 21 

I1  0.4 V1

I2  -0.2 V1

To get y 22 and y 12 , refer to the circuit in Fig. (b).

0.2 V 1

I1

V1

I2

+ 5

V1 = 0

10 

+ 

V2



(b) Since V1  0 , the dependent current source can be replaced with an open circuit. V2  10 I 2

y 12 

  y 22 

I1 0 V2

Thus,

 0.4 0 [y ]   S  - 0.2 0.1

I2 1   0 .1 V2 10

Consequently, the y parameter equivalent circuit is shown in Fig. (c).

I1

I2

+ V1

+ 0.2 V 1

0.4 S



0.1 S

V2 

(c)

Chapter 19, Solution 22. To obtain y 11 and y 21 , consider the circuit below. 5

I2

+

+

I1

5

0.5V 2

2

V 2 =0

V1 – – The 2- resistor is short-circuited. I I 2 V1  5 1   y 11 1   0.4 V1 5 2 1 I1 1 I I2  I1   y21  2  2  0.2 2 V1 2.5I1 To obtain y 12 and y 22 , consider the circuit below. 5

I1 +

+ 5

0.5V 2

2

V2

V 1 =0 – – At the top node, KCL gives V V I2  0.5V2  2  2  1.2V2 2 5 I1  

V2  0.2V2 5

  y22 

  y12 

I1  0.2 V2

Hence,  0.4 0.2  [y ]    S  0.2 1.2 

I2  1.2 V2

I2


(a) 1 /(  y 12 )  1 //

1 1  s s1

y 11  y 12  1

 

y 22  y 12  s

 

y 12   (s  1)

y 11  1  y 12  1  (s  1)  s  2

 

y 22  s  y 12

1 s2  s  1   (s  1)  s s

 (s  1)   s2  [y ]  s2  s  1   (s  1)  s   (b) Consider the network below.

I1

I2

1 +

+

+ [y]

Vs -

V1 -

V2 -

2

Vs  I1  V1

(1)

V2  2I 2

(2)

I1  y11V1  y12 V2

(3)

I 2  y 21V1  y 22 V2

(4)

From (1) and (3)

Vs  V1  y11V1  y12 V2

 

Vs  (1  y11 )V1  y12 V2

(5)

From (2) and (4),

 0.5V2  y 21V1  y 22 V2



V1  

1 (0.5  y 22 )V2 y 21

(6)


Vs   

2 s

(1  y 11 )(0.5  y 22 ) V2  y 12 V2 y 21  

2/s   1 (1  y 11 )(0.5  y 22 )  y 12  y 21   2/s

V2   (s  1)  

V2 

2   1 1  s  2 0.5  s  s  1  s s1  



2/s  s  s  s  s  (s  3)(0.5s  s 2  s  1) s(s  1) 3

2

2

2(s  1) 2(s  1) 0.8(s  1)  2  2 2 2  s  2s  s  s  1.5s  s  3s  4.5s  3 2.5s  4.5s  3 s 1.8s  1.2 3

2

3

Chapter 19, Solution 24. Since this is a reciprocal network, a  network is appropriate, as shown below.

Y2

Y1

Y3 (a) 4

1/4 S

1/4 S

1/8 S

4

8

(b) Y1  y 11  y 12 

Y2  - y 12 

(c) 1 1 1   S, 2 4 4

1 S, 4

Y3  y 22  y 21 

Z1  4 

Z2  4 

3 1 1   S, 8 4 8

Z3  8 


This is a reciprocal network and is shown below.

0.5 S

0.5S

1S

Chapter 19, Solution 26. To get y 11 and y 21 , consider the circuit in Fig. (a). 4 2

1 +

V1

+ 

Vx

I2

2

1

+

2 Vx

V2 = 0 

 (a) At node 1, V1  Vx V V  2 Vx  x  x 2 1 4

But

I1 

Also,

I2  y 21

  2 V1  -Vx

V1  Vx V1  2 V1   1.5 V1 2 2

(1)

  y 11 

I1  1.5 V1

Vx  2 Vx   I 2  1.75 Vx  -3.5 V1 4 I2   -3.5 V1

To get y 22 and y 12 , consider the circuit in Fig.(b). 4 2 I1

1

2

+ Vx

1

2 Vx

I2

+ 

V2

 (b) At node 2, I 2  2 Vx 

V2  Vx 4

(2)

At node 1, 2 Vx 

V2  Vx Vx Vx 3    Vx 4 2 1 2

  V2  -Vx

Substituting (3) into (2) gives 1 I 2  2 Vx  Vx  1.5 Vx  -1.5 V2 2 I2  -1.5 y 22  V2 I1 

- Vx V2  2 2

  y 12 

I1  0.5 V2

Thus,  1.5 0.5  [y ]   S  - 3.5 - 1.5 

(3)

Chapter 19, Solution 27. Consider the circuit in Fig. (a).

4

I1

I2 +

+ 

V1

0.1 V 2

 +

10 

20 I 1

V2 = 0 

(a) V1  4 I 1

  y 11 

I 2  20 I 1  5 V1

I1 I1   0.25 V1 4 I 1

  y 21 

I2 5 V1

Consider the circuit in Fig. (b).

I1

4

I2

+ V1 = 0

0.1 V 2

 +

10 

20 I 1

+ 

V2



(b) 4 I 1  0.1 V2

I 2  20 I 1 

  y 12 

I 1 0.1   0.025 V2 4

V2  0.5 V2  0.1 V2  0.6 V2 10

Thus,

 0.25 0.025  [y ]   S 0.6   5 Alternatively, from the given circuit, V1  4 I 1  0.1 V2

  y 22 

I2  0.6 V2

I 2  20 I 1  0.1 V2

Comparing these with the equations for the h parameters show that h11  4 , h12  -0.1, h 21  20 , h 22  0.1 Using Table 18.1,

as above.

1 1   0.25 S , h 11 4 h 20  21   5S , h 11 4

- h 12 0.1   0.025 S h 11 4  0.4  2  h   0 .6 S h 11 4

y 11 

y 12 

y 21

y 22

Chapter 19, Solution 28. We obtain y 11 and y 21 by considering the circuit in Fig.(a).

2

8

I2

+ I1

+ 12 

V1

4



V2 = 0 

(a) Z in = 2 + (12||8) = 6.8 Ω I 1  147.06 mS y 11  1  V1 Z in

I 2 = (–6/10)I 1 = (–0.6)(V 1 /6.8) = –0.08824 I y 21  2  –88.24 mS V1 To get y 22 and y 12 , consider the circuit in Fig. (b).

I1

2

8

Io

+

+

V1 = 0

12 

4



V2

I2



(b) (1/y 22 ) = [4||(8+(12||2))] = [4||(8+(1.714286))] = 2.833333 = V 2 /I 2

y 22 = 352.9 mS I 1 = (–12/14)I o = –0.857143I o and I o = [4/(4+(8+1.714286))]I 2 = 0.29166667I 2 = V 2 /9.714286 Thus, I 1 = [(–0.857143)/9.714286]V 2 = –0.088235V 2 or

y 12 = I 1 /V 2 = –88.24 mS

Thus,

147.06 - 88.24 [y ]    mS - 88.24 352.9 

We note that I = YV, I 1 = 1 A, and –I 2 = V 2 /2. We now have the following equations, 1 = 0.14706V 1 – 0.08824V 2 and I 2 = –0.08824V 1 + 0.3529V 2 or –V 2 /2 = –0.08824V 1 + 0.3529V 2 or 0.08824V 1 = 0.8529V 2 which leads to V 1 = 9.6657V 2 . Substituting this into the first equation we get, 1 = (1.42144–0.08824)V 2 or V 2 = 0.75 V. Finally we get, P 2Ω = (0.75)2/2 = 281.2 mW. Now to check our answer. The equivalent circuit is shown in Fig. (c). After transforming the current source to a voltage source, we have the circuit in Fig. (d). 88.24mS

58.82 mS

1A

264.7 mS

2

(c) 17 

11.333 

+ 17 V

+ 

V  (d)

3.778 

2

V

(2 || 3.778)(17) (1.3077)(17)   0.75 V (2 || 3.778)  17  11.333 1.3077  28.333 P

V 2 (0.75) 2   281.2 mW R 2


(a)

Transforming the  subnetwork to Y gives the circuit in Fig. (a).

1

1

Vo

+ 10 A

+ 2

V1 

-4 A

V2 

(a) It is easy to get the z parameters z 12  z 21  2 , z 11  1  2  3 ,

z 22  3

 z  z 11 z 22  z 12 z 21  9  4  5

y 11 

z 22 3   y 22 , z 5

y 12  y 21 

- z 12 - 2  z 5

Thus, the equivalent circuit is as shown in Fig. (b).

2/5 S

I1

I2

+ 10 A

V1

+ 1/5 S

1/5 S



V2

-4 A



(b) I 1  10 

3 2 V1  V2 5 5

  50  3 V1  2 V2

(1)

-2 3 V1  V2   - 20  -2 V1  3 V2 5 5 10  V1  1.5 V2   V1  10  1.5 V2 (2) I 2  -4 

Substituting (2) into (1), 50  30  4.5 V2  2 V2

  V2  8 V

V1  10  1.5 V2  22 V

(b)

For direct circuit analysis, consider the circuit in Fig. (a). For the main non-reference node, Vo 10  4    Vo  12 2 10 

V1  Vo 1

  V1  10  Vo  22 V

-4

V2  Vo 1

  V2  Vo  4  8 V


(a)

Convert to z parameters; then, convert to h parameters using Table 18.1. z 11  z 12  z 21  60  , z 22  100   z  z 11 z 22  z 12 z 21  6000  3600  2400

 z 2400   24 , z 22 100 - z 21   -0.6 , z 22

h 11 

h12 

z 12 60   0.6 z 22 100

h 21

h 22 

1  0.01 z 22

Thus,  24  0.6  [h]     - 0.6 0.01 S 

(b)

Similarly, z 11  30 

z 12  z 21  z 22  20 

 z  600  400  200

h11 

200  10 20

h 21  -1

Thus,  10  1  [h]     - 1 0.05 S 

20 1 20 1   0.05 20

h12  h 22

Chapter 19, Solution 31. We get h11 and h 21 by considering the circuit in Fig. (a).

1

2

V3

V4

1

I2

+ I1

2

V1

4 I1




V3 V3  V4  2 2

  2 I 1  2 V3  V4

(1)

At node 2, V3  V4 V  4 I1  4 2 1 8 I 1  -V3  3 V4   16 I 1  -2 V3  6 V4

(2)

Adding (1) and (2), 18 I 1  5 V4   V4  3.6 I 1 V3  3 V4  8 I 1  2.8 I 1 V1  V3  I 1  3.8 I 1 V h11  1  3.8  I1 I2 

- V4  -3.6 I 1 1

  h 21 

I2  -3.6 I1

To get h 22 and h 12 , refer to the circuit in Fig. (b). The dependent current source can be replaced by an open circuit since 4 I 1  0 . I1

1

1

2

I2

+ V1

2

4 I1 = 0



(b)

+ 

V2

V1 

2 2 V2  V2 2  2 1 5

I2 

V2 V2  2  2 1 5

  h 12 

  h 22 

V1  0 .4 V2

I2 1   0 .2 S V2 5

Thus,

 38  0.4  [h]     - 3.6 0.2 S 

Chapter 19, Solution 32. Using Fig. 19.90, design a problem to help other students to better understand how to find the h and g parameters for a circuit in the s-domain. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the h and g parameters of the two-port network in Fig.19.90 as functions of s.


