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Chapter 10, Problem 1.
Determine i in the circuit of Fig. 10.50.
Figure 10.50
For Prob. 10.1.
Chapter 10, Solution 1.
We first determine the input impedance.
1H
⎯⎯
→
jω L = j1x10 = j10
1F
⎯⎯
→
1
jω C
=
1
= − j 0.1
j10 x1
−1
⎛ 1
1
1⎞
Zin = 1+ ⎜
+
+ ⎟ = 1.0101− j0.1 = 1.015 < −5.653o
⎝ j10 − j 0.1 1⎠
I=
2 < 0o
= 1.9704 < 5.653o
1.015 < −5.653o
i(t) = 1.9704 cos(10t + 5.653o ) A = 1.9704cos(10t+5.65˚) A
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Chapter 10, Problem 2.
Solve for V o in Fig. 10.51, using nodal analysis.
Figure 10.51
For Prob. 10.2.
Chapter 10, Solution 2.
Consider the circuit shown below.
2
Vo
+
4∠0o V- _
–j5
j4
At the main node,
4 − Vo
V
V
= o + o
⎯⎯
→ 40 = Vo (10 + j )
2
− j5 j 4
40
Vo =
= 3.98 < 5.71o A
10 − j
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Chapter 10, Problem 3.
Determine v o in the circuit of Fig. 10.52.
Figure 10.52
For Prob. 10.3.
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
-j16 V
-j3 Ω
+
−
Vo
1Ω
j8 Ω
6Ω
2∠0° A
Applying nodal analysis,
- j16 − Vo
Vo
Vo
+2=
+
4 − j3
1 6 + j8
⎛
- j16
1
1 ⎞
⎟V
+ 2 = ⎜1 +
+
4 − j3
⎝ 4 − j3 6 + j8 ⎠ o
Vo =
Therefore,
3.92 − j2.56 4.682∠ - 33.15°
=
= 3.835∠ - 35.02°
1.22 + j0.04
1.2207 ∠1.88°
v o ( t ) = 3.835 cos(4t – 35.02°) V
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Chapter 10, Problem 4.
Determine i1 in the circuit of Fig. 10.53.
Figure 10.53
For Prob. 10.4.
Chapter 10, Solution 4.
⎯⎯
→
0.5H
jω L = j 0.5 x103 = j 500
1
= − j 500
jω C j10 x2 x10 −6
Consider the circuit as shown below.
2µ F
1
⎯⎯
→
I1
50∠0o V
=
3
2000
V1
+
_
-j500
j500
+
–
30I1
At node 1,
50 − V1 30I1 − V1
V
+
= 1
− j500
2000
j 500
50 − V1
But I1 =
2000
50 − V1 + j 4 x30(
I1 =
50 − V1
=0
2000
50 − V1
) + j 4V1 − j 4V1 = 0
2000
→ V1 = 50
i1(t) = 0 A
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Chapter 10, Problem 5.
Find io in the circuit of Fig. 10.54.
Figure 10.54
For Prob. 10.5.
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Chapter 10, Solution 5.
0.25H
2µ F
jω L = j 0.25 x4 x103 = j1000
⎯⎯
→
⎯⎯
→
1
jω C
=
1
= − j125
j 4 x10 x2 x10 −6
3
Consider the circuit as shown below.
Io
2000
Vo
25∠0o V +
_
-j125
j1000
+
–
10Io
At node Vo,
Vo − 25 Vo − 0 Vo − 10I o
=0
+
+
2000
j1000
− j125
Vo − 25 − j2Vo + j16Vo − j160I o = 0
(1 + j14)Vo − j160I o = 25
But Io = (25–Vo)/2000
(1 + j14)Vo − j2 + j0.08Vo = 25
Vo =
25 + j2
25.08∠4.57°
1.7768∠ − 81.37°
=
1 + j14.08 14.115∠58.94°
Now to solve for io,
25 − Vo 25 − 0.2666 + j1.7567
=
= 12.367 + j0.8784 mA
2000
2000
= 12.398∠4.06°
Io =
io = 12.398cos(4x103t + 4.06˚) mA.
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Chapter 10, Problem 6.
Determine V x in Fig. 10.55.
Figure 10.55
For Prob. 10.6.
Chapter 10, Solution 6.
Let Vo be the voltage across the current source. Using nodal analysis we get:
Vo
Vo − 4Vx
20
−3+
= 0 where Vx =
Vo
20
20 + j10
20 + j10
Combining these we get:
Vo
Vo
4Vo
−
−3+
= 0 → (1 + j0.5 − 3)Vo = 60 + j30
20 20 + j10
20 + j10
Vo =
60 + j30
20(3)
or Vx =
= 29.11∠–166˚ V.
− 2 + j0.5
− 2 + j0.5
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Chapter 10, Problem 7.
Use nodal analysis to find V in the circuit of Fig. 10.56.
Figure 10.56
For Prob. 10.7.
Chapter 10, Solution 7.
At the main node,
120∠ − 15 o − V
V
V
= 6∠30 o +
+
40 + j20
− j30 50
⎯
⎯→
115.91 − j31.058
− 5.196 − j3 =
40 + j20
⎛
1
j
1⎞
V⎜⎜
+
+ ⎟⎟
⎝ 40 + j20 30 50 ⎠
V=
− 3.1885 − j4.7805
= 124.08∠ − 154 o V
0.04 + j0.0233
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Chapter 10, Problem 8.
Use nodal analysis to find current io in the circuit of Fig. 10.57. Let
i s = 6 cos(200t + 15°) A.
Figure 10.57
For Prob. 10.8.
Chapter 10, Solution 8.
ω = 200,
100mH
50µF
⎯
⎯→
⎯
⎯→
jωL = j200x 0.1 = j20
1
1
=
= − j100
jωC j200x 50x10 − 6
The frequency-domain version of the circuit is shown below.
0.1 Vo
40 Ω
V1
6∠15
o
20 Ω
+
Vo
-
Io
V2
-j100 Ω
j20 Ω
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At node 1,
or
V
V1
V − V2
6∠15 o + 0.1V1 = 1 +
+ 1
20 − j100
40
5.7955 + j1.5529 = (−0.025 + j 0.01)V1 − 0.025V2
(1)
At node 2,
V1 − V2
V
= 0.1V1 + 2
40
j20
From (1) and (2),
⎯⎯→
0 = 3V1 + (1 − j2)V2
⎡(−0.025 + j0.01) − 0.025⎤⎛ V1 ⎞ ⎛ (5.7955 + j1.5529) ⎞
⎜ ⎟=⎜
⎟⎟
⎢
3
(1 − j2) ⎥⎦⎜⎝ V2 ⎟⎠ ⎜⎝
0
⎠
⎣
or
(2)
AV = B
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(200t − 82.17 o ) A
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Chapter 10, Problem 9.
Use nodal analysis to find v o in the circuit of Fig. 10.58.
Figure 10.58
For Prob. 10.9.
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Chapter 10, Solution 9.
10 cos(10 3 t ) ⎯
⎯→ 10 ∠0°, ω = 10 3
10 mH ⎯
⎯→
50 µF ⎯
⎯→
jωL = j10
1
1
=
= - j20
3
jωC j (10 )(50 × 10 -6 )
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,
At node 2,
10 − V1 V1 V1 − V2
=
+
20
20
- j20
10 = (2 + j) V1 − jV2
(1)
V1 − V2
V
V2
V1
, where I o =
has been substituted.
= (4) 1 +
20
- j20
20 30 + j10
(-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
Vo =
Therefore,
30
3
170
V2 =
⋅
= 6.154∠70.26°
30 + j10
3 + j 0.6 − j26.2
v o ( t ) = 6.154 cos(103 t + 70.26°) V
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Chapter 10, Problem 10.
Use nodal analysis to find v o in the circuit of Fig. 10.59. Let ω = 2 krad/s.
Figure 10.59
For Prob. 10.10.
Chapter 10, Solution 10.
⎯
⎯→
50 mH
2µF
⎯
⎯→
jωL = j2000x50 x10 − 3 = j100,
1
1
=
= − j250
jωC j2000x 2x10 − 6
ω = 2000
Consider the frequency-domain equivalent circuit below.
V1
36<0o
2k Ω
-j250
j100
V2
0.1V1 4k Ω
At node 1,
36 =
V1
V
V − V2
+ 1 + 1
2000 j100 − j250
⎯
⎯→
36 = (0.0005 − j0.006)V1 − j0.004V2
(1)
At node 2,
V1 − V2
V
= 0.1V1 + 2
− j250
4000
⎯⎯→
0 = (0.1 − j0.004)V1 + (0.00025 + j0.004)V2 (2)
Solving (1) and (2) gives
Vo = V2 = −535.6 + j893.5 = 8951.1∠93.43o
vo (t) = 8.951 sin(2000t +93.43o) kV
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Chapter 10, Problem 11.
Apply nodal analysis to the circuit in Fig. 10.60 and determine I o .
Figure 10.60
For Prob. 10.11.
Chapter 10, Solution 11.
Consider the circuit as shown below.
–j5 Ω
Io
2Ω
2Ω
V1
o
4∠0 V
+
_
V2
j8 Ω
2Io
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At node 1,
V1 − 4
V − V2
=0
− 2I o + 1
2
2
V1 − 0.5V2 − 2I o = 2
But, Io = (4–V2)/(–j5) = –j0.2V2 + j0.8
Now the first node equation becomes,
V1 – 0.5V2 + j0.4V2 – j1.6 = 2 or
V1 + (–0.5+j0.4)V2 = 2 + j1.6
At node 2,
V2 − V1 V2 − 4 V2 − 0
+
+
=0
− j5
2
j8
–0.5V1 + (0.5 + j0.075)V2 = j0.8
Using MATLAB to solve this, we get,
>> Y=[1,(-0.5+0.4i);-0.5,(0.5+0.075i)]
Y=
1.0000
-0.5000
-0.5000 + 0.4000i
0.5000 + 0.0750i
>> I=[(2+1.6i);0.8i]
I=
2.0000 + 1.6000i
0 + 0.8000i
>> V=inv(Y)*I
V=
4.8597 + 0.0543i
4.9955 + 0.9050i
Io = –j0.2V2 + j0.8 = –j0.9992 + 0.01086 + j0.8 = 0.01086 – j0.1992
= 199.5∠86.89˚ mA.
