Chapter 11, Problem 1.
If v(t) = 160 cos 50t V and i(t) = –20 sin(50t – 30°) A, calculate the instantaneous power
and the average power.
Chapter 11, Solution 1.
v( t ) = 160 cos(50t )
i( t ) = 20 sin(50t − 30°) = 2 cos(50t − 30° + 180° − 90°)
i( t ) = 20 cos(50t + 60°)
p( t ) = v( t ) i( t ) = (160)(20) cos(50t ) cos(50t + 60°)
p( t ) = 1600 [ cos(100 t + 60°) + cos(60°) ] W
p( t ) = 800 + 1600 cos(100t + 60°) W
1
1
Vm I m cos(θ v − θi ) = (160)(20) cos(60°)
2
2
P = 800 W
P=
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Chapter 11, Problem 2.
Given the circuit in Fig. 11.35, find the average power supplied or absorbed by each
element.
Figure 11.35
For Prob. 11.2.
Chapter 11, Solution 2.
Using current division,
j1 Ω
I2
I1
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.
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Chapter 11, Problem 3.
A load consists of a 60 Ω resistor in parallel with a 90 µ F capacitor. If the load is
connected to a voltage source v s (t) = 40 cos 2000t, find the average power delivered to
the load.
Chapter 11, Solution 3.
I
+
–
90 µ F
˚
40∠0
⎯⎯
→
C
R
1
1
=
= − j 5.5556
−6
jω C j 90 x10 x 2 x103
I = 40/60 = 0.6667A or Irms = 0.6667/1.4142 = 0.4714A
The average power delivered to the load is the same as the average power absorbed by
the resistor which is
Pavg = Irms260 = 13.333 W.
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Chapter 11, Problem 4.
Find the average power dissipated by the resistances in the circuit of Fig. 11.36.
Additionally, verify the conservation of power.
Figure 11.36
For Prob. 11.4.
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Chapter 11, Solution 4.
We apply nodal analysis. At the main node,
I1
20∠30o V
+
–
5 Ω I2
Vo
j4 Ω
8Ω
–j6 Ω
20 < 30o − Vo Vo
V
=
+ o
⎯⎯
→ Vo = 5.152 + j10.639
5
j 4 8 − j6
For the 5Ω resistor,
20 < 30o − Vo
= 2.438 < −3.0661o A
I1 =
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 8Ω resistor,
V
I 2 = o = 1.466 < 71.29o
8− j
The average power dissipated by the resistor is
1
1
P2 =  I 2 2 R2 = x1.4662 x8 = 8.5966 W
2
2
The complex power supplied is
1
1
S = Vs I1* = (20 < 30o )(2.438 < 3.0661o ) = 20.43 + j13.30 VA
2
2
Adding P1 and P2 gives the real part of S, showing the conservation of power.
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Chapter 11, Problem 5.
Assuming that v s = 8 cos(2t – 40º) V in the circuit of Fig. 11.37, find the average power
delivered to each of the passive elements.
Figure 11.37
For Prob. 11.5.
Chapter 11, Solution 5.
Converting the circuit into the frequency domain, we get:
1Ω
8∠–40˚
I1Ω =
P1Ω =
+
−
2Ω
j6
–j2
8∠ − 40°
= 1.6828∠ − 25.38°
j6(2 − j2)
1+
j6 + 2 − j2
1.6828 2
1 = 1.4159 W
2
P3H = P0.25F = 0
I 2Ω =
P2Ω =
j6
1.6828∠ − 25.38° = 2.258
j6 + 2 − j2
2.258 2
2 = 5.097 W
2
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Chapter 11, Problem 6.
For the circuit in Fig. 11.38, i s = 6 cos 10 3 t A. Find the average power absorbed by the
50 Ω resistor.
Figure 11.38
For Prob. 11.6.
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Chapter 11, Solution 6.
