Chapter 10, Solution 1.
ω=1
10 cos( t − 45°)
→ 10∠ - 45°
5 sin( t + 30°)
→ 5∠ - 60°
1H
→ jωL = j
1
1F
→ = -j
jωC
The circuit becomes as shown below.
3Ω Vo jΩ
+ +
10∠-45° V 2 Io 5∠-60° V
− −
Applying nodal analysis,
(10∠ - 45°) − Vo (5∠ - 60°) − Vo Vo
+ =
3 j -j
j10∠ - 45° + 15∠ - 60° = j Vo
Vo = 10 ∠ - 45° + 15∠ - 150° = 15.73∠247.9°
Therefore, v o ( t ) = 15.73 cos(t + 247.9°) V
Chapter 10, Solution 2.
ω = 10
4 cos(10t − π 4)
→ 4∠ - 45°
20 sin(10 t + π 3)
→ 20 ∠ - 150°
1H
→ jωL = j10
1 1
0.02 F
→ = = - j5
jωC j 0.2
The circuit becomes that shown below.
10 Ω Vo
Io
+
20∠-150° V j10 Ω -j5 Ω 4∠-45° A
−
Applying nodal analysis,
(20∠ - 150°) − Vo Vo Vo
+ 4∠ - 45° = +
10 j10 - j5
20 ∠ - 150° + 4∠ - 45° = 0.1(1 + j) Vo
Vo 2 ∠ - 150° + 4 ∠ - 45°
Io = = = 2.816 ∠150.98°
j10 j (1 + j)
Therefore, i o ( t ) = 2.816 cos(10t + 150.98°) A
Chapter 10, Solution 3.
ω= 4
2 cos(4t )
→ 2∠0°
16 sin(4 t )
→ 16∠ - 90° = -j16
2H
→ jωL = j8
1 1
1 12 F
→ = = - j3
jωC j (4)(1 12)
The circuit is shown below.
4Ω -j3 Ω Vo j8 Ω 6Ω
+
-j16 V 1Ω 2∠0° A
−
Applying nodal analysis,
- j16 − Vo Vo Vo
+2= +
4 − j3 1 6 + j8
- j16 1 1
+ 2 = 1 + + V
4 − j3 4 − j3 6 + j8 o
3.92 − j2.56 4.682∠ - 33.15°
Vo = = = 3.835∠ - 35.02°
1.22 + j0.04 1.2207 ∠1.88°
Therefore, v o ( t ) = 3.835 cos(4t – 35.02°) V
Chapter 10, Solution 4.
16 sin(4 t − 10°)
→ 16∠ - 10°, ω = 4
1H
→ jωL = j4
1 1
0.25 F
→ = = -j
jωC j (4)(1 4)
Ix j4 Ω V1 -j Ω
+
+
16∠-10° V 0.5 Ix 1Ω Vo
−
−
(16∠ - 10°) − V1 1 V
+ Ix = 1
j4 2 1− j
But
(16∠ - 10°) − V1
Ix =
j4
3 ((16∠ - 10°) − V1 ) V
So, = 1
j8 1− j
48∠ - 10°
V1 =
- 1 + j4
Using voltage division,
1 48∠ - 10°
Vo = V1 = = 8.232∠ - 69.04°
1− j (1 - j)(-1 + j4)
Therefore, v o ( t ) = 8.232 sin(4t – 69.04°) V
Chapter 10, Solution 5.
Let the voltage across the capacitor and the inductor be Vx and we get:
Vx − 0.5I x − 10∠30° Vx Vx
+ + =0
4 − j2 j3
Vx
(3 + j6 − j4)Vx − 1.5I x = 30∠30° but I x = = j0.5Vx
− j2
Combining these equations we get:
30∠30°
(3 + j2 − j0.75)Vx = 30∠30° or Vx =
3 + j1.25
30∠30°
I x = j0.5 = 4.615∠97.38° A
3 + j1.25
Chapter 10, Solution 6.
Let Vo be the voltage across the current source. Using nodal analysis we get:
Vo − 4Vx Vo 20
−3+ = 0 where Vx = Vo
20 20 + j10 20 + j10
Combining these we get:
Vo 4Vo Vo
− −3+ = 0 → (1 + j0.5 − 3)Vo = 60 + j30
20 20 + j10 20 + j10
60 + j30 20(3)
Vo = or Vx = = 29.11∠–166˚ V.