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Chapter 10, Problem 12.
By nodal analysis, find io in the circuit of Fig. 10.61.
Figure 10.61
For Prob. 10.12.
Chapter 10, Solution 12.
20 sin(1000t ) ⎯
⎯→ 20 ∠0°, ω = 1000
⎯→
10 mH ⎯
50 µF ⎯
⎯→
jωL = j10
1
1
=
= - j20
3
jωC j (10 )(50 × 10 -6 )
The frequency-domain equivalent circuit is shown below.
2 Io
V1
10 Ω
V2
Io
20∠0° A
20 Ω
-j20 Ω
j10 Ω
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At node 1,
20 = 2 I o +
V1 V1 − V2
+
,
20
10
where
V2
j10
2V2 V1 V1 − V2
20 =
+
+
j10 20
10
400 = 3V1 − (2 + j4) V2
(1)
Io =
At node 2,
or
V
V
2V2 V1 − V2
+
= 2 + 2
j10
10
- j20 j10
j2 V1 = (-3 + j2) V2
V1 = (1 + j1.5) V2
(2)
Substituting (2) into (1),
400 = (3 + j4.5) V2 − (2 + j4) V2 = (1 + j0.5) V2
Therefore,
V2 =
400
1 + j0.5
Io =
V2
40
=
= 35.74 ∠ - 116.6°
j10 j (1 + j0.5)
i o ( t ) = 35.74 sin(1000t – 116.6°) A
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Chapter 10, Problem 13.
Determine V x in the circuit of Fig. 10.62 using any method of your choice.
Figure 10.62
For Prob. 10.13.
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.
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Chapter 10, Problem 14.
Calculate the voltage at nodes 1 and 2 in the circuit of Fig. 10.63 using nodal analysis.
Figure 10.63
For Prob. 10.14.
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
At node 2,
V2 V2 V2 − V1
+
+
= 20∠30°
j2 - j5
j4
- j5.5 V2 + j2.5 V1 = 173.2 + j100
(1)
(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
j2.5
- j5.5
∆1 =
= 20 − j5.5 = 20.74∠ - 15.38°
- 200 ∠30° j2.5
= j3 (200∠30°) = 600∠120°
200 ∠30° - j5.5
1 + j2.5 - 200∠30°
= (200 ∠30°)(1 + j5) = 1020∠108.7°
j2.5
200∠30°
∆1
V1 =
= 28.93∠135.38°
∆
∆2
V2 =
= 49.18∠124.08°
∆
∆2 =
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Chapter 10, Problem 15.
Solve for the current I in the circuit of Fig. 10.64 using nodal analysis.
Figure 10.64
For Prob. 10.15.
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Chapter 10, Solution 15.
We apply nodal analysis to the circuit shown below.
5A
2Ω
jΩ
V1
V2
I
-j20 V
+
−
-j2 Ω
2I
4Ω
At node 1,
V
V − V2
- j20 − V1
= 5+ 1 + 1
2
- j2
j
- 5 − j10 = (0.5 − j0.5) V1 + j V2
(1)
At node 2,
V1 − V2 V2
,
=
j
4
V
where I = 1
- j2
5
V2 =
V1
0.25 − j
5 + 2I +
(2)
Substituting (2) into (1),
j5
= 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°
- 5 − j10 −
V1
= (0.5∠90°)(15.81∠313.5°)
- j2
I = 7.906∠43.49° A
I=
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Chapter 10, Problem 16.
Use nodal analysis to find V x in the circuit shown in Fig. 10.65.
Figure 10.65
For Prob. 10.16.
Chapter 10, Solution 16.
Consider the circuit as shown in the figure below.
j4 Ω
V1
V2
+ Vx –
2∠0o A
5Ω
–j3 Ω
3∠45o A
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At node 1,
V − 0 V1 − V2
−2+ 1
+
=0
5
j4
(0.2 − j0.25)V1 + j0.25V2 = 2
(1)
At node 2,
V2 − V1 V2 − 0
+
− 3∠45° = 0
j4
− j3
j0.25V1 + j0.08333V2 = 2.121 + j2.121
In matrix form, (1) and (2) become
⎡0.2 − j0.25
⎢ j0.25
⎣
(2)
j0.25 ⎤ ⎡ V1 ⎤ ⎡
2
⎤
=⎢
⎢
⎥
⎥
j0.08333⎦ ⎣V2 ⎦ ⎣2.121 + j2.121⎥⎦
Solving this using MATLAB, we get,
>> Y=[(0.2-0.25i),0.25i;0.25i,0.08333i]
Y=
0.2000 - 0.2500i
0 + 0.2500i
0 + 0.2500i
0 + 0.0833i
>> I=[2;(2.121+2.121i)]
I=
2.0000
2.1210 + 2.1210i
>> V=inv(Y)*I
V=
5.2793 - 5.4190i
9.6145 - 9.1955i
Vs = V1 – V2 = –4.335 + j3.776 = 5.749∠138.94˚ V.
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Chapter 10, Problem 17.
By nodal analysis, obtain current I o in the circuit of Fig. 10.66.
Figure 10.66
For Prob. 10.17.
Chapter 10, Solution 17.
Consider the circuit below.
j4 Ω
100∠20° V
+
−
1Ω
Io
V1
3Ω
2Ω
V2
-j2 Ω
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At node 1,
100∠20° − V1 V1 V1 − V2
=
+
j4
3
2
V1
100 ∠20° =
(3 + j10) − j2 V2
3
(1)
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
∆1 =
∆2 =
100∠20° 1.5 + j0.5
100∠20°
- j2
= -55.45 − j286.2
- 0.5
100∠20°
= -26.95 − j364.5
1 + j10 3 100∠20°
∆1
= 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° A
V1 =
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Chapter 10, Problem 18.
Use nodal analysis to obtain V o in the circuit of Fig. 10.67 below.
Figure 10.67
For Prob. 10.18.
Chapter 10, Solution 18.
Consider the circuit shown below.
8Ω
V1
j6 Ω
4Ω
V2
j5 Ω
+
4∠45° A
2Ω
Vx
−
+
2 Vx
-j Ω
-j2 Ω
Vo
−
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At node 1,
V1 V1 − V2
+
2
8 + j6
200 ∠45° = (29 − j3) V1 − (4 − j3) V2
(1)
4∠45° =
At node 2,
V1 − V2
V
V2
,
+ 2Vx = 2 +
8 + j6
- j 4 + j5 − j2
(104 − j3) V1 = (12 + j41) V2
12 + j41
V1 =
V
104 − j3 2
(2)
where Vx = V1
Substituting (2) into (1),
(12 + j41)
V − (4 − j3) V2
104 − j3 2
200 ∠45° = (14.21∠89.17°) V2
200∠45°
V2 =
14.21∠89.17°
200∠45° = (29 − j3)
- j2
- j2
- 6 − j8
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
Vo =
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Chapter 10, Problem 19.
Obtain V o in Fig. 10.68 using nodal analysis.
Figure 10.68
For Prob. 10.19.
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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
At node 3,
V1 − V3 V3 − V2
0.2V1 +
=
j2
4
(0.8 − j2) V1 + V2 + (-1 + j2) V3 = 0
(1)
(2)
Subtracting (2) from (1),
0 = 1.2V1 + j V2
(3)
But at the supernode,
V1 = 12 ∠0° + V2
V2 = V1 − 12
or
(4)
Substituting (4) into (3),
0 = 1.2V1 + j (V1 − 12)
j12
V1 =
= Vo
1.2 + j
12∠90°
1.562∠39.81°
Vo = 7.682∠50.19° V
Vo =
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Chapter 10, Problem 20.
Refer to Fig. 10.69. If v s (t ) = Vm sin ωt and vo (t ) = A sin (ωt + φ ) derive the expressions
for A and φ
Figure 10.69
For Prob. 10.20.
Chapter 10, Solution 20.
The circuit is converted to its frequency-domain equivalent circuit as shown below.
R
+
Vm∠0°
+
−
jωL
Vo
1
jωC
−
Let
1
=
Z = jωL ||
jωC
L
C
=
jωL
1 − ω2 LC
1
jωC
jωL
jωL
Z
1 − ω2 LC
V
Vm =
Vm =
Vo =
jωL
R (1 − ω2 LC) + jωL m
R+Z
R+
1 − ω2 LC
⎛
⎞
ωL Vm
ωL
⎜90° − tan -1
⎟
∠
Vo =
R (1 − ω2 LC) ⎠
R 2 (1 − ω2 LC) 2 + ω2 L2 ⎝
jωL +
If
Vo = A∠φ , then
ωL Vm
A=
R 2 (1 − ω 2 LC) 2 + ω 2 L2
and
φ = 90° − tan -1
ωL
R (1 − ω 2 LC)
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Chapter 10, Problem 21.
For each of the circuits in Fig. 10.70, find V o /V i for ω = 0, ω → ∞, and ω 2 = 1 / LC .
Figure 10.70
For Prob. 10.21.
Chapter 10, Solution 21.
(a)
Vo
=
Vi
1
jωC
R + jωL +
1
jωC
As ω → ∞ ,
(b)
Vo
=
Vi
1
LC
Vo
=
Vi
,
jωL
R + jωL +
As ω → ∞ ,
1
LC
1
jωC
1
jRC ⋅
1
=
-j L
R C
LC
− ω2 LC
=
1 − ω2 LC + jωRC
Vo
= 0
Vi
Vo 1
= = 1
Vi 1
At ω = 0 ,
At ω =
1
1 − ω LC + jωRC
2
Vo 1
= = 1
Vi 1
Vo
= 0
Vi
At ω = 0 ,
At ω =
=
,
Vo
=
Vi
−1
jRC ⋅
1
=
j L
R C
LC
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Chapter 10, Problem 22.