⎯⎯
→ jω L = j103 x 20 x10−3 = j 20
1
1
40µF →
=
= − j25
jωC j10 3 x 40x10 − 6
20 mH
We apply nodal analysis to the circuit below.
Vo
+
20Ix
–
Ix
j20
6∠0o
50
–j25
10
V − 20I x
V −0
−6+ o
+ o
=0
10 + j20
50 − j25
Vo
But I x =
. Substituting this and solving for Vo 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)Vo = 6 or Vo = 6/(0.03225∠–43.99˚ = 186.05∠43.99˚
For power, all we need is the magnitude of the rms value of Ix.
Ix = 186.05/55.9 = 3.328 and Ixrms = 3.328/1.4142 = 2.353
We can now calculate the average power absorbed by the 50Ω resistor.
Pavg = (2.353)2x50 = 276.8 W.
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Chapter 11, Problem 7.
Given the circuit of Fig. 11.39, find the average power absorbed by the 10 Ω resistor.
Figure 11.39
For Prob. 11.7.
Chapter 11, Solution 7.
Applying KVL to the lefthand 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
Vo =
V1 ⎯
⎯→ V1 =
Vo
But,
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 ⎠
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Chapter 11, Problem 8.
In the circuit of Fig. 11.40, determine the average power absorbed by the 40 Ω resistor.
Figure 11.40
For Prob. 11.8.
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Chapter 11, Solution 8.
We apply nodal analysis to the following circuit.
V1 Io j20 Ω
V2
I2
j10 Ω
6∠0° A
0.5 Io
40 Ω
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
2
1⎛ 9 ⎞
1
2
⎟ (40) = 43.78 W
P = I2 R = ⎜
2 ⎝ 37 ⎠
2
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Chapter 11, Problem 9.
For the op amp circuit in Fig. 11.41, Vs = 10∠30° V rms . Find the average power
absorbed by the 20k Ω resistor.
Figure 11.41
For Prob. 11.9.
Chapter 11, Solution 9.
This is a noninverting 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 20kς resistor is
Vo
Io =
= 0.1411 + j1.491 mA
20k − j12k
P = I o 2 R = (1.4975) 2 x10−6 x 20 x103 = 44.85 mW
Chapter 11, Problem 10.
In the op amp circuit in Fig. 11.42, find the total average power absorbed by the resistors.
Figure 11.42
For Prob. 11.10.
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.
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Chapter 11, Problem 11.
For the network in Fig. 11.43, assume that the port impedance is
Z ab =
R
1+ ω R C
2
2
2
−1
∠ − tan ωRC
Find the average power consumed by the network when R = 10 kΩ , C = 200 nF , and
i = 2 sin(377t + 22º) mA.
Figure 11.43
For Prob. 11.11.
Chapter 11, Solution 11.
ω = 377 ,
R = 10 4 ,
C = 200 × 10 9
ωRC = (377)(10 4 )(200 × 10 9 ) = 0.754
tan 1 (ωRC) = 37.02°
Z ab =
10k
1 + (0.754) 2
∠  37.02° = 7.985∠  37.02° kΩ
i( t ) = 2 sin(377 t + 22°) = 2 cos(377 t − 68°) mA
I = 2 ∠  68°
2
⎛ 2 × 10  3 ⎞
⎟ (7.985∠  37.02°) × 10 3
S=
=⎜
⎜
2 ⎟⎠
⎝
S = 15.97∠  37.02° mVA
I 2rms Z ab
P = S cos(37.02) = 12.751 mW
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Chapter 11, Problem 12.
For the circuit shown in Fig. 11.44, determine the load impedance Z for maximum power
transfer (to Z). Calculate the maximum power absorbed by the load.
Figure 11.44
For Prob. 11.12.
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 5Ω resistor and the combination is in
series with the capacitor. That whole combination is in parallel with the 4Ω resistor.
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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°
ZThev = 0.8233 – j0.8642 or ZL = 0.8233 + j0.8642Ω.
