37151512 Network Analysis by Van Valkenburg Solution CHAP 6

Muha Muhamm mmad ad Irfa Irfan n Yous Yousuf uf (Peo (Peon n of Holy Holy Proph Prophet et (P.B (P.B.U .U.H) .H)))

20002000-EE-41 41

118 118

cos t: A + [1/CR 1]B = 0 A = -B/ CR 1 sin t: - B + [1/CR 1]A = I0/C - B + [1/CR 1][-B/ CR 1] = I0/C B = I0/C[A = -[I0/C[-

- 1/C2R 12] - 1/C2R 12]]/

CR 1

Complete solution V(t) = Ke-(1/CR1)t + [-[I0/C[- - 1/C2R 12]]/ CR 1]sin t + [I0/C[- - 1/C2R 12]]cos t At t = 0+ V(0+) = Ke-(1/CR1)0+ + [-[I0/C[- - 1/C2R 12]]/ CR 1]sin (0+) + [I0/C[- - 1/C2R 12]]cos (0+) I0sin t [R 1 + R 2] = K(1) + [I0/C[- - 1/C2R 12]] I0sin t [R 1 + R 2] = K + [I0/C[- - 1/C2R 12]] K = I0sin

- 1/C2R 12]]

t [R 1 + R 2] - [I0/C[-

- 1/C2R 12]]]e-(1/CR1)t + [-[I0/C[- 1/C2R 12]]cos t

V(t) = [I0sin t [R 1 + R 2] - [I0/C[2 2 1/C R 1 ]]/ CR 1]sin t + [I0/C[-

-

Q#6.29: Consider a series RLC network which is excited by a voltage source. 1. Determi Determine ne the the charac characteri teristi sticc equati equation. on. 2. Locus Locus of of the the roots roots of the equatio equation. n. 3. Plot Plot the the roo roots ts of the equ equati ation. on. Solution:

R

L C

V(t) i(t)

For t 0 According to KVL di 1 L + idt + Ri = V(t) dt C Differentiating with respect to ‘t’ d2i i di

Muha Muhamm mmad ad Irfa Irfan n Yous Yousuf uf (Peo (Peon n of Holy Holy Proph Prophet et (P.B (P.B.U .U.H) .H))) L

+

+R

20002000-EE-41 41

119 119

=0

2

dt C dt Dividing both sides by ‘L’ d2i i Rdi + + =0 (i) 2 dt LC Ldt The characteristic equation can be found by substituting the trial solution i = est or by the equivalent of substituting s2 for (d2i/dt2), and s for (di/dt); thus 1 2

s +

R

+ LC L

s=0

2) =0

j j

n

=1

-j

n

=0 1 2

s +

R +

s=0

LC L Characteristic equation: as2 + bs + c = 0 Here a

1 R

b L 1 c LC -b

b2 – 4ac

s1, s2 = 2a

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H))

R

R

2

-

120

1 -

L

2000-E-41

4(1)

L

LC

s1, s2 = 2(1)

R

R

2

-

1 -

L

4(1)

L

LC

s1, s2 = 2

2

R

R

2

-

1 -

L

4(1)

L

LC

s1, s2 = 2

4

R =

=

R

2

-

1 -

2L

2L

R

R

-

4(1) 4LC

2

1 -

2L

2L

LC

radical term

(ii)

Hint: 4 = 2 To convert equation (i) to a standard form, we define the value of resistance that causes the radical (pertaining to the root) term in the a bove equation as the critical resistance, R cr . This value is found by solving the equation

2 R

1 -

2L

=0 LC

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) R = R cr 2 R cr

1 -

=0

2L

LC 2

R cr

1 =

2L

LC

Taking square root of both the sides 2 R cr

1 =

2L

LC

R cr

1 =

2L

LC

Using cross multiplication L R cr = 2 C

Hint:

1

=1

R = R cr

R

C

2

L

=

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121


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1 n

= LC

R n = L

2

1 2 n

=

LC Substituting the corresponding values in equation (i) we get 2 s2 + 2 ns + n = 0 roots of the characteristic equation are Characteristic equation: as2 + bs + c = 0 Here a b c

1 2

b2 – 4ac

-b s1, s2 =

2a -2

)2 – 4(1)(

(2

n

n

s1, s2 = 2(1) -2

4

n

2

2 n

s1, s2 = 2

2

Simplifying we get

s1, s2 = -

when

n

n

n

n

2

–1

=0

s1, s2 = -(0)

n 2 n

(0)2 – 1

–4

2 n

2 n

)

