9

Sinusoidal Steady State Analysis

Assessment Problems

AP 9.1 [a] V = 170/−40◦ V.

[b] 10 sin(1000t + 20◦ ) = 10 cos(1000t − 70◦ ; )

.· . I = 10/−70◦ A.

[c] I = 5/36.87◦ + 10/−53.13◦

= 4 + j3 + 6 − j8 = 10 − j5 = 11.18/−26.57◦ A.
[d] sin(20,000πt + 30◦ ) = cos(20,000πt − 60◦ ).
Thus,

V = 300/45◦ − 100/−60◦ = 212.13 + j212.13 − (50 − j86.60)

= 162.13 + j298.73 = 339.90/61.51◦ mV.

AP 9.2 [a] v = 18.6 cos(ωt − 54◦ ) V.

[b] I = 20/45◦ − 50/ − 30◦ = 14.14 + j14.14 − 43.3 + j25

= −29.16 + j39.14 = 48.81/126.68◦ .

Therefore i = 48.81 cos(ωt + 126.68◦ ) mA.

[c] V = 20 + j80 − 30/15◦ = 20 + j80 − 28.98 − j7.76

= −8.98 + j72.24 = 72.79/97.08◦ .

v = 72.79 cos(ωt + 97.08◦ ) V.

AP 9.3 [a] ωL = (104 )(20 × 10−3 ) = 200 Ω.

[b] ZL = jωL = j200 Ω.

9–1
 9–2 CHAPTER 9. Sinusoidal Steady State Analysis

[c] VL = IZL = (10/30◦ )(200/90◦ ) × 10−3 = 2/120◦ V.

[d] vL = 2 cos(10,000t + 120◦ ) V.
−1 −1
AP 9.4 [a] XC = = = −50 Ω.
ωC 4000(5 × 10−6 )
[b] ZC = jXC = −j50 Ω.
V 30/25◦
[c] I = = = 0.6/115◦ A.
ZC 50/−90◦
[d] i = 0.6 cos(4000t + 115◦ ) A.

AP 9.5 I1 = 100/25◦ = 90.63 + j42.26;

I2 = 100/145◦ = −81.92 + j57.36;

I3 = 100/−95◦ = −8.72 − j99.62;

I4 = −(I1 + I2 + I3 ) = (0 + j0) A, therefore i4 = 0 A.

125/−60◦ 125
AP 9.6 [a] I = = /(−60 − θZ )◦ ;
|Z|/θz |Z|

But −60 − θZ = −105◦ .·. θZ = 45◦ ;

Z = 90 + j160 + jXC ;
1
.·. XC = −70 Ω; XC = − = −70;
ωC
1
.· . C = = 2.86 µF.
(70)(5000)
Vs 125/−60◦
[b] I = = = 0.982/−105◦ A; .·. |I| = 0.982 A.
Z (90 + j90)
AP 9.7 [a]

ω = 2000 rad/s;
−1
ωL = 10 Ω, = −20 Ω;
ωC
 Problems 9–3

20(j10)
Zxy = 20kj10 + 5 + j20 = + 5 − j20;
(20 + j10)
= 4 + j8 + 5 − j20 = (9 − j12) Ω.
−1
[b] ωL = 40 Ω, = −5 Ω;
ωC
" #
(20)(j40)
Zxy = 5 − j5 + 20kj40 = 5 − j5 +
20 + j40
= 5 − j5 + 16 + j8 = (21 + j3) Ω.
j106
" # !
20(jωL)
[c] Zxy = + 5−
20 + jωL 25ω
20ω 2 L2 j400ωL j106
= + + 5 − .
400 + ω 2 L2 400 + ω 2 L2 25ω
The impedance will be purely resistive when the j terms cancel, i.e.,
400ωL 106
= .
400 + ω 2 L2 25ω
Solving for ω yields ω = 4000 rad/s.
20ω 2 L2
[d] Zxy = + 5 = 10 + 5 = 15 Ω.
400 + ω 2 L2
AP 9.8 The frequency 4000 rad/s was found to give Zxy = 15 Ω in Assessment
Problem 9.7. Thus,
V 150/0◦
V = 150/0◦ , Is = = = 10/0◦ A.
Zxy 15
Using current division,
20
IL = (10) = 5 − j5 = 7.07/−45◦ A;
20 + j20

iL = 7.07 cos(4000t − 45◦ ) A, Im = 7.07 A.

AP 9.9 After replacing the wye made up of the j40 Ω, 50 Ω, and 40 Ω impedances with
its equivalent delta, the circuit becomes
 9–4 CHAPTER 9. Sinusoidal Steady State Analysis

where
j40(50) + 50(40) + 40(j40)
Za = = 90 − j50 Ω;
j40

j40(50) + 50(40) + 40(j40)

Zb = = 40 + j72 Ω;
50

j40(50) + 50(40) + 40(j40)

Zc = = 50 + j90 Ω.
40

The circuit is further simplified by combining the impedances to the right of

the 14 Ω resistor into a single equivalent impedance:

Zeq = Zb k[(Zc k − j15) + (Za k10)] = (16 − j16) Ω.

136/0◦
Therefore I = = 4/28.07◦ A.
14 + 16 − j16

AP 9.10
V1 = 240/53.13◦ = 144 + j192 V;

V2 = 96/−90◦ = −j96 V;

jωL = j(4000)(15 × 10−3 ) = j60 Ω;

1 6 × 106
= −j = −j60 Ω.
jωC (4000)(25)

Perform a source transformation:

V1 144 + j192
= = 3.2 − j2.4 A;
j60 j60

V2 96
= −j = −j4.8 A.
20 20

Combine the parallel impedances:

1 1 1 1 j5 1
Y = + + + = = ;
j60 30 −j60 20 j60 12
 Problems 9–5

1
Z= = 12 Ω.
Y

Vo = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48/36.87◦ V;

vo = 48 cos(4000t + 36.87◦ ) V.

AP 9.11 Use the lower node as the reference node. Let V1 = node voltage across the
20 Ω resistor and VTh = node voltage across the capacitor. Writing the node
voltage equations gives us

V1 V1 − 10Ix −j10
− 2/45◦ + = 0 and VTh = (10Ix ).
20 j10 10 − j10

We also have
V1
Ix = .
20

Solving these equations for VTh gives VTh = 10/45◦ V. To find the Thévenin
impedance, we remove the independent current source and apply a test
voltage source at the terminals a, b. Thus

It follows from the circuit that

10Ix = (20 + j10)Ix .

Therefore
VT VT
Ix = 0 and IT = +
−j10 10.

VT
ZTh = , therefore ZTh = (5 − j5) Ω.
IT
 9–6 CHAPTER 9. Sinusoidal Steady State Analysis

AP 9.12 The phasor domain circuit is as shown in the following diagram:

The node voltage equation is

V V V V − 100/−90◦
−10 + + + + = 0;
5 −j(20/9) j5 20

Solving, V = 10 − j30 = 31.62/−71.57◦ .

Therefore v = 31.62 cos(50,000t − 71.57◦ ) V.

AP 9.13 Let Ia , Ib , and Ic be the three clockwise mesh currents going from left to
right. Summing the voltages around meshes a and b gives

33.8 = (1 + j2)Ia + (3 − j5)(Ia − Ib )

and

0 = (3 − j5)(Ib − Ia ) + 2(Ib − Ic ).

But

Vx = −j5(Ia − Ib ),

therefore

Ic = −0.75[−j5(Ia − Ib )].

Solving for I = Ia = 29 + j2 = 29.07/3.95◦ A.

√
AP 9.14 [a] M = 0.4 0.0625 = 0.1 H, ωM = 80 Ω;

Z22 = 40 + j800(0.125) + 360 + j800(0.25) = (400 + j300) Ω.

∗
Therefore |Z22 | = 500 Ω, Z22 = (400 − j300) Ω.
2
80

Zr = (400 − j300) = (10.24 − j7.68) Ω.
500
 Problems 9–7

245.20
[b] I1 = = 0.50/ − 53.13◦ A;
184 + 100 + j400 + Zτ
i1 = 0.5 cos(800t − 53.13◦ ) A.
jωM j80
 
[c] I2 = I1 = (0.5/ − 53.13◦ ) = 0.08/0◦ A;
Z22 500/36.87◦
i2 = 80 cos 800t mA.