For the circuit in Fig. 10.71, determine V o /V s .
Figure 10.71
For Prob. 10.22.
Chapter 10, Solution 22.
Consider the circuit in the frequency domain as shown below.
R1
R2
Vs
+
−
1
jωC
jωL
Let
Z = (R 2 + jωL) ||
+
Vo
−
1
jωC
1
(R + jωL)
R 2 + jωL
jωC 2
Z=
=
1
1 + jωR 2 − ω2 LC
R 2 + jωL +
jωC
R 2 + jωL
Vo
1 − ω2 LC + jωR 2 C
Z
=
=
R 2 + jωL
Vs Z + R 1
R1 +
1 − ω2 LC + jωR 2 C
Vo
R 2 + jωL
=
2
Vs R 1 + R 2 − ω LCR 1 + jω (L + R 1 R 2 C)
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Chapter 10, Problem 23.
Using nodal analysis obtain V in the circuit of Fig. 10.72.
Figure 10.72
For Prob. 10.23.
Chapter 10, Solution 23.
V − Vs
V
+
+ jωCV = 0
1
R
jωL +
jω C
V+
jωRCV
− ω2LC + 1
+ jωRCV = Vs
⎛ 1 − ω2LC + jωRC + jωRC − jω3RLC2 ⎞
⎜
⎟ V = Vs
2
⎟
⎜
1
−
ω
LC
⎝
⎠
V=
(1 − ω2 LC)Vs
1 − ω2LC + jωRC(2 − ω2LC)
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Chapter 10, Problem 24.
Use mesh analysis to find V o in the circuit of Prob. 10.2.
Chapter 10, Solution 24.
Consider the circuit as shown below.
2Ω
+
o
4∠0 V
I1
+
_
j4 Ω
–j5 Ω
Vo
I2
–
For mesh 1,
4 = (2 − j 5) I1 + j 5 I1
For mesh 2,
(1)
0 = j 5I 1 + ( j 4 − j 5) I 2
⎯⎯
→ I1 =
1
I2
5
(2)
Substituting (2) into (1),
1
4 = (2 − j 5) I 2 + j 5I 2
5
Vo = j 4 I 2 =
⎯⎯
→ I2 =
1
0.1 + j
j4
= 3.98 < 5.71o V
0.1 + j
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Chapter 10, Problem 25.
Solve for io in Fig. 10.73 using mesh analysis.
Figure 10.73
For Prob. 10.25.
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Chapter 10, Solution 25.
ω= 2
10 cos(2t ) ⎯
⎯→ 10∠0°
6 sin(2t ) ⎯
⎯→ 6 ∠ - 90° = -j6
2H ⎯
⎯→
jωL = j4
1
1
0.25 F ⎯
⎯→
=
= - j2
jωC j (2)(1 4)
The circuit is shown below.
4Ω
j4 Ω
Io
10∠0° V
+
−
I1
-j2 Ω
I2
+
−
6∠-90° V
For loop 1,
- 10 + (4 − j2) I 1 + j2 I 2 = 0
5 = (2 − j) I 1 + j I 2
(1)
For loop 2,
j2 I 1 + ( j4 − j2) I 2 + (- j6) = 0
I1 + I 2 = 3
(2)
In matrix form (1) and (2) become
⎡ 2 − j j ⎤ ⎡ I 1 ⎤ ⎡ 5⎤
⎢ 1 1 ⎥ ⎢ I ⎥ = ⎢ 3⎥
⎣
⎦⎣ 2 ⎦ ⎣ ⎦
∆ = 2 (1 − j) ,
∆ 1 = 5 − j3 ,
∆ 2 = 1 − j3
∆1 − ∆ 2
4
=
= 1 + j = 1.414 ∠45°
2 (1 − j)
∆
i o ( t ) = 1.4142 cos(2t + 45°) A
I o = I1 − I 2 =
Therefore,
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Chapter 10, Problem 26.
Use mesh analysis to find current io in the circuit of Fig. 10.74.
Figure 10.74
For Prob. 10.26.
Chapter 10, Solution 26.
0.4H
⎯⎯
→
1µ F
⎯⎯
→
jω L = j103 x0.4 = j400
1
jωC
=
1
= − j1000
j10 x10−6
3
20sin103 t = 20 cos(103 t − 90o )
⎯⎯
→ 20 < −90 = − j 20
The circuit becomes that shown below.
2 kΩ
–j1000
Io
10∠0o
+
_
+
_
I1
j400
–j20
I2
For loop 1,
−10 + (12000 + j 400) I1 − j 400 I 2 = 0
⎯⎯
→ 1 = (200 + j 40) I1 − j 40 I 2 (1)
For loop 2,
− j 20 + ( j 400 − j1000) I 2 − j 400 I1 = 0 ⎯⎯
→ −12 = 40 I1 + 60 I 2
(2)
In matrix form, (1) and (2) become
⎡ 1 ⎤ ⎡ 200 + j 40 − j 40 ⎤ ⎡ I1 ⎤
⎢ −12 ⎥ = ⎢
40
60 ⎥⎦ ⎢⎣ I 2 ⎥⎦
⎣
⎦ ⎣
Solving this leads to
I1 =0.0025-j0.0075, I2 = -0.035+j0.005
I o = I1 − I 2 = 0.0375 − j 0.0125 = 39.5 < −18.43 mA
io = 39.5cos(103 t − 18.43o ) mA
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Chapter 10, Problem 27.
Using mesh analysis, find I 1 and I 2 in the circuit of Fig. 10.75.
Figure 10.75
For Prob. 10.27.
Chapter 10, Solution 27.
For mesh 1,
- 40 ∠30° + ( j10 − j20) I 1 + j20 I 2 = 0
4 ∠30° = - j I 1 + j2 I 2
(1)
50 ∠0° + (40 − j20) I 2 + j20 I 1 = 0
5 = - j2 I 1 − (4 − j2) I 2
(2)
For mesh 2,
From (1) and (2),
⎡ 4∠30°⎤ ⎡ - j
j2 ⎤⎡ I 1 ⎤
⎢ 5 ⎥ = ⎢ - j2 - (4 − j2) ⎥⎢ I ⎥
⎦⎣ 2 ⎦
⎣
⎦ ⎣
∆ = -2 + 4 j = 4.472∠116.56°
∆ 1 = -(4 ∠30°)(4 − j2) − j10 = 21.01∠211.8°
∆ 2 = - j5 + 8∠120° = 4.44 ∠154.27°
I1 =
∆1
= 4.698∠95.24° A
∆
I2 =
∆2
= 0.9928∠37.71° A
∆
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Chapter 10, Problem 28.
In the circuit of Fig. 10.76, determine the mesh currents i1 and i2 . Let v1 = 10 cos 4t V
and v 2 = 20 cos(4t − 30°) V.
Figure 10.76
For Prob. 10.28.
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Chapter 10, Solution 28.
1
1
=
= − j0.25
jωC j1x 4
The frequency-domain version of the circuit is shown below, where
1H
⎯
⎯→
V1 = 10∠0 o ,
jωL = j4,
⎯
⎯→
1F
V2 = 20∠ − 30 o .
1
j4
j4
1
-j0.25
+
+
V1
-
I1
V1 = 10∠0 o ,
1
I2
V2
-
V2 = 20∠ − 30 o
Applying mesh analysis,
10 = (2 + j3.75)I1 − (1 − j0.25)I 2
(1)
− 20∠ − 30 o = −(1 − j0.25)I1 + (2 + j3.75)I 2
(2)
From (1) and (2), we obtain
10
⎛
⎞ ⎛ 2 + j3.75 − 1 + j0.25 ⎞⎛ I1 ⎞
⎜⎜
⎟⎟ = ⎜⎜
⎟⎟⎜⎜ ⎟⎟
⎝ − 17.32 + j10 ⎠ ⎝ − 1 + j0.25 2 + j3.75 ⎠⎝ I 2 ⎠
Solving this leads to
I1 = 2.741∠ − 41.07 o ,
I 2 = 4.114∠92 o
Hence,
i1(t) = 2.741cos(4t–41.07˚)A, i2(t) = 4.114cos(4t+92˚)A.
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Chapter 10, Problem 29.
By using mesh analysis, find I 1 and I 2 in the circuit depicted in Fig. 10.77.
Figure 10.77
For Prob. 10.29.
Chapter 10, Solution 29.
For mesh 1,
(5 + j5) I 1 − (2 + j) I 2 − 30 ∠20° = 0
30 ∠20° = (5 + j5) I 1 − (2 + j) I 2
(1)
For mesh 2,
(5 + j3 − j6) I 2 − (2 + j) I 1 = 0
0 = - (2 + j) I 1 + (5 − j3) I 2
(2)
From (1) and (2),
⎡30∠20°⎤ ⎡ 5 + j5 - (2 + j) ⎤⎡ I 1 ⎤
⎢ 0 ⎥ = ⎢ - (2 + j) 5 - j3 ⎥⎢ I ⎥
⎣
⎦ ⎣
⎦⎣ 2 ⎦
∆ = 37 + j6 = 37.48∠9.21°
∆ 1 = (30 ∠20°)(5.831∠ - 30.96°) = 175∠ - 10.96°
∆ 2 = (30 ∠20°)(2.356 ∠26.56°) = 67.08∠46.56°
I1 =
∆1
= 4.67∠–20.17° A
∆
I2 =
∆2
= 1.79∠37.35° A
∆
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Chapter 10, Problem 30.
Use mesh analysis to find vo in the circuit of Fig. 10.78. Let v s1 = 120 cos(100t + 90°) V,
v s 2 = 80 cos 100t V.
Figure 10.78
For Prob. 10.30.
Chapter 10, Solution 30.
300mH
⎯⎯
→
jω L = j100 x300 x10 −3 = j30
200mH
⎯⎯
→
jω L = j100 x200 x10 −3 = j20
400mH
⎯⎯
→
jω L = j100 x400 x10 −3 = j40
1
= − j200
j100 x50 x10−6
The circuit becomes that shown below.
