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
P= Vm I m cos(θ v − θi ) = (160)(20) cos(60°)
2 2
P = 800 W
Chapter 11, Solution 2.
First, transform the circuit to the frequency domain.
30 cos(500t ) → 30 ∠0° , ω = 500
0.3 H
→ jωL = j150
1 -j
20µF
→ = = - j100
jωC (500)(20)(10 -6 )
I I2 -j100 Ω
I1
+
30∠0° V j150 Ω 200 Ω
−
30∠0°
I1 = = 0.2∠ − 90° = - j0.2
j150
i1 ( t ) = 0.2 cos(500 t − 90°) = 0.2 sin(500 t )
30∠0° 0.3
I2 = = = 0.1342∠26.56° = 0.12 + j0.06
200 − j100 2 − j
i 2 ( t ) = 0.1342 cos(500 t + 25.56°)
I = I 1 + I 2 = 0.12 − j0.14 = 0.1844 ∠ - 49.4°
i( t ) = 0.1844 cos(500t − 35°)
For the voltage source,
p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.1844 cos(500t − 35°) ]
At t = 2 s , p = 5.532 cos(1000) cos(1000 − 35°)
p = (5.532)(0.5624)(0.935)
p = 2.91 W
For the inductor,
p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.2 sin(500t ) ]
At t = 2 s , p = 6 cos(1000) sin(1000)
p = (6)(0.5624)(0.8269)
p = 2.79 W
For the capacitor,
Vc = I 2 (- j100) = 13.42∠ - 63.44°
p( t ) = v( t ) i( t ) = [13.42 cos(500 − 63.44°) ] × [ 0.1342 cos(500t + 25.56°)
At t = 2 s , p = 18 cos(1000 − 63.44°) cos(1000 + 26.56°)
p = (18)(0.991)(0.1329)
p = 2.37 W
For the resistor,
VR = 200 I 2 = 26.84 ∠25.56°
p( t ) = v( t ) i( t ) = [ 26.84 cos(500t + 26.56°) ] × [ 0.1342 cos(500t + 26.56°) ]
At t = 2 s , p = 3.602 cos 2 (1000 + 25.56°)
p = (3.602)(0.1329 2
p = 0.0636 W
Chapter 11, Solution 3.
10 cos(2t + 30°)
→ 10∠30° , ω= 2
1H
→ jωL = j2
1
0.25 F
→ = -j2
jωC
I 4Ω I1 2Ω
I2
+
10∠30° V j2 Ω -j2 Ω
−
( j2)(2 − j2)
j2 || (2 − j2) = = 2 + j2
2
10 ∠30°
I= = 1.581∠11.565°
4 + 2 + j2
j2
I1 = I = j I = 1.581∠101.565°
2
2 − j2
I2 = I = 2.236 ∠56.565°
2
For the source,
1
S = V I* = (10∠30°)(1.581∠ - 11.565°)
2
S = 7.905∠18.43° = 7.5 + j2.5
The average power supplied by the source = 7.5 W
For the 4-Ω resistor, the average power absorbed is
1 2 1
P = I R = (1.581) 2 (4) = 5 W
2 2
For the inductor,
1 2 1
S = I 2 Z L = (2.236) 2 ( j2) = j5
2 2
The average power absorbed by the inductor = 0 W
For the 2-Ω resistor, the average power absorbed is
1 2 1
P = I 1 R = (1.581) 2 (2) = 2.5 W
2 2
For the capacitor,
1 2 1
S= I 1 Z c = (1.581) 2 (- j2) = - j2.5
2 2
The average power absorbed by the capacitor = 0 W
Chapter 11, Solution 4.
20 Ω 10 Ω
+ I1 I2
50 V -j10 Ω j5 Ω
−
For mesh 1,
50 = (20 − j10) I 1 + j10 I 2
5 = (2 − j) I 1 + j I 2 (1)
For mesh 2,
0 = (10 + j5 − j10) I 2 + j10 I 1
0 = (2 − j) I 2 + j2 I 1 (2)
In matrix form,
5 2 − j j I 1
0 = j2 2 − j I
2
∆ = 5 − j4 , ∆ 1 = 5 (2 − j) , ∆ 2 = -j10
∆ 1 5 (2 − j)
I1 = = = 1.746∠12.1°
∆ 5 − j4
∆ 2 - j10
I2 = = = 1.562 ∠128.66°
∆ 5 - j4
For the source,
1
S= V I 1 = 43.65∠ - 12.1°
*
2
The average power supplied = 43.65 cos(12.1°) = 42.68 W
For the 20-Ω resistor,
1 2
P = I 1 R = 30.48 W
2
For the inductor and capacitor,
P=0W
For the 10-Ω resistor,
1 2
P = I 2 R = 12.2 W
2
Chapter 11, Solution 5.