AP 9.15
Vs 25 × 103 /0◦
I1 = =
Z1 + 2s2 Z2 1500 + j6000 + (25)2 (4 − j14.4)

= 4 + j3 = 5/36.87◦ A;

V1 = Vs − Z1 I1 = 25,000/0◦ − (4 + j3)(1500 + j6000)

= 37,000 − j28,500;

1
V2 = − V1 = −1480 + j1140 = 1868.15/142.39◦ V;
25

V2 1868.15/142.39◦
I2 = = = 125/216.87◦ A.
Z2 4 − j14.4
 9–8 CHAPTER 9. Sinusoidal Steady State Analysis

Problems

ω
P 9.1 [a] ω = 2πf = 600 rad/s, f= = 95.54 Hz
2π
[b] T = 1/f = 10.47 ms
[c] Im = 100 mA
[d] i(0) = 100 cos(45◦ ) = 70.72 mA
π
[e] φ = 45◦ ; = 0.7857 rad
4
[f ] i = 0 when 600t + 45◦ = 90◦
45◦
(600 rad/s)t = 45 ⇒ t = = 0.785 rad
57.3◦ /rad
i.e. 600t = 0.7853 ⇒ t = 1.308 m sec

[g] (di/dt) = (−100)600 sin(600t + 45◦ )

(di/dt) = 0 when 600t + 45◦ = 180◦

135◦
i.e. 600t = = 2.356 rad
57.3◦ /rad
⇒ t = 3.926 ms

P 9.2

[a] Left as φ becomes more positive.

[b] Right.

P 9.3 [a] 25 V
[b] 2πf = 400π; f = 200 Hz
[c] ω = 400π = 1256.64 rad/s
 Problems 9–9

π π
 
[d] θ(rad) = 60◦ = = 1.0472 rad
180◦ 3
[e] θ = 60◦
1 1
[f ] T = = = 5 ms
f 200
π π π
[g] 400πt + = ; .·. 400πt =
3 2 6
1
.· . t= = 416.67 µs
2400
0.005 π
   
[h] v = 25 cos 400π t − +
6 3
= 25 cos[400πt − (π/3) + (π/3)]
= 25 cos 400πt V
[i] 400π(t + to ) + (π/3) = 400πt + (3π/2)
7π 7
.·. 400πto = ; to = = 2.92 ms
6 2400
P 9.4 W = 2πf = 6.25 × 108 rad/sec

W KT : Imax = W CVc,max = 3.0A

Then
q
vmax = Imax R2 + (1/W C)2
√
= 62 + 242 = 24.74 V

T
P 9.5 [a] = 25 − 5 = 20 ms; T = 40 ms
2
1 1
f= = = 25 Hz
T 40 × 10−3
[b] i = Im sin(ωt + θ)

ω = 2πf = 50π rad/s

−π
50π(5 × 10−3 ) + θ = 0; .· . θ = rad = −45◦
4
i = Im sin[50πt − 45◦ ]

0.5 = Im sin −45◦ ; Im = −70.71 mA

i = −70.71 sin[50πt − 45◦ ] = 70.71 cos[50πt + 45◦ ] mA

√ √
P 9.6 Vm = 2Vrms = 2(240) = 339.41 V
 9–10 CHAPTER 9. Sinusoidal Steady State Analysis

s
1 Z T /2 2 2 2π
P 9.7 Vrms = Vm sin t dt;
T 0 T

Z T /2
2π V 2 Z T /2 4π V 2T
   
Vm2 sin2 t dt = m 1 − cos t dt = m ;
0 T 2 0 T 4
s
1 Vm2 T Vm
Therefore Vrms = = .
T 4 2
Z to +T Z to +T
1 1
P 9.8 Vm2 2
cos (ωt + φ) dt = Vm2 + cos(2ωt + 2φ) dt
to t
(o
2 2
Vm2 Z to +T )
R to +T
= to dt + cos(2ωt + 2φ) dt
2 to
Vm2 1 h
 i 
= sin(2ωt + 2φ) |ttoo +T
T+
2 2ω
Vm2 1
 
= T+ [sin(2ωto + 4π + 2φ) − sin(2ωto + 2φ)]
2 2ω
2 T 1 2 T
   
= Vm + (0) = Vm .
2 2ω 2
P 9.9 [a] The numerical values of the terms in Eq. 9.8 are
Vm = 75, R/L = 5333.33, ωL = 300
√
R2 + ω 2 L2 = 500

φ = −60◦ , θ = tan−1 300/400, θ = 36.87◦

Substitute these values into Equation 9.9:

h i
i = −17.94e−5333.33t + 150 cos(4000t − 96.87◦ ) mA, t≥0

[b] Transient component = −17.94e−5333.33t mA

Steady-state component = 150 cos(4000t − 96.87◦ ) mA
[c] By direct substitution into Eq 9.9 in part (a), i(750 µs) = 38.44 mA
[d] 150 mA, 4000 rad/s, −96.87◦
[e] The current lags the voltage by 36.87◦ .

P 9.10 [a] From Eq. 9.7 we have

di Vm R cos(φ − θ) −(R/L)t ωLVm sin(ωt + φ − θ)
L = √ 2 e − √ ;
dt R + ω 2 L2 R 2 + ω 2 L2
−Vm R cos(φ − θ)e−(R/L)t Vm R cos(ωt + φ − θ)
Ri = √ + √ ;
R 2 + ω 2 L2 R2 + ω 2 L2
 Problems 9–11

" #
di R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)
L + Ri = Vm √ .
dt R2 + ω 2 L2
But
R ωL
√ = cos θ and √ = sin θ.
R + ω 2 L2
2 R2 + ω 2 L2
Therefore the right-hand side reduces to
Vm cos(ωt + φ).
At t = 0, Eq. 9.7 reduces to
−Vm cos(φ − θ) Vm cos(φ − θ)
i(0) = √ + √ 2 = 0.
R 2 + ω 2 L2 R + ω 2 L2
Vm
[b] iss = √ 2 cos(ωt + φ − θ).
R + ω 2 L2
Therefore
diss −ωLVm
L =√ 2 sin(ωt + φ − θ)
dt R + ω 2 L2
and
Vm R
Riss = √ 2 cos(ωt + φ − θ).
R + ω 2 L2
" #
diss R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)
L + Riss = Vm √
dt R 2 + ω 2 L2
= Vm cos(ωt + φ).

P 9.11 [a] Y = 30/ − 160◦ + 15/70◦ = 29.38/170.56◦

y = 28.38 cos(200t + 170.56◦ )

[b] Y = 90/ − 110◦ + 60/ − 70◦ = 141.33/ − 94.16◦

y = 141.33 cos(50t − 94.16◦ )

[c] Y = 50/ − 60◦ + 25/20◦ − 75/ − 30◦ = 16.7/170.52◦

y = 16.7 cos(5000t + 170.52◦ )

[d] Y = 10/30◦ + 10/ − 90◦ + 10/150◦ = 0

y=0

P 9.12 [a] 200 Hz.

[b] θv = 0◦ ;
1006 0◦ 100
I= = 6 −90◦ ; θi = −90◦ .
jωL ωL
 9–12 CHAPTER 9. Sinusoidal Steady State Analysis

[c] ωL = 2π(200)(2mH) = 2.5 Ω.

1006 0 1006 0
[d] i = = = 406 − 90
jωL j2.5
[e] ZL = jωL = j2.5 Ω.

P 9.13 [a] ω = 2πf = 160π × 103 = 502.65 krad/s = 502,654.82 rad/s

25 × 10−3 /0◦
[b] I = = jωC(25 × 10−3 )/0◦ = 25 × 10−3 ωC /90◦
1/jωC
.·. θi = 90◦

[c] 628.32 × 10−6 = 25 × 10−3 ωC

1 25 × 10−3
= = 39.79 Ω, .·. XC = −39.79 Ω
ωC 628.32 × 10−6
1 1
[d] C = =
39.79(ω) (39.79)(160π × 103 )

C = 0.05 × 10−6 = 0.05 µF

−1
 
[e] Zc = j = −j39.79 Ω
ωC
P 9.14 [a] Vg = 1006 0◦ ; Ig = 26 40◦ ;
Vg 1006 0◦
.· . Z = = = 506 −40◦ Ω.
Ig 26 40◦
[b] ig lags vg by 40◦ :
2πf = 1000π; f = 500 Hz; T = 1/f = 2 ms;
40◦
.·. ig lags vg by = 0.222 ms.
360◦
P 9.15 [a] ZL = j(500)(50 × 10−3 ) = j25 Ω
−j
ZC = = −j62.5 Ω
(500)(32 × 10−6 )

25/ − 60◦
[b] I = = 554.7/ − 3.69◦ mA
25 + j25 − j62.5
 Problems 9–13

[c] i = 554.7 cos(500t − 3.69◦ ) mA

P 9.16 [a] jωL = j(2 × 104 )(600 × 10−6 ) = j12 Ω;
jωL = j(2 × 104 )(0.5 × 10−3 ) = j10 Ω
Ig = 9226 30circ

[b] Vo = 9226 30circ Ze ;

1 1 1 1
Ze = ; Ye = + + ;
Ye 20 j10 16 + j12
Ye = 0.09 + j(−0.13)S = (0.09 − j0.13) S;
1
Ze = = 3.6 + j5.21Ω = 6.326 55◦ Ω;
(0.09 − j0.13)
Vo = (9226 30◦ )(6.326 55◦ ) = 5827.046 85◦
[c] vo = 5827.04 cos(2 × 104 t + 85◦ ) V.
1 1 1
P 9.17 [a] Y = + +
3 + j4 16 − j12 −j4
= 0.12 − j0.16 + 0.04 + j0.03 + j0.25
= 160 + j120 = 200/36.87◦ mS.
[b] G = 160 mS.
[c] B = 120 mS.
I 8
[d] I = 8/0◦ A, V= = = 40/−36.87◦ V;
Y 0.2/36.87◦
V 40/−36.87◦
IC = = ◦
= 10/53.13◦ A;
ZC 4/−90
iC = 10 cos(ωt + 53.13◦ ) A, Im = 10 A.