We obtain VTh using the circuit below. We apply nodal analysis.
j2 Ω
I
4Ω
–j3 Ω
V2
+
o
40∠0 V
+
–
VTh
5Ω
–
V2 − 40 V2 − 40 V2 − 0
=0
+
+
4 − j3
j2
5
(0.16 + j0.12 − j0.5 + 0.2)V2 = (0.16 + j0.12 − j0.5)40
(0.5235∠ − 46.55°)V2 = (0.4123∠ − 67.17°)40
Thus,
V2 = 31.5∠–20.62˚V = 29.48 – j11.093V
I = (40 – V2)/(4 – j3) = (40 – 29.48 + j11.093)/(4 – j3)
= 15.288∠46.52˚/5∠–36.87˚ = 3.058∠83.39˚ = 0.352 + j3.038
VThev = 40 – 4I = 40 – 1.408 – j12.152 = 38.59 – j12.152V
= 40.46∠–17.479˚V
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We can check our value of VThev by letting V1 = VThev. Now we can use nodal analysis to
solve for V1.
At node 1,
V1 − 40 V1 − V2 V2 − 0
+
+
= 0 → (0.25 + j0.3333)V1 + (0.2 − j0.3333)V2 = 10
4
− j3
5
At node 2,
V2 − V1 V2 − 40
+
= 0 → − j0.3333V1 + (− j0.1667)V2 = − j20
− j3
j2
>> Z=[(0.25+0.3333i),0.3333i;0.3333i,(0.20.1667i)]
Z=
0.2500 + 0.3333i
0  0.3333i
0  0.3333i 0.2000  0.1667i
>> I=[10;20i]
I=
10.0000
0 20.0000i
>> V=inv(Z)*I
V=
38.5993 12.1459i
29.4890 11.0952i
Please note, these values check with the ones obtained above.
To calculate the maximum power to the load,
ILrms = (40.46/(2x0.8233))/1.4141 = 17.376A
Pavg = (ILrms)20.8233 = 248.58 W.
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Chapter 11, Problem 13.
The Thevenin impedance of a source is Z Th = 120 + j 60 Ω , while the peak Thevenin
voltage is VTh = 110 + j 0 V . Determine the maximum available average power from the
source.
Chapter 11, Solution 13.
For maximum power transfer to the load, ZL = 120 – j60Ω.
ILrms = 110/(240x1.4142) = 0.3241A
Pavg = ILrms2120 = 12.605 W.
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Chapter 11, Problem 14.
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 i s = 5cos 40t A .
Figure 11.45
For Prob. 11.14.
Chapter 11, Solution 14.
We find the Thevenin equivalent at the terminals of Z.
40 mF
7.5 mH
1
1
=
= j 0.625
jωC j 40 x 40 x10−3
⎯⎯
→ jω L = j 40 x7.5 x10−3 = j 0.3
⎯⎯
→
To find ZTh, consider the circuit below.
j0.3
j0.625
8Ω
12 Ω
ZTh = 8 − j 0.625 + 12 // j 0.3 = 8 − j 0.625 +
ZTh
12 x0.3
= 8.0075 − j 0.3252
12 + 0.3
ZL = (ZThev)* = 8.008 + j0.3252Ω.
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To find VTh, consider the circuit below.
j0.625
8Ω
I1
5∠0o
j0.3
12 Ω
+
VTh
–
By current division,
I1 = 5(j0.3)/(12+j0.3) = 1.5∠90˚/12.004∠1.43˚ = 0.12496∠88.57˚
= 0.003118 + j0.12492A
VThev rms = 12I1/ 2 = 1.0603∠88.57˚V
ILrms = 1.0603∠88.57˚/2(8.008) = 66.2∠88.57˚mA
Pavg = ILrms28.008 = 35.09 mW.
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Chapter 11, Problem 15.
In the circuit of Fig. 11.46, find the value of ZL that will absorb the maximum power and
the value of the maximum power.
Figure 11.46
For Prob. 11.15.