− 2 + j0.5 − 2 + j0.5
Chapter 10, Solution 7.
At the main node,
120∠ − 15 o − V V V 115.91 − j31.058
= 6∠30 o + +
→ − 5.196 − j3 =
40 + j20 − j30 50 40 + j20
1 j 1
V
40 + j20 + 30 + 50
− 3.1885 − j4.7805
V= = 124.08∠ − 154 o V
0.04 + j0.0233
Chapter 10, Solution 8.
ω = 200,
100mH
→ jωL = j200x 0.1 = j20
1 1
50µF
→ = = − j100
jωC j200x 50x10 − 6
The frequency-domain version of the circuit is shown below.
0.1 Vo
40 Ω
V1 Io V2
+ -j100 Ω
6∠15 o
20 Ω Vo j20 Ω
-
At node 1,
V V1 V − V2
6∠15 o + 0.1V1 = 1 + + 1
20 − j100 40
or 5.7955 + j1.5529 = (−0.025 + j 0.01)V1 − 0.025V2 (1)
At node 2,
V1 − V2 V
= 0.1V1 + 2 → 0 = 3V1 + (1 − j2)V2 (2)
40 j20
From (1) and (2),
(−0.025 + j0.01) − 0.025 V1 (5.7955 + j1.5529)
= or AV = B
3 (1 − j2) V2
0
Using MATLAB,
V = inv(A)*B
leads to V1 = −70.63 − j127.23, V2 = −110.3 + j161.09
V − V2
Io = 1 = 7.276∠ − 82.17 o
40
Thus,
i o ( t ) = 7.276 cos(200 t − 82.17 o ) A
Chapter 10, Solution 9.
10 cos(10 3 t )
→ 10 ∠0°, ω = 10 3
10 mH
→ jωL = j10
1 1
50 µF
→ = = - j20
jωC j (10 )(50 × 10 -6 )
3
Consider the circuit shown below.
20 Ω V1 -j20 Ω V2 j10 Ω
Io
+
+
10∠0° V 20 Ω 4 Io 30 Ω Vo
−
−
At node 1,
10 − V1 V1 V1 − V2
= +
20 20 - j20
10 = (2 + j) V1 − jV2 (1)
At node 2,
V1 − V2 V V2 V
= (4) 1 + , where I o = 1 has been substituted.
- j20 20 30 + j10 20
(-4 + j) V1 = (0.6 + j0.8) V2
0.6 + j0.8
V1 = V2 (2)
-4+ j
Substituting (2) into (1)
(2 + j)(0.6 + j0.8)
10 = V2 − jV2
-4+ j
170
or V2 =
0.6 − j26.2
30 3 170
Vo = V2 = ⋅ = 6.154 ∠70.26°
30 + j10 3 + j 0.6 − j26.2
Therefore, v o ( t ) = 6.154 cos(103 t + 70.26°) V
Chapter 10, Solution 10.
50 mH
→ jωL = j2000x50 x10 − 3 = j100, ω = 2000
1 1
2µF
→ = = − j250
jωC j2000 x 2x10 − 6
Consider the frequency-domain equivalent circuit below.
V1 -j250 V2
36At node 1,
V1 V V − V2
36 = + 1 + 1
→ 36 = (0.0005 − j0.006)V1 − j0.004V2 (1)
2000 j100 − j250
At node 2,
V1 − V2 V
= 0.1V1 + 2 → 0 = (0.1 − j0.004)V1 + (0.00025 + j0.004)V2 (2)
− j250 4000
Solving (1) and (2) gives
Vo = V2 = −535.6 + j893.5 = 8951.1∠93.43o
vo (t) = 8.951 sin(2000t +93.43o) kV
Chapter 10, Solution 11.
cos(2t )
→ 1∠0°, ω = 2
8 sin( 2t + 30°)
→ 8∠ - 60°
1 1
1H
→ jωL = j2 12F
→ = = -j
jωC j (2)(1 2)
1 1
2H
→ jωL = j4 14F
→ = = - j2
jωC j (2)(1 4)
Consider the circuit below.