P 9.18 [a] Z1 = R1 + jωL1 ;

R2 (jωL2 ) ω 2 L22 R2 + jωL2 R22
Z2 = = ;
R2 + jωL2 R22 + ω 2 L22
ω 2 L22 R2 R22 L2
Z1 = Z2 when R1 = 2 and L1 = 2 .
R2 + ω 2 L22 R2 + ω 2 L22
 9–14 CHAPTER 9. Sinusoidal Steady State Analysis

(4000)2 (1.25)2 (5000)

[b] R1 = = 2500 Ω;
50002 + 40002 (1.25)2
(5000)2 (1.25)
L1 = = 625 mH.
50002 + 40002 (1.25)2

1 j
P 9.19 [a] Y2 = − ;
R2 ωL2
1 R1 − jωL1
Y1 = = 2 .
R1 + jωL1 R1 + ω 2 L21
Therefore Y2 = Y1 when
R12 + ω 2 L21 R12 + ω 2 L21
R2 = and L2 = .
R1 ω 2 L1
80002 + 10002 (4)2
[b] R2 = = 10 kΩ;
8000
80002 + 10002 (4)2
L2 = = 20 H.
10002 (4)

1
P 9.20 [a] Z1 = R1 − j ;
ωC1
R2 /jωC2 R2 R2 − jωR22 C2
Z2 = = = ;
R2 + (1/jωC2 ) 1 + jωR2 C2 1 + ω 2 R22 C22
R2
Z1 = Z2 when R1 = and
1 + ω 2 R22 C22
1 ωR22 C2 1 + ω 2 R22 C22
= or C1 = .
ωC1 1 + ω 2 R22 C22 ω 2 R22 C2
1000
[b] R1 = = 200 Ω;
1 + (40 × 103 )2 (1000)2 (50 × 10−4 )2
1 + (40 × 103 )2 (1000)2 (50 × 10−9 )2
C1 = = 62.5 nF.
(40 × 103 )2 (1000)2 (50 × 10−9 )

1
P 9.21 [a] Y2 = + jωC2 ;
R2
1 jωC1 ω 2 R1 C12 + jωC1
Y1 = = = ;
R1 + (1/jωC1 ) 1 + jωR1 C1 1 + ω 2 R12 C12
Therefore Y1 = Y2 when
1 + ω R12 C12
2
C1
R2 = and C2 = .
ω 2 R1 C12 1 + ω 2 R12 C12
 Problems 9–15

1 + (50 × 103 )2 (1000)2 (40 × 10−9 )2

[b] R2 = = 1250 Ω;
(50 × 103 )2 (1000)(40 × 10−9 )2

40 × 10−9
C2 = = 8 nF.
1 + (50 × 103 )2 (1000)2 (40 × 10−9 )2

P 9.22 Zab = 5 + j8 + 10k − j20 + (8 + j16)k(40 − j80)

= 5 + j8 + 8 − j4 + 12 + j16 = 25 + j20 Ω = 32.02/38.66◦ Ω

P 9.23 First find the admittance of the parallel branches

1 1 1 1
Yp = + + + = (0.213 − 0.132j) S;
12 − j4 8 + j12 10 j10

1 1
Zp = = = (3.37 + 2.10j) Ω;
Yp 0.213 − j0.132

Zab = −j12.8 + 3.37 + j2.10 + 13.6 = (16.97 − j10.69) Ω;

1 1
Yab = = = (0.042 + j0.0265) S
Zab (16.97 − j10.69)

= (42 + j26.5) mS = 49.666 32.2◦ .

1 1 jωRL
P 9.24 [a] + RkjωL = +
jωC jωC jωL + R
jωL + R − ω 2 RLC
=
jωC(jωL + R)

(R − ω 2 RLC + jωL)(−ω 2 LC − jωRC)

=
(−ω 2 LC + jωRC)(−ω 2 LC − jωRC)
The denominator in the expression above is purely real; set the imaginary
part of the numerator in the above expression equal to zero and solve for
ω:
−ω 3 L2 C − ωR2 C + ω 3 R2 C 2 L = 0

−ω 2 L2 − R2 + ω 2 R2 LC = 0

2 R2 2002
ω = 2 = = 250,000
R LC − L2 2002 (0.4)(20 × 10−6 ) − (0.4)2
.· . ω = 500 rad/s
 9–16 CHAPTER 9. Sinusoidal Steady State Analysis

(200)(j200)
[b] Zab (500) = −j100 + = 100 Ω
200 + j200
P 9.25 [a] R = 300 Ω = 120 Ω + 180 Ω;
1 1
ωL − = −400 so 10,000L − = −400.
ωC 10,000C
Choose L = 10 mH. Then,
1 1
= 100 + 400 so C = = 0.2 µF.
10,000C 10,000(500)
We can achieve the desired capacitance by combining two 0.1 µF
capacitors in parallel. The final circuit is shown here:

1 1
[b] 0.01ω = so ω 2 = = 5 × 108 ;
ω(0.2 × 10−6 ) −6
0.01(0.2 × 10 )
.·. ω = 22,360.7 rad/s.

P 9.26 [a] Using the notation and results from Problem 9.19:
20
RkL = 40 + j20 so R1 = 40, L1 = = 4 mH;
5000
402 + 50002 (0.004)2
R2 = = 50 Ω;
40
402 + 50002 (0.004)2
L2 = = 20 mH;
50002 (0.004)
R2 kjωL2 = 50kj100 = 40 + j20 Ω. (checks)

The circuit, using combinations of components from Appendix H, is

shown here:
 Problems 9–17

[b] Using the notation and results from Problem 9.21:

RkC = 40 − j20 so R1 = 40, C1 = 10 µF;
1 + 50002 (40)2 (10 µ)2
R2 = = 50 Ω;
50002 (40)(10 µ)2
10 µ
C2 = = 2 µF;
1+ 50002 (40)2 (10 µ)2
R2 k(−j/ωC2 ) = 50k(−j100) = 40 − j20 Ω. (checks)

The circuit, using combinations of components from Appendix H, is

shown here:

P 9.27 [a] (40 + j20)k(−j/ωC) = 50kj100k(−j/ωC).

To cancel out the j100 Ω impedance, the capacitive impedance must be
−j100 Ω:
−j 1
= −j100 so C = = 2 µF.
5000C (100)(5000)
Check:
RkjωLk(−j/ωC) = 50kj100k(−j100) = 50 Ω.
Create the equivalent of a 2 µF capacitor from components in Appendix
H by combining two 1 µF capacitors in parallel.
[b] (40 − j20)k(jωL) = 50k(−j100)k(jωL).
To cancel out the −j100 Ω impedance, the inductive impedance must be
j100 Ω:
100
j5000L = j100 so L = = 20 mH.
5000
Check:
RkjωLk(−j/ωC) = 50kj100k(−j100) = 50 Ω.
Create the equivalent of a 20 mH inductor from components in Appendix
H by combining two 10 mH inductors in series.
1
P 9.28 Z = 4000 + j(2000)(0.5) − j
(2000)(100 × 10−9 )

= 3000 + j1000 − j5000 = 3000 − j4000 Ω

 9–18 CHAPTER 9. Sinusoidal Steady State Analysis

V 80/0◦
Io = = = 9.6 + j12.8 = 16/53.13◦ mA
Z 3000 − j4000

io (t) = 16 cos(2000t + 53.13◦ ) mA

1 1
P 9.29 = −6
= −j25 Ω
jωC j(5 × 10 )(8000)

jωL = j8000(3.125 × 10−3 ) = j25 Ω

Vg = 60/ − 90◦ V

Ze = j25 + (50k − j25) = 10 + j5 Ω

60/ − 90◦
Ig = = −j2.4 − j4.8 A
10 + j5

Vo = (50k − j25)Ig = (10 − j20)(−j2.4 − j4.8) = −120 V

vo = −120 cos 8000t V

P 9.30 Vs = 256 −90◦ V;

1
= −j40 Ω;
jωC

jωL = j10 Ω.

Zeq = 10 + j10k(20 + 10k − j40) = 13.18 + j9.17 Ω;

 Problems 9–19

Vo 256 −90◦
Io = = = −0.88 − j1.278 = 1.556 −124.82◦
Zeq 13.18 + j9.17

io = 1.55 cos(2000t − 124.82) A.

1
P 9.31 [a] = −j2.5 Ω;
jωC
jωL = j20 Ω;

Ze = j20k(50 + 200k − j2.5) = 7.12 + 17.5j

Ig = 0.0256 0◦ ;

Vg = Ig Ze = 0.025(7.12 + j17.5) = 0.178 + j0.437 V.