Converting the circuit into the frequency domain, we get:
1Ω 2Ω
+ –j2
8∠–40˚ − j6
8∠ − 40°
I1Ω = = 1.6828∠ − 25.38°
j6(2 − j2)
1+
j6 + 2 − j2
1.6828 2
P1Ω = 1 = 1.4159 W
2
P3H = P0.25F = 0
j6
I 2Ω = 1.6828∠ − 25.38° = 2.258
j6 + 2 − j2
2.258 2
P2Ω = 2 = 5.097 W
2
Chapter 11, Solution 6.
20 Ω 10 Ω
+ I1 I2
50 V -j10 Ω j5 Ω
−
For mesh 1,
(4 + j2) I 1 − j2 (4 ∠60°) + 4 Vo = 0 (1)
Vo = 2 (4 ∠60° − I 2 ) (2)
For mesh 2,
(2 − j) I 2 − 2 (4∠60°) − 4Vo = 0 (3)
Substituting (2) into (3),
(2 − j) I 2 − 8∠60° − 8 (4 ∠60° − I 2 ) = 0
40∠60°
I2 =
10 − j
Hence,
40∠60° - j8∠60°
Vo = 2 4 ∠60° − =
10 − j 10 − j
Substituting this into (1),
j32 ∠60° 14 − j
(4 + j2) I 1 = j8∠60° + = ( j8∠60°)
10 − j 10 − j
(4∠60°)(1 + j14)
I1 = = 2.498∠125.06°
21 + j8
1 2 1
P4 = I 1 R = (2.498) 2 (4) = 12.48 W
2 2
Chapter 11, Solution 7.
20 Ω 10 Ω
+ I1 I2
50 V -j10 Ω j5 Ω
−
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 10 − j5
But, Vo = V
→ V1 = Vo
10 − j5 1 10
10 − j5 Vo
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
Vo 10
I1 = = ∠ - 25°
10 2
1 2 1 100
P= I 1 R = (10) = 250 W
2 2 2
Chapter 11, Solution 8.
We apply nodal analysis to the following circuit.
V1 Io -j20 Ω V2
I2
6∠0° A j10 Ω 0.5 Io 40 Ω
At node 1,
V1 V1 − V2
6= + V1 = j120 − V2 (1)
j10 - j20
At node 2,
V2
0 .5 I o + I o =
40
V1 − V2
But, Io =
- j20
1.5 (V1 − V2 ) V2
Hence, =
- j20 40
3V1 = (3 − j) V2 (2)
Substituting (1) into (2),
j360 − 3V2 − 3V2 + j V2 = 0
j360 360
V2 = = (-1 + j6)
6 − j 37
V2 9
I2 = = (-1 + j6)
40 37
2
1 2 1 9
P = I2 R = (40) = 43.78 W
2 2 37
Chapter 11, Solution 9.
6
Vo = 1 + Vs = (4)(2) = 8 V rms
2
Vo2 64
P10 = = mW = 6.4 mW
R 10
The current through the 2 -kΩ resistor is
Vs
= 1 mA
2k
P2 = I 2 R = 2 mW
Similarly,
P6 = I 2 R = 6 mW
Chapter 11, Solution 10.
No current flows through each of the resistors. Hence, for each resistor,
P = 0 W.
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°
10k
Z ab = ∠ - 37.02° = 6.375∠ - 37.02° kΩ
1 + (0.754) 2
i( t ) = 2 sin(377 t + 22°) = 2 cos(377 t − 68°) mA
I = 2 ∠ - 68°
2
2 × 10 -3
S= I 2
Z ab = (6.375∠ - 37.02°) × 10 3
rms
2
S = 12.751∠ - 37.02° mVA
P = S cos(37.02) = 10.181 mW
Chapter 11, Solution 12.
(a) We find Z Th using the circuit in Fig. (a).
Zth
8Ω -j2 Ω
(a)
(8)(-j2) 8
Z Th = 8 || -j2 = = (1 − j4) = 0.471 − j1.882
8 − j2 17
Z L = Z * = 0.471 + j1.882 Ω
Th
We find VTh using the circuit in Fig. (b).