(a)

We obtain h11 and h 21 by referring to the circuit in Fig. (a).

1

s

s

+ I1

I2 +

1/s

V1 

V2 = 0 

(a)   s  1 I V1  1  s  s ||  I 1  1  s  2   s  1 1 s V s h11  1  s  1  2 I1 s 1 By current division, - I1 I2 -1 s -1   h 21   2 I2  I1  s 1 s s 1 I1 s  1

To get h 22 and h 12 , refer to Fig. (b).

I1 = 0

1

s

s

I2

+ + 

1/s

V1

V2

 (b) V1 

V2 V1 1s 1   h12   2 V2  2 s 1 s s 1 V2 s  1

 1 V2  s   I 2  s

  h 22 

I2 1 s   2 V2 s  1 s s  1

Thus, s   s  1  s2  1 [h]   -1  2 s 1 

(b)

1  s 1  s   s2  1  2

To get g11 and g 21 , refer to Fig. (c). I1

1

s

s

I2 = 0 +

V1

+ 

1/s

V2 

(c)  1 V1  1  s   I 1  s V2 

  g 11 

I1 1 s   2 V1 1  s  1 s s  s  1

V1 V2 1s 1 V1  2   g 21   2 1 s 1 s s  s 1 V1 s  s  1

To get g 22 and g 12 , refer to Fig. (d).

I1

1

s

s

I2

+

+ 1/s

V1 = 0

V2

I2



 (d)   1 (s  1) s   I 2 V2  s  || (s  1)  I 2  s   s   1 s 1 s  g 22 

I1 

V2 s 1 s 2 I2 s  s 1

-I I1 -1 s -1 I2  2 2   g 12   2 1 s 1 s s  s 1 I2 s  s 1

Thus,   2 [g ]   s  2 s

s -1  2 s1 s s1  1 s1   s 2 s1 s s1 


To get h 11 and h 21 , consider the circuit below. 4

j6 

+ I1

V1  5 //( 4  j6)I1 

Also, I 2  

-j3 

5

V1 -

5 I1 9  j6

I2

5(4  j6)I1 9  j6

+ V 2 =0 -

V h11  1  3.0769  j1.2821 I1

I h 21  2  0.3846  j0.2564 I1

 

To get h 22 and h 12 , consider the circuit below.

4

j6 

I2

I1 +

-j3 

5

V1 -

V1 

5 V2 9  j6

+ + V2

-

 

V2   j3 //(9  j6)I 2

V 5 h12  1   0.3846  j0.2564 V2 9  j6

 

I 1 9  j3 h 22  2   V2  j3 //(9  j6)  j3(9  j6)  0.0769  j0.2821

Thus, ( 3.077  j1.2821)  0.3846  j0.2564  [h]      0.3846  j0.2564 (0.0769  j0.2821) S 

Chapter 19, Solution 34. Refer to Fig. (a) to get h11 and h 21 .

300 

10 

50 

2

1 + I1

V1

Vx



I2

+

+

 +

100 

10 V x

V2 = 0 



(a) At node 1, Vx Vx  0    300 I 1  4 Vx 100 300 300 Vx  I  75 I 1 4 1

I1 

V1  10 I 1  Vx  85 I 1

But

  h11 

(1)

V1  85  I1

At node 2, 0  10 Vx Vx Vx Vx 75 75     I1  I  14.75 I 1 50 300 5 300 5 300 1 I2   14.75 I1

I2  h 21

To get h 22 and h 12 , refer to Fig. (b). 300  I 1 = 0 10 

50  1

+ V1 

+ Vx

 +

100 



(b)

10 V x

2

I2

+ 

V2

At node 2, V2 V2  10 Vx    400 I 2  9 V2  80 Vx 400 50 V2 100 Vx  V2  400 4 400 I 2  9 V2  20 V2  29 V2 I2 29   0.0725 S h 22  V2 400 I2 

But Hence,

V1  Vx 

V2 4

  h 12 

V1 1   0.25 V2 4

 85  0.25  [h]     14.75 0.0725 S 

To get g 11 and g 21 , refer to Fig. (c). 300 

I1

10 

50  1

+ 

Vx

I2 = 0 +

+ V1

2

100 

 +

10 V x

V2 

 (c)

At node 1, I1 

But or

Vx Vx  10 Vx  100 350

  350 I 1  14.5 Vx

V1  Vx   10 I 1  V1  Vx 10 Vx  V1  10 I 1

(2)

I1 

Substituting (3) into (2) gives 350 I 1  14.5 V1  145 I 1

  495 I 1  14.5 V1

(3)

g 11 

I 1 14.5   0.02929 S V1 495

At node 2,   11 V2  (50)  Vx   10 Vx  -8.4286 Vx  350   14.5   -8.4286 V1  84.286 I 1  -8.4286 V1  (84.286)   V1  495  V2  -5.96 V1

  g 21 

V2  -5.96 V1

To get g 22 and g 12 , refer to Fig. (d).

300 

I1 + V1 = 0

Io

10 

+

+ Vx



Io

50 

 +

100 

10 V x

I2

V2 

 (d)

10 || 100  9.091 I2 

But

V2  10 Vx V2  50 300  9.091

309.091 I 2  7.1818 V2  61.818 Vx

(4)

9.091 V  0.02941 V2 309.091 2

(5)

Vx 

Substituting (5) into (4) gives 309.091 I 2  9 V2 V2 g 22   34.34  I2

Io 

34.34 I 2 V2  309.091 309.091

- 34.34 I 2 - 100 Io  110 (1.1)(309.091) I1   -0.101 I2

I1  g 12

Thus,  0.02929 S - 0.101  [g ]   34.34    - 5.96

Chapter 19, Solution 35. To get h11 and h 21 consider the circuit in Fig. (a). 1

I1

1:2

4

I2

+

+

V1

V2 = 0



 (a)

ZR 

4 4 1 2  n 4

V1  (1  1) I 1  2 I 1

  h11 

V1  2 I1

I 2 -1 I1 - N 2   -2   h 21    -0.5 I1 2 I2 N1

To get h 22 and h 12 , refer to Fig. (b). I1 = 0 1 

4

1:2

+

 +

V1  (b)

Since I 1  0 , I 2  0 . h 22  0 . Hence,

I2

V2

At the terminals of the transformer, we have V1 and V2 which are related as V1 1 V2 N 2  n2   h12    0 .5 V2 2 V1 N 1 Thus,

 2  0.5  [h]     - 0.5 0 

Chapter 19, Solution 36. We replace the two-port by its equivalent circuit as shown below.

4

I1

16 

2 I1

+ 10 V

+ 

V1

I2 +

3 V2

+ 

-2 I 1



100  V 2 

100 || 25  20  V2  (20)(2 I 1 )  40 I 1

(1)

- 10  20 I 1  3 V2  0 10  20 I 1  (3)(40 I 1 )  140 I 1

I1 

1 , 14

V1  16 I 1  3 V2 

V2  136 14

-8  100  I2    (2 I 1 )  70  125  (a)

V2 40   0.2941 V1 136

(b)

I2  - 1.6 I1

(c)

I1 1   7.353  10 -3 S V1 136

(d)

V2 40   40  1 I1

40 14

25 


We first obtain the h parameters. To get h11 and h 21 refer to Fig. (a). 6

3

I2

+ I1

+ 6

V1

3

V2 = 0



 (a)

3 || 6  2 V1  (6  2) I 1  8 I 1

  h11 

V1 8 I1

-6 -2 I1  I 3 6 3 1

  h 21 

I2 - 2  3 I1

I2 

To get h 22 and h 12 , refer to the circuit in Fig. (b). I1 = 0 6 

3

I2

+ 6

V1

3

 (b)

3 || 9 

V2 

9 4

9 I 4 2

  h 22 

I2 4  V2 9

+ 

V2

6 2 V2  V2 63 3

V1 

  h 12 

V1 2  V2 3

 2  8  3  [h]   -2 4  S  3 9 

The equivalent circuit of the given circuit is shown in Fig. (c). 8

I1

I2 +

10 V

+ 

2/3 V 2

+ 

9/4  V 2

-2/3 I 1

5



(c)

8 I1 

2 V  10 3 2

(1)

2  9  2  45  30 I 5 ||   I    I 3 1  4  3 1  29  29 1 29 I1  V 30 2 V2 

(2)

Substituting (2) into (1),  29  2 (8)   V2  V2  10  30  3 300 V2   1.19 V 252 (b)

By direct analysis, refer to Fig.(d).

6

3 +

10 V

+ 

6

3

V2 

(d)

5

10 -A current source. Since 6 6 || 6  3  , we combine the two 6- resistors in parallel and transform 10 the current source back to  3  5 V voltage source shown in Fig. (e). 6 Transform the 10-V voltage source to a

3

3 +

+ 

5V

V2 

(e) 3 || 5 

V2 

(3)(5) 15  8 8

15 8 75  1.1905 V (5)  63 6  15 8

3 || 5 

Chapter 19, Solution 38. From eq. (19.75), h h R h h R 0.04 x30 x 400 Z in  hie  re fe L  h11  12 21 L  600   333.33  1  hoe RL 1  h22 RL 1  2 x103 x 400 From eq. (19.79), Rs  hie Rs  h11 2, 000  600 Z out     650  ( Rs  hie )h0 e  hre h fe ( Rs  h11 )h22  h21h12 2600 x 2 x103  30 x0.04

Chapter 19, Solution 39. We obtain g 11 and g 21 using the circuit below. R1

I1

R3

I 2 =0 +

V1

R2

+ _

V2 –

I1 

V1 R1  R2

  g11 

By voltage division, R2 V2  V1 R1  R2

 

I1 1  V1 R1  R2

g 21 

V2 R2  V1 R1  R2

We obtain g 12 and g 22 using the circuit below. I1

R1

R3

+

+ R2

I2

V 1 =0

V2

–

–

By current division, R2 I1   I2 R1  R2

  g12 

I1 R2  I2 R1  R2

Also,  V RR RR  V2  I 2 ( R3  R1 // R2 )  I 2  R3  1 2  g 22  2  R3  1 2 I2 R1  R2 R1  R2  

g 11 

R2 1 , g 12   R1  R 2 R1  R 2

g 21 

R2 R1R 2 , g 22  R 3  R1  R 2 R1  R 2

Chapter 19, Solution 40. Using Fig. 19.97, design a problem to help other students to better understand how to find g parameters in an ac circuit. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the g parameters for the circuit in Fig.19.97.