(0.0312 − j2.49)(0.178 + j0.437)

Vo = = 0.02 − j7.48 = 0.0236 −18.59.
(50 + 0.0312 − j2.49)
vo = 0.023 cos(50,000t − 18.59) V.
25,000
[b] ω = 2πf = 50,000; f= ;
π
1 π
T = = = 40π µs;
f 25,000
26.57
.· . (40π µs) = 9.27 µs;
360
.·. vo leads ig by 9.27 µs.
 9–20 CHAPTER 9. Sinusoidal Steady State Analysis

P 9.32

V1 = j5(−j2) = 10 V;

15
−25 + 10 + (4 − j3)I1 = 0; .· . I1 = = 2.4 + j1.8 A;
4 − j3

Ib = I1 − j5 = (2.4 + j1.8) − j5 = 2.4 − j3.2 A;

VZ = −j5I2 + (4 − j3)I1 = −j5(2.4 − j3.2) + (4 − j3)(2.4 + j1.8) = −1 − j12 V;

−25 + (1 + j3)I3 + (−1 − j12) = 0 .· . I3 = 6.2 − j6.6 A;

IZ = I3 − I2 = (6.2 − j6.6) − (2.4 − j3.2) = 3.8 − j3.4 A;

VZ −1 − j12
Z= = = 1.42 − j1.88 Ω.
IZ 3.8 − j3.4

P 9.33

Va − (100 − j50) Va Va − (140 + j30)

+ + = 0.
20 j5 12 + j16
 Problems 9–21

Solving,

Va = 40 + j30 V;

140 + j30 (40 + j30) − (140 + j30)

IZ + (30 + j20) − + = 0.
−j10 12 + j16

Solving,

IZ = −30 − j10 A;

(100 − j50) − (140 + j30)

Z= = 2 + j2 Ω.
−30 − j10

P 9.34 ZL = j(10,000)(10 × 10−3 ) = j100 Ω

−j
ZC = = −j50 Ω
(10,000)(2 × 10−6 )

Construct the phasor domain equivalent circuit:

Using current division:

(50 + j100)
I= (0.06) = 30 + j30 mA
50 + j100 + 100 − j50

Vo = 100I = 3 + j3 = 4.24/45◦

vo = 4.24 cos(10,000t + 45◦ ) V

P 9.35 Vg = 40/ − 15◦ V; Ig = 40/ − 68.13◦ mA

Vg
Z= = 1000/53.13◦ Ω = 600 + j800 Ω
Ig
 9–22 CHAPTER 9. Sinusoidal Steady State Analysis

0.4 × 106
!
Z = 600 + j 3.2ω −
ω

0.4 × 106
.·. 3.2ω − = 800
ω

.·. ω 2 − 250ω − 125,000 = 0

Solving,

ω = 500 rad/s

P 9.36 [a]

Vb = (2000 − j1000)(0.025) = 50 − j25 V;

50 − j25
Ia = = 120 − j160 mA = 2006 −53.13◦ mA;
250 + j125
50 − j25 + j100
Ic = = 100 + j150 mA = 180.26 56.3◦ mA;
500
Ig = Ia + Ib + Ic = 245 − j10 mA = 245.196 −2.35◦ mA.

[b] ia = 200 cos(2000t − 53.13◦ ) mA;

ic = 180.2 cos(2000t + 56.3◦ ) mA;

ig = 245.19 cos(2000t − 2.35◦ ) mA.

P 9.37 [a] In order for vg and ig to be in phase, the impedance to the right of the
500 Ω resistor must be purely real:
jωL(R + 1/jωC)
Zeq = jωLk(R + 1/ωC) =
jωL + R + 1/jωC
jωL(jωRC + 1)
=
jωLRC − ω 2 LC + 1
 Problems 9–23

(−ω 2 RLC + jωL)(1 − ω 2 LC − jωRC)

= .
(1 − ω 2 LC + jωRC)(1 − ω 2 LC − jωRC)
The denominator of the above expression is purely real. Now set the
imaginary part of the numerator in that expression to zero and solve for
ω:
ωL(1 − ω 2 LC) + ω 3 R2 LC 2 = 0.
1 1
So ω2 = = = 6,250,000;
2
LC − R C 2 (0.1)(2 × 10 ) − 1002 (2 × 10−6 )2
−6

.· . ω = 2500 rad/s and f = 397.9 Hz.

[b] Zeq = 300 + j250k(100 − j200) = 800 Ω;
Ig = 45/1006 0c irc = 56.256 0c irc mA;
ig (t) = 56.25 cos 2500t mA.
[c] As derived in [a], the circuit will have real value of resonant frequency
when
LC − R2 C 2 ≥ 0
L − R2 C ≥ 0
s s
L 100 × 10−3
R≤ ⇒R≤
C 2 × 10−6
R ≤ 223.6Ω
Thus, R should be less than 223.6 Ω for real value of resonant frequency.
P 9.38 [a] For ig and vg to be in phase, the impedance to the right of the 480 Ω
resistor must be purely real.
(1/jωC)(300) 300
jωL + = ωL +
(1/jωC) + 300 1 + 300jωC
300(1 − 300jωC)
= jωL +
1 + 1002 ω 2 C 2
jωL(1 + 3002 ω 2 C 2 ) + 300(1 − 300jωC)
= .
1 + 3002 ω 2 C 2
In the above expression the denominator is purely real. So set the
imaginary part of the numerator to zero and solve for ω:
ωL(1 + 3002 ω 2 C 2 ) − 3002 ωC = 0;
3002 C − L 3002 (5 × 10−6 ) − 0.2
ω2 = = = 5.55 × 105 ;
3002 C 2 L 3002 (5 × 10−6 )2 (0.2)
.· . ω = 745.35 rad/s.
 9–24 CHAPTER 9. Sinusoidal Steady State Analysis

[b] When ω = 745.35 rad/s

Zg = 480k[j149 + (300k − j268.33)] = 104.33
.·. Vg = Zg Ig = 104.34(0.08) = 8.34 V;
vo = 8.34 cos 745.34 V.
[c] R2 C − L > 0
s
L
R≥
C
R ≥ 200Ω
Thus, R should always be greater than 200 Ω for real value of resonant
frequency.
P 9.39 [a] The voltage and current are in phase when the impedance to the left of
the 1 kΩ resistor is purely real:
1 R + jωL
Zeq = k(R + jωL) =
jωC 1 + jωRC − ω 2 LC
(R + jωL)(1 − ω 2 LC − jωRC)
= .
(1 + ω 2 LC)2 + ω 2 R2 C 2
The denominator in the above expression is purely real, so set the
imaginary part of the expression’s numerator to zero and solve for ω:
−ωR2 C + ωL − ω 3 L2 C = 0;
L − R2 C (0.02) − 1252 (135 × 10−9 )
ω2 = = = 3.31 × 108 ;
L2 C (0.01)2 (62.5 × 10−9 )
.· . ω = 1.82 × 104 rad/s.
[b] Zt = 1000 + (−j407)k(125 + j364) = 2184.9 Ω;
Vo 156 0◦
Io = = = 6.86 0◦ mA;
ZT 2184.9
io = 6.8 cos 18,200t mA.
[c] As derived in [a], the circuit will have real value of resonant frequency
when
L − R2 C ≥ 0
s s
L L 20 × 10−3
≥ R2 ⇒ R ≤ ⇒≤ .
C C 13.5 × 10−3
Thus, the maximum value of R for which circuit will have real value of
resonant frequency is 384 Ω.
 Problems 9–25

106
P 9.40 [a] Z1 = 400 − j = 400 − j800 Ω;
500(2.5)
j106 L
Z2 = 2000kj500L = ;
2000 + j500L
j106 L
ZT = Z1 + Z2 = 400 − j800 +
2000 + j500L
500 × 106 L2 2 × 109 L
= 400 − − j800 + j .
20002 + 5002 L2 20002 + 5002 L2
ZT is resistive when
2 × 109 L
2 2 2
= 800 or 5002 L2 − 25 × 105 L + 20002 = 0.
2000 + 500 L
Solving, L1 = 8 H and L2 = 2 H.
[b] When L = 8 H:
500 × 106 (8)2
ZT = 400 + = 2000 Ω;
20002 + 5002 (8)2
200/0◦
Ig = = 100/0◦ mA;
2000
ig = 100 cos 500t mA.

When L = 2 H:
500 × 106 (2)2
ZT = 400 + = 800 Ω;
20002 + 500(2)2
200/0◦
Ig = = 250/0◦ mA;
800
ig = 250 cos 500t mA.

