Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 11, Problem 7 (6).
Given the circuit of Fig. 11.40, find the average power absorbed by the 10-Ω
resistor.
Figure 11.40
Chapter 11, Solution 7 (6).
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
V
10 − j5 1
But,
Vo =
Hence,
8Io +

→ V1 =
10 − j5
Vo
10
Vo
10 − j5
Vo +
=0
j50
10
I o = j0.025 Vo
Substituting (2) into (1),
8∠20° = 0.1 Vo (1 + j)
Vo =
80∠20°
1+ j
I1 =
Vo 10
=
∠ - 25°
10
2
P=
 1 100 
1
2
(10) = 250 W
I 1 R =  
 2  2 
2
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
1
Chapter 11, Problem 12(11).
For each of the circuits in Fig. 11.45, determine the value of load Z for maximum
power transfer and the maximum average power transferred.
Figure 11.45
Chapter 11, Solution 12(11).
We find Z Th using the circuit in Fig. (a).
Zth
8Ω
Z Th
-j2 Ω
(a)
(8)(-j2) 8
= 8 || -j2 =
= (1 − j4) = 0.471 − j1.882
8 − j2 17
Z L = Z *Th = 0.471 + j1.882 Ω
We find VTh using the circuit in Fig. (b).
Io
8Ω
+
-j2 Ω
Vth
(b)
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4∠0° A
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Io =
- j2
(4 ∠0°)
8 − j2
VTh = 8 I o =
- j64
8 − j2
2
Pmax =
VTh
2
8RL
 64 


 68 
=
= 15.99 W
(8)(0.471)
We obtain Z Th from the circuit in Fig. (c).
5Ω
-j3 Ω
j2 Ω
4Ω
Zth
(c)
Z Th = j2 + 5 || (4 − j3) = j2 +
(5)(4 − j3)
= 2.5 + j1.167
9 − j3
Z L = Z *Th = 2.5 − j1.167 Ω
From Fig.(d), we obtain VTh using the voltage division principle.
5Ω
-j3 Ω
j2 Ω
10∠30° V
+
-
4Ω
+
Vth
(d)
 4 − j3 
 4 − j3  10

(10 ∠ 30 °) = 
 ∠ 30 ° 
V Th = 

 3 − j  3
 9 − j3 
2
Pmax =
VTh
8RL
2
 5 10 

⋅ 
 10 3 
=
= 1.389 W
(8)(2.5)
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 11, Problem 20(17).
The load resistance RL in Fig. 11.53 is adjusted until it absorbs the maximum
average power. Calculate the value of RL and the maximum average power.
Figure 11.53
Chapter 11, Solution 20(17).
Combine j20 W and -j10 W 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
40 W
V1
-j20 W
4 Io
V2
+ -j10 W
1A
(a)
At the supernode,
1=
V1
V
V
+ 1 + 2
40 - j20 - j10
40 = (1 + j2) V1 + j4 V2
Also,
V1 = V2 + 4 I o ,
1.1 V1 = V2