2 Io 2 Io
2 -j Ω
2 Io
2
2
2 Io 2 2 Io
I -j Ω
At node 1,
(8∠ - 60°) − V1 V1 V1 − V2
= +
2 -j j2
8∠ - 60° = (1 + j) V1 + j V2 (1)
At node 2,
V1 − V2 (8∠ - 60°) − V2
1+ + =0
j2 j4 − j2
V2 = 4 ∠ - 60° + j + 0.5 V1 (2)
Substituting (2) into (1),
1 + 8∠ - 60° − 4 ∠30° = (1 + j1.5) V1
1 + 8∠ - 60° − 4∠30°
V1 =
1 + j1.5
V1 1 + 8∠ - 60° − 4 ∠30°
Io = = = 5.024∠ - 46.55°
-j 1.5 − j
Therefore, i o ( t ) = 5.024 cos(2t – 46.55°)
Chapter 10, Solution 12.
20 sin(1000t )
→ 20 ∠0°, ω = 1000
10 mH
→ jωL = j10
1 1
50 µF
→ = = - j20
jωC j (10 )(50 × 10 -6 )
3
The frequency-domain equivalent circuit is shown below.
2 Io
V1 10 Ω V2
Io
20∠0° A 20 Ω -j20 Ω j10 Ω
At node 1,
V1 V1 − V2
20 = 2 I o + + ,
20 10
where
V2
Io =
j10
2V2 V1 V1 − V2
20 = + +
j10 20 10
400 = 3V1 − (2 + j4) V2 (1)
At node 2,
2V2 V1 − V2 V V
+ = 2 + 2
j10 10 - j20 j10
j2 V1 = (-3 + j2) V2
or V1 = (1 + j1.5) V2 (2)
Substituting (2) into (1),
400 = (3 + j4.5) V2 − (2 + j4) V2 = (1 + j0.5) V2
400
V2 =
1 + j0.5
V2 40
Io = = = 35.74 ∠ - 116.6°
j10 j (1 + j0.5)
Therefore, i o ( t ) = 35.74 sin(1000t – 116.6°) A
Chapter 10, Solution 13.
Nodal analysis is the best approach to use on this problem. We can make our work easier
by doing a source transformation on the right hand side of the circuit.
–j2 Ω 18 Ω j6 Ω
+
+ +
40∠30º V − Vx 3Ω 50∠0º V −
−
Vx − 40∠30° Vx Vx − 50
+ + =0
− j2 3 18 + j6
which leads to Vx = 29.36∠62.88˚ A.
Chapter 10, Solution 14.
At node 1,
0 − V1 0 − V1 V2 − V1
+ + = 20∠30°
- j2 10 j4
- (1 + j2.5) V1 − j2.5 V2 = 173.2 + j100 (1)
At node 2,
V2 V2 V2 − V1
+ + = 20∠30°
j2 - j5 j4
- j5.5 V2 + j2.5 V1 = 173.2 + j100 (2)
Equations (1) and (2) can be cast into matrix form as
1 + j2.5 j2.5 V1 - 200 ∠30°
=
j2.5
- j5.5 V2 200 ∠30°
1 + j2.5 j2.5
∆= = 20 − j5.5 = 20.74∠ - 15.38°
j2.5 - j5.5
- 200 ∠30° j2.5
∆1 = = j3 (200∠30°) = 600∠120°
200 ∠30° - j5.5
1 + j2.5 - 200∠30°
∆2 = = (200 ∠30°)(1 + j5) = 1020∠108.7°
j2.5 200∠30°
∆1
V1 = = 28.93∠135.38°
∆
∆2
V2 = = 49.18∠124.08°
∆
Chapter 10, Solution 15.
We apply nodal analysis to the circuit shown below.
5A
2Ω V1 jΩ V2
I
+
-j20 V -j2 Ω 2I 4Ω
−
At node 1,
- j20 − V1 V V − V2
= 5+ 1 + 1
2 - j2 j
- 5 − j10 = (0.5 − j0.5) V1 + j V2 (1)
At node 2,
V1 − V2 V2
5 + 2I + = ,
j 4
V1
where I =
- j2
5
V2 = V1
0.25 − j
(2)
Substituting (2) into (1),
j5
- 5 − j10 − = 0.5 (1 − j) V1
0.25 − j
j40
(1 − j) V1 = -10 − j20 −
1 − j4
160 j40
( 2 ∠ - 45°) V1 = -10 − j20 + −
17 17
V1 = 15.81∠313.5°
V1
I= = (0.5∠90°)(15.81∠313.5°)
- j2
I = 7.906∠43.49° A
Chapter 10, Solution 16.