Io
+
8Ω Vth -j2 Ω 4∠0° A
−
(b)
- j2
Io = (4∠0°)
8 − j2
- j64
VTh = 8 I o =
8 − j2
2
64
2
VTh 68
Pmax = = = 15.99 W
8RL (8)(0.471)
(b) We obtain Z Th from the circuit in Fig. (c).
5Ω -j3 Ω
j2 Ω
Zth 4Ω
(c)
(5)(4 − j3)
Z Th = j2 + 5 || (4 − j3) = j2 + = 2.5 + j1.167
9 − j3
Z L = Z * = 2.5 − j1.167 Ω
Th
Chapter 11, Solution 13.
(a) We find Z Th at the load terminals using the circuit in Fig. (a).
j100 Ω
Zth
80 Ω -j40 Ω
(a)
(-j40)(80 + j100)
Z Th = -j40 || (80 + j100) = = 51.2 − j1.6
80 + j60
Z L = Z * = 51.2 + j1.6 Ω
Th
(b) We find VTh at the load terminals using Fig. (b).
Io j100 Ω
+
3∠20° A 80 Ω -j40 Ω Vth
−
(b)
80 (8)(3∠20°)
Io = (3∠20°) =
80 + j100 − j40 8 + j6
(- j40)(24∠20°)
VTh = - j40 I o =
8 + j6
2
40
2 ⋅ 24
VTh 10
Pmax = = = 22.5 W
8RL (8)(51.2)
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
VTh = (10∠30°) = ∠30°
9 − j3 3 − j 3
2
5 10
2 ⋅
VTh 10 3
Pmax = = = 1.389 W
8RL (8)(2.5)
Chapter 11, Solution 14.
j24 Ω –j10 Ω
I
+
16 Ω
VTh ZTh
40∠90º A 10 Ω
j8 Ω _
(10 + j24)(16 + j8)
Z Th = − j10 + = − j10 + 8.245 + j7.7 = 8.245 − j2.3Ω
10 + j24 + 16 + j8
Z = Z∗ = 8.245 + j2.3Ω
Th
10
VTh = I(16 + j8) = j40(16 + j8)
10 + j24 + 16 + j8
= 173.55∠65.66° = 71.53 + j158.12 V
2
VTh
2
Pmax = I 2 8.245 = 2 8.245 = 456.6 W
rms
(2x8.245) 2
Chapter 11, Solution 15.
To find Z Th , insert a 1-A current source at the load terminals as shown in Fig. (a).
1Ω 1 -j Ω 2
+
Vo jΩ 2 Vo 1A
−
(a)
At node 1,
Vo Vo V2 − Vo
+ =
→ Vo = j V2 (1)
1 j -j
At node 2,
V2 − Vo
1 + 2 Vo =
→ 1 = j V2 − (2 + j) Vo (2)
-j
Substituting (1) into (2),
1 = j V2 − (2 + j)( j) V2 = (1 − j) V2
1
V2 =
1− j
V2 1 + j
VTh = = = 0.5 + j0.5
1 2
Z L = Z * = 0.5 − j0.5 Ω
Th
We now obtain VTh from Fig. (b).
1Ω -j Ω
+ +
+
12∠0° V Vo jΩ 2 Vo Vth
−
− −
(b)
12 − Vo Vo
2 Vo + =
1 j
- 12
Vo =
1+ j
Vo − (- j × 2 Vo ) + VTh = 0
(12)(1 + j2)
VTh = -(1 + j2)Vo =
1+ j
2
12 5
2
VTh 2
Pmax = = = 90 W
8RL (8)(0.5)
Chapter 11, Solution 16.
1 1
ω = 4, 1H
→ jωL = j 4, 1 / 20F
→ = = − 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
2Ω V1 4Ω V2
+
+ +
10At node 1,
10 − V1 V V − V2
= 1 + 0.25V1 + 1
→ 5 = V1 (1 + j 0.2) − 0.25V2 (1)
2 − j5 4
At node 2,
V1 − V2 V
+ 0.25V1 = 2
→ 0 = 0.5V1 + V2 (−0.25 + j 0.25) (2)
4 j4
Solving (1) and (2) leads to
VTh = V2 = 6.1947 + j 7.0796 = 9.4072∠48.81o
Chapter 11, Solution 17.
We find R Th at terminals a-b following Fig. (a).