Figure 19.97 Solution To get g 11 and g 21 , consider the circuit in Fig. (a). -j6 

I1

j10 

I2 = 0 +

+ 

V1

12 

V2 

(a) V1  (12  j6) I 1 g 21 

  g 11 

I1 1   0.0667  j0.0333 S V1 12  j6

V2 12 I 1 2    0.8  j0.4 V1 (12  j6) I 1 2  j

To get g 12 and g 22 , consider the circuit in Fig. (b). I1

-j6 

j10 

I2

+ V1 = 0 R

R

12 

 (b)

I2 R

I1 

- 12 I 12 - j6 2

  g 12 

I1 - 12   - g 21  -0.8  j0.4 I 2 12 - j6

V2  ( j10  12 || -j6) I 2 V2 (12)(-j6)  j10   2.4  j5.2  g 22  12 - j6 I2

 0.0667  j0.0333 S - 0.8  j0.4  [g ]   0.8  j0.4 2.4  j5.2   

Chapter 19, Solution 41. For the g parameters I 1  g 11 V1  g 12 I 2 V2  g 21 V1  g 22 I 2 But

or

and V1  Vs  I 1 Z s V2  - I 2 Z L  g 21 V1  g 22 I 2 0  g 21 V1  (g 22  Z L ) I 2 - (g 22  Z L ) I2 V1  g 21

Substituting this into (1), (g g  Z L g 11  g 21 g 12 ) I 1  22 11 I2 - g 21 I2 - g 21 or  I 1 g 11 Z L   g Also,

V2  g 21 (Vs  I 1 Z s )  g 22 I 2  g 21 Vs  g 21 Z s I 1  g 22 I 2  g 21 Vs  Z s (g 11 Z L   g ) I 2  g 22 I 2

But

I2 

- V2 ZL

V  V2  g 21 Vs  [ g 11 Z s Z L   g Z s  g 22 ] 2   ZL  V2 [ Z L  g 11 Z s Z L   g Z s  g 22 ]  g 21 Vs ZL V2 g 21 Z L  Vs Z L  g 11 Z s Z L   g Z s  g 22 V2 g 21 Z L  Vs Z L  g 11 Z s Z L  g 11 g 22 Z s  g 21 g 12 Z s  g 22 V2 g 21 Z L  Vs (1  g 11 Z s )(g 22  Z L )  g 12 g 21 Z s

(1) (2)

Chapter 19, Solution 42. With the help of Fig. 19.20, we obtain the circuit model below. I1

600 

I2

+ V1 –

+ 10-3 V 2

+ –

120I 1

500 k

V2 –


(a)

To find A and C , consider the network in Fig. (a).

Z

I1

I2 +

V1

+ 

V2 

(a) V1  V2

  A 

 C  I1  0 

V1 1 V2

I1 0 V2

To get B and D , consider the circuit in Fig. (b).

Z

I1

I2 +

V1

+ 

V2 = 0 

(b) V1  Z I 1 , B

- V1 - Z I 1  Z - I1 I2

D

- I1 1 I2

I 2  - I1

Hence,

1 Z [T]    0 1 

(b)

To find A and C , consider the circuit in Fig. (c).

I1

I2 +

V1

+ 

Z

V2 

(c) V1  V2

  A 

V1  Z I 1  V2

V1 1 V2

  C 

I1 1  Y V2 Z

To get B and D , refer to the circuit in Fig.(d).

I2 + I1

+ Y

V1

V2 = 0



 (d)

V1  V2  0 B

- V1  0, I2

I 2  - I1 D

- I1 1 I2

Thus,

 1 0 [T]    Y 1

Chapter 19, Solution 44. Using Fig. 19.99, design a problem to help other students to better understand how to find the transmission parameters of an ac circuit. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine the transmission parameters of the circuit in Fig.19.99.

Figure 19.99 Solution To determine A and C , consider the circuit in Fig.(a).

j15 

Io

I1

-j10 

-j20  Io'

V1

+ 

I2 = 0 Io

20 

+ V2 

(a) V1  [ 20  (- j10) || ( j15  j20) ] I 1

  10  (-j10)(-j5)  V1   20  I 1   20  j  I 1   3 - j15   I o  I1 '

 - j10  2  I 1    I 1 I o   3  - j10  j5 

V2  (-j20) I o  20 I o ' =  j

V1 (20  j10 3) I 1  0.7692 + j0.3461  40  V2   20  j I1 3  

A

C

40  40  I1  20I1 =  20  j I1 3  3 

I1  V2

1 40 20  j 3

 0.03461 + j0.023

To find B and D , consider the circuit in Fig. (b). j15 

I1

-j10 

-j20 

I2 +

V1

+ 

20 

V2 = 0

 (b)

We may transform the  subnetwork to a T as shown in Fig. (c). ( j15)(-j10)  j10 j15  j10  j20 40 (-j10)(-j20) Z2   -j 3 - j15 Z1 

Z3 

( j15)(-j20)  j20 - j15 I1

j10 

j20 

I2 +

V1

+ 

20 – j40/3 

V2 = 0 

(c) - I2 

D

20  j40 3 3  j2 I1  I 20  j40 3  j20 3 j 1

- I1 3 j   0.5385  j0.6923 I2 3  j2

 ( j20)(20  j40 3)  V1   j10  I 20  j40 3  j20  1  V1  [ j10  2 (9  j7) ] I 1  j I 1 (24  j18)

- V1 - j I 1 (24  j18) 6   (-15  j55) - (3 - j2) I2 13 I1 3 j B  -6.923  j25.385 

B

(-6.923  j25.38)    0.7692  j0.3461 [T]     (0.03461  j0.023) S 0.5385  j0.6923 

Chapter 19, Solution 45. To determine A and C consider the circuit below. I1

-j2 

I 2 =0 +

V1

4

+ _

V2 –

V1  (4  j 2)I1, V2  4I1 V 4  j2 A 1   1 j 0.5 V2 4

C

I1 I  1  0.25 V2 4I1

To determine B and D, consider the circuit below. I1

-j2 

I2 +

V1

4 V 2 =0

+ _

– The 4- resistor is short-circuited. Hence, I I2  I1, D   1  1 I2 V1   j 2I1  j 2I2

B

V1 j 2I   2  2 j I2 I2

Hence,

1  j0.5  j2   [T] =  1   0.25 S

Chapter 19, Solution 46. To get A and C , refer to the circuit in Fig.(a).

1

I1

V1

I2 = 0

2

Ix

+ + 

1

1

+

2

Vo

4 Ix

V2 




Vo Vo  V2  2 1

  2 I 1  3 Vo  2 V2

(1)

At node 2, Vo  V2 4 Vo  4Ix   2 Vo 1 2

  Vo  -V2

(2)

From (1) and (2), 2 I 1  -5 V2

But

I1 

  C 

V1  Vo  V1  V2 1

- 2.5 V2  V1  V2 A

I1 - 5   -2.5 S V2 2

  V1  -3.5 V2

V1  -3.5 V2

To get B and D , consider the circuit in Fig. (b). I1

1 +

V1

+ 

1

1

Vo

I2

2

Ix 2

+ 4 Ix

V2 = 0 



(b)

At node 1, I1 

Vo Vo  2 1

  2 I 1  3 Vo

(3)

At node 2, Vo  4Ix  0 1 – I 2  Vo  2 Vo  0   I 2  -3 Vo I2 

Adding (3) and (4), 2 I1  I 2  0   I 1  -0.5 I 2

But

D

- I1  0.5 I2

I1 

V1  Vo 1

  V1  I 1  Vo

Substituting (5) and (4) into (6), -1 -1 -5 V1  I 2  I 2  I 2 3 6 2 B

- V1 5   0.8333  I2 6

Thus,  - 3.5 0.8333   [T]   - 0.5   - 2.5 S

(4)

(5)

(6)

Chapter 19, Solution 47. To get A and C, consider the circuit below. 6

I1

1

+

4

+ Vx

V1 -

V1  Vx Vx Vx  5Vx   1 2 10

2

-

 

V2  4(0.4Vx )  5Vx  3.4Vx

V  Vx I1  1  1.1Vx  Vx  0.1Vx 1

I 2 =0

+ 5V x

+ V2

-

-

V1  1.1Vx

 



A

V1  1.1 / 3.4  0.3235 V2

I C  1  0.1 / 3.4  0.02941 V2

To get B and D, consider the circuit below. 6 1

I1

4

I2 0V

+

+ Vx

V1 -

-

V1  Vx Vx Vx   1 6 2

I2  

2

 

+ 5V x -

V1 

10 Vx 6

5Vx Vx 17    Vx 4 6 12

-

(1)

(2)

V1  I1  Vx

(3)

From (1) and (3) I1  V1  Vx 

+ V 2 =0

4 Vx 6



I 4 12 D   1  ( )  0.4706 I 2 6 17

V 10 12 B   1  ( )  1.176 I2 6 17 S Ω

 0.3235 1.176   [T]    0.02941 S 0.4706 


(a)

Refer to the circuit below. I1

I2 +

V1

+ 

[T]

V2

ZL



V1  4 V2  30 I 2 I 1  0.1 V2  I 2

(1) (2)

When the output terminals are shorted, V2  0 . So, (1) and (2) become V1  -30I 2 and I1  - I 2 Hence, Z in 

(b)

V1  30  I1

When the output terminals are open-circuited, I 2  0 . So, (1) and (2) become V1  4 V2 I 1  0.1 V2 or V2  10 I 1 V1  40 I 1 Z in 

(c)

V1  40  I1

When the output port is terminated by a 10- load, V2  -10 I 2 . So, (1) and (2) become V1  -40 I 2  30 I 2  -70 I 2 I 1  - I 2  I 2  -2 I 2 V1  35 I 1

Z in 

V1  35  I1

Alternatively, we may use Z in 

A ZL  B CZL  D

Chapter 19, Solution 49. To get A and C , refer to the circuit in Fig.(a).

1/s

I1

I2 = 0 +

V1

+ 

1

1/s

1/s

1

V2 

(a)

1 ||

1s 1 1   s 11 s s 1

V2 

1 || 1 s V 1 s  1 || 1 s 1

1 V2 s s 1 A   1 V1 1 2s  1  s s 1  1   2s  1   1  1 1    ||    I1  ||   V1  I 1   s  1   s (s  1)   s  1  s s  1

 1   2s  1     V1  s  1   s (s  1)  2s  1   1 2s  1 I1 (s  1)(3s  1)  s  1 s (s  1) But Hence,

2s  1 s V2 2s  1 2s  1   I1 s (s  1)(3s  1)

V1  V2 

C

V2 (s  1)(3s  1)  s I1

To get B and D , consider the circuit in Fig. (b).