→ V1 =
(1)
where I o =
- V1
40
V2
1 .1
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Substituting (2) into (1),
V 
40 = (1 + j2)  2  + j4 V2
 1 .1 
Z Th =
V2 =
44
1 + j6.4
V2
= 1.05 − j6.71 Ω
1
R L = Z Th = 6.792 Ω
To find VTh , consider the circuit in Fig. (b).
40 W
Io
V1
4 Io
V2
+ -
+
120Ð0° V
+
-
-j20 W
-j10 W
Vth
-
(b)
At the supernode,
120 − V1
V
V
= 1 + 2
40
- j20 - j10
120 = (1 + j2) V1 + j4 V2
Also,
V1 = V2 + 4 I o ,
V1 =
(3)
where I o =
120 − V1
40
V2 + 12
1 .1
(4)
Substituting (4) into (3),
109.09 − j21.82 = (0.9091 + j5.818) V2
VTh = V2 =
Pmax =
109.09 − j21.82
= 18.893∠ - 92.43°
0.9091 + j5.818
VTh
8RL
2
(18.893) 2
=
= 6.569 W
(8)(6.792)
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 11, Problem 51(38).
For the entire circuit in Fig. 11.71, calculate:
(a) the power factor
(b) the average power delivered by the source
(c) the reactive power
(d) the apparent power
(e) the complex power
Figure 11.71
Chapter 11, Solution 51(38).
Z T = 2 + (10 − j5) || (8 + j6)
ZT = 2 +
(10 − j5)(8 + j6)
110 + j20
= 2+
18 + j
18 + j
Z T = 8.152 + j0.768 = 8.188∠5.382°
pf = cos(5.382°) = 0.9956 (lagging)
2
V
1
(16) 2
S = V I* =
=
2
2 Z * (2)(8.188∠ - 5.382°)
S = 15.63∠5.382°
P = S cos θ = 15.56 W
Q = S sin θ = 1.466 VAR
S = S = 15.63 VA
S = 15.63∠5.382° = 15.56 + j1.466 VA
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 11, Problem 57(42).
For the circuit in Fig. 11.77, find the average, reactive, and complex power
delivered by the dependent voltage source.
Figure 11.77
Chapter 11, Solution 57(42).
4Ω
24∠0° V
+
-
Vo
-j1 Ω
V1
2Ω
+
1Ω
j2 Ω
2 Vo
V2
-
At node o,
At node 1,
24 − Vo Vo Vo − V1
=
+
4
1
-j
24 = (5 + j4) Vo − j4 V1
Vo − V1
V
+ 2 Vo = 1
-j
j2
V1 = (2 − j4) Vo
(1)
(2)
Substituting (2) into (1),
24 = (5 + j4 − j8 − 16) Vo
Vo =
- 24
,
11 + j4
V1 =
(-24)(2 - j4)
11 + j4
The voltage across the dependent source is
V2 = V1 + (2)(2 Vo ) = V1 + 4 Vo
- 24
(-24)(6 − j4)
V2 =
⋅ (2 − j4 + 4) =
11 + j4
11 + j4
1
1
(-24)(6 − j4) - 24  576 
(6 − j4)
S=
⋅
=
S = V2 I * = V2 (2 Vo* )
2
2
11 + j4
11 - j4  137 
S = 25.23 − j16.82 VA
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 11, Problem 60(45).
For the circuit in Fig. 11.80, find Vo and the input power factor.
Figure 11.70
Chapter 11, Solution 60(45).
S1 = 20 + j
20
sin(cos -1 (0.8)) = 20 + j15
0.8
S 2 = 16 + j
16
sin(cos -1 (0.9)) = 16 + j7.749
0.9
S = S1 + S 2 = 36 + j22.749 = 42.585∠32.29°
But
Vo =
S = Vo I * = 6 Vo
S
= 7.098 ∠ 32.29°
6
pf = cos(32.29°) = 0.8454 (lagging)
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 11, Problem 62(47).
For the circuit in Fig. 11.82, find Vs.
Figure 11.72
Chapter 11, Solution 62(47).
0.2 + j0.04 Ω
I
I2
0.3 + j0.15 Ω
I1
Vs
+
-
S 2 = 15 − j
But
But
+
+
V1
V2
-
-
15
sin(cos -1 (0.8)) = 15 − j11.25
0.8
S 2 = V2 I *2
S 2 15 − j11.25
I *2 =
=
V2
120
I 2 = 0.125 + j0.09375
V1 = V2 + I 2 (0.3 + j0.15)
V1 = 120 + (0.125 + j0.09375)(0.3 + j0.15)
V1 = 120.02 + j0.0469
10
S1 = 10 + j
sin(cos -1 (0.9)) = 10 + j4.843
0.9
S1 11.111∠25.84°
S1 = V1 I 1*
I 1* =
=
V1 120.02 ∠0.02°
I 1 = 0.093∠ - 25.82° = 0.0837 − j0.0405
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
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 11, Problem 74(57).
A 120-V rms 60-Hz source supplies two loads connected in parallel, as shown in
Fig. 11.90.
(a) Find the power factor of the parallel combination.
(b) Calculate the value of the capacitance connected in parallel that will raise the
power factor to unity.
Figure 11.90
Chapter 11, Solution 74(57).
P1
24
=
= 30 kVA
cos θ1 0.8
Q1 = S1 sin θ1 = (30)(0.6) = 18 kVAR
S1 = 24 + j18 kVA
θ1 = cos -1 (0.8) = 36.87°
S1 =
θ 2 = cos -1 (0.95) = 18.19° S 2 =
P2
40
=
= 42.105 kVA
cos θ 2 0.95
Q 2 = S 2 sin θ 2 = 13.144 kVAR
S 2 = 40 + j13.144 kVA
S = S1 + S 2 = 64 + j31.144 kVA
 31.144 
 = 25.95°
θ = tan -1 
pf = cos θ = 0.8992
 64 
θ 2 = 25.95° ,
θ1 = 0°
Q c = P [ tan θ 2 − tan θ1 ] = 64 [ tan(25.95°) − 0 ] = 31.144 kVAR
Qc
31,144
C=
=
= 5.74 mF
2
ω Vrms (2π )(60)(120) 2
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