-j10 Ω 30 Ω
a b
40 Ω j20 Ω
(a)
(30)( j20) (40)(-j10)
Z Th = 30 || j20 + 40 || (- j10) = +
30 + j20 40 − j10
Z Th = 9.23 + j13.85 + 2.353 − j9.41
Z Th = 11.583 + j4.44 Ω
Z L = Z * = 11.583 − j4.44 Ω
Th
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
2
VTh 5000
Pmax = = = 53.96 W
8RL (8)(11.583)
Chapter 11, Solution 18.
We find Z Th at terminals a-b as shown in the figure below.
40 Ω -j10 Ω
40 Ω 80 Ω
a Zth
j20 Ω
b
(80)(-j10)
Z Th = j20 + 40 || 40 + 80 || (-j10) = j20 + 20 +
80 − j10
Z Th = 21.23 + j10.154
Z L = Z * = 21.23 − j10.15 Ω
Th
Chapter 11, Solution 19.
At the load terminals,
(6)(3 + j)
Z Th = - j2 + 6 || (3 + j) = -j2 +
9+ j
Z Th = 2.049 − j1.561
R L = Z Th = 2.576 Ω
To get VTh , let Z = 6 || (3 + j) = 2.049 + j0.439 .
By transforming the current sources, we obtain
VTh = (4 ∠0°) Z = 8.196 + j1.756
2
VTh 70.258
Pmax = = = 3.409 W
8RL 20.608
Chapter 11, Solution 20.
Combine j20 Ω and -j10 Ω 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).
4 Io
Io 40 Ω V1 V2
+ −
-j20 Ω -j10 Ω 1A
(a)
At the supernode,
V1 V V
1= + 1 + 2
40 - j20 - j10
40 = (1 + j2) V1 + j4 V2 (1)
- V1
Also, V1 = V2 + 4 I o , where I o =
40
V2
1.1 V1 = V2
→ V1 = (2)
1 .1
Substituting (2) into (1),
V
40 = (1 + j2) 2 + j4 V2
1 .1
44
V2 =
1 + j6.4
V2
Z Th = = 1.05 − j6.71 Ω
1
R L = Z Th = 6.792 Ω
To find VTh , consider the circuit in Fig. (b).
4 Io
Io 40 Ω V1 V2
+ −
+
+
120∠0° V -j20 Ω -j10 Ω Vth
−
−
(b)
At the supernode,
120 − V1 V V
= 1 + 2
40 - j20 - j10
120 = (1 + j2) V1 + j4 V2 (3)
120 − V1
Also, V1 = V2 + 4 I o , where I o =
40
V2 + 12
V1 = (4)
1 .1
Substituting (4) into (3),
109.09 − j21.82 = (0.9091 + j5.818) V2
109.09 − j21.82
VTh = V2 = = 18.893∠ - 92.43°
0.9091 + j5.818
2
VTh (18.893) 2
Pmax = = = 6.569 W
8RL (8)(6.792)
Chapter 11, Solution 21.
We find Z Th at terminals a-b, as shown in the figure below.
100 Ω -j10 Ω
a
40 Ω
Zth
50 Ω
j30 Ω
b
Z Th = 50 || [ - j10 + 100 || (40 + j30) ]
(100)(40 + j30)
where 100 || (40 + j30) = = 31.707 + j14.634
140 + j30
(50)(31.707 + j4.634)
Z Th = 50 || (31.707 + j4.634) =
81.707 + j4.634
Z Th = 19.5 + j1.73
R L = Z Th = 19.58 Ω
Chapter 11, Solution 22.
i (t ) = 4 sin t , 0Chapter 11, Solution 23.
15, 0 < t < 2
v( t ) =
5, 2 < t < 6
Vrms =
2
1
6
[ ∫ 15
0
2
2
dt + ∫2 5 2 dt =
6
] 550
6
Vrms = 9.574 V
Chapter 11, Solution 24.
5, 0 < t < 1
T = 2, v( t ) =
- 5, 1 < t < 2
Vrms =
2
1
2
[∫ 5
0
1
2
dt + ∫1 (-5) 2 dt =
2
] 25
2
[1 + 1] = 25
Vrms = 5 V
Chapter 11, Solution 25.
2 1 T 2
f rms =
T
1 1 2
[2 3 2
∫ 0 f ( t )dt = 3 ∫ 0 (−4) dt + ∫ 1 0dt + ∫2 4 dt ]
1 32
= [16 + 0 + 16] =
3 3
32
f rms = = 3.266
3
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