1/s

I1

I2 +

V1

+ 

1

1/s

1

1/s

V2 = 0 

(b) I  1 1  1 V1  I 1 1 || ||   I 1 1 ||   1  s s  2s  2s  1 -1 I -s s 1 1 I2   I 1 1 2s  1 1  s 1 s D

- I 1 2s  1 1   2 s I2 s

I  1  2s  1  I2  2  V1    2s  1  - s  -s

  B 

- V1 1  s I2

 2  2s  1 [T]   (s  1)( 3s  1)  s

  1 2  s

Thus, 1 s

Chapter 19, Solution 50. To get a and c, consider the circuit below. 2 I 1 =0

s

I2

+

+

V1

4/s

V2

-

V1 

-

4/s 4 V2  V2 s  4/s s2  4

a  V2

 

V1

 1  0.25s 2

V2  (s  4 / s)I 2 or

I2 

V2 (1  0.25s 2 )V1  s  4/s s  4/s

 

I s  0.25s3 c 2  V1 s2  4

To get b and d, consider the circuit below. I1

2

s

I2

+

+

V 1 =0

4/s

V2

-

I1 

 4/s 2I I2   2 2  4/s s2

-

 

I d   2  1  0.5s I1

4 (s 2  2s  4) V2  (s  2 // )I2  I2 s s2 

(s 2  2s  4)( s  2) I1 s2 2

 

V b   2  0.5s 2  s  2 I1

 0.25s 2  1 0.5s 2  s  2  [t ]   0.25s 2  s  2 0.5s  1    s  4

Chapter 19, Solution 51. To get a and c , consider the circuit in Fig. (a).

j I1 = 0

1

-j3 

I2

+ j2 

V1

+ 

j

V2



(a) V2  I 2 ( j  j3)  -j2 I 2 V1  -jI 2

a

V2 - j2 I 2  2 V1 - jI 2

c

I2 1  j V1 - j

To get b and d , consider the circuit in Fig. (b).

j I1

1

-j3 

I2

+ V1 = 0

j2 

j



(b) For mesh 1, or

0  (1  j2) I1  j I 2 I 2 1  j2   2 j I1 j d

- I2  -2  j I1

+ 

V2

For mesh 2, V2  I 2 ( j  j3)  j I 1 V2  I 1 (2  j)(- j2)  j I 1  (-2  j5) I 1 b

- V2  2  j5 I1

Thus,  2 2  j5  [t ]     j -2 j

Chapter 19, Solution 52. It is easy to find the z parameters and then transform these to h parameters and T parameters.  R1  R 2 [z ]    R2

R2  R 2  R 3 

 z  (R 1  R 2 )(R 2  R 3 )  R 22  R 1R 2  R 2 R 3  R 3 R 1

(a)

 z z [h]   22 -z  21  z 22

z 12   R 1 R 2  R 2 R 3  R 3 R 1 z 22   R2  R3  - R2 1   z 22   R2  R3

R2  R2  R3   1  R2  R3 

Thus, h 11  R 1 

R 2R 3 , R2  R3

h 12 

R2  - h 21 , R2  R3

h 22 

1 R2  R3

as required.

(b)

 z 11 z [T]   21 1   z 21

 z   R1  R 2 z 21   R 2  z 22 1   z 21   R 2

R 1R 2  R 2 R 3  R 3 R 1   R2  R2  R3  R2 

Hence, A  1

as required.

R3 R1 R1 1 , B  R3  (R 2  R 3 ) , C  , D  1 R2 R2 R2 R2

Chapter 19, Solution 53. For the z parameters, V1  z11 I1  z12 I 2 V2  z12 I1  z 22 I 2

(1) (2)

For ABCD parameters, V1  A V2  B I 2 I1  C V2  D I 2

(3) (4)

From (4),

V2 

I1 D  I C C 2

Comparing (2) and (5), 1 z 21  , C

(5)

z 22 

D C

Substituting (5) into (3),  AD  A  B I 2 V1  I1    C  C A AD  BC  I1  I2 C C Comparing (6) and (1), A z11  C

(6)

z 12 

AD  BC  T  C C

Thus, A  [Z] =  C 1  C

T  C D  C

Chapter 19, Solution 54. For the y parameters I 1  y 11 V1  y 12 V2 I 2  y 21 V1  y 22 V2

(1) (2)

From (2), I 2 y 22  V y 21 y 21 2 -y 1 V1  22 V2  I y 12 y 21 2 V1 

or

(3)

Substituting (3) into (1) gives - y 11 y 22 y 11 I V2  y 12 V2  I1  y 21 2 y 21 - y y 11 or I1  V2  I y 21 y 21 2

(4)

Comparing (3) and (4) with the following equations V1  A V2  B I 2 I 1  C V2  D I 2 clearly shows that A

as required.

- y 22 , y 21

B

-1 , y 21

C

- y y 21

,

D

- y 11 y 21

Chapter 19, Solution 55. For the z parameters V1  z11 I1  z12 I 2 V2  z 21 I1  z 22 I 2

(1) (2)

From (1),

I1 

z 1 V1  12 I 2 z11 z11

(3)

Substituting (3) into (2) gives  z z z  V2  21 V1   z 22  21 12  I 2 z11 z11   z  V2  21 V1  z I 2 or z11 z11

(4)

Comparing (3) and (4) with the following equations I1  g11 V1  g12 I 2 V2  g 21 V1  g 22 I 2 indicates that g 11 

as required.

1 , z 11

g 12 

- z 12 , z 11

g 21 

z 21 , z 11

g 22 

z z 11

Chapter 19, Solution 56. Using Fig. 19.20, we obtain the equivalent circuit as shown below. I1

Rs

h 11

I2

+ Vs

V +1 _

V1

+ h 12 V o + –

h 21 I 1

h 22

–

Vo

RL

–

We can solve this using MATLAB. First, we generate 4 equations from the given circuit. It may help to let V s = 10 V. –10 + R s I 1 + V 1 = 0 or V 1 + 1000I 1 = 10 –10 + R s I 1 + h 11 I 1 + h 12 V o = 0 or 0.0001V s + 1500 = 10 I 2 = –V o /R L or V o + 2000I 2 = 0 h 21 I 1 + h 22 V o – I 2 = 0 or 2x10–6V o + 100I 1 – I 2 = 0 >> A=[1,0,1000,0;0,0.0001,1500,0;0,1,0,2000;0,(2*10^-6),100,-1] A= 1.0e+003 * 0.0010 0 1.0000 0 0 0.0000 1.5000 0 0 0.0010 0 2.0000 0 0.0000 0.1000 -0.0010 >> U=[10;10;0;0] U= 10 10 0 0 >> X=inv(A)*U X= 1.0e+003 * 0.0032 -1.3459 0.0000 0.0007 Gain = V o /V s = –1,345.9/10 = –134.59. There is a second approach we can take to check this problem. First, the resistive value of h 22 is quite large, 500 kΩ versus R L so can be ignored. Working on the right side of the circuit we obtain the following,

I 2 = 100I 1 which leads to V o = –I 2 x2k = –2x105I 1 . Now the left hand loop equation becomes, –V s + (1000 + 500 + 10–4(–2x105))I 1 = 1480I 1 . Solving for V o /V s we get, Our answer checks!

V o /V s = –200,000/1480 = –134.14.

Chapter 19, Solution 57.  T  (3)(7)  (20)(1)  1 A  [z ]   C 1  C

T  C   3 1 D   1 7   C

D  [y ]   -B1  B

- T   B   A     B  

B  [h]   -D1  D

1   T   20    D  7 7   C -1 1  S   D  7 7 

C  [g ]   A 1  A

- T  A  B   A 

D  [t ]   CT   T

7 20 -1 20

1 3S  1  3

-1 20 3 20

  S 

-1  3  20    3

B   T   7 20   A    1 S 3   T 

Chapter 19, Solution 58. Design a problem to help other students to better understand how to develop the y parameters and transmission parameters, given equations in terms of the hybrid parameters. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A two-port is described by V 1 = I 1 + 2V 2 , I 2 = -2I 1 + 0.4V 2 Find: (a) the y parameters, (b) the transmission parameters. Solution The given set of equations is for the h parameters. 1  2   h  (1)(0.4)  (2)(-2)  4.4 [h]     - 2 0.4 S

(a)

 1 h [y ]   11 h  21  h11

(b)

  [T ]    

- h h 21 - h 22 h 21

- h 12 h 11 h h 11

   1 -2   S   - 2 4.4  

- h11 h 21 -1 h 21

   2.2 0.5        0.2 S 0.5  

Chapter 19, Solution 59.  g  (0.06)(2)  (-0.4)(0.2)  0.12  0.08  0.2

(a)

  [z]    

1 g 11 g 21 g 11

g

(b)

  [y ]    

(c)

 g 22   g [h]   - g  21   g

- g 12  2   g   10   g 11   - 1 0.3 S   g 

(d)

  [T]    

g 22  10   g 21   5   g 1    0.3 S g 21 

g 22 - g 21 g 22

1 g 21 g 11 g 21

- g 12  g 11   16.667 6.667    g   3.333 3.333    g 11 

g 12 g 22 1 g 22

   0.1 - 0.2   S   - 0.1 0.5  

Chapter 19, Solution 60. Comparing this with Fig. 19.5, z11  z12  4  j 3  2  2  j3k

z 22 – z 12 = 5 – j – 2 = 3 – j kΩ XL  3 x103   L

  L

3 x103  3mH 106

X C = 1x103 = 1/(ωC) or C = 1/(103x106) = 1 nF Hence, the resulting T network is shown below. 2 k

3 mH

3 k

2 k

1 nF


To obtain z 11 and z 21 , consider the circuit in Fig. (a).

1

Io 1

1

I2 = 0

+ I1

+ 1

V1

V2 

 (a)  2 5 V1  I 1 [1  1 || (1  1) ]  I 1 1    I 1  3 3 V 5 z 11  1  I1 3

Io 

1 1 I1  I1 1 2 3

- V2  I o  I 1  0 1 4 V2  I 1  I 1  I 1 3 3 z 21 

V2 4  3 I1

To obtain z 22 and z 12 , consider the circuit in Fig. (b). 1

I1

1

1

+ V1

+ 1



V2 

(b)

I2

Due to symmetry, this is similar to the circuit in Fig. (a). 5 4 z 22  z 11  , z 21  z 12  3 3   [z ]    

(b)

  [h]     

z z 22 - z 21 z 22

(c)

  [T]     

z 11 z 21 1 z 21

z 12 z 22 1 z 22 z z 21 z 22 z 21

                   

3 4   5 5  -4 3  S 5 5 

5 4 3 S 4

3   4  5   4 

5 3 4 3

4 3  5 3


I1

10 k

a

40 k

+ 

+

I2 +

50 k b

30 k

V1

Ib 20 k



Since no current enters the input terminals of the op amp, V1  (10  30)  10 3 I 1 But

Va  Vb 

V2



(1)

30 3 V1  V1 40 4

Vb 3 V 3  20  10 80  10 3 1 which is the same current that flows through the 50-k resistor. Ib 

Thus,

V2  40  10 3 I 2  (50  20)  10 3 I b 3 V2  40  10 3 I 2  70  10 3  V 80  10 3 1 21 V2  V1  40  10 3 I 2 8 V2  105  10 3 I 1  40  10 3 I 2

From (1) and (2),  40 0  [z ]    k  105 40   z  z 11 z 22  z 12 z 21  16  10 8

(2)

 A B  [T]    C D  

z 11 z 21 1 z 21

z z 21 z 22 z 21

   0.381 15.24 k   0.381    9.52 S 

Chapter 19, Solution 63. To get z 11 and z 21 , consider the circuit below. I1

1:3

I 2 =0

 +

4

V1

 +

+ V’ 1

V’ 2

-

-

+

9

-

V2 -

ZR 

9 n2

 1,

V1  (4 // ZR )I1 

n  3 4 I1 5

 

V2  V2 '  nV1'  nV1  3(4 / 5)I1

V z11  1  0.8 I1  

z 21 

V2  2.4 I1

To get z 21 and z 22 , consider the circuit below. I 1 =0

1:3 

+ V1

4

I2  +

+ V’ 1

V’ 2

-

-

+ 9

-

-

Z R '  n 2 ( 4 )  36 , V2  (9 // ZR ' )I 2 

V2

9x36 I2 45

n 3 

z 22 

V2  7.2 I2

V1 

V2 V2   2.4I 2 n 3



V z 21  1  2.4 I2

Thus,  0.8 2.4 [z ]     2.4 7.2

Chapter 19, Solution 64. 1 -j   - j k 3 jC (10 )(10 -6 )

1 F  

Consider the op amp circuit below.