R
jωC R
P 9.41 [a] Zp = =
R + (1/jωC) 1 + jωRC

10,000 10,000
= =
1 + j(5000)(10,000)C 1 + j50 × 106 C
10,000(1 − j50 × 106 C)
=
1 + 25 × 1014 C 2
10,000 5 × 1011 C
= − j
1 + 25 × 1014 C 2 1 + 25 × 1014 C 2
jωL = j5000(0.8) = j4000
 9–26 CHAPTER 9. Sinusoidal Steady State Analysis

5 × 1011 C
.·. 4000 =
1 + 25 × 1014 C 2
.·. 1014 C 2 − 125 × 106 C + 1 = 0

.·. C 2 − 5 × 10−8 C + 4 × 10−16 = 0

Solving,
C1 = 40 nF C2 = 10 nF
10,000
[b] Re =
1 + 25 × 1014 C 2
When C = 40 nF Re = 2000 Ω;
80/0◦
Ig = = 40/0◦ mA; ig = 40 cos 5000t mA
2000
When C = 10 nF Re = 8000 Ω;
80/0◦
Ig = = 10/0◦ mA; ig = 10 cos 5000t mA
8000
P 9.42 Simplify the top triangle using series and parallel combinations:

(1 + j1)k(1 − j1) = 1 Ω.

Convert the lower left delta to a wye:

(j1)(1)
Z1 = = j1 Ω;
1 + j1 − j1

(−j1)(1)
Z2 = = −j1 Ω;
1 + j1 − j1

(j1)(−j1)
Z3 = = 1 Ω.
1 + j1 − j1

Convert the lower right delta to a wye:

 Problems 9–27

(−j1)(1)
Z4 = = −j1 Ω;
1 + j1 − j1

(−j1)(j1)
Z5 = = 1 Ω;
1 + j1 − j1

(j1)(1)
Z6 = = j1 Ω.
1 + j1 − j1

The resulting circuit is shown below:

Simplify the middle portion of the circuit by making series and parallel
combinations:

(1 + j1 − j1)k(1 + 1) = 1k2 = 2/3 Ω;

Zab = −j1 + 2/3 + j1 = 2/3 Ω.

P 9.43 [a] jωL = j(400)(400) × 10−3 = j160 Ω;

1 −j
= = −j80 Ω.
jωC (400)(31.25 × 10−6 )

Using voltage division,

−j80kj160
Vab = (−j50) = −20 − j10 V;
320 + (−j80kj160)
VTh = Vab = −20 − j10 V.
 9–28 CHAPTER 9. Sinusoidal Steady State Analysis

[b] Remove the voltage source and combine impedances in parallel to find
ZTh = Zab :
ZTh = Zab = 320k − j80kj160 = 64 − j128 Ω.
[c]

P 9.44 Step 1 to Step 2:

180/90◦
= 6/0◦ A
j30

Step 2 to Step 3:

(j30)k15 = 12 + j6 Ω; (6/0◦ )(12 + j6) = 72 + j36 V

Step 3 to Step 4:

72 + j36
ZN = 12 + j6 − j30 = 12 − j24 Ω; IN = = j3 A
12 − j24

P 9.45 Step 1 to Step 2:

(0.12/0◦ )(250) = 30/0◦ V

Step 2 to Step 3:

30/0◦
250 − j400 + j150 = 250 − j250 Ω; = 60 − j60 mA
250 − j250
 Problems 9–29

Step 3 to Step 4:

(250 − j250)k500 = 200 − j100 Ω; (200 − j100)(0.06 − j0.06) = 18 − j6 V

P 9.46 Open circuit voltage:

−j20Ia + 40Ia + 20(Ia − 0.4 − j0.2) = 0.

Solving,

20(0.4 + j0.2)
Ia = = 0.1 + j0.1 A;
60 − j20

Voc = 40Ia + j16(0.4 + j0.2) = 0.8 + j10.4 V.

Short circuit current:

 9–30 CHAPTER 9. Sinusoidal Steady State Analysis

−j20Ia + 40(Ia − Isc ) + 20(Ia − 0.4 − j0.2) = 0;

40(Isc − Ia ) + j16(Isc − 0.4 − j0.2) = 0.

Solving,

IN = Isc = 0.3 + j0.5 A;

VTh 0.8 + j10.4

ZTh = = = 16 + j8 Ω.
IN 0.3 + j0.5

P 9.47

5 − j15
VTh = + (1 − j3) mA, ZN in kΩ.
ZN

−18 − j13.5
IN = + +4.5 − j6 mA, ZN in kΩ;
ZN
 Problems 9–31

5 − j15 −18 − j13.5

+ 1 − j3 = + (4.5 − j6);
ZN ZN

23 − j15
= 3.5 − j3 .· . ZN = 4 + j3 kΩ;
ZN

5 − j15
IN = + 1 − j3 = −j6 mA.
4 + j3

P 9.48 Short circuit current

Vx − j20(I1 − Isc ) = −40 + j40;

−2Vx − j20(Isc − I1 ) = 0;

Vx = 20I1 .

Solving,

IrmN = Isc = −2 + j2 A.

Open circuit voltage

 9–32 CHAPTER 9. Sinusoidal Steady State Analysis

−40 + j40
I= = −2 A;
20 − j20

Vx = 20I = −40 V;

Voc = 2Vx − j20I = −80 + j40 V;

−80 + j40V
ZrmN = = 30 + j10 Ω.
−2 + j2

P 9.49 Open circuit voltage:

V1 − 250 V1
− 0.03Vo + = 0;
20 + j10 50 − j100

−j100
. · . Vo = V1 .
50 − j100

V1 j3V1 V1 250
+ + = ;
20 + j10 50 − j100 50 − j100 20 + j10

V1 = 500 − j250 V; Vo = 300 − j400 V = VTh .

Short circuit current:

 Problems 9–33

250/0◦
Isc = = 3.5 − j0.5 A;
70 + j10

VTh 300 − j400

ZTh = = = 100 − j100 Ω.
Isc 3.5 − j0.5

The Thévenin equivalent circuit:

P 9.50 Open circuit voltage:

V2 V2 − 51 V2
+ 88Iφ + = 0;
10 −j50

5 − (V2 /5)
Iφ = .
200

Solving,

V2 = −66 + j88 = 110/126.87◦ V = VTh .

Find the Thévenin (Norton) equivalent impedance using a test source:

 9–34 CHAPTER 9. Sinusoidal Steady State Analysis

VT 0.8VT
IT = + 88Iφ + ;
10 −j50
−VT /5
Iφ = ;
200
!
1 1/5 0.8
IT = VT − 88 + ;
10 200 −j50

VT
.· . ZN = = 30 − j40.
IT
VTh −66 + j88
IN = = = −2.2 + j0 A.
ZN 30 − j40
The Norton equivalent circuit:

P 9.51 [a]

VT VT − αVT
IT = + ;
1000 −j1000
IT 1 (1 − α) j−1+α
= − =
VT 1000 j1000 j1000;
VT j1000
.·. ZTh = = .
IT α−1+j
ZTh is real when α = 1.
 Problems 9–35

[b] ZTh = 1000 Ω.

j1000
[c] ZTh = 500 − j500 =
α−1+j
1000 1000(α − 1)
= 2
+j .
(α − 1) + 1 (α − 1)2 + 1
Equate the real parts:
1000
= 500 .·. (α − 1)2 + 1 = 2;
(α − 1)2 + 1
.· . (α − 1)2 = 1 so α = 0.
Check the imaginary parts:
(α − 1)1000
= −500.
(α − 1)2 + 1 α=1

Thus, α = 0.
1000 1000(α − 1)
[d] ZTh = 2
+j .
(α − 1) + 1 (α − 1)2 + 1
For Im(ZTh ) > 0, α must be greater than 1. So ZTh is inductive for
1 < α ≤ 10.
P 9.52 jωL = j100 × 103 (1.2 × 10−3 ) = j120 Ω;

1 −j
= = −j12.5 Ω.
jωC (100 × 10 )(0.8 × 10−6 )
3

VT = −j12.5IT + 5I∆ − 60I∆ ;

−j120
I∆ = IT ;
60 + j120

j120
VT = −j12.5IT + 55 IT ;
60 + j120

VT
= Zab = 44 + j9.5 = 45.016 12.18
IT
 9–36 CHAPTER 9. Sinusoidal Steady State Analysis

1 109
P 9.53 = = 8 kΩ
ωC1 50,000(2.5)

1 109
= = 4 kΩ
ωC2 50,000(5)

VT = (2400 − j8000)IT + 40IT (90)

VT
ZTh = = 6000 − j8000 Ω
IT
P 9.54

V1 − 240 V1 V1
+ + = 0.
j20 50 60 + j10

Solving for V1 yields

V1 = 154.7 − j106.26

= 187.6896 −34.85◦
60
Vo = (V1 ) = 137.46 − 103.38j = 172.06 − 36.94
60 + j10

P 9.55 Make a source transformation on the left side:

 Problems 9–37

Write a KCL equation on the right hand side:

Vo Vo
16Io + + = 0.
50 −j25
Write a KVL equation on the left hand side:
Vo
−1 + (40 + j20)Io + = 0.
8
Solving,

Vo = 4 + j4 = 5.66/45◦ V;

1 − Vo /8
Io = = −25 − j25 mA = 35.36/ − 135◦ mA.
j20

P 9.56 jωL = j(2500)(1.6 × 10−3 ) = j4 Ω;

1 −j
= = −j4 Ω;
jωCtop (2500)(100 × 10−6 )

1 −j
= = −j16 Ω;
jωCmid (2500)(25 × 10−6 )

Ig = 56 0◦ A; Vg = 206 90◦ V.