122

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) s1, s2 =

n

s1, s2 =

j

2000-E-41

–1

n

Hint: –1 = j 3) R L C

500 1H 1 10-6 F

Substituting the corresponding values in equation (ii)

500 =

500

2

-

2(1)

= -250 = -250 = -250

1 (1)(10-6)

2(1)

(ii)

62500 - 1000000 -937500 937500 -1

s1, s2 = -250

j968.246 R L C

1000 1H 1 10-6 F


1000 =

-

2

1 -

2(1)

= -500 = -500 = -500

1000 2(1)

250000 - 1000000 -750000 750000 -1

(1)(10-6)

(ii)

123


s1, s2 = -500

2000-E-41

j 866.025

R L C

3000 1H 1 10-6 F


3000 =

3000

2

-

1 -

2(1)

(1)(10-6)

2(1)

(ii)

= -1500 2250000 - 1000000 = -1500 1250000 = -1500 1118.034 = (-1500 + 1118.034), (-1500 - 1118.034) s1, s2 = -381.966, -2618.034 R L C

5000 1H 1 10-6 F


5000 =

5000

-

2

1 -

2(1)

2(1)

= -2500 6250000 - 1000000 = -2500 5250000 = -2500 2291.288 = (-2500 + 2291.288), (-2500 - 2291.288) s1, s2 = -208.712, -4791.288

(1)(10-6)

(ii)

124


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125

Q#6.31: Analyze the network given in the figure on the loop basis, and determine the characteristic equation for the currents in the network as a function of k 1. Find the values of k 1 for which the roots of the characteristic equation are on the imaginary axis of the s plane. Find the range of values of k 1 for which the roots of the characteristic equation have positive real parts. Solution: 1H

+

i2 1

K 1i1

+ -

1 1

1

V1(t) i1

i3 1F

Loop i1: For t 0 According to KVL V1(t) = (i1)(1 ) + (i1 – i2)(1 1 XC = j2π fc ω = 2π f j2π fc = jω c jω = s 1 XC = sc c =1F

) + (i1 – i3)(1

) + (i1 – i3)(XC)

1 XC = s(1 F) 1 XC = s V1(t) = (i1)(1

1 ) + (i1 – i2)(1

) + (i1 – i3)(1

) + (i1 – i3)

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) Simplifying

2000-E-41

s 1

V1(t) = i1 + i1 – i2 + i1 – i3 + i1

1 - i3

s 1 V1(t) = (3 +

s

1 )i1 – i2 – (1 +

s

)

(i)

s

Loop i2: For t 0 According to KVL (i2 – i1)(1 ) + i2(XL) = 0 XL = jω L s = jω XL = s(1 H) XL = s Substituting (i2 – i1)(1 ) + i2(s) = 0 Simplifying i2 – i1 + si2 = 0 (1 + s)i2 – i1 = 0

(ii)

Loop i3: For t 0 According to KVL Sum of voltage rise = sum of voltage drop Sum of voltage rise = k 1i1

(a) 1

Sum of voltage drop = (i3 – i1)(1

) + (i3 – i1)

+ (i3)(1

s Substituting in (a) 1 (i3 – i1)(1

) + (i3 – i1)

+ (i3)(1

) = k 1i1

+ (i3)(1

) - k 1i1 = 0

s Simplifying 1 (i3 – i1)(1

) + (i3 – i1)

s 1 i3 – i1 + i3

1 - i1

s

+ i3 – k 1i1 = 0 s

)

126

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 1

2000-E-41

127

1

-

+ k 1 + 1 i1 + 2 +

i3 = 0

s

(iii)

s

Equations (i), (ii) & (iii) can be written in matrix form

1

1

3+

-

-1

1+

i1

s

V1

s

-1

(1 + s)

0

i2

=

1 -

0

1

1 + k 1 +

0

2+

s

i3

0

s

A

X

B

Determinant of A =

1 3+

1 (1 + s) 2 +

s

1 - (0)(0)

+ (-)

(-1)

2+

s

1+

1 (-1)0 – (-)

s After simplifying Characteristic equation: (5 – k 1)s2 + (6 – 2k 1)s + (2 – k 1) = 0 When k 1 = 0 (5 – 0)s2 + (6 – 2(0))s + (2 – 0) = 0 5s2 + 6s + 2 = 0 as2 + bs + c = 0 Here

s

1 (0)