40 k I1

20 k

Vx

10 k

1

+

 +

2

I2 +

-j k

V1

V2 

 At node 1,

V1  Vx Vx Vx  0   20 -j 10 V1  (3  j20) Vx

(1)

At node 2, Vx  0 0  V2  10 40

But

I1 

  Vx 

-1 V 4 2

V1  Vx 20  10 3

Substituting (2) into (3) gives V1  0.25 V2 I1   50  10 -6 V1  12.5  10 -6 V2 3 20  10 Substituting (2) into (1) yields -1 V1  (3  j20) V2 4 0  V1  (0.75  j5) V2 or

Comparing (4) and (5) with the following equations I 1  y 11 V1  y 12 V2 I 2  y 21 V1  y 22 V2

(2)

(3)

(4)

(5)

indicates that I 2  0 and that  50  10 -6 [y ]   1 

12.5  10 -6  S 0.75  j5 

 y  (77.5  j25.  12.5)  10 -6  (65  j250)  10 -6

  [h]    

1 y 11 y 21 y 11

- y 12  - 0.25  y 11   2  10 4    4 y 1.3  j5 S    2  10 y 11 

Chapter 19, Solution 65. The network consists of two two-ports in series. It is better to work with z parameters and then convert to y parameters. It is obvious that the upper 1 Ω resistor is shorted out by the top circuit so we are essentially left with 2 Ω connected to 3 Ω. This then produces the Z parameters  5 3  [z ]     3 3 

 z  15  9  6  z 22   [y ]   z -z  21  z

- z 12  z   z 11  z 

 0.5  0.5  5 S   - 0.5 6 

Chapter 19, Solution 66. Since we have two two-ports in series, it is better to convert the given y parameters to z parameters.  y  y 11 y 22  y 12 y 21  (2  10 -3 )(10  10 -3 )  0  20  10 -6   [z a ]    

y 22 y - y 21 y

- y 12 y y 11 y

   500  0   100     0 

 500 0   100 100   600 100  [z ]       0 100   100 100   100 200  i.e.

V1  z 11 I 1  z 12 I 2 V2  z 21 I 1  z 22 I 2

or

V1  600 I 1  100 I 2 V2  100 I 1  200 I 2

(1) (2)

But, at the input port, Vs  V1  60 I 1

(3)

and at the output port, V2  Vo  -300I 2

(4)

From (2) and (4), 100 I 1  200 I 2  -300 I 2 I 1  -5 I 2

(5)

Substituting (1) and (5) into (3), Vs  600 I 1  100 I 2  60 I 1  (660)(-5) I 2  100 I 2  -3200 I 2

(6)

From (4) and (6), Vo - 300 I 2   0.09375 V2 - 3200 I 2

Chapter 19, Solution 67. We first the y parameters, To find y 11 and y 21 consider the circuit below. 30 

I1

40  I 2

+ 10 

1A

V 2 =0

V1 -

V1  I1(30  10 // 40)  38I1

 

y11 

I1 1  V1 38

By current division, 10 I 0.2I1 1 I2  I1  0.2I   y21  2   50 V1 38I1 190 1 To find y 22 and y 12 consider the circuit below. I1

30 

40  I2

+

+ 10 

V 1 =0 -

V2

1A

-

V2  (40  10 // 30)I2  47.5I2

 

y22 

I2 2  y 22 = 2/95 V2 93

By current division, I1  

10 I I2   2 30  10 4

 

1  I2 I 1 y12  1  4   V2 47.5I2 190

1/190   1/ 38 [y ]     1/190 2 / 95  For three copies cascaded in parallel, we can use MATLAB. >> Y=[1/38,-1/190;-1/190,2/95]

Y= 0.0263 -0.0053 -0.0053 0.0211 >> Y3=3*Y Y3 = 0.0789 -0.0158 -0.0158 0.0632 >> DY=0.0789*0.0632-0.0158*0.158 DY = 0.0025 >> T=[0.0632/0.0158,1/0.0158;DY/0.0158,0.0789/0.0158] T= 4.0000 63.2911 0.1576 4.9937

63.29    4 T=   0.1576 S 4.994 


 4 -2 For the upper network N a , [y a ]    -2 4 

 2 -1  and for the lower network N b , [y b ]    1 2  For the overall network,  6 -3 [y ]  [y a ]  [y b ]     -3 6   y  36  9  27

  [h]    

1 y 11 y 21 y 11

- y 12   1 y 11   6   y   1   y 11   2

1  2  9  S 2 

Chapter 19, Solution 69. We first determine the y parameters for the upper network N a . To get y 11 and y 21 , consider the circuit in Fig. (a).

n

1 , 2

ZR 

1s 4  n2 s

 2s  4   4 I V1  (2  Z R ) I 1   2   I 1    s  1  s I s y 11  1  V1 2 (s  2) - I1 - s V1  -2 I 1  n s2 I2 -s   V1 s  2

I2  y 21

To get y 22 and y 12 , consider the circuit in Fig. (b).

I1

2

2:1

1/s

I2

+

+

V 1 =0

V2





(b) 1 1 Z R '  (n 2 )(2)    (2)  4 2

1   1 1 s  2 I V2    Z R '  I 2     I 2   s   s 2  2s  2 y 22 

I2 2s  V2 s  2

 - 1  2s   -s   V2   V I 1  - n I 2     2  s  2  s  2 2

I2

y 12 

I1 -s  V2 s  2

 s  2 (s  2) [y a ]   -s   s2

-s s2 2s s2

    

For the lower network N b , we obtain y 11 and y 21 by referring to the network in Fig. (c). 2

I1

I2 +

+ 

V1

s

V2 = 0 

(c) V1  2 I 1

  y 11 

I 2  - I1 

- V1 2

I1 1  V1 2 I 2 -1  V1 2

  y 21 

To get y 22 and y 12 , refer to the circuit in Fig. (d). I1

2

I2

+

+ s

V1 = 0 

V2

I2

 (d)

V2  (s || 2) I 2 

I1  - I 2 

2s I s2 2

  y 22 

I2 s  2  V2 2s

- V2  - s  s  2  -s   V2   s  2  s  2  2s  2

y 12 

I1 - 1  V2 2

12 -1 2  [y b ]     - 1 2 (s  2) 2s   s1  s2 [y ]  [y a ]  [y b ]    - (3s  2)  2 (s  2)

- (3s  2) 2 (s  2) 5s 2  4s  4 2s (s  2)

    

Chapter 19, Solution 70. We may obtain the g parameters from the given z parameters.  25 20   z a  250  100  150 [z a ]   ,  5 10 

 50 25  [z b ]   ,  25 30    [g ]    

1 z 11 z 21 z 11

- z 12 z 11 z z 11

 z b  1500  625  875     

 0.04 - 0.8  [g a ]   , 6   0.2

 0.02 - 0.5  [g b ]     0.5 17.5 

 0.06 S - 1.3  [g ]  [g a ]  [ g b ]   23.5    0.7

Chapter 19, Solution 71. This is a parallel-series connection of two two-ports. We need to add their g parameters together and obtain z parameters from there. For the transformer,

V1 

1 V2 , I1  2I 2 2

Comparing this with V1  AV2  BI2 ,

I1  CV2  DI 2

shows that 0.5 0 [Tb1]     0 2

To get A and C for T b2 , consider the circuit below. I1

4

+

5

V1 -

V1  9I1,

A

I 2 =0

2

V2  5I1

V1  9 / 5  1.8, V2

I C  1  1 / 5  0.2 V2

+ V2 -

We obtain B and D by looking at the circuit below. 4

I1

I 2 =0

+ V1 -

5 I 2   I1 7

+

5 2

 

V 2 =0 -

I D   1  7 / 5  1.4 I2

7 38 V1  4I1  2I 2  4( I 2 )  2I 2   I 2 5 5

 

1.8 7.6 [Tb 2 ]    0.2 1.4  0.9 3.8 [T ]  [Tb1][Tb 2 ]   , 0.4 2.8

T  1

C / A   T / A  0.4444  1.1111  [g b ]   B / A  1.1111 4.2222   1/ A

From Prob. 19.52, 1.8 18.8 [Ta ]    0.1 1.6  C / A   T / A  0.05555  0.5555  [g a ]   B / A   0.5555 10.4444   1/ A 0.4999  1.6667 [g ]  [ g a ]  [ g b ]    1.6667 14.667  Thus,

V B   1  7.6 I2

 1 / g 11 [z ]   g 21 / g 11

 g 21 / g 11   2  3.334     g / g 11   3.334 20.22 

Chapter 19, Solution 72. Consider the network shown below. I1

I a1 + V a1

+

V1

I a2

Na

I b1 + V b1



I2

+ V a2

+

V2

I b2

Nb

Va1  25 I a1  4 Va 2 I a 2  - 4 I a1  Va 2 Vb1  16 I b1  Vb 2 I b 2  - I b1  0.5 Vb 2

+ V b2



(1) (2) (3) (4)

V1  Va1  Vb1 V2  Va 2  Vb 2 I 2  I a 2  I b2 I 1  I a1 Now, rewrite (1) to (4) in terms of I 1 and V2 Va1  25 I 1  4 V2 I a 2  - 4 I 1  V2 Vb1  16 I b1  V2 I b 2  - I b1  0.5 V2

(5) (6) (7) (8)

Adding (5) and (7), V1  25 I 1  16 I b1  5 V2

(9)

Adding (6) and (8), I 2  - 4 I 1  I b1  1.5 V2

(10)

I b1  I a1  I 1

(11)

Because the two networks N a and N b are independent, I 2  - 5 I 1  1.5 V2 or V2  3.333 I 1  0.6667 I 2

(12)

Substituting (11) and (12) into (9), 25 5 V1  41I 1  I1  I 1.5 1.5 2 V1  57.67 I 1  3.333 I 2

(13)

Comparing (12) and (13) with the following equations V1  z 11 I 1  z 12 I 2 V2  z 21 I 1  z 22 I 2 indicates that  57.67 3.333  [z ]     3.333 0.6667  Alternatively,  25 4  [h a ]   , -4 1

 16 1  [h b ]     - 1 0.5 

 41 5  [h]  [h a ]  [h b ]     - 5 1.5    [z ]    

as obtained previously.

h h 22 - h 21 h 22

 h  61.5  25  86.5

h12 h 22 1 h 22

   57.67 3.333    3 . 333 0 . 6667    

Chapter 19, Solution 73. From Problem 19.6,  25 20  [ z]   z  25 x30  20 x24  270 , 24 30  A C

z11 25  , z21 24 1 z 21



1 ,D 24

B

z 270  z21 24

z 22 30  z 21 24

The overall ABCD parameters can be found using MATLAB. >> T=[25/24,270/24;1/24,30/24] T= 1.0417 11.2500 0.0417 1.2500 >> T3=T*T*T T3 = 2.6928 49.7070 0.1841 3.6133 >> Z=[2.693/0.1841,(2.693*3.613-0.1841*49.71)/0.1841;1/0.1841,3.613/0.1841] Z= 14.6279 3.1407 5.4318 19.6252