50 µ F
⎯⎯
→
1
jω C
=
j40
j20
20 Ω
+
120∠90o
+
_
10 Ω
I1
j30
–j200
vo
I2
-
I3`
+
_
80∠0o
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you are using it without permission.
For mesh 1,
−120 < 90o + (20 + j30) I1 − j 30 I 2 = 0
⎯⎯
→ j120 = (20 + j 30) I1 − j 30 I 2
For mesh 2,
− j 30 I1 + ( j 30 + j 40 − j 200) I 2 + j 200 I 3 = 0
⎯⎯
→ 0 = −3I1 − 13I 2 + 20 I 3
For mesh 3,
80 + j200I 2 + (10 − j180)I 3 = 0 → −8 = j20I 2 + (1 − j18)I 3
(3)
(1)
(2)
We put (1) to (3) in matrix form.
0 ⎤ ⎡ I1 ⎤ ⎡ j12⎤
⎡2 + j3 − j3
⎢ − 3 − 13
20 ⎥⎥ ⎢⎢I 2 ⎥⎥ = ⎢⎢ 0 ⎥⎥
⎢
⎢⎣ 0
j20 1 − j18⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ − 8⎥⎦
This is an excellent candidate for MATLAB.
>> Z=[(2+3i),-3i,0;-3,-13,20;0,20i,(1-18i)]
Z=
2.0000 + 3.0000i
0 - 3.0000i
0
-3.0000
-13.0000
20.0000
0
0 +20.0000i 1.0000 -18.0000i
>> V=[12i;0;-8]
V=
0 +12.0000i
0
-8.0000
>> I=inv(Z)*V
I=
2.0557 + 3.5651i
0.4324 + 2.1946i
0.5894 + 1.9612i
Vo = –j200(I2 – I3) = –j200(–0.157+j0.2334) = 46.68 + j31.4 = 56.26∠33.93˚
vo = 56.26cos(100t + 33.93˚ V.
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Chapter 10, Problem 31.
Use mesh analysis to determine current I o in the circuit of Fig. 10.79 below.
Figure 10.79
For Prob. 10.31.
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Chapter 10, Solution 31.
Consider the network shown below.
80 Ω
100∠120° V
+
−
I1
Io
-j40 Ω
j60 Ω
I2
-j40 Ω
20 Ω
I3
+
−
60∠-30° V
For loop 1,
- 100∠120° + (80 − j40) I1 + j40 I 2 = 0
10 ∠20° = 4 (2 − j) I 1 + j4 I 2
(1)
j40 I 1 + ( j60 − j80) I 2 + j40 I 3 = 0
0 = 2 I1 − I 2 + 2 I 3
(2)
60∠ - 30° + (20 − j40) I 3 + j40 I 2 = 0
- 6∠ - 30° = j4 I 2 + 2 (1 − j2) I 3
(3)
For loop 2,
For loop 3,
From (2),
2 I 3 = I 2 − 2 I1
Substituting this equation into (3),
- 6 ∠ - 30° = -2 (1 − j2) I 1 + (1 + j2) I 2
(4)
From (1) and (4),
⎡ 10∠120° ⎤ ⎡ 4 (2 − j)
j4 ⎤⎡ I 1 ⎤
=
⎢ - 6∠ - 30°⎥ ⎢ - 2 (1 − j2) 1 + j2⎥⎢ I ⎥
⎣
⎦ ⎣
⎦⎣ 2 ⎦
∆=
∆2 =
8 − j4
- j4
= 32 + j20 = 37.74∠32°
- 2 + j4 1 + j2
8 − j4 10∠120°
= -4.928 + j82.11 = 82.25∠93.44°
- 2 + j4 - 6∠ - 30°
Io = I2 =
∆2
= 2.179∠61.44° A
∆
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Chapter 10, Problem 32.
Determine V o and I o in the circuit of Fig. 10.80 using mesh analysis.
Figure 10.80
For Prob. 10.32.
Chapter 10, Solution 32.
Consider the circuit below.
j4 Ω
Io
+
4∠-30° V
2Ω
Vo
I1
+
3 Vo
I2
-j2 Ω
−
For mesh 1,
where
(2 + j4) I 1 − 2 (4∠ - 30°) + 3 Vo = 0
Vo = 2 (4∠ - 30° − I 1 )
Hence,
(2 + j4) I 1 − 8∠ - 30° + 6 (4 ∠ - 30° − I 1 ) = 0
4 ∠ - 30° = (1 − j) I 1
I 1 = 2 2 ∠15°
or
Io =
3 Vo
3
=
(2)(4∠ - 30° − I 1 )
- j2 - j2
I o = j3 (4 ∠ - 30° − 2 2 ∠15°)
I o = 8.485∠15° A
Vo =
- j2 I o
= 5.657∠-75° V
3
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Chapter 10, Problem 33.
Compute I in Prob. 10.15 using mesh analysis.
Chapter 10, Solution 33.
Consider the circuit shown below.
5A
I4
2Ω
jΩ
I
-j20 V
+
−
I1
-j2 Ω
I2
2I
I3
4Ω
For mesh 1,
j20 + (2 − j2) I 1 + j2 I 2 = 0
(1 − j) I 1 + j I 2 = - j10
For the supermesh,
( j − j2) I 2 + j2 I 1 + 4 I 3 − j I 4 = 0
Also,
I 3 − I 2 = 2 I = 2 (I 1 − I 2 )
I 3 = 2 I1 − I 2
(1)
(2)
(3)
For mesh 4,
I4 = 5
Substituting (3) and (4) into (2),
(8 + j2) I 1 − (- 4 + j) I 2 = j5
(4)
(5)
Putting (1) and (5) in matrix form,
⎡ 1− j
j ⎤⎡ I 1 ⎤ ⎡ - j10 ⎤
⎢8 + j2 4 − j⎥⎢ I ⎥ = ⎢ j5 ⎥
⎦
⎣
⎦⎣ 2 ⎦ ⎣
∆ = -3 − j5 ,
∆ 1 = -5 + j40 ,
∆ 2 = -15 + j85
∆ − ∆ 2 10 − j45
I = I1 − I 2 = 1
=
= 7.906∠43.49° A
∆
- 3 − j5
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Chapter 10, Problem 34.
Use mesh analysis to find I o in Fig. 10.28 (for Example 10.10).
Chapter 10, Solution 34.
The circuit is shown below.
Io
I2
5Ω
3A
20 Ω
8Ω
40∠90° V
+
−
-j2 Ω
I3
10 Ω
I1
j15 Ω
j4 Ω
For mesh 1,
- j40 + (18 + j2) I 1 − (8 − j2) I 2 − (10 + j4) I 3 = 0
For the supermesh,
(13 − j2) I 2 + (30 + j19) I 3 − (18 + j2) I 1 = 0
(1)
(2)
Also,
I2 = I3 − 3
(3)
Adding (1) and (2) and incorporating (3),
- j40 + 5 (I 3 − 3) + (20 + j15) I 3 = 0
3 + j8
I3 =
= 1.465∠38.48°
5 + j3
I o = I 3 = 1.465∠38.48° A
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Chapter 10, Problem 35.
Calculate I o in Fig. 10.30 (for Practice Prob. 10.10) using mesh analysis.
Chapter 10, Solution 35.
Consider the circuit shown below.
4Ω
j2 Ω
I3
8Ω
1Ω
-j3 Ω
10 Ω
20 V
+
−
I1
-j4 A
I2
-j5 Ω
For the supermesh,
- 20 + 8 I 1 + (11 − j8) I 2 − (9 − j3) I 3 = 0
(1)
I 1 = I 2 + j4
(2)
Also,
For mesh 3,
(13 − j) I 3 − 8 I 1 − (1 − j3) I 2 = 0
Substituting (2) into (1),
(19 − j8) I 2 − (9 − j3) I 3 = 20 − j32
Substituting (2) into (3),
- (9 − j3) I 2 + (13 − j) I 3 = j32
From (4) and (5),
⎡ 19 − j8 - (9 − j3) ⎤⎡ I 2 ⎤ ⎡ 20 − j32 ⎤
⎢ - (9 − j3) 13 − j ⎥⎢ I ⎥ = ⎢ j32 ⎥
⎦
⎣
⎦⎣ 3 ⎦ ⎣
∆ = 167 − j69 ,
(3)
(4)
(5)
∆ 2 = 324 − j148
∆ 2 324 − j148 356.2∠ - 24.55°
=
=
∆
167 − j69 180.69∠ - 22.45°
I 2 = 1.971∠-2.1° A
I2 =
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Chapter 10, Problem 36.
Compute V o in the circuit of Fig. 10.81 using mesh analysis.
Figure 10.81
For Prob. 10.36.
Chapter 10, Solution 36.
Consider the circuit below.
j4 Ω
-j3 Ω
+
4∠90° A
I1
2Ω
Vo
I2
+
−
12∠0° V
−
2Ω
2Ω
I3
2∠0° A
Clearly,
I 1 = 4 ∠90° = j4
and
I 3 = -2
For mesh 2,
(4 − j3) I 2 − 2 I 1 − 2 I 3 + 12 = 0
(4 − j3) I 2 − j8 + 4 + 12 = 0
- 16 + j8
I2 =
= -3.52 − j0.64
4 − j3
Thus,
Vo = 2 (I 1 − I 2 ) = (2)(3.52 + j4.64) = 7.04 + j9.28
Vo = 11.648∠52.82° V
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Chapter 10, Problem 37.
Use mesh analysis to find currents I 1 , I 2 , and I 3 in the circuit of Fig. 10.82.
Figure 10.82
For Prob. 10.37.
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Chapter 10, Solution 37.
