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[1/C2R 12]]/ 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

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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


R

R

2

-

120 120

1 -

L

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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 eq uation

2 R

1 -

2L

=0 LC

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))) R = R cr cr 2 R cr cr

1 -

=0

2L

LC 2

R cr cr

1 =

2L

LC

Taking square root of both the sides 2 R cr cr

1 =

2L

LC

R cr cr

1 =

2L

LC

Using cross multiplication L R cr cr = 2 C

Hint:

1

=1

R = R cr cr

R

C

2

L

=

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121 121


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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

)

122 122

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))) s1, s2 =

n

s1, s2 =

j

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–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)

123 123


s1, s2 = -500

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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)

124 124


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125 125

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)

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))) Simplifying

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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

)

126 126

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))) 1

20002000-EE-41 41

127 127

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

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))) a b c -b

20002000-EE-41 41 5 6 2

b2 – 4ac

s1, s2 = 2a -6

62 – 4(5)(2)

s1, s2 = 2(5) -6

36 – 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

128 128


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42 – 4(4)(1)

-4 s1, s2 =

2(4) -4

16 – 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

129 129


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 – 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

130 130


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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 vo ltage V with a series RL circuit. (a) (a) Solu Soluti tion on:: 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

131 131

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))) R P= L V Q= L Hence the solution of this equation i = e-Pt QePtdt + ke-Pt V -(R/L)t

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

i=e

L V i = e-(R/L)t

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

20002000-EE-41 41

132 132

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))) 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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133 133

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))) (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

20002000-EE-41 41

134 134


20002000-EE-41 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)

135 135


20002000-EE-41 41

136 136

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 1000 100000 00co coss 1000 1000tt = + (100 (1000) 0) + 1000 100000 000i 0i 2 dt dt The trial solution for the particular integral is i p = A cos 1000t + B sin 1000t d2i p di p 1000 100000 00co coss 1000 1000tt = + (100 (1000) 0) + 1000 100000 000i 0i p 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

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137 137


THE END

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138 138

network analysis by van valkenburg solution CHAP#6

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