14.628 3.141  [Z] =    5.432 19.625

Chapter 19, Solution 74. From Prob. 18.35, the transmission parameters for the circuit in Figs. (a) and (b) are

1 Z [Ta ]   , 0 1 

 1 0 [Tb ]    1 Z 1 

Z

Z (a)

(b)

We partition the given circuit into six subcircuits similar to those in Figs. (a) and (b) as shown in Fig. (c) and obtain [T] for each.

s

s

1

1/s

T1

T2

T3

1

T4

T5

1/s

T6

1 0  [T1 ]   , 1 1 

1 s [T2 ]   ,  0 1

1 0  [T3 ]    s 1

[T4 ]  [T2 ] ,

[T5 ]  [T1 ] ,

[T6 ]  [T3 ]

 1 0  1 0  [T]  [T1 ][T2 ][T3 ][T4 ][T5 ][T6 ]  [T1 ][T2 ][T3 ][T4 ]    1 1  s 1 

0 0  1 1 s 1 T T T  [T1 ][T2 ][T3 ][T4 ]   [ ] [ ] [ ] 1 2 3   0 1   s 1 1   s 1 1      1 0   s2  s 1 s   [T1 ][T2 ]    1  s 1   s 1

s   1 s   s2  s 1  [T1 ]   3   2 2  0 1   s  s  2s  1 s  1   1 0   s 4  s 3  3s 2  2s  1 s 3  2s     3 2 s2 1   1 1   s  s  2s  1  s 4  s 3  3s 2  2s  1  s 3  2s [T]   4   s  2s 3  4s 2  4s  2 s 3  s 2  2s  1 

Note that AB  CD  1 as expected.

Chapter 19, Solution 75. (a) We convert [z a ] and [z b ] to T-parameters. For N a ,  z  40  24  16 .  z / z 21   2 4  z / z  [Ta ]   11 21    1 / z 21 z 22 / z 21  0.25 1.25

For N b ,  y  80  8  88 .

 y 22 / y 21  1 / y 21    5  0.5 [Tb ]        y / y 21  y11 / y 21   44  4   17    186 [T]  [Ta ][Tb ]     56.25  5.125

We convert this to y-parameters.  T  AD  BC  3.  D / B   T / B  0.3015  0.1765 [y ]    S A / B  0.0588 10.94   1 / B

(b)

The equivalent z-parameters are A / C  T / C  3.3067 0.0533 [z]      1 / C D / C   0.0178 0.0911

Consider the equivalent circuit below. I1

z 11

z 22

+

I2

+

+ +

Vi

z 12 I 2

ZL z 21 I 1

-

Vo -

-

-

Vi  z11I1  z12 I 2

(1)

Vo  z 21I1  z 22 I 2 But Vo  I 2 ZL



(2) I 2  Vo / ZL

(3)

From (2) and (3) , V Vo  z 21I1  z 22 o ZL



 1  z  22  I1  Vo   z 21 ZL z 21 

(4)

Substituting (3) and (4) into (1) gives Vi  z 11 z 11 z 22   Vo  z 21 z 21 Z L

 z 12    194.3 Z L 

 

Vo .  0.0051 Vi


To get z 11 and z 21 , we open circuit the output port and let I 1 = 1A so that V V z11  1  V1, z 21  2  V2 I1 I1

The schematic is shown below. After it is saved and run, we obtain z11  V1  3.849,

z 21  V2  1.122

Similarly, to get z 22 and z 12 , we open circuit the input port and let I 2 = 1A so that V z12  1  V1, I2

z 22 

V2  V2 I2


z 22  V2  3.849

Thus,  3.949 1.122  [z ]    1.122 3.849

Chapter 19, Solution 77. We follow Example 19.15 except that this is an AC circuit. (a) We set V 2 = 0 and I 1 = 1 A. The schematic is shown below. In the AC Sweep Box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, the output file includes FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

3.163 E–.01

–1.616 E+02

FREQ

VM($N_0001)

VP($N_0001)

1.592 E–01

9.488 E–01

–1.616 E+02

From this we obtain h 11 = V 1 /1 = 0.9488–161.6 h 21 = I 2 /1 = 0.3163–161.6.

(b) In this case, we set I 1 = 0 and V 2 = 1V. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain an output file which includes

FREQ

VM($N_0001)

VP($N_0001)

1.592 E–01

3.163 E–.01

1.842 E+01

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

9.488 E–01

–1.616 E+02

From this, h 12 = V 1 /1 = 0.316318.42 h 21 = I 2 /1 = 0.9488–161.6.

Thus,

0.316318.42  0.9488  161.6  [h] =    0.3163  161.6 0.9488  161.6 S 

Chapter 19, Solution 78 For h 11 and h 21 , short-circuit the output port and let I 1 = 1A. f   / 2  0.6366 . The schematic is shown below. When it is saved and run, the output file contains the following: FREQ


6.366E-01 FREQ

1.202E+00

1.463E+02

VM($N_0003) VP($N_0003)

6.366E-01

3.771E+00

-1.350E+02

From the output file, we obtain I 2  1.202146.3o ,

V1  3.771  135o

so that V h11  1  3.771  135o , 1

I h 21  2  1.202146.3o 1

For h 12 and h 22 , open-circuit the input port and let V 2 = 1V. The schematic is shown below. When it is saved and run, the output file includes: FREQ 6.366E-01

VM($N_0003) VP($N_0003) 1.202E+00

-3.369E+01

FREQ


6.366E-01

3.727E-01

-1.534E+02

From the output file, we obtain I 2  0.3727  153.4o ,

V1  1.202  33.69o

so that V h12  1  1.202  33.69o , 1

I h 22  2  0.3727  153.4o 1

Thus,

 3.771  135 o  1.202  33.69 o  [h]    0.3727  153.4 o S   1.202146.3

Chapter 19, Solution 79 We follow Example 19.16. (a) We set I 1 = 1 A and open-circuit the output-port so that I 2 = 0. The schematic is shown below with two VPRINT1s to measure V 1 and V 2 . In the AC Sweep box, we enter Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes

FREQ

VM(1)

VP(1)

3.183 E–01

4.669 E+00

–1.367 E+02

FREQ

VM(4)

VP(4)

3.183 E–01

2.530 E+00

–1.084 E+02

From this,

z 11 = V 1 /I 1 = 4.669–136.7/1 = 4.669–136.7 z 21 = V 2 /I 1 = 2.53–108.4/1 = 2.53–108.4.

(b) In this case, we let I 2 = 1 A and open-circuit the input port. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes

FREQ

VM(1)

VP(1)

3.183 E–01

2.530 E+00

–1.084 E+02

FREQ

VM(2)

VP(2)

3.183 E–01

1.789 E+00

–1.534 E+02

From this, z 12 = V 1 /I 2 = 2.53–108.4/1 = 2.53–108..4 z 22 = V 2 /I 2 = 1.789–153.4/1 = 1.789–153.4. Thus, 4.669  136.7 2.53  108.4  [z] =  Ω  2.53  108.4 1.789  153.4

Chapter 19, Solution 80 To get z 11 and z 21 , we open circuit the output port and let I 1 = 1A so that

V z11  1  V1, I1

z 21 

V2  V2 I1


z 21  V2  70.37

Similarly, to get z 22 and z 12 , we open circuit the input port and let I 2 = 1A so that V z12  1  V1, I2

z 22 

V2  V2 I2


z 22  V2  11.11

Thus,  29.88 3.704 [z ]      70.37 11.11

Chapter 19, Solution 81 (a) We set V 1 = 1 and short circuit the output port. The schematic is shown below. After simulation we obtain y 11 = I 1 = 1.5, y 21 = I 2 = 3.5

(b) We set V 2 = 1 and short-circuit the input port. The schematic is shown below. Upon simulating the circuit, we obtain y 12 = I 1 = –0.5, y 22 = I 2 = 1.5 1.5  0.5 [Y] =  S  3.5 1.5 

Chapter 19, Solution 82 We follow Example 19.15. (a) Set V 2 = 0 and I 1 = 1A. The schematic is shown below. After simulation, we obtain h 11 = V 1 /1 = 3.8, h 21 = I 2 /1 = 3.6

(b) Set V 1 = 1 V and I 1 = 0. The schematic is shown below. After simulation, we obtain h 12 = V 1 /1 = 0.4, h 22 = I 2 /1 = 0.25 Hence,

0 .4   3 .8  [h] =    3.6 0.25 S 

Chapter 19, Solution 83 To get A and C, we open-circuit the output and let I 1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V 1 = 11 and V 2 = 34.

A

V1  0.3235, V2

I 1 C 1   0.02941 V2 34

Similarly, to get B and D, we open-circuit the output and let I 1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V 1 = 2.5 and I 2 = -2.125.

V 2.5 B 1   1.1765, I2 2.125

I 1 D 1   0.4706 I 2 2.125

Thus,  0.3235 1.1765   [T]    0.02941 S 0.4706 


(a)

Since A =

V1 V2

and C = I 2 0

I1 V2

, we open-circuit the output port and let V 1 I 2 0

= 1 V. The schematic is as shown below. After simulation, we obtain A = 1/V 2 = 1/0.7143 = 1.4 C = I 2 /V 2 = 1.0/0.7143 = 1.4

(b) To get B and D, we short-circuit the output port and let V 1 = 1. The schematic is shown below. After simulating the circuit, we obtain B = –V 1 /I 2 = –1/1.25 = –0.8 D = –I 1 /I 2 = –2.25/1.25 = –1.8 Thus

A B   1.4  0.8    C D = 1.4 S  1.8     


(a)

Since A =

V1 V2

and C = I 2 0

I1 V2

, we let V 1 = 1 V and I 2 0

open-circuit the output port. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain an output file which includes

FREQ 1.592 E–01

IM(V_PRINT1) 6.325 E–01

IP(V_PRINT1) 1.843 E+01

FREQ 1.592 E–01

VM($N_0002) 6.325 E–01

VP($N_0002) –7.159 E+01

From this, we obtain A =

1 1   1.58171.59 V2 0.6325  71.59

C =

I1 0.632518.43   190 = j V2 0.6325  71.59

(b)

Similarly, since B =

V1 I2

and D =  V2  0

I1 I2

, we let V 1 = 1 V and shortV2  0

circuit the output port. The schematic is shown below. Again, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. After simulation, we get an output file which includes the following results:

FREQ 1.592 E–01

IM(V_PRINT1) 5.661 E–04

IP(V_PRINT1) 8.997 E+01

FREQ 1.592 E–01

IM(V_PRINT3) 9.997 E–01

IP(V_PRINT3) –9.003 E+01

From this, B = 

1 1   190   j I2 0.9997   90

D = 

I1 5.661x10 4 89.97  = 5.561x10–4 I2 0.9997  90  j  A B  1.58171.59  C D =  jS 5.661x10 4    


By definition, g 11 =

I1 V1

, g 21 = I 2 0

V1 V2

. I 2 0

We let V 1 = 1 V and open-circuit the output port. The schematic is shown below. After simulation, we obtain g 11 = I 1 = 2.7 g 21 = V 2 = 0.0

(b)

Similarly, g 12 =

I1 I2

, g 22 = V1  0

V2 I2

V1  0

We let I 2 = 1 A and short-circuit the input port. The schematic is shown below. After simulation, g 12 = I 1 = 0 g 22 = V 2 = 0 Thus

 2.727S 0 [g] =  0  0


(a)

Since

a =

V2 V1

and c = I1  0

I2 V1

, I1  0

we open-circuit the input port and let V 2 = 1 V. The schematic is shown below. In the AC Sweep box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain an output file which includes FREQ 1.592 E–01

IM(V_PRINT2) 5.000 E–01

IP(V_PRINT2) 1.800 E+02

FREQ 1.592 E–01

VM($N_0001) 5.664 E–04

VP($N_0001) 8.997 E+01

From this, a =

c =

(b)

1 5.664 x10  4 89.97

 1765  89.97

0.5180  882.28  89.97 5.664x10  4 89.97

Similarly, b = 

V2 I1

and d =  V1  0

I2 I1

V1  0

We short-circuit the input port and let V 2 = 1 V. The schematic is shown below. After simulation, we obtain an output file which includes

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

5.000 E–01

1.800 E+02

FREQ 1.592 E–01

IM(V_PRINT3) 5.664 E–04

IP(V_PRINT3) –9.010 E+01

From this, we get b =  d = 

Thus

1 4

5.664x10   90.1

= –j1765

0.5180 = j888.28 5.664x10  4   90.1   j1765  j1765   [t] =  j888.2   j888.2 S

Chapter 19, Solution 88 To get Z in , consider the network in Fig. (a).