- (-1)

1 + k 1 +

(1 + s) s

1 1 + k 1 + s

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) a b c -b

2000-E-41 5 6 2

b2 – 4ac

s1, s2 = 2a -6

62 – 4(5)(2)

s1, s2 = 2(5) -6

36 – 40

s1, s2 = 10 -6

-4

s1, s2 = 10 -6

-1 4

s1, s2 = 10 -6

j2

s1, s2 = 10 s1, s2 = -0.6

j0.2

s1, s2 = (-0.6 + j0.2), (-0.6 - j0.2) When k 1 = 1 (5 – 1)s2 + (6 – 2(1))s + (2 – 1) = 0 4s2 + 4s + 1 = 0 as2 + bs + c = 0 Here a b c -b

b2 – 4ac

s1, s2 = 2a

4 4 1

128


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42 – 4(4)(1)

-4 s1, s2 =

2(4) -4

16 – 16

s1, s2 = 8 -4

0

s1, s2 = 8 -4

0

s1, s2 = 8 s1, s2 = -0.5, -0.5 When k 1 = 2 (5 – 2)s2 + (6 – 2(2))s + (2 – 2) = 0 3s2 + 2s + 0 = 0 as2 + bs + c = 0 Here a b c b2 – 4ac

-b s1, s2 =

2a 22 – 4(3)(0)

-2 s1, s2 =

2(3) -2

4 – 0

-2

6 4

s1, s2 =

s1, s2 = 6 -2

2

s1, s2 = 6 s1, s2 = 0, 0.667

3 2 0

129


When k 1 = -1 (5 – (-1))s2 + (6 – 2(-1))s + (2 – (-1)) = 0 6s2 + 8s + 3 = 0 as2 + bs + c = 0 Here a b c b2 – 4ac

-b s1, s2 =

2a 82 – 4(6)(3)

-8 s1, s2 =

2(6) -8

64 – 72

-8

12 -8

s1, s2 =

s1, s2 = 6 -8

-1 8

-8

6 j2.828

s1, s2 =

s1, s2 = 6 s1, s2 = (-1.334 + j0.472), (-1.334 - j0.472)

Q#6.32: Show that equation 6-121 can be written in the form ζω i = ke- nt cos (ω n√1 - ζ 2 t + φ )

Give the values for k and φ in terms of k 5 and k 6 of Eq. (6-121). Solution: Let

k 5 = kcosφ (i) k 6 = -ksinφ (ii) 2 k = (kcosφ ) + (-ksinφ )2 k = k 2cos2φ + k 2sin2φ

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6 8 3

130


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k = k 2(cos2φ + sin2φ ) k = k 2(1) k = √k 2 = √k 52 + k 62 Dividing Eq. (i) by (ii) kcosφ

k 5 = -cot φ =

-ksinφ

φ

k 6

-1 k 5 = cot k 6

Using the trigonometric identity cos (x + y) = cos x cos y – sin x sin y Q#6.33: A switch is closed at t = 0 connecting a battery of voltage V with a series RL circuit. (a) Solution: sw t=0 R

L

V

i

For t 0 According to KVL

di V = iR + L dt Dividing both sides by ‘L’ di R V + i= dt L L This is a linear non-homogeneous equation of the first order and its solution is, Thus

131

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) R P= L V Q= L Hence the solution of this equation i = e-Pt QePtdt + ke-Pt V i = e-(R/L)t

e(R/L)tdt + ke-(R/L)t L V

-(R/L)t

e(R/L)tdt + ke-(R/L)t

i=e

L e(R/L)t e(R/L)tdt = d dt

(R/L)t

L e(R/L)t e(R/L)tdt = R Substituting V

L e(R/L)t

i = e-(R/L)t

+ ke-(R/L)t L

R

V + ke-(R/L)t

i=

R i(0-) = i(0+) = 0 Substituting i = 0 at t = 0 V + ke-(R/L)(0)