V1 − V2 V1 − j20
−5 + + = 0;
−j16 −j4

V2 − V1 V2 V2 − j20
+ + = 0.
−j16 j4 24

Solving,
V2
V2 = +3.64 + j0.868 Io = = 0.217 + j(−0.911) = 0.9366 −76.6
j4

io = 0.936 cos(2500t − 76.6) A.

 9–38 CHAPTER 9. Sinusoidal Steady State Analysis

P 9.57 jωL = j(400)(100 × 10−3 ) = j40 Ω;

1 −j
= = −j100 Ω;
jωC (400)(25 × 10−6 )

Vg1 = 206 53.13◦ = 12 + j16 V;

10
Vg2 = 6.016 33.69◦ = 5 + j V.
3

 
Vo − (12 + j16) Vo Vo − 5 + j 10
3
+ + = 0.
−j100 300 j40

Solving,

Vo = −0.765 − j4.94 = 5.06 − 98.80◦

vo (t) = 5.0[cos(400t − 98.8)] = 5.0 cos(400t − 98.80)

P 9.58 Transform the circuit to the phasor domain; the sources are

Va = 50/ − 90◦ = −j50 V;

Vb = 25/90◦ = j25 V.

The impedances are

jωL = j106 (10 × 10−6 ) = j10 Ω;

1 −j
= = −j10 Ω.
jωC (10 )(0.1 × 10−6 )
6

The phasor domain circuit is

 Problems 9–39

The KVL equations are

(10 − j10)I1 + j10I2 − 10I3 = −j50;

j10I1 + 10I2 − 10I3 = 0;

−10I1 − 10I2 + 20I3 = j25.

Solving,

I1 = 0.5 − j1.5 A; I3 = −1 + j0.5 A;

Ia = −I1 = −0.5 + j1.5 = 1.58/108.43◦ A;

Ib = −I3 = 1 − j0.5 = 1.12/ − 26.57◦ A.

Inverse phasor-transform back to the time domain to get

ia = 1.58 cos(106 t + 108.43◦ ) A;

ib = 1.12 cos(106 t − 26.57◦ ) A.

P 9.59

Vo Vo
+ + 20Io = 0
50 −j25
 9–40 CHAPTER 9. Sinusoidal Steady State Analysis

(2 + j4)Vo = −2000Io

Vo = (−200 + j400)Io

V1 − (Vo /10)
Io =
j25

.·. V1 = (−20 + j65)Io

V1
0.006 + j0.013 = + Io = (−0.4 + j1.3)Io + Io = (0.6 + j1.3)Io
50

0.6 + j1.3(10 × 10−3 )

.·. Io = = 10/0◦ mA
(0.6 + j1.3)

Vo = (−200 + j400)Io = −2 + j4 = 4.47/116.57◦ V

P 9.60

(12 − j12)Ia − 12Ig − 5(−j8) = 0

−12Ia + (12 + j4)Ig + j20 − 5(j4) = 0

Solving,

Ig = 4 − j2 = 4.47/ − 26.57◦ A

P 9.61
 Problems 9–41

(12 − j12)Ia − 12Ig − 5(−j8) = 0;

−12Ia + (12 + j4)Ig + j20 − 5(j4) = 0.

Solving,

Ig = 4 − j2 = 4.47/ − 26.57◦ A so Io = 5 − Ig = 1 + j2 = 2.24/63.43◦ A;

io = 2.24 cos(2500t + 63.43◦ ) A.

P 9.62

10/0◦ = (1 − j1)I1 − 1I2 + j1I3 ;

−5/0◦ = −1I1 + (1 + j1)I2 − j1I3 ;

1 = j1I1 − j1I2 + I3 .
 9–42 CHAPTER 9. Sinusoidal Steady State Analysis

Solving,

I1 = 11 + j10 A; I2 = 11 + j5 A; I3 = 6 A;

Ia = I3 − 1 = 5 A;

Ib = I1 − I3 = 5 + j10 A;

Ic = I2 − I3 = 5 + j5 A;

Id = I1 − I2 = j5 A.

P 9.63

(10 + j5)Ia − j5Ib = 100/0◦

−j5Ia − j5Ib = j100

Solving,

Ia = −j10 A; Ib = −20 + j10 A

Io = Ia − Ib = 20 − j20 = 28.28/ − 45◦ A

io (t) = 28.28 cos(50,000t − 45◦ ) A

P 9.64 jωL1 = j5000(4 × 10−3 ) = j20 Ω; jωL2 = j5000(110 × 10−3 ) = j550 Ω

1 −j
= = −j50 Ω
jωC (5000)(4 × 10−6 )

75/0◦ = j500I∆ − 100I∆ − j550Ia

 Problems 9–43

0 = (10 + j20)Ia + 100I∆ + j550(Ia − I∆ )

Solving,

Ia = j2.5 A

Vo = 10Ia = j25 = 25/90◦

vo = 25 cos(5000t − 90◦ ) = 25 sin 5000t V

Zo 200 − j1000
P 9.65 Vo = Vg = (1006 0◦ ) = 199.46 −0.229 V;
ZT 300 + j1600 + 500 − j1000

vo = 199.4 cos(8000t − 0.229◦ ) V.

1 −j
P 9.66 = = −j400 Ω;
jωC (20,000)(125 × 10−9 )

jωL = j(20,000)(0.03) = j600 Ω;

Let Z1 = 100 − j400 Ω; Z2 = 120 + j600 Ω;

Ig = 4006 0◦ mA;

Z1 kZ2 (100 − j400)k(120 + j600)

Io = (Ig ) = (0.4);
Z1 (100 − j400)

= 48.54 + j0.597 mA = 48.546 0.70◦ mA;

io = 48.54 cos(20,000t + 0.70◦ ) mA.

P 9.67 [a] Superposition must be used because the frequencies of the two sources are
different.
[b] For ω = 2000 rad/s:

10k − j2.5 = 0.588 − j2.35

 9–44 CHAPTER 9. Sinusoidal Steady State Analysis

0.588 − j2.35
so Vo1 = (206 36.87◦ ) = 70.76 −81.8 V.
0.588 − j2.35 + j2
For ω = 5000 rad/s:

j10k10 = 5 + j5 Ω;
5 + j5
Vo2 = (106 10.26◦ ) = 11.046 22.60 V.
5 + j5 − j2
Thus,
vo (t) = [70.7 cos(2000t − 81.8) + 11.04 cos(5000t + 22.6)] V.

P 9.68 [a] Superposition must be used because the frequencies of the two sources are
different.
[b] For ω = 16,000 rad/s:

V0o − 10 V0o V0
+ + o = 0;
−j100 j400 400
!
1 1 1 10
V0o + + = ;
−j100 j400 400 −j100

.·. V0o = 12 + j4 = 12.65/18.435◦ V.

For ω = 4000 rad/s:

V00o V00 V00 − 20

+ o + o = 0;
−j400 j100 400
 Problems 9–45

20
V00o (j − j4 + 1) = 20 so V00o = = 2 + j6 = 6.32/71.565◦ V.
1 − j3
Thus,
vo (t) = [12.65 cos(16,000t + 18.435◦ ) + 6.32 cos(4000t + 71.565◦ )] V.
1
P 9.69 Vg = 20/0◦ V; = −j400 Ω
jωC
Let Va = voltage across the capacitor, positive at upper terminal
Then:

Va − 20/0◦ Va Va
+ + = 0; .·. Va = (8 − j4) V
400 −j400 400

0 − V a 0 − Vo Va
+ = 0; Vo = −
400 200 2

.·. Vo = −4 + j2 = 4.47/153.43◦ V

vo = 4.47 cos(5000t + 153.43◦ ) V

P 9.70 [a]

Va − 20/0◦ Va
+ jωCo Va + =0
400 400
20
Va =
2 + j400ωCo
Va
Vo = −
2
−10 10/180◦
Vo = =
2 + j2 × 106 Co 2 + j2 × 106 Co
.·. denominator angle = 45◦

so 2 × 106 Co = 2 .·. C = 1 µF
 9–46 CHAPTER 9. Sinusoidal Steady State Analysis

10/180◦
[b] Vo = = 3.54/135◦ V
2 + j2
vo = 3.54 cos(5000t + 135◦ ) V

P 9.71 [a] Vg = 25/0◦ V;

20
Vp = Vg = 5/0◦ ; Vn = Vp = 5/0◦ V;
100
5 5 − Vo
+ = 0;
80,000 Zp
Zp = −j80,000k40,000 = 32,000 − j16,000 Ω;
5Zp
Vo = + 5 = 7 − j = 7.07/ − 8.13◦ ;
80,000
vo = 7.07 cos(50,000t − 8.13◦ ) V.
[b] Vp = 0.2Vm /0◦ ; Vn = Vp = 0.2Vm /0◦ ;
0.2Vm 0.2Vm − Vo
+ = 0;
80,000 32,000 − j16,000
32,000 − j16,000
.·. Vo = 0.2Vm + Vm (0.2) = Vm (0.28 − j0.04);
80,000
.·. |Vm (0.28 − j0.04)| ≤ 10;
.·. Vm ≤ 35.36 V.
1
P 9.72 = −j10 kΩ;
jωC1

1
= −j100 kΩ.
jωC2

Va − 2 Va Va Va − Vo
+ + + = 0;
5000 −j10,000 20,000 100,000
 Problems 9–47

20Va − 40 + j10Va + 5Va + Va − Vo = 0;

.· . (26 + j10)Va − Vo = 40.