At node 1,
V1 V1 − V2 V1 − V2
j2 = + +
20 10 - j5
j40 = (3 + j4) V1 − (2 + j4) V2
At node 2,
V1 − V2 V1 − V2 V
+ +1+ j = 2
10 - j5 j10
10 (1 + j) = - (1 + j2) V1 + (1 + j) V2
Thus,
j40 3 + j4 - 2 (1 + j2) V1
=
10 (1 +
j) - (1 + j2)
1 + j V2
3 + j4 - 2 (1 + j2)
∆= = 5 − j = 5.099 ∠ - 11.31°
- (1 + j2) 1+ j
j40 - 2 (1 + j2)
∆1 = = −60 + j100 = 116.62 ∠120.96°
10 (1 + j) 1+ j
3 + j4 j40
∆2 = = -90 + j110 = 142.13∠129.29°
- (1 + j2) 10 (1 + j)
∆1
V1 = = 22.87∠132.27° V
∆
∆2
V2 = = 27.87∠140.6° V
∆
Chapter 10, Solution 17.
Consider the circuit below.
j4 Ω 1Ω
Io 2Ω
+
100∠20° V V1 V2
−
3Ω -j2 Ω
At node 1,
100∠20° − V1 V1 V1 − V2
= +
j4 3 2
V1
100 ∠20° = (3 + j10) − j2 V2 (1)
3
At node 2,
100∠20° − V2 V1 − V2 V2
+ =
1 2 - j2
100 ∠20° = -0.5 V1 + (1.5 + j0.5) V2 (2)
From (1) and (2),
100∠20° - 0.5 0.5 (3 + j) V1
=
100∠20° 1 + j10 3
- j2 V2
- 0.5 1.5 + j0.5
∆= = 0.1667 − j4.5
1 + j10 3 - j2
100∠20° 1.5 + j0.5
∆1 = = -55.45 − j286.2
100∠20° - j2
- 0.5 100∠20°
∆2 = = -26.95 − j364.5
1 + j10 3 100∠20°
∆1
V1 = = 64.74 ∠ - 13.08°
∆
∆2
V2 = = 81.17 ∠ - 6.35°
∆
V1 − V2 ∆ 1 − ∆ 2 - 28.5 + j78.31
Io = = =
2 2∆ 0.3333 − j 9
I o = 9.25∠-162.12°
Chapter 10, Solution 18.
Consider the circuit shown below.
V1 8Ω j6 Ω V 4Ω j5 Ω
2
+ +
4∠45° A 2Ω Vx 2 Vx -j Ω -j2 Ω Vo
− −
At node 1,
V1 V1 − V2
4∠45° = +
2 8 + j6
200 ∠45° = (29 − j3) V1 − (4 − j3) V2 (1)
At node 2,
V1 − V2 V V2
+ 2Vx = 2 + , where Vx = V1
8 + j6 - j 4 + j5 − j2
(104 − j3) V1 = (12 + j41) V2
12 + j41
V1 = V (2)
104 − j3 2
Substituting (2) into (1),
(12 + j41)
200∠45° = (29 − j3) V − (4 − j3) V2
104 − j3 2
200 ∠45° = (14.21∠89.17°) V2
200∠45°
V2 =
14.21∠89.17°
- j2 - j2 - 6 − j8
Vo = V2 = V2 = V2
4 + j5 − j2 4 + j3 25
10∠233.13° 200∠45°
Vo = ⋅
25 14.21∠89.17°
Vo = 5.63∠189° V
Chapter 10, Solution 19.
We have a supernode as shown in the circuit below.
j2 Ω
V1 V2 4Ω
V3
+
2Ω Vo -j4 Ω 0.2 Vo
−
Notice that Vo = V1 .