I1
+
120∠ − 90 o V
-
120∠ − 30 V
+
Ix
Z
Z=80-j35 Ω
I2
Iz
Iy
o
Z
I3
For mesh x,
ZI x − ZI z = − j120
(1)
ZI y − ZI z = −120∠30 o = −103.92 + j60
(2)
− ZI x − ZI y + 3ZI z = 0
(3)
For mesh y,
For mesh z,
Putting (1) to (3) together leads to the following matrix equation:
− j120
0
(−80 + j35) ⎞⎛ I x ⎞ ⎛
⎛ (80 − j35)
⎞
⎜
⎟⎜ ⎟ ⎜
⎟
0
(80 − j35) (−80 + j35) ⎟⎜ I y ⎟ = ⎜ − 103.92 + j60 ⎟
⎜
⎜ (−80 + j35) (−80 + j35) (240 − j105) ⎟⎜ I ⎟ ⎜
⎟
0
⎝
⎠⎝ z ⎠ ⎝
⎠
⎯
⎯→
AI = B
Using MATLAB, we obtain
⎛ - 0.2641 − j2.366 ⎞
⎜
⎟
I = inv(A) * B = ⎜ - 2.181 - j0.954 ⎟
⎜ - 0.815 − j1.1066 ⎟
⎝
⎠
I1 = I x = −0.2641 − j2.366 = 2.38∠ − 96.37 o A
I 2 = I y − I x = −1.9167 + j1.4116 = 2.38∠143.63o A
I 3 = − I y = 2.181 + j0.954 = 2.38∠23.63o A
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Chapter 10, Problem 38.
Using mesh analysis, obtain I o in the circuit shown in Fig. 10.83.
Figure 10.83
For Prob. 10.38.
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Chapter 10, Solution 38.
Consider the circuit below.
Io
I1
2∠0° A
2Ω
j2 Ω
1Ω
I2
+
−
-j4 Ω
4∠0° A
I3
I4
10∠90° V
1Ω
A
Clearly,
I1 = 2
(1)
(2 − j4) I 2 − 2 I 1 + j4 I 4 + 10 ∠90° = 0
(2)
For mesh 2,
Substitute (1) into (2) to get
(1 − j2) I 2 + j2 I 4 = 2 − j5
For the supermesh,
(1 + j2) I 3 − j2 I 1 + (1 − j4) I 4 + j4 I 2 = 0
j4 I 2 + (1 + j2) I 3 + (1 − j4) I 4 = j4
At node A,
I3 = I4 − 4
Substituting (4) into (3) gives
j2 I 2 + (1 − j) I 4 = 2 (1 + j3)
From (2) and (5),
⎡1 − j2 j2 ⎤⎡ I 2 ⎤ ⎡ 2 − j5⎤
⎢ j2 1 − j⎥⎢ I ⎥ = ⎢ 2 + j6⎥
⎦
⎣
⎦⎣ 4 ⎦ ⎣
∆ = 3 − j3 ,
(3)
(4)
(5)
∆ 1 = 9 − j11
- ∆ 1 - (9 − j11) 1
=
= (-10 + j)
∆
3 − j3
3
I o = 3.35∠174.3° A
Io = -I2 =
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Chapter 10, Problem 39.
Find I 1 , I 2 , I 3 , and I x in the circuit of Fig. 10.84.
Figure 10.84
For Prob. 10.39.
Chapter 10, Solution 39.
For mesh 1,
(28 − j15)I1 − 8I 2 + j15I 3 = 12∠64 o
(1)
− 8I1 + (8 − j9)I 2 − j16I 3 = 0
(2)
j15I1 − j16I 2 + (10 + j)I 3 = 0
(3)
For mesh 2,
For mesh 3,
In matrix form, (1) to (3) can be cast as
j15 ⎞⎛ I1 ⎞ ⎛⎜12∠64 o ⎞⎟
−8
⎛ (28 − j15)
⎟⎜ ⎟
⎜
(8 − j9) − j16 ⎟⎜ I 2 ⎟ = ⎜ 0 ⎟
⎜ −8
⎜
⎟
⎜
j15
− j16 (10 + j) ⎟⎠⎜⎝ I 3 ⎟⎠ ⎜ 0 ⎟
⎝
⎝
⎠
Using MATLAB,
or
AI = B
I = inv(A)*B
I1 = −0.128 + j0.3593 = 0.3814∠109.6 o A
I 2 = −0.1946 + j0.2841 = 0.3443∠124.4 o A
I 3 = 0.0718 − j0.1265 = 0.1455∠ − 60.42 o A
I x = I1 − I 2 = 0.0666 + j0.0752 = 0.1005∠48.5 o A
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Chapter 10, Problem 40.
Find io in the circuit shown in Fig. 10.85 using superposition.
Figure 10.85
For Prob. 10.40.
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Chapter 10, Solution 40.
Let i O = i O1 + i O 2 , where i O1 is due to the dc source and i O 2 is due to the ac source. For
i O1 , consider the circuit in Fig. (a).
4Ω
2Ω
iO1
+
−
8V
(a)
Clearly,
i O1 = 8 2 = 4 A
For i O 2 , consider the circuit in Fig. (b).
4Ω
2Ω
IO2
10∠0° V
+
−
j4 Ω
(b)
If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2 = 4 3 Ω .
IO2
2.5∠0° A
4Ω
2Ω
j4 Ω
(c)
By the current division principle,
43
I O2 =
(2.5∠0°)
4 3 + j4
I O 2 = 0.25 − j0.75 = 0.79∠ - 71.56°
Thus,
i O 2 = 0.79 cos(4t − 71.56°) A
Therefore,
i O = i O1 + i O 2 = 4 + 0.79 cos(4t – 71.56°) A
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Chapter 10, Problem 41.
Find v o for the circuit in Fig. 10.86, assuming that v s = 6 cos 2t + 4 sin 4t V.
Figure 10.86
For Prob. 10.41.
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Chapter 10, Solution 41.
We apply superposition principle. We let
vo = v1 + v2
where v1 and v2 are due to the sources 6cos2t and 4sin4t respectively. To find v1,
consider the circuit below.
-j2
+
+
_
6∠0o
2Ω
V1
–
1/ 4F
⎯⎯
→
1
jω C
=
1
= − j2
j 2 x1/ 4
2
(6) = 3 + j 3 = 4.2426 < 45o
2 − j2
Thus,
v1 = 4.2426cos(2t + 45o )
To get v2, consider the circuit below
–j
V1 =
+
4∠0o
+
_
2Ω
V2
–
1/ 4F
⎯⎯
→
1
jω C
=
1
= − j1
j 4 x1/ 4
2
(4) = 3.2 + j11.6 = 3.578 < 26.56o
2− j
v2 = 3.578 sin(4t + 26.56o )
V2 =
Hence,
vo = 4.243cos(2t + 45˚) + 3.578sin(4t + 25.56˚) V.
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Chapter 10, Problem 42.
Solve for I o in the circuit of Fig. 10.87.
Figure 10.87
For Prob. 10.42.
Chapter 10, Solution 42.
Let I o = I1 + I 2
where I1 and I2 are due to 20<0o and 30<45o sources respectively. To get I1, we use the
circuit below.
I1
j10 Ω
60 Ω
o
20∠0 V
+
_
50 Ω
–j40 Ω
Let Z1 = -j40//60 = 18.4615 –j27.6927, Z2 = j10//50=1.9231 + j9.615
Transforming the voltage source to a current source leads to the circuit below.
I1
Z2
Z1
–j2
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Using current division,
Z2
I1 =
(− j 2) = 0.6217 + j 0.3626
Z1 + Z 2
To get I2, we use the circuit below.
j10 Ω
I2
50 Ω
60 Ω
–j40 Ω
+
_
30∠45o V
After transforming the voltage source, we obtain the circuit below.
I2
Z2
Z1
0.5∠45o
Using current division,
− Z1
I2 =
(0.5 < 45o ) = −0.5275 − j 0.3077
Z1 + Z 2
Hence,
I o = I1 + I 2 = 0.0942 + j 0.0509 = 0.109 < 30o A
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Chapter 10, Problem 43.
Using the superposition principle, find i x in the circuit of Fig. 10.88.
Figure 10.88
For Prob. 10.43.
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Chapter 10, Solution 43.
Let I x = I 1 + I 2 , where I 1 is due to the voltage source and I 2 is due to the current
source.
ω= 2
5 cos(2t + 10°) ⎯
⎯→ 5∠10°
10 cos(2t − 60°) ⎯
⎯→ 10 ∠ - 60°
4H ⎯
⎯→ jωL = j8
1
1
1
F ⎯
⎯→
=
= -j4
8
jωC j (2)(1 / 8)
For I 1 , consider the circuit in Fig. (a).
-j4 Ω
3Ω
I1
+
−
j8 Ω
10∠-60° V
(a)
I1 =
10∠ - 60° 10 ∠ - 60°
=
3 + j8 − j4
3 + j4
For I 2 , consider the circuit in Fig. (b).
-j4 Ω
5∠10° A
3Ω
I2
j8 Ω
(b)
I2 =
- j8
- j40 ∠10°
(5∠10°) =
3 + j8 − j4
3 + j4
1
(10∠ - 60° − j40∠10°)
3 + j4
49.51∠ - 76.04°
Ix =
= 9.902∠ - 129.17°
5∠53.13°
i x = 9.902 cos(2t – 129.17°) A
I x = I1 + I 2 =
Therefore,
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Chapter 10, Problem 44.
Use the superposition principle to obtain v x in the circuit of Fig. 10.89. Let
v s = 50 sin 2t V and i s = 12 cos(6t + 10°) A.
Figure 10.89
For Prob. 10.44.
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Chapter 10, Solution 44.
Let v x = v1 + v 2 , where v1 and v2 are due to the current source and voltage source
respectively.
For v1 , ω = 6 , 5 H
⎯
⎯→
jωL = j30
The frequency-domain circuit is shown below.
20 Ω
16 Ω
Is
Let Z = 16 //(20 + j30) =
+
V1
-
16(20 + j30)
= 11.8 + j3.497 = 12.31∠16.5 o
36 + j30
V1 = I s Z = (12∠10 o )(12.31∠16.5 o ) = 147.7∠26.5 o
For v2 , ω = 2 , 5 H
j30
⎯
⎯→
v1 = 147.7 cos(6 t + 26.5 o ) V
jωL = j10
The frequency-domain circuit is shown below.