Rs

I1

I2

+ Vs

+ 

+ Two-Port

V1

RL

V2



 (a)

Z in I 1  y 11 V1  y 12 V2 I 2  y 21 V1  y 22 V2

But

(1) (2)

- V2  y 21 V1  y 22 V2 RL - y 21 V1 V2  y 22  1 R L

I2 

(3)

Substituting (3) into (1) yields  - y 21 V1  , I 1  y 11 V1  y 12    y 22  1 R L 

  y  y 11 YL   V1 , I1    y 22  YL 

1 RL

 y  y 11 y 22  y 12 y 21

Z in 

or

YL 

V1 y 22  YL  I 1  y  y 11 YL

 y   - y 21 V1  I2 y V  y 22 V2   21 1  y 21 Z in   22   I1 I1  I 1   y 22  YL   y  YL  y y Z  y 21  y 22 y 21  y 21 Z in  22 21 in   22   y 22  YL y 22  YL   y  y 11 YL 

Ai 

Ai 

From (3),

y 21 YL  y  y 11 YL

  

Av 

V2 - y 21  V1 y 22  YL

To get Z out , consider the circuit in Fig. (b). I1

I2

+ Rs

V1

+ 

Two-Port

V2

 (b) Z out 

But

Z out

V2 V2  I 2 y 21 V1  y 22 V2

(4)

V1  - R s I 1

Substituting this into (1) yields I 1  - y 11 R s I 1  y 12 V2 (1  y 11 R s ) I 1  y 12 V2 y 12 V2 - V1 I1   1  y 11 R s Rs - y 12 R s V1  or V2 1  y 11 R s Substituting this into (4) gives 1 Z out  y 12 y 21 R s y 22  1  y 11 R s 1  y 11 R s  y 22  y 11 y 22 R s  y 21 y 22 R s Z out 

y 11  Ys  y  y 22 Ys


Av 

- h fe R L h ie  (h ie h oe  h re h fe ) R L

- 72  10 5 Av  2640  (2640  16  10 -6  2.6  10 -4  72)  10 5 - 72  10 5 Av   - 1613 2640  1824

dc gain  20 log A v  20 log (1613)  64.15 dB


(a)

Z in  h ie 

h re h fe R L 1  h oe R L

1500  2000 

10 -4  120 R L 1  20  10 -6 R L

12  10 -3 500  1  2  10 -5 R L

500  10 -2 R L  12  10 -3 R L 500  10 2  0.2 R L R L  250 k (b)


Av 

- 120  250  10 3 2000  (2000  20  10 -6  120  10 -4 )  250  10 3 - 30  10 6 Av   - 3333 2  10 3  7  10 3

Av 

Ai 

h fe 120   20 1  h oe R L 1  20  10 -6  250  10 3

R s  h ie 600  2000  (R s  h ie ) h oe  h re h fe (600  2000)  20  10 -6  10 -4  120 2600  k  65 k 40

Z out 

Z out

(c)

Av 

Vc Vc  Vb Vs

  Vc  A v Vs  -3333  4  10 -3  - 13.33 V

Chapter 19, Solution 91 R s  1.2 k , (a)

At 

R L  4 k


- 80  4  10 3 1200  (1200  20  10 -6  1.5  10 -4  80)  4  10 3 - 32000 At   - 25.64 This is just the gain for the transistor. If we 1248 calculate the gain for the circuit we get A t = V o /V be and V be = V s [1.2k/(1.2k+2k)] = 0.375, thus, V A = (0.375)(–25.64) = –9.615. At 

h fe 80   74.07 1  h oe R L 1  20  10 -6  4  10 3

(b)

Ai 

(c)

Z in  h ie  h re A i Z in  1200  1.5  10 -4  74.074  1.2 k

(d)

R s  h ie (R s  h ie ) h oe  h re h fe 1200  1200 2400    51.28 k -6 -4 2400  20  10  1.5  10  80 0.0468

Z out 

Z out

(a) –25.64 for the transistor and –9.615 for the circuit, (b) 74.07, (c) 1.2 kΩ, (d) 51.28 kΩ

Chapter 19, Solution 92 Due to the resistor R E  240  , we cannot use the formulas in section 18.9.1. We will need to derive our own. Consider the circuit in Fig. (a).

Rs

Ib

h ie

Ic

+

+ h re V c

Vs

+ 

+ 

h fe I b

Vb

Vc IE RE





(a)

Z in IE  Ib  Ic

(1)

Vb  h ie I b  h re Vc  (I b  I c ) R E

(2)

Vc RE  1

(3)

I c  h fe I b 

But

h oe

h oe

Vc  - I c R L

(4)


I c  h fe I b 

or

Ai 

RL RE  1

Ic h oe

I c h fe (1  R E h oe )  Ib 1  h oe (R L

100(1  240x 30 x10 6 ) 1  30  10 -6 (4,000  240) A i  79.18

Ai 

From (3) and (5),

(5)

RL

Ic 

h fe (1  R E )h oe Vc I b  h fe I b  1  h oe (R L  R E ) RE  1

(6) h oe

Substituting (4) and (6) into (2), Vb  (h ie  R E ) I b  h re Vc  I c R E Vb 

Vc (h ie  R E ) V  h re Vc  c R E RL   1   h fe (1  R E h oe )    R E   h fe  h oe  1  h oe (R L  R E )  

V 1  b  A v Vc  1  R E  h oe 

(h ie  R E )    h fe (1  R E h oe )    h fe    1  h oe (R L  R E )

 h re 

RE RL

(7)

1 (4000  240) 240 -4 10    6 Av  4000  1  100(1  240 x30x10 )  100  240   6  -6 30 x10   1  30  10  4240   1  6.06x10 3  10 -4  0.06  -0.066 Av

A v  –15.15 From (5), Ic 

h fe I 1  h oe R L b

We substitute this with (4) into (2) to get Vb  (h ie  R E ) I b  (R E  h re R L ) I c  h (1  R E h oe )  Vb  (h ie  R E ) I b  (R E  h re R L )  fe I b   1  h oe (R L  R E )  Z in 

h (R  h re R L )(1  R E h oe ) Vb  h ie  R E  fe E 1  h oe (R L  R E ) Ib (8)

Z in  4000  240  Z in  12.818 k

(100)(240  10 -4  4  10 3 )(1  240x 30x10 6 ) 1  30  10 -6  4240

To obtain Z out , which is the same as the Thevenin impedance at the output, we introduce a 1-V source as shown in Fig. (b). Rs

h ie

Ib

Ic

+

+ + 

h re V c

h fe I b

Vb

h oe + 

Vc IE RE



1V

 (b)

Z out

From the input loop, I b (R s  h ie )  h re Vc  R E (I b  I c )  0 But

Vc  1

So, I b (R s  h ie  R E )  h re  R E I c  0

(9)

From the output loop, Ic 

or

Vc

RE 

1 h oe

 h fe I b 

h oe h fe Ic Ib   h fe 1  R E h oe

h oe  h fe I b R E h oe  1

(10)

Substituting (10) into (9) gives h  (R s  R E  h ie ) oe  h  Ic  fe     h re  R E I c  (R s  R E  h ie )  0 h 1  R h fe E oe   R s  R E  h ie R  R E  h ie  h oe    h re  Ic  R E Ic  s h fe 1  R E h oe  h fe 

 R  R E  h ie  (h oe h fe )  s   h re 1  R E h oe   Ic  R E  (R s  R E  h ie ) h fe Z out 

240  100  (1200  240  4000) 1200  240  4000  -6 -4  1  240 x 30 x10 6   30  10  10  100   24000  5440   193.7 k 0.152

Z out 

Z out

R E h fe  R s  R E  h ie 1  I c  R s  R E  h ie   h oe  h re h fe   1  R E h oe 

Chapter 19, Solution 93 We apply the same formulas derived in the previous problem. (h ie  R E ) R 1   h re  E Av  RL  1   h fe (1  R E h oe )  R E     h fe  h oe  1  h oe (R L  R E )   1  Av

(2000  200) 200  2.5  10 -4  3800  150(1  0.002) (200  10 5 )   150   1  0.04

1  0.004  2.5  10- 4  0.05263  -0.05638 Av A v  –17.74 Ai 

h fe (1  R E h oe ) 150(1  200x10 5 )   144.5 1  h oe (R L  R E ) 1  10 -5  (200  3800)

Z in  h ie  R E 

h fe (R E  h re R L )(1  R E h oe ) 1  h oe (R L  R E )

(150)(200  2.5  10 -4  3.8  10 3 )(1.002) Z in  2000  200  1.04 Z in  2200  28966 Z in  31.17 k Z out 

Z out 

R E h fe  R s  R E  h ie  R s  R E  h ie    h oe  h re h fe  1  R E h oe  200  150  1000  200  2000 33200  - 0.0055  3200  10 -5  -4    2.5  10  150  1.002 

Z out  –6.148 M

Chapter 19, Solution 94 We first obtain the ABCD parameters.  200 0  Given [h]   ,  100 10 -6    [T]    

h h 21 - h 22 h 21

- h11 h 21 -1 h 21

 h  h11 h 22  h12 h 21  2  10 -4

   - 2  10 -6  -8   - 10 

-2   - 10 -2 

The overall ABCD parameters for the amplifier are  - 2  10 -6 - 2  - 2  10 -6 -2 [T]   -8 -2  -8 - 10  - 10 - 10 -2  - 10

  2  10 -8    10 -10

 T  2  10 -12  2  10 -12  0 B  [h]   D -1  D

Thus,

h ie  200 ,

Av 

T D C D

 0    200  4 -6    - 10 10  

h re  0 ,

h fe  -10 4 ,

h oe  10 -6

(10 4 )(4  10 3 ) 5 -4 3  2  10 200  (2  10  0)  4  10

Z in  h ie 

h re h fe R L  200  0  200  1  h oe R L

2  10 -2   10 -4 

Chapter 19, Solution 95 1 s 4  10s 2  8  Let Z A  y 22 s 3  5s

Using long division, ZA  s 

i.e.