0= R 0

e =1

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132

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) V k=R Substituting V

-V

i=

e-(R/L)t

+ R

R

V (1 - e-(R/L)t )

i= R P = i2R t

i2R dt

WR = 0 t

V

2 (1 - e-(R/L)t )2Rdt

WR =

0 R (a - b)2 = a2 + b2 – 2ab V2

t

(1 + e-2(R/L)t – 2(1)(e-(R/L)t))Rdt

WR = 0

R 2

t

V2 (1 + e-2(R/L)t – 2e-(R/L)t)dt

WR = 0

R

V2

t

t

WR =

t

(1)dt +

R 0 Simplifying V2 WR =

(b) Li2 WL = 2 LV2

0

2L

L e

R

(-2e-(R/L)tdt)

dt +

0

t+ R

e

-2(R/L)t

-(R/L)t

-

3L -2(R/L)t

e 2R

2R

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133

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) (1 - e-(R/L)t )2

WL = 2R 2 (c) At t = 0 V2 WR = R

2L (0) +

e

2L

R

e R

2L

WR =

L (1) -

R

R

V2 WR =

0 R

WR = 0 joules At t = 0 LV2 WL = (1 - e-(R/L)0)2 2R 2 LV2 WL = (1 – e0)2 2R 2 LV2 WL = (1 – 1)2 2R 2

WL = 0 joules At t = LV2

∞

(1 - e-(R/L) )2

WL = 2R 2 LV2

∞

(1 – e- )2

WL = 2R 2

-

L 0

(0) +

V2

L e-2(R/L)(0) 2R

R

V2 WR =

-(R/L)(0)

3L 0

e 2R 3L

(1) 2R 2R

2R

3L 2R

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134


2000-E-41

LV2 (1 – 0)2

WL = 2R 2 LV2 WL =

joules 2

2R

(d) In steady state total energy supply W = WR + WL

V2 W =

2L e-(R/L)t -

t+ R

L

R

3L

e-2(R/L)t 2R

+ 2R

LV2 (1 – e-(R/L)t)2 2R2

Q#6.34: In the series RLC circuit shown in the accompanying diagram, the frequency of the driving force voltage is (1) = n (2) = n 1- 2 Solution:

1000 Ω

1H

+ -

100 sin ω t

For t 0 According to KVL di 100 sin t = L + iR + dt Here = n di 100 sin nt = L + iR + dt 1 n =

i(t) 1µ F

1 idt C

1 idt C

… (i)

135


2000-E-41

136

LC L=1H C=1 10-6 F 1 n = (1 H)( 1

10-6 F)

After simplifying n = 1000 rad/sec Substituting in (i) we get di 1 100 sin 1000t = L + iR + idt … (i) dt C Differentiating both the sides & substituting the values of L & C we get d2i di i 100 (1000) cos 1000t = (1) + (1000) + 2 dt dt 10-6 Simplifying we get d2i di 100000cos 1000t = + (1000) + 1000000i 2 dt dt The trial solution for the particular integral is i p = A cos 1000t + B sin 1000t d2i p di p 100000cos 1000t = + (1000) + 1000000i p 2 dt dt (i p)′ = -1000A sin 1000t + B 1000cos 1000t (i p)′ ′ = -1000000A cos 1000t - B 1000000sin 1000t (i p)′ = Ist derivative (i p)′ ′ = 2nd derivative 100000cos 1000t = -1000000A cos 1000t - B 1000000sin 1000t + 1000(-1000A sin 1000t + B 1000cos 1000t) + 1000000(A cos 1000t + B sin 1000t) Simplifying 100000cos 1000t = -1000000A cos 1000t – 1000000B sin 1000t - 1000000A sin 1000t + 1000000B cos 1000t + 1000000A cos 1000t + 1000000B sin 1000t Simplifying Equating the coefficients Cos: 100000 = 1000000B

100000 B= 1000000 B = 0.1


Sin: 0 = - 1000000B – 1000000A + 1000000B 0 = –1000000A A=0 i p = A cos 1000t + B sin 1000t Substituting the values of A & B i p = (0) cos 1000t + (0.1) sin 1000t i p = 0.1 sin 1000t e j

ωt

– e-j

ωt

sin ω t = 2j = 1000

Here ω

e j1000t – e-j1000t sin 1000t = 2j e j1000t – e-j1000t ip = 0.1

Transient response 2j

In steady state At resonance XL = XC In a series RLC circuit Z = R + j(XL - XC) Z = R + j(XC - XC) Z = R V Im = Z 100 Im = 1000 Im = 0.1 A (2) = n 1- 2 Determine the values of Do yourself.

n

&

substitute & simplify

2000-E-41

137


THE END

2000-E-41

138

37151512 Network Analysis by Van Valkenburg Solution CHAP 6

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