20 Ω
16 Ω
⎯
⎯→
j10
+
V2
+
Vs
-
-
Using voltage division,
16(50∠0 o )
16
V2 =
Vs =
= 21.41∠ − 15.52 o
16 + 20 + j10
36 + j10
⎯
⎯→
v 2 = 21.41sin(2t − 15.52 o ) V
Thus,
v x = 147.7 cos(6 t + 26.5 o ) + 21.41sin( 2 t − 15.52 o ) V
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Chapter 10, Problem 45.
Use superposition to find i (t ) in the circuit of Fig. 10.90.
Figure 10.90
For Prob. 10.45.
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Chapter 10, Solution 45.
Let i = i1 + i2 , where i1 and i2 are due to 16cos(10t +30o) and 6sin4t sources respectively.
To find i1 , consider the circuit below.
I1
16<30o V
20 Ω
+
_
jX
X = ω L = 10 x300 x10 −3 = 3
16 < 30o
I1 =
= 0.7911
20 + j 3
i1 = 0.7911cos(10t + 21.47o ) A
To find i2 , consider the circuit below.
I2
20 Ω
+
_
6∠0o V
jX
X = ω L = 4 x300 x10 −3 = 1.2
6 < 0o
I2 = −
= 0.2995 < 176.6o
20 + j1.2
i1 = 0.2995 sin(4t + 176.6o ) A
Thus,
i = i1 + i2 = 0.7911cos(10t + 21.47o ) + 0.2995 sin(4t + 176.6o ) A
= 791.1cos(10t+21.47˚)+299.5sin(4t+176.6˚) mA
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Chapter 10, Problem 46.
Solve for vo (t ) in the circuit of Fig. 10.91 using the superposition principle.
Figure 10.91
For Prob. 10.46.
Chapter 10, Solution 46.
Let v o = v1 + v 2 + v 3 , where v1 , v 2 , and v 3 are respectively due to the 10-V dc source,
the ac current source, and the ac voltage source. For v1 consider the circuit in Fig. (a).
6Ω
2H
+
1/12 F
+
−
v1
10 V
−
(a)
The capacitor is open to dc, while the inductor is a short circuit. Hence,
v1 = 10 V
For v 2 , consider the circuit in Fig. (b).
ω= 2
2H ⎯
⎯→ jωL = j4
1
1
1
F ⎯
⎯→
=
= - j6
12
jωC j (2)(1 / 12)
+
6Ω
-j6 Ω
4∠0° A
V2
j4 Ω
−
(b)
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Applying nodal analysis,
V
V
V ⎛1 j j ⎞
4 = 2 + 2 + 2 = ⎜ + − ⎟ V2
6 - j6 j4 ⎝ 6 6 4 ⎠
V2 =
Hence,
24
= 21.45∠26.56°
1 − j0.5
v 2 = 21.45 sin( 2 t + 26.56°) V
For v 3 , consider the circuit in Fig. (c).
ω=3
2H ⎯
⎯→ jωL = j6
1
1
1
F ⎯
⎯→
=
= - j4
12
jωC j (3)(1 / 12)
6Ω
j6 Ω
+
12∠0° V
+
−
-j4 Ω
V3
−
(c)
At the non-reference node,
12 − V3 V3 V3
=
+
6
- j4 j6
12
V3 =
= 10.73∠ - 26.56°
1 + j0.5
Hence,
v 3 = 10.73 cos(3t − 26.56°) V
Therefore,
v o = 10 + 21.45 sin(2t + 26.56°) + 10.73 cos(3t – 26.56°) V
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Chapter 10, Problem 47.
Determine io in the circuit of Fig. 10.92, using the superposition principle.
Figure 10.92
For Prob. 10.47.
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Chapter 10, Solution 47.
Let i o = i1 + i 2 + i 3 , where i1 , i 2 , and i 3 are respectively due to the 24-V dc source, the
ac voltage source, and the ac current source. For i1 , consider the circuit in Fig. (a).
1Ω
24 V
1/6 F
2H
− +
i1
2Ω
4Ω
Since the capacitor is an open circuit to dc,
24
i1 =
=4A
4+2
For i 2 , consider the circuit in Fig. (b).
ω=1
2H ⎯
⎯→ jωL = j2
1
1
F ⎯
⎯→
= - j6
6
jωC
1Ω
j2 Ω
-j6 Ω
I2
10∠-30° V
+
−
I1
2Ω
I2
4Ω
(b)
For mesh 1,
- 10 ∠ - 30° + (3 − j6) I 1 − 2 I 2 = 0
10 ∠ - 30° = 3 (1 − 2 j) I 1 − 2 I 2
(1)
0 = -2 I 1 + (6 + j2) I 2
I 1 = (3 + j) I 2
(2)
For mesh 2,
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Substituting (2) into (1)
10 ∠ - 30° = 13 − j15 I 2
I 2 = 0.504 ∠19.1°
i 2 = 0.504 sin( t + 19.1°) A
Hence,
For i 3 , consider the circuit in Fig. (c).
ω=3
2H ⎯
⎯→ jωL = j6
1
1
1
F ⎯
⎯→
=
= - j2
6
jωC j (3)(1 / 6)
1Ω
j6 Ω
-j2 Ω
I3
2Ω
2∠0° A
4Ω
(c)
2 || (1 − j2) =
2 (1 − j2)
3 − j2
Using current division,
2 (1 − j2)
⋅ (2∠0°)
2 (1 − j2)
3 − j2
=
I3 =
2 (1 − j2)
13 + j3
4 + j6 +
3 − j2
I 3 = 0.3352 ∠ - 76.43°
Hence
i 3 = 0.3352 cos(3t − 76.43°) A
Therefore,
i o = 4 + 0.504 sin(t + 19.1°) + 0.3352 cos(3t – 76.43°) A
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Chapter 10, Problem 48.
Find io in the circuit of Fig. 10.93 using superposition.
Figure 10.93
For Prob. 10.48.
Chapter 10, Solution 48.
Let i O = i O1 + i O 2 + i O 3 , where i O1 is due to the ac voltage source, i O 2 is due to the dc
voltage source, and i O3 is due to the ac current source. For i O1 , consider the circuit in
Fig. (a).
ω = 2000
50 cos(2000t ) ⎯
⎯→ 50∠0°
40 mH ⎯
⎯→
20 µF ⎯
⎯→
jωL = j (2000)(40 × 10 -3 ) = j80
1
1
=
= - j25
jωC j (2000)(20 × 10 -6 )
80 || (60 + 100) = 160 3
50
30
=
I=
160 3 + j80 − j25 32 + j33
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Using current division,
- 80 I
-1
10∠180°
= I=
80 + 160 3
46∠45.9°
I O1 = 0.217 ∠134.1°
Hence,
i O1 = 0.217 cos(2000 t + 134.1°) A
For i O 2 , consider the circuit in Fig. (b).
I O1 =
i O2 =
24
= 0.1 A
80 + 60 + 100
For i O3 , consider the circuit in Fig. (c).
ω = 4000
2 cos(4000t ) ⎯
⎯→ 2∠0°
40 mH ⎯
⎯→
jωL = j (4000)(40 × 10 -3 ) = j160
20 µF ⎯
⎯→
1
1
=
= - j12.5
jωC j (4000)(20 × 10 -6 )
For mesh 1,
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I1 = 2
(1)
For mesh 2,
(80 + j160 − j12.5) I 2 − j160 I 1 − 80 I 3 = 0
Simplifying and substituting (1) into this equation yields
(8 + j14.75) I 2 − 8 I 3 = j32
(2)
For mesh 3,
240 I 3 − 60 I 1 − 80 I 2 = 0
Simplifying and substituting (1) into this equation yields
I 2 = 3 I 3 − 1.5
(3)
Substituting (3) into (2) yields
(16 + j44.25) I 3 = 12 + j54.125
12 + j54.125
I3 =
= 1.1782∠7.38°
16 + j44.25
Hence,
I O 3 = - I 3 = -1.1782 ∠7.38°
i O 3 = -1.1782 sin( 4000t + 7.38°) A
Therefore,
i O = 0.1 + 0.217 cos(2000t + 134.1°) – 1.1782 sin(4000t + 7.38°) A
Chapter 10, Problem 49.
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Using source transformation, find i in the circuit of Fig. 10.94.
Figure 10.94
For Prob. 10.49.
Chapter 10, Solution 49.
8 sin( 200t + 30°) ⎯
⎯→ 8∠30°, ω = 200
5 mH ⎯
⎯→
jωL = j (200)(5 × 10 -3 ) = j
1 mF ⎯
⎯→
1
1
=
= - j5
jωC j (200)(1 × 10 -3 )
After transforming the current source, the circuit becomes that shown in the figure below.
5Ω
40∠30° V
I=
3Ω
I
+
−
jΩ
-j5 Ω
40 ∠30°
40 ∠30°
=
= 4.472∠56.56°
5 + 3 + j − j5
8 − j4
i = 4.472 sin(200t + 56.56°) A
Chapter 10, Problem 50.
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Use source transformation to find vo in the circuit of Fig. 10.95.
Figure 10.95
For Prob. 10.50.
Chapter 10, Solution 50.
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5 cos(10 5 t )
⎯⎯→ 5∠0°, ω = 10 5
0.4 mH ⎯
⎯→
0.2 µF ⎯
⎯→
jωL = j (10 5 )(0.4 × 10 -3 ) = j40
1
1
=
= - j50
5
jωC j (10 )(0.2 × 10 -6 )
After transforming the voltage source, we get the circuit in Fig. (a).
j40 Ω
+
20 Ω
0.25∠0°
-j50 Ω
80 Ω
Vo
−
(a)
Let
Z = 20 || - j 50 =
- j100
2 − j5
- j25
2 − j5
With these, the current source is transformed to obtain the circuit in Fig.(b).
and
Vs = (0.25∠0°) Z =
j40 Ω
Z
+
Vs
+
−
80 Ω
Vo
−
(b)
By voltage division,
80
80
- j25
Vs =
⋅
j
100
Z + 80 + j40
2 − j5
+ 80 + j40
2 − j5
8 (- j25)
Vo =
= 3.615∠ - 40.6°
36 − j42
v o = 3.615 cos(105 t – 40.6°) V
Vo =
Therefore,
Chapter 10, Problem 51.