0 − Va 0 − Vo
+ = 0;
20,000 −j100,000

j5Va − Vo = 0.

Solving,

Vo = 1.43 + j7.42 = 7.56/79.09◦ V;

vo (t) = 7.56 cos(106 t + 79.09◦ ) V.

1 −j109
P 9.73 [a] = = −j100 Ω
jωC (106 )(10)
Vg = 30/0◦ V

Vg (1/jωCo ) 30/0◦
Vp = = = Vn
25 + (1/jωCo ) 1 + j25ωCo
Vn Vn − Vo
+ =0
100 −j100
1 + j1 30(1 − j1)
Vo = Vn = (1 − j1)Vn =
j 1 + j25ωCo
√
30 2
|Vo | = q =6
1 + 625ω 2 Co2

Solving,
Co = 280 nF
30(1 − j1)
[b] Vo = = 6/ − 126.87◦
1 + j7
vo = 6 cos(106 t − 126.87◦ ) V

P 9.74 jωL1 = j100 Ω;

jωL2 = j64 Ω;

1
= −j40 Ω;
jωC
 9–48 CHAPTER 9. Sinusoidal Steady State Analysis

q
jωM = j(4 × 103 )k (25)(16) × 10−3 = j80k Ω;

Z22 = 5 + j64 − j40 = 5 + j24 Ω;

∗
Z22 = 5 − j24 Ω;
" #2
80k
Zr = (5 − j24) = 53.26k 2 − 255.68k 2 ;
|5 + j24|

Zab = 20 + j100 + 53.26k 2 − 255.68k 2 = (20 + 53.26k 2 ) + j(100 − 255.689k 2 ).

Zab is resistive when

100 − 255.68k 2 = 0 or k 2 = 0.391 so k = 0.625;

.·. Zab = 20 + (53.267)(0.391) = 40.83 Ω.

P 9.75 Remove the voltage source to find the equivalent impedance:

!2
20
ZTh = 45 + j150 + (5 − j6) = 77.78 + j110.65 Ω.
|5 + j6|
Using voltage division:
!
425
VTh = Vcd = j30I1 = j30 = 1254.09 + j1045.08 V.
5 + j6

P 9.76 [a] jωL1 = j(800)(100 × 10−3 ) = j80 Ω

jωL2 = j(800)(400 × 10−3 ) = j320 Ω
jωM = j80 Ω

168 = (80 + j80)Ig + j80IL

 Problems 9–49

0 = j80Ig + (240 + j320)IL

Solving,
Ig = 1.2 − j0.9 A; IL = −0.3 A

ig = 1.5 cos(800t − 36.87◦ ) A

iL = 0.3 cos(5000t − 180◦ ) A

M 0.1
[b] k = √ =q = 0.5
L1 L2 (0.1)(0.4)
[c] When t = 625π µs,
800t = (800)(625π) × 10−6 = 0.5π = π/2 rad = 90◦

ig (625πµs) = 1.5 cos(53.13◦ ) = 0.9 A

iL (625πµs) = 0.3 cos(−90◦ ) = 0 A

1 1 1
w = L1 i2g + L2 i2L + M ig iL = (100 × 10−3 )(0.81) + 0 + 0 = 40.5 mJ
2 2 2
When t = 1250π µs,
800t = π rad = 180◦

ig (1250πµs) = 1.5 cos(180 − 36.87) = −1.2 A

iL (1250πµs) = 0.3 cos(180 − 180) = 0.3 A

1 1
w = (100 × 10−3 )(1.44) + (400 × 10−3 )(0.09)
2 2
+100 × 10−3 (−1.2)(0.3) = 54 mJ

P 9.77 [a] jωL1 = j(200 × 103 )(10−3 ) = j200 Ω;

jωL2 = j(200 × 103 )(4 × 10−3 ) = j800 Ω;

1 −j
= = −j400 Ω;
jωC (200 × 10 )(12.5 × 10−9 )
3

.·. Z22 = 100 + 200 + j800 − j400 = 300 + j400 Ω;

.·. Z22
∗
= 300 − j400 Ω.
q
M = k L1 L2 = 2k × 10−3 ;

ωM = (200 × 103 )(2k × 10−3 ) = 400k;

" #2
400k
Zr = (300 − j400) = k 2 (192 − j256) Ω;
500
 9–50 CHAPTER 9. Sinusoidal Steady State Analysis

Zin = 200 + j200 + 192k 2 − j256k ; 2

1
|Zin | = [(200 + 192k)2 + (200 − 256k)2 ] 2 ;
d|Zin | 1 1
= [(200 + 192k)2 + (200 − 256k)2 ]− 2 ×
dk 2

[2(200 + 192k 2 )384k + 2(200 − 256k 2 )(−512k)];

d|Zin |
= 0 when
dk
768k(200 + 192k 2 ) − 1024k(200 − 256k 2 ) = 0.
√
.·. k 2 = 0.125; .·. k = 0.125 = 0.3536.

[b] Zin (min) = 200 + 192(0.125) + j[200 − 0.125(256)]

= 224 + j168 = 280/36.87◦ Ω.

560/0◦
I1 (max) = = 2/ − 36.87◦ A;
224 + j168
.·. i1 (peak) = 2 A.

Note — You can test that the k value obtained from setting d|Zin |/dt = 0
leads to a minimum by noting 0 ≤ k ≤ 1. If k = 1,
Zin = 392 − j56 = 395.98/ − 8.13◦ Ω.
Thus,
|Zin |k=1 > |Zin |k=√0.125 .
If k = 0,
Zin = 200 + j200 = 282.84/45◦ Ω.
Thus,
|Zin |k=0 > |Zin |k=√0.125 .

P 9.78 [a] jωLL = j100 Ω

jωL2 = j500 Ω

Z22 = 300 + 500 + j100 + j500 = 800 + j600 Ω

∗
Z22 = 800 − j600 Ω

ωM = 270 Ω
2
270

Zr = [800 − j600] = 58.32 − j43.74 Ω
1000
 Problems 9–51

[b] Zab = R1 + jωL1 + Zr = 41.68 + j180 + 58.32 − j43.74 = 100 + j136.26 Ω

P 9.79 In Eq. 9.34 replace ω 2 M 2 with k 2 ω 2 L1 L2 and then write Xab as

k 2 ω 2 L1 L2 (ωL2 + ωLL )
Xab = ωL1 − 2
R22 + (ωL2 + ωLL )2
k 2 ωL2 (ωL2 + ωLL )
(
= ωL1 1 − 2
}
R22 + (ωL2 + ωLL )2

For Xab to be negative requires

2
R22 + (ωL2 + ωLL )2 < k 2 ωL2 (ωL2 + ωLL ),

or

2
R22 + (ωL2 + ωLL )2 − k 2 ωL2 (ωL2 + ωLL ) < 0,

which reduces to

2
R22 + ω 2 L22 (1 − k 2 ) + ωL2 ωLL (2 − k 2 ) + ω 2 L2L < 0.

But k ≤ 1, so it is impossible to satisfy the inequality. Therefore Xab can

never be negative if XL is an inductive reactance.

P 9.80 [a]

Vab V1 + V 2
Zab = = ;
I1 I1
V1 V2 N2
= , V2 = V1 ;
N1 N2 N1
N1
N1 I1 = N2 I2 , I2 = I1 ;
N2
N1
 
V2 = (I1 + I2 )ZL = I1 1+ ZL ;
N2
 9–52 CHAPTER 9. Sinusoidal Steady State Analysis

2
N1 N1
  
V1 + V2 = + 1 V2 = 1 + ZL I1 ;
N2 N2
(1 + N1 /N2 )2 ZL I1
.·. Zab = .
I1
2
N1

Zab = 1 + ZL .
N2
[b] Assume dot on N2 is moved to the lower terminal, then
V1 −V2 −N1
= , V1 = V2 ;
N1 N2 N2
−N1
N1 I1 = −N2 I2 , I2 = I1 .
N2
As in part [a]
V 1 + V2
V2 = (I2 + I1 )ZL and Zab = ;
I1
(1 − N1 /N2 )V2 (1 − N1 /N2 )(1 − N1 /N2 )ZL I1
Zab = = ;
I1 I1
Zab = [1 − (N1 /N2 )]2 ZL .