Chapter 11, Solution 15.
To find Z Th , insert a 1A current source at the load terminals as shown in Fig. (a).
1Ω
j Ω
1
2
+
2 Vo
jΩ
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 Th = 2 =
= 0.5 + j0.5
1
2
Z L = Z *Th = 0.5 − j0.5 Ω
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We now obtain VTh from Fig. (b).
1Ω
12∠0° V
j Ω
+
+
+
−
Vo
jΩ
2 Vo
Vth
−
−
(b)
12 − Vo Vo
=
1
j
 12
Vo =
1+ j
2 Vo +
– Vo − ( j × 2 Vo ) + VTh = 0
(−12)(1 − j2)
VTh = (1  j2) Vo =
1+ j
2
Pmax =
VTh
8RL
2
⎛12 5 ⎞
⎟⎟
⎜⎜
⎝ 2 ⎠
= 90 W
=
(8)(0.5)
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Chapter 11, Problem 16.
For the circuit of Fig. 11.47, find the maximum power delivered to the load ZL.
Figure 11.47
For Prob. 11.16.
Chapter 11, Solution 16.
1
1
=
= − j5
jωC j 4 x1 / 20
We find the Thevenin equivalent at the terminals of ZL. To find VTh, we use the circuit
shown below.
0.5Vo
ω = 4,
1H
jωL = j 4,
⎯
⎯→
⎯
⎯→
1 / 20F
2Ω
4Ω
V1
V2
+
+
10<0o

+
Vo

j5
j4
VTh

At node 1,
10 − V1
V
V − V2
= 1 + 0.5V1 + 1
2
− j5
4
⎯⎯→
5 = V1 (1.25 + j0.2) − 0.25V2
(1)
At node 2,
V1 − V2
V
+ 0.25V1 = 2
j4
4
⎯
⎯→
0 = 0.5V1 + V2 (−0.25 + j 0.25)
(2)
PROPRIETARY MATERIAL. © 2007 The McGrawHill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGrawHill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
Solving (1) and (2) leads to
VTh = V2 = 6.1947 + j 7.0796 = 9.4072∠48.81o
To obtain RTh, consider the circuit shown below. We replace ZL by a 1A current source.
0.5V1
2Ω
4Ω
V1
j5
V2
j4
1A
At node 1,
V − V2
V1
V
+ 1 + 0.25V1 + 1
=0 ⎯
⎯→
2 − j5
4
(3)
0 = V1 (1 + j 0.2) − 0.25V2
At node 2,
1+
V1 − V2
V
+ 0.25V1 = 2
4
j4
⎯
⎯→
− 1 = 0.5V1 + V2 (−0.25 + j 0.25)
(4)
Solving (1) and (2) gives
V
Z Th = 2 = 1.9115 + j 3.3274 = 3.8374∠60.12 o
1
Pmax =
 VTh  2
9.4072 2
=
= 5.787 W
8RTh
8 x1.9115
PROPRIETARY MATERIAL. © 2007 The McGrawHill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGrawHill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
Chapter 11, Problem 17.
Calculate the value of ZL in the circuit of Fig. 11.48 in order for ZL to receive maximum
average power. What is the maximum average power received by ZL?
Figure 11.48
For Prob. 11.17.
PROPRIETARY MATERIAL. © 2007 The McGrawHill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGrawHill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
Chapter 11, Solution 17.
We find R Th at terminals ab following Fig. (a).
j10 Ω
30 Ω
a
b
40 Ω
j20 Ω
(a)
Z Th = − j10 + 30  j20 + 40 =
(30 − j10)(40 + j20)
= 20 Ω = ZL
70 + j10
We obtain VTh from Fig. (b).
I1
I2
j10 Ω
30 Ω
j5 A
+ VTh −
40 Ω
j20 Ω
(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
8RL
2
=
5000
= 31.25 W
(8)(20)
PROPRIETARY MATERIAL. © 2007 The McGrawHill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGrawHill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.