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Use source transformation to find I o in the circuit of Prob. 10.42.
Chapter 10, Solution 51.
Transforming the voltage sources into current sources, we have the circuit as
shown below.
–j2
j10
50
60
-j40
0.5∠45o
j10 x50
= 1.9231 + j 9.615
50 + j10
V1 = − j 2 Z1 = 19.231 − j 3.846
− j 40 x60
Let Z 2 = − j 40 // 60 =
= 18.4615 − j 27.6923
60 − j 40
V2 = Z 2 x0.5 < 45o = 16.315 − 3.263
Let Z1 = j10 // 50 =
Transforming the current sources to voltage sources leads to the circuit below.
Io
Z2
Z1
V1
+
_
+
_
V2
Applying KVL to the loop gives
−V1 + I o ( Z1 + Z 2 ) + V2 = 0
Io =
⎯⎯
→ Io =
V1 − V2
Z1 + Z 2
19.231 − j 3.846 − 16.316 + j 3.263
= 0.1093 < 30o A = 109.3∠30˚ mA
1.9231 + j 9.615 + 18.4615 − j 27.6923
Chapter 10, Problem 52.
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Use the method of source transformation to find I x in the circuit of Fig. 10.96.
Figure 10.96
For Prob. 10.52.
Chapter 10, Solution 52.
We transform the voltage source to a current source.
60∠0°
Is =
= 6 − j12
2 + j4
The new circuit is shown in Fig. (a).
-j2 Ω
Ix
2Ω
Is = 6 – j12 A
6Ω
j4 Ω
4Ω
5∠90° A
-j3 Ω
(a)
Let
6 (2 + j4)
= 2.4 + j1.8
8 + j4
Vs = I s Z s = (6 − j12)(2.4 + j1.8) = 36 − j18 = 18 (2 − j)
Z s = 6 || (2 + j4) =
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With these, we transform the current source on the left hand side of the circuit to a
voltage source. We obtain the circuit in Fig. (b).
Zs
-j2 Ω
Ix
Vs
4Ω
+
−
j5 A
-j3 Ω
(b)
Let
Z o = Z s − j2 = 2.4 − j0.2 = 0.2 (12 − j)
Vs
18 (2 − j)
=
= 15.517 − j6.207
Io =
Z o 0.2 (12 − j)
With these, we transform the voltage source in Fig. (b) to a current source. We obtain the
circuit in Fig. (c).
Ix
Io
4Ω
Zo
j5 A
-j3 Ω
(c)
Using current division,
Zo
2.4 − j0.2
(I o + j5) =
(15.517 − j1.207)
Ix =
6.4 − j3.2
Z o + 4 − j3
I x = 5 + j1.5625 = 5.238∠17.35° A
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Chapter 10, Problem 53.
Use the concept of source transformation to find V o in the circuit of Fig. 10.97.
Figure 10.97
For Prob. 10.53.
Chapter 10, Solution 53.
We transform the voltage source to a current source to obtain the circuit in Fig. (a).
-j3 Ω
j4 Ω
+
5∠0° A
4Ω
j2 Ω
2Ω
Vo
-j2 Ω
−
(a)
Let
j8
= 0.8 + j1.6
4 + j2
Vs = (5∠0°) Z s = (5)(0.8 + j1.6) = 4 + j8
Z s = 4 || j2 =
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With these, the current source is transformed so that the circuit becomes that shown in
Fig. (b).
-j3 Ω
Zs
j4 Ω
+
Vs
+
−
2Ω
-j2 Ω
Vo
−
(b)
Z x = Z s − j3 = 0.8 − j1.4
V
4 + j8
= −3.0769 + j4.6154
Ix = s =
Z s 0.8 − j1.4
With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c).
Let
j4 Ω
+
Ix
2Ω
Zx
-j2 Ω
Vo
−
(c)
Let
1.6 − j2.8
= 0.8571 − j0.5714
2.8 − j1.4
Vy = I x Z y = (−3.0769 + j4.6154) ⋅ (0.8571 − j0.5714) = j5.7143
Z y = 2 || Z x =
With these, we transform the current source to obtain the circuit in Fig. (d).
Using current division,
j4 Ω
Zy
+
Vy
+
−
-j2 Ω
Vo
−
(d)
Vo =
- j2 ( j5.7143)
- j2
Vy =
= (3.529 – j5.883) V
Z y + j4 − j2
0.8571 − j0.5714 + j4 − j2
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Chapter 10, Problem 54.
Rework Prob. 10.7 using source transformation.
Chapter 10, Solution 54.
50 x(− j 30)
= 13.24 − j 22.059
50 − j 30
We convert the current source to voltage source and obtain the circuit below.
50 //( − j 30) =
40 Ω
+
Vs =115.91 –j31.06V
13.24 – j22.059 Ω
j20 Ω
+
-
I
134.95-j74.912 V
V
-
+
-
Applying KVL gives
-115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0
or I =
− 250.86 + j105.97
= −4.7817 + j1.8055
53.24 − j 2.059
But − Vs + (40 + j20)I + V = 0
⎯
⎯→
V = Vs − (40 + j20)I
V = 115.91 − j31.05 − (40 + j20)(−4.7817 + j1.8055) = 124.06∠ − 154 o V
which agrees with the result in Prob. 10.7.
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Chapter 10, Problem 55.
Find the Thevenin and Norton equivalent circuits at terminals a-b for each of the circuits
in Fig. 10.98.
Figure 10.98
For Prob. 10.55.
Chapter 10, Solution 55.
(a)
To find Z th , consider the circuit in Fig. (a).
j20 Ω
10 Ω
Zth
-j10 Ω
(a)
( j20)(- j10)
j20 − j10
= 10 − j20 = 22.36∠-63.43° Ω
To find Vth , consider the circuit in Fig. (b).
Z N = Z th = 10 + j20 || (- j10) = 10 +
j20 Ω
10 Ω
+
50∠30° V
+
−
-j10 Ω
Vth
−
(b)
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Vth =
IN =
(b)
- j10
(50∠30°) = -50∠30° V
j20 − j10
Vth
- 50 ∠30°
=
= 2.236∠273.4° A
Z th 22.36 ∠ - 63.43°
To find Z th , consider the circuit in Fig. (c).
-j5 Ω
8Ω
Zth
j10 Ω
(c)
Z N = Z th = j10 || (8 − j5) =
( j10)(8 − j5)
= 10∠26° Ω
j10 + 8 − j5
To obtain Vth , consider the circuit in Fig. (d).
-j5 Ω
Io
4∠0° A
8Ω
j10 Ω
+
Vth
−
(d)
By current division,
8
32
(4∠0°) =
Io =
8 + j10 − j5
8 + j5
Vth = j10 I o =
IN =
j320
= 33.92∠58° V
8 + j5
Vth 33.92 ∠58°
=
= 3.392∠32° A
10 ∠26°
Z th
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Chapter 10, Problem 56.
For each of the circuits in Fig. 10.99, obtain Thevenin and Norton equivalent circuits at
terminals a-b.
Figure 10.99
For Prob. 10.56.
Chapter 10, Solution 56.
(a)
To find Z th , consider the circuit in Fig. (a).
j4 Ω
6Ω
-j2 Ω
Zth
(a)
( j4)(- j2)
Z N = Z th = 6 + j4 || (- j2) = 6 +
= 6 − j4
j4 − j2
= 7.211∠-33.69° Ω
By placing short circuit at terminals a-b, we obtain,
I N = 2∠0° A
Vth = Z th I th = (7.211∠ - 33.69°) (2∠0°) = 14.422∠-33.69° V
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(b)
To find Z th , consider the circuit in Fig. (b).
j10 Ω
30 Ω
60 Ω
-j5 Ω
Zth
(b)
30 || 60 = 20
(- j5)(20 + j10)
20 + j5
= 5.423∠-77.47° Ω
Z N = Z th = - j5 || (20 + j10) =
To find Vth and I N , we transform the voltage source and combine the 30 Ω
and 60 Ω resistors. The result is shown in Fig. (c).
j10 Ω
4∠45° A
20 Ω
a
IN
-j5 Ω
(c)
b
20
2
(4∠45°) = (2 − j)(4∠45°)
20 + j10
5
= 3.578∠18.43° A
IN =
Vth = Z th I N = (5.423∠ - 77.47°) (3.578∠18.43°)
= 19.4∠-59° V
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Chapter 10, Problem 57.
Find the Thevenin and Norton equivalent circuits for the circuit shown in Fig. 10.100.
Figure 10.100
For Prob. 10.57.
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Chapter 10, Solution 57.
To find Z th , consider the circuit in Fig. (a).
5Ω
-j10 Ω
2Ω
Zth
j20 Ω
(a)
( j20)(5 − j10)
5 + j10
= 18 − j12 = 21.63∠-33.7° Ω
Z N = Z th = 2 + j20 || (5 − j10) = 2 +
To find Vth , consider the circuit in Fig. (b).
5Ω
-j10 Ω
2Ω
+
60∠120° V
+
−
j20 Ω
Vth
−
(b)
j20
j4
(60 ∠120°) =
(60∠120°)
5 − j10 + j20
1 + j2
= 107.3∠146.56° V
Vth =
IN =
Vth
107.3∠146.56°
=
= 4.961∠-179.7° A
Z th 21.633∠ - 33.7°
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Chapter 10, Problem 58.
For the circuit depicted in Fig. 10.101, find the Thevenin equivalent circuit at terminals
a-b.
Figure 10.101
For Prob. 10.58.
Chapter 10, Solution 58.