P 9.81 [a]

N1
N1 I1 = N2 I2 , I2 = I1 ;
N2
Vab V2 V2
Zab = = = ;
I1 + I2 I1 + I2 (1 + N1 /N2 )I1
V1 N1 N1
= , V1 = V2 ;
V2 N2 N2
N1
 
V1 + V2 = ZL I1 = + 1 V2 ;
N2
I1 ZL
Zab = ;
(N1 /N2 + 1)(1 + N1 /N2 )I1
 Problems 9–53

ZL
.·. Zab = .
[1 + (N1 /N2 )]2
[b] Assume dot on the N2 coil is moved to the lower terminal. Then
N1 N1
V1 = − V2 and I2 = − I1 .
N2 N2
As before
V2
Zab = and V1 + V2 = ZL I1 ;
I1 + I2
V2 ZL I1
.·. Zab = = .
(1 − N1 /N2 )I1 [1 − (N1 /N2 )]2 I1
ZL
Zab = .
[1 − (N1 /N2 )]2

P 9.82

V3
ZL = ;
I3

V2 −V3
= ; 1I2 = −20I3 ;
1 20

V1 V2
= ; 50I1 = 1I2 ;
50 1

V1
Zab = .
I1

Substituting,

V1 50V2 502 V2
Zab = = =
I1 I2 /50 I2

502 (−V3 /20) (50)2 V3

= = 2
= 6.25ZL = 6.25(200/ − 45◦ ) = 1250/ − 45◦ Ω.
−20I3 (20) I3
 9–54 CHAPTER 9. Sinusoidal Steady State Analysis

P 9.83 The phasor domain equivalent circuit is

Vm Vm
Vo = − IRx ; I= .
2 Rx − jXC

As Rx varies from 0 to ∞, the amplitude of vo remains constant and its phase

angle increases from 0◦ to −180◦ , as shown in the following phasor diagram:

240 240
P 9.84 [a] I = + = (10 − j7.5) A
24 j32
Vs = 240/0◦ + (0.1 + j0.8)(10 − j7.5) = 247 + j7.25 = 247.11/1.68◦ V
[b] Use the capacitor to eliminate the j component of I, therefore
240
Ic = j7.5 A, Zc = = −j32 Ω
j7.5
Vs = 240 + (0.1 + j0.8)10 = 241 + j8 = 241.13/1.90◦ V
[c] Let Ic denote the magnitude of the current in the capacitor branch. Then
I = (10 − j7.5 + jIc ) = 10 + j(Ic − 7.5) A

Vs = 240/α = 240 + (0.1 + j0.8)[10 + j(Ic − 7.5)]

= (247 − 0.8Ic ) + j(7.25 + 0.1Ic )

It follows that
240 cos α = (247 − 0.8Ic ) and 240 sin α = (7.25 + 0.1Ic )
Now square each term and then add to generate the quadratic equation
Ic2 − 605.77Ic + 5325.48 = 0; Ic = 302.88 ± 293.96
 Problems 9–55

Therefore
Ic = 8.92 A (smallest value) and Zc = 240/j8.92 = −j26.90 Ω.
Therefore, the capacitive reactance is −26.90 Ω.

P 9.85 [a]

120 120
I` = + = 16 − j10 A;
7.5 j12
V` = (0.15 + j6)(16 − j10) = 62.4 + j94.5 = 113.24/56.52◦ V(rms);

Vs = 120/0◦ + V` = 205.43/27.39◦ V.

[b]

120 120
[c] I` = + = 48 − j30 A;
2.5 j4
V` = (0.15 + j6)(48 − j30) = 339.73/56.56◦ ;

Vs = 120 + V` = 418.02/42.7◦ .

The amplitude of Vs must be increased from 205.43 to 418.02 (more than

doubled) to maintain the load voltage at 120 V.
120 120 120
[d] I` = + + = 48 + j30 A;
2.5 j4 −j2
V` = (0.15 + j6)(48 + j30) = 339.73/120.57◦ ;
 9–56 CHAPTER 9. Sinusoidal Steady State Analysis

Vs = 120 + V` = 297.23/100.23◦ .

The amplitude of Vs must be increased from 205.43 to 297.23 to

maintain the load voltage at 120 V.

P 9.86 [a] Let N1 = primary winding turns and 2N2 = secondary winding turns.
Then
14,000 250 N2 1
= ; .· . = = a.
N1 2N2 N1 112
In part c),
IP = 2aIa ;
2N2 Ia 1
.·. IP = = Ia
N1 56
1
= (31.656 − j0.16);
56

IP = 565.3 − j2.9 mA.

In part d),
IP N1 = I1 N2 + I2 N2 ;
N2
.·. IP = (I1 + I2 )
N1
1
= (34.19 − j0.182 + 31.396 − j0.164)
112
1
= (65.586 − j0.346);
112

IP = 585.6 − j3.1 mA.

[b] Yes, because the neutral conductor carries non-zero current whenever the
load is not balanced.

P 9.87 [a] The circuit is redrawn, with mesh currents identified:

 Problems 9–57

The mesh current equations are:

120/0◦ = 23Ia − 2Ib − 20Ic ;

120/0◦ = −2Ia + 43Ib − 40Ic ;

0 = −20Ia − 40Ib + 70Ic .

Solving,
Ia = 24/0◦ A; Ib = 21.96/0◦ A; Ic = 19.40/0◦ A.
The branch currents are:
I1 = Ia = 24/0◦ A;

I2 = Ib − Ia = 2.04/180◦ A;

I3 = Ib = 21.96/0◦ A;

I4 = Ic = 19.40/0◦ A;

I5 = Ia − Ic = 4.6/0◦ A;

I6 = Ib − Ic = 2.55/0◦ A.

[b] Let N1 be the number of turns on the primary winding; because the
secondary winding is center-tapped, let 2N2 be the total turns on the
secondary. Therefore,
13,200 240 N2 1
= or = .
N1 2N2 N1 110
The ampere turn balance requires
N1 IP = N2 I1 + N2 I3 .
Therefore,
N2 1
IP = (I1 + I3 ) = (24 + 21.96) = 0.42/0◦ A.
N1 110
 9–58 CHAPTER 9. Sinusoidal Steady State Analysis

P 9.88 [a]

The three mesh current equations are

120/0◦ = 23Ia − 2Ib − 20Ic ;

120/0◦ = −2Ia + 23Ib − 20Ic ;

0 = −20Ia − 20Ib + 50Ic .

Solving,
Ia = 24/0◦ A; Ib = 24/0◦ A; Ic = 19.2/0◦ A;

.·. I2 = Ib − Ia = 0 A.
N2 N2
[b] IP = (I1 + I3 ) = (Ia + Ib )
N1 N1
1
= (24 + 24) = 0.436/0◦ A.
110
[c] Yes; when the two 120 V loads are equal, there is no current in the
“neutral” line, so no power is lost to this line. Since you pay for power,
the cost is lower when the loads are equal.

P 9.89 [a]

125 = (R + 0.05 + j0.05)I1 − (0.03 + j0.03)I2 − RI3 ;

125 = −(0.03 + j0.03)I1 + (R + 0.05 + j0.05)I2 − RI3 .

 Problems 9–59

Subtracting the above two equations gives

0 = (R + 0.08 + j0.08)I1 − (R + 0.08 + j0.08)I2 ;

.·. I1 = I2 so In = I2 − I1 = 0 A.

[b] V1 = R(I1 − I3 ); V2 = R(I2 − I3 ).

Since I1 = I2 (from part [a]) V1 = V2 .

[c]

250 = (440.04 + j0.04)Ia − 440Ib ;

0 = −440Ia + 448Ib .

Solving,
Ia = 31.656207 − j0.160343 A;

Ib = 31.090917 − j0.157479 A;

I = Ia − Ib = 0.56529 − j0.002864 A;

V1 = 40I = 22.612 − j0.11456 = 22.612/ − 0.290282◦ V;

V2 = 400I = 226.116 − j1.1456 = 226.1189/ − 0.290282◦ V.

[d]

125 = (40.05 + j0.05)Ix − (0.03 + j0.03)Iy − 40Iz ;

 9–60 CHAPTER 9. Sinusoidal Steady State Analysis

125 = −(0.03 + j0.03)Ix + (400.05 + j0.05)Iy − 400Iz ;

0 = −40Ix − 400Iy + 448Iz .

Solving,
Ix = 34.19 − j0.182 A;

Iy = 31.396 − j0.164 A;

Iz = 31.085 − j0.163 A;

V1 = 40(Ix − Iz ) = 124.2/ − 0.35◦ V;

V2 = 400(Iy − Iz ) = 124.4/ − 0.18◦ V.

[e] Because an open neutral can result in severely unbalanced voltages across
the 125 V loads.

P 9.90 No, the motor current drops to 5 A, well below its normal running value of
22.86 A.

P 9.91 After fuse A opens, the current in fuse B is only 15 A.