5s 2  8  s L1  Z B s 3  5s

L1  1 H

and

5s 2  8 ZB  3 s  5s

as shown in Fig (a).

L1 ZB

y 22 = (a) 1 s 3  5s YB   Z B 5s 2  8

Using long division,

YB  0.2s 

where

C 2  0.2 F

3.4s  sC 2  YC 5s 2  8 and

YC 

3.4s 5s 2  8

as shown in Fig. (b). L1 C2 Y c = 1/Z C (b)

1 5s 2  8 5s 8 1 ZC      s L3  YC 3.4s 3.4 3.4s s C4

i.e. an inductor in series with a capacitor 5 L3   1.471 H and 3.4

C4 

3.4  0.425 F 8

Thus, the LC network is shown in Fig. (c). 425 mF

1.471 H

1H

200 mF

(c)

Chapter 19, Solution 96 This is a fourth order network which can be realized with the network shown in Fig. (a).

L1

L3

C2

1

C4

(a)  (s)  (s 4  3.414s 2  1)  (2.613s 3  2.613s) 1 2.613s  2.613s H(s)  s 4  3.414s 2  1 1 2.613s 3  2.613s 3

which indicates that

-1 2.613s  2.613s s 4  3.414s  1  2.613s 3  2.613s

y 21  y 22

3

We seek to realize y 22 . By long division, 2.414s 2  1 y 22  0.383s   s C 4  YA 2.613s 3  2.613s i.e.

C 4  0.383 F

and

2.414s 2  1 YA  2.613s 3  2.613s

as shown in Fig. (b). L1

YA

L3

C2

C4

(b)

y 22

ZA 

1 2.613s 3  2.613s  YA 2.414s 2  1

By long division, Z A  1.082s 

i.e.

1.531s  s L3  Z B 2.414s 2  1

L 3  1.082 H

and

ZB 

1.531s 2.414s 2  1

as shown in Fig.(c). L1

ZB

L3

C2

C4

(c) YB  i.e.

1 1 1  1.577s   s C2  s L1 1.531s ZB

C 2  1.577 F

and

L1  1.531 H

Thus, the network is shown in Fig. (d). 1.531 H

1.577 F

1.082 H

0.383 F

(d)

1


s3 s s 3  12s H(s)  3  6s 2  24 (s  12s)  (6s 2  24) 1 3 s  12s 3

Hence, y 22

6s 2  24 1  3   ZA s  12s s C 3

(1)

where Z A is shown in the figure below.

C1

C3

L2

ZA

y 22

We now obtain C 3 and Z A using partial fraction expansion. Let

6s 2  24 A Bs  C   2 2 s (s  12) s s  12

6s 2  24  A (s 2  12)  Bs 2  Cs Equating coefficients : 24  12A   A  2 s0 : 1 0C s : 2 s : 6 AB   B  4

Thus, 6s 2  24 2 4s   2 2 s (s  12) s s  12

(2)

Comparing (1) and (2), 1 1 C3   F A 2

But

3 1 s 2  12 1   s ZA s 4s 4

(3)

1 1  sC1  s L2 ZA

(4)

Comparing (3) and (4), 1 C1  F 4

and

L2 

1 H 3

Therefore, C1  250 mF ,

L 2  333.3 mH ,

C 3  500 mF

Chapter 19, Solution 98  h  1  0.8  0.2    h / h 21  h11 / h 21    0.001 [Ta ]  [Tb ]    6  h 22 / h 21  1 / h 21   2.5x10

 10   0.005

2.6x105 0.06  [T]  [Ta ][Tb ]    8 5x105  1.5x10 We now convert this to z-parameters

A / C  T / C 1.733x103 [z]    7  1 / C D / C  6.667 x10

1000

I1

0.0267   3.33x103 

z 11

+

z 22 +

I2 +

+ Vs

z 12 I 2

ZL z 21 I 1

-

-

Vo -

-

Vs  (1000  z11)I1  z12 I 2

(1)

Vo  z 22 I 2  z 21I1

(2)

But Vo  I 2 ZL



I 2  Vo / ZL

(3)

Substituting (3) into (2) gives  1  z  22  I1  Vo   z 21 z 21ZL 

We substitute (3) and (4) into (1)

(4)

 1  z z  22  Vo  12 Vo Vs  (1000  z 11 ) ZL  z 11 z 21 Z L   7.653x10  4  2.136x10  5  744V


Z ab  Z1  Z 3  Z c || (Z b  Z a ) Z c (Z a  Z b ) Z1  Z 3  Za  Zb  Zc Z cd  Z 2  Z 3  Z a || (Z b  Z c ) Z a (Z b  Z c ) Z2  Z3  Za  Zb  Zc Z ac  Z1  Z 2  Z b || (Z a  Z c ) Z b (Z a  Z c ) Z1  Z 2  Za  Zb  Zc

(1)

(2)

(3)

Subtracting (2) from (1), Z1  Z 2 

Z b (Z c  Z a ) Za  Zb  Zc

(4)

Adding (3) and (4), Z1 

ZbZc Za  Zb  Zc

(5)

Subtracting (5) from (3), Z2 

ZaZb Za  Zb  Zc

(6)

Subtracting (5) from (1), Z3 

ZcZa Za  Zb  Zc

(7)

Using (5) to (7)

Z a Z b Z c (Z a  Z b  Z c ) (Z a  Z b  Z c ) 2 Za ZbZc Z1Z 2  Z 2 Z 3  Z 3 Z1  Za  Zb  Zc Z1Z 2  Z 2 Z 3  Z 3 Z1 

Dividing (8) by each of (5), (6), and (7),

(8)

Za 

Z1Z 2  Z 2 Z 3  Z 3 Z1 Z1

Zb 

Z1Z 2  Z 2 Z 3  Z 3 Z1 Z3

Zc 

Z1Z 2  Z 2 Z 3  Z 3 Z1 Z2

as required. Note that the formulas above are not exactly the same as those in Chapter 9 because the locations of Z b and Z c are interchanged in Fig. 18.122.

Below are answers for the Network Analysis Tutorials. Some of the tutorial pages have random parameters. For these pages, there are no fixed right answers, and formulas are provided instead. Introductory Tutorial (Tut22) 1. Orange 2. 12 3. 3.14159 The Physics of Electricity (Tut1A) 1. 36000 2. 32.04 3. 60 4. 4.32 5. 18000 Basic Elements and Circuit Laws (Tut1) 1. 3 2. -2 3. 50 4. 50 5. -60 6. Answerless page Resistors in Series and Parallel (Tut2) 1. 4 2. 1 3. 10 4. 17 5. 11 6. Formula: R1×R2/(R1 + R2) + (R3 + R4)×R5/(R3 + R4 + R5) + R7 Voltage Dividers and Current Dividers (Tut2A) 1. 40 2. 160 3. 12 4. 20 Circuit Solving with Kirchhoff's Laws (Tut3) 1. 3 2. 2 3. 3 4. 5 5. 4 6. 6 7. I6 8. V9

9. Answerless page The Node Voltage Method (Tut4) 1. a 2. g 3. -6 4. 5 5. 4 6. g 7. 9 8. Ve 9. 0 10. Answerless page 11. -2 12. 4 13. Answerless page 14. I4 15. I4 16. Answerless page The Mesh Current Method (Tut5) 1. c 2. -2 3. -4 4. -2 5. Ib 6. 3A 7. Ia 8. Answerless page 9. 5 Thevenin Laboratory (Tut6) 1. Formula: 1000×Voc×Vr/[R×(Voc - Vr)] Maximum Power Transfer (DC) (Tut6A) 1. Formula: R2×R3/(R2 + R3) 2. Formula: E×R3/(R2 + R3) 3. Formula: VT2/(4×RT) where VT = E×R3/(R2 + R3) Superposition (Tut6B) 1. 2 2. 5 3. -3 4. 4 Inductors and Capacitors (Tut7)

1. 0.8 2. 0.2 3. 0 4. 12t 5. -20e-4t 6. 16cos(8t) 7. -80 8. 9.79992 9. 3t 10. 2t3 11. -1.25e-4t 12. -0.25cos[8t] 13. 1.5 14. 29.532 15. Answerless page First Order Systems (Tut8) 1. 0 2. 0 3. 120 4. 600 5. 120 6. 34.016 7. 0 8. 0 9. 90 10. 450 11. 0.45 12. -200 13. -200 14. 40 15. 450 16. 240 17. Answerless page Second Order Systems (Tut9) 1. 0 2. 0 3. -80 4. capacitor 5. 80 6. 20000 7. 0 8. series 9. 6000 10. 5000

11. over 12. -2683 13. A1 14. -A2 15. s 16. 3.0148 17. 188 18. 1.297 19. Answerless page The Properties of Sinusoids (Tut10) 1. 86 2. -170 3. 220 4. 23.4 5. -220 6. 440 7. 0 8. 155.6 9. -67 10. 50 11. 7.958 12. 125.7 13. 24.2 Root-mean-square (Tut10A) 1. 212 2. 50 3. 40 4. 0 5. 3 6. 10 7. 2500 8. 0 9. 5 10. -150 11. 25000 12. 450000 13. 900000 14. 466667 15. 32.27 Complex Numbers (Tut10B) 1. -1 2. -1 3. -j

4. 85.9 5. 383 6. 321 7. 655 8. -459 9. 85 10. -47.86 11. 138 12. 141.5 13. 92 14. 35 15. -15 16. 73 17. 17.0 18. 14.13 19. Formula: The equation is in the form M/θ = (-5 + jA)(B/-152°) - (C - j100)/(2/D°). The answer is √(X2 + Y2)where X = √(52 + A2)×B×cos[arctan(-A/5) + 28°] - ( √(C2 + 1002)/2)×cos[arctan(-100/C) - D] Y = √(52 + A2)×B×sin[arctan(-A/5) + 28°] - ( √(C2 + 1002)/2)×sin[arctan(-100/C) - D] AC Circuits (Tut11) 1. 98 2. 120 3. 140 4. -125 5. 25 6. 30 7. 14.97 8. 200 AC Power (Tut12) 1. 378 2. 466 3. 441 4. B 5. 0 6. 466 7. 578 8. .779 9. 3373 10. D 11. B 12. 166 13. 2732

Maximum Power Transfer (AC) (Tut12A) 1. Formula: |Z1|2×R/(R2 + |Z1|2) 2. Formula: -(R2×|Z1| - R2×|Z2| - |Z1|2×|Z2|)/(R2 + |Z1|2) 3. Formula: |E|×R/√ R2 + |Z1|2 4. Formula: Divide the result from page 3 by twice the result from page 1. 5. Formula: Square the result from page 4 and multiply by the result from page 1. Balanced Three-Phase Circuits (Tut13) 1. 133.3 2. 866 3. 866 4. 173.2 5. 200 6. 346.4 7. 400

Solucionário Fundamentos Circuitos Elétricos-Sadiku 5ed

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