At the supernode,
V3 − V2 V2 V1 V1 − V3
= + +
4 - j4 2 j2
0 = (2 − j2) V1 + (1 + j) V2 + (-1 + j2) V3 (1)
At node 3,
V1 − V3 V3 − V2
0.2V1 + =
j2 4
(0.8 − j2) V1 + V2 + (-1 + j2) V3 = 0 (2)
Subtracting (2) from (1),
0 = 1.2V1 + j V2 (3)
But at the supernode,
V1 = 12 ∠0° + V2
or V2 = V1 − 12 (4)
Substituting (4) into (3),
0 = 1.2V1 + j (V1 − 12)
j12
V1 = = Vo
1.2 + j
12∠90°
Vo =
1.562∠39.81°
Vo = 7.682∠50.19° V
Chapter 10, Solution 20.
The circuit is converted to its frequency-domain equivalent circuit as shown below.
R
+
+ 1
Vm∠0° jωL Vo
− jωC
−
L
1 C jωL
Let Z = jωL || = =
jωC 1 1 − ω2 LC
jωL +
jωC
jωL
Z 1 − ω2 LC jωL
Vo = Vm = Vm = V
R+Z jωL R (1 − ω2 LC) + jωL m
R+
1 − ω2 LC
ωL Vm ωL
Vo = ∠90° − tan -1
R 2 (1 − ω2 LC) 2 + ω2 L2 R (1 − ω LC)
2
If Vo = A∠φ , then
ωL Vm
A=
R 2 (1 − ω 2 LC) 2 + ω 2 L2
ωL
and φ = 90° − tan -1
R (1 − ω 2 LC)
Chapter 10, Solution 21.
1
Vo jωC 1
(a) = =
Vi 1 1 − ω LC + jωRC
2
R + jωL +
jωC
Vo 1
At ω = 0 , = = 1
Vi 1
Vo
As ω → ∞ , = 0
Vi
1 Vo 1 -j L
At ω = , = =
LC Vi 1 R C
jRC ⋅
LC
Vo jωL − ω2 LC
(b) = =
Vi 1 1 − ω2 LC + jωRC
R + jωL +
jωC
Vo
At ω = 0 , = 0
Vi
Vo 1
As ω → ∞ , = = 1
Vi 1
1 Vo −1 j L
At ω = , = =
LC Vi 1 R C
jRC ⋅
LC
Chapter 10, Solution 22.
Consider the circuit in the frequency domain as shown below.
R1
R2 +
+ 1
Vs Vo
− jωC
jωL −
1
Let Z = (R 2 + jωL) ||
jωC
1
(R + jωL)
jωC 2 R 2 + jωL
Z= =
1 1 + jωR 2 − ω2 LC
R 2 + jωL +
jωC
R 2 + jωL
Vo Z 1 − ω2 LC + jωR 2 C
= =
Vs Z + R 1 R 2 + jωL
R1 +
1 − ω2 LC + jωR 2 C
Vo R 2 + jωL
=
Vs R 1 + R 2 − ω LCR 1 + jω (L + R 1 R 2 C)
2
Chapter 10, Solution 23.
V − Vs V
+ + jωCV = 0
R 1
jωL +
jω C
jωRCV
V+ + jωRCV = Vs
− ω2LC + 1
1 − ω2LC + jωRC + jωRC − jω3RLC2
V = Vs
1 − ω2LC
(1 − ω2 LC)Vs
V=
1 − ω2LC + jωRC(2 − ω2LC)
Chapter 10, Solution 24.
For mesh 1,
1 1 1
Vs = + I1 − I (1)
jωC1 jωC 2 jωC 2 2
For mesh 2,
−1 1
0= I 1 + R + jωL + I (2)
jωC 2 jωC 2 2
Putting (1) and (2) into matrix form,
1 1 −1
Vs jωC + jωC jωC 2 I1
= 1 2
0
−1 1 I 2
R + jωL +
jω C 2 jωC 2
1 1 1 1
∆ = + R + jωL + + 2
jωC1 jωC 2 jωC 2 ω C1C 2
1 Vs
∆ 1 = Vs R + jωL + and ∆2 =
jωC 2 jωC 2
1
Vs R + jωL +
∆1 jωC 2
I1 = =
∆ 1 1 1 1
+ R + jωL + + 2
jωC 1 jωC 2 jωC 2 ω C1 C 2
Vs
∆2 jωC 2
I2 = =
∆ 1 1 1 1
+ R + jωL + + 2
jωC 1 jωC 2 jωC 2 ω C1 C 2
Chapter 10, Solution 25.
ω= 2
10 cos(2t )
→ 10∠0°
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