Consider the circuit in Fig. (a) to find Z th .
8Ω
Zth
j10 Ω
-j6 Ω
(a)
( j10)(8 − j6)
Z th = j10 || (8 − j6) =
= 5 (2 + j)
8 + j4
= 11.18∠26.56° Ω
Consider the circuit in Fig. (b) to find Vth .
Io
+
8Ω
j10 Ω
5∠45° A
Vth
-j6 Ω
(b)
Io =
8 − j6
4 − j3
(5∠45°) =
(5∠45°)
8 − j6 + j10
4 + j2
Vth = j10 I o =
( j10)(4 − j3)(5∠45°)
= 55.9∠71.56° V
(2)(2 + j)
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Chapter 10, Problem 59.
Calculate the output impedance of the circuit shown in Fig. 10.102.
Figure 10.102
For Prob. 10.59.
Chapter 10, Solution 59.
Insert a 1-A current source at the output as shown below.
-j2 Ω
10 Ω
V1
+
–
Vo
j40 Ω
0.2 Vo
+
Vin
1A
–
v1
j 40
But vo = −1(− j 2) = j 2
0.2vo + 1 =
j 2 x0.2 + 1 =
V1
j 40
⎯⎯
→ V1 = −16 + j 40
Vin = V1 – Vo + 10 = –6 + j38 = 1xZin
Zin = –6 + j38 Ω.
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Chapter 10, Problem 60.
Find the Thevenin equivalent of the circuit in Fig. 10.103 as seen from:
(b) terminals c-d
(a) terminals a-b
Figure 10.103
For Prob. 10.60.
Chapter 10, Solution 60.
(a)
To find Z th , consider the circuit in Fig. (a).
10 Ω
-j4 Ω
a
j5 Ω
Zth
4Ω
b
(a)
Z th = 4 || (- j4 + 10 || j5) = 4 || (- j4 + 2 + j4)
Z th = 4 || 2 = 1.333 Ω
To find Vth , consider the circuit in Fig. (b).
10 Ω
V1
-j4 Ω
V2
+
20∠0° V
+
−
j5 Ω
4∠0° A
4Ω
Vth
−
(b)
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At node 1,
20 − V1 V1 V1 − V2
=
+
10
j5
- j4
(1 + j0.5) V1 − j2.5 V2 = 20
(1)
At node 2,
V1 − V2 V2
=
- j4
4
V1 = (1 − j) V2 + j16
4+
(2)
Substituting (2) into (1) leads to
28 − j16 = (1.5 − j3) V2
28 − j16
V2 =
= 8 + j5.333
1.5 − j3
Therefore,
Vth = V2 = 9.615∠33.69° V
(b)
To find Z th , consider the circuit in Fig. (c).
Zth
c
d
10 Ω
-j4 Ω
j5 Ω
4Ω
(c)
⎛
j10 ⎞
⎟
Z th = - j4 || (4 + 10 || j5) = - j4 || ⎜ 4 +
2 + j⎠
⎝
- j4
Z th = - j4 || (6 + j4) =
(6 + j4) = 2.667 – j4 Ω
6
To find Vth ,we will make use of the result in part (a).
V2 = 8 + j5.333 = (8 3 ) (3 + j2)
V1 = (1 − j) V2 + j16 = j16 + (8 3) (5 − j)
Vth = V1 − V2 = 16 3 + j8 = 9.614∠56.31° V
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Chapter 10, Problem 61.
Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 10.104.
Figure 10.104
For Prob. 10.61.
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Chapter 10, Solution 61.
To find VTh, consider the circuit below
4Ω
Vo
a
Ix
+
-j3 Ω
o
2∠0 A
1.5Ix
VTh
–
b
2 + 1.5Ix = Ix
But
Ix = –4
Vo = –j3Ix = j12
VTh = Vo + 6Ix = j12 − 24 V
To find ZTh, consider the circuit shown below.
4Ω
Vo
Ix
-j3 Ω
1+1.5 Ix = Ix
1A
Ix = -2
−Vo + Ix(4 − j 3) = 0
ZTh =
1.5Ix
⎯⎯
→ Vo = −8 + j6
Vo
= −8 + j 6 Ω
1
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Chapter 10, Problem 62.
Using Thevenin’s theorem, find vo in the circuit of Fig. 10.105.
Figure 10.105
For Prob. 10.62.
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Chapter 10, Solution 62.
First, we transform the circuit to the frequency domain.
12 cos( t ) ⎯
⎯→ 12∠0°, ω = 1
2H ⎯
⎯→
1
F ⎯
⎯→
4
1
F ⎯
⎯→
8
jωL = j2
1
= - j4
jωC
1
= - j8
jωC
To find Z th , consider the circuit in Fig. (a).
3 Io
Io
4Ω
Vx
j2 Ω
1
Ix
2
-j4 Ω
-j8 Ω
+
−
1V
(a)
At node 1,
Vx Vx
1 − Vx
+
+ 3Io =
,
4 - j4
j2
Thus,
where I o =
- Vx
4
Vx 2 Vx 1 − Vx
−
=
- j4
4
j2
Vx = 0.4 + j0.8
At node 2,
I x + 3Io =
1 1 − Vx
+
- j8
j2
I x = (0.75 + j0.5) Vx − j
3
8
I x = -0.1 + j0.425
Z th =
1
= -0.5246 − j2.229 = 2.29∠ - 103.24° Ω
Ix
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To find Vth , consider the circuit in Fig. (b).
3 Io
Io
4Ω
j2 Ω
V1
V2
1
12∠0° V
+
−
2
-j4 Ω
-j8 Ω
+
Vth
−
(b)
At node 1,
V
V − V2
12 − V1
= 3Io + 1 + 1
,
4
- j4
j2
24 = (2 + j) V1 − j2 V2
where I o =
12 − V1
4
(1)
At node 2,
V1 − V2
V
+ 3Io = 2
j2
- j8
72 = (6 + j4) V1 − j3 V2
(2)
From (1) and (2),
⎡ 24⎤ ⎡ 2 + j - j2⎤ ⎡ V1 ⎤
⎢ 72 ⎥ = ⎢ 6 + j4 - j3⎥ ⎢ V ⎥
⎣ ⎦ ⎣
⎦⎣ 2⎦
∆ = -5 + j6 ,
Vth = V2 =
Thus,
∆ 2 = - j24
∆2
= 3.073∠ - 219.8°
∆
2
(2)(3.073∠ - 219.8°)
Vth =
2 + Z th
1.4754 − j2.229
6.146∠ - 219.8°
Vo =
= 2.3∠ - 163.3°
2.673∠ - 56.5°
Vo =
Therefore,
v o = 2.3 cos(t – 163.3°) V
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Chapter 10, Problem 63.
Obtain the Norton equivalent of the circuit depicted in Fig. 10.106 at terminals a-b.
Figure 10.106
For Prob. 10.63.
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Chapter 10, Solution 63.
Transform the circuit to the frequency domain.
4 cos(200t + 30°) ⎯
⎯→ 4∠30°, ω = 200
10 H ⎯
⎯→
5 µF ⎯
⎯→
jωL = j (200)(10) = j2 kΩ
1
1
=
= - j kΩ
jωC j (200)(5 × 10 -6 )
Z N is found using the circuit in Fig. (a).
-j kΩ
j2 kΩ
ZN
2 kΩ
(a)
Z N = - j + 2 || j2 = - j + 1 + j = 1 kΩ
We find I N using the circuit in Fig. (b).
-j kΩ
4∠30° A
j2 kΩ
2 kΩ
IN
(b)
j2 || 2 = 1 + j
By the current division principle,
1+ j
IN =
(4 ∠30°) = 5.657 ∠75°
1+ j − j
Therefore,
i N = 5.657 cos(200t + 75°) A
Z N = 1 kΩ
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Chapter 10, Problem 64.
For the circuit shown in Fig. 10.107, find the Norton equivalent circuit at terminals a-b.
Figure 10.107
For Prob. 10.64.
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Chapter 10, Solution 64.
Z N is obtained from the circuit in Fig. (a).
60 Ω
40 Ω
ZN
-j30 Ω
j80 Ω
(a)
Z N = (60 + 40) || ( j80 − j30) = 100 || j50 =
(100)( j50)
100 + j50
Z N = 20 + j40 = 44.72∠63.43° Ω
To find I N , consider the circuit in Fig. (b).
60 Ω
3∠60° A
I1
40 Ω
I2
-j30 Ω
IN
Is
j80 Ω
(b)
I s = 3∠60°
For mesh 1,
100 I 1 − 60 I s = 0
I 1 = 1.8∠60°
For mesh 2,
( j80 − j30) I 2 − j80 I s = 0
I 2 = 4.8∠60°
IN = I2 – I1 = 3∠60° A
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Chapter 10, Problem 65.
Compute io in Fig. 10.108 using Norton’s theorem.
Figure 10.108
For Prob. 10.65.
Chapter 10, Solution 65.
5 cos(2 t ) ⎯
⎯→ 5∠0°, ω = 2
4H ⎯
⎯→
1
F ⎯
⎯→
4
1
F ⎯
⎯→
2
jωL = j (2)(4) = j8
1
1
=
= - j2
jωC j (2)(1 / 4)
1
1
=
= -j
jωC j (2)(1 / 2)
To find Z N , consider the circuit in Fig. (a).
2Ω
ZN
-j2 Ω
-j Ω
(a)
Z N = - j || (2 − j2) =
- j (2 − j2) 1
= (2 − j10)
2 − j3
13
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To find I N , consider the circuit in Fig. (b).
5∠0° V
2Ω
+ −
-j2 Ω
-j Ω
IN
(b)
IN =
5∠0°
= j5
-j
The Norton equivalent of the circuit is shown in Fig. (c).
Io
ZN
IN
j8 Ω
(c)
Using current division,
Io =
ZN
(1 13)(2 − j10)( j5) 50 + j10
IN =
=
(1 13)(2 − j10) + j8 2 + j94
Z N + j8
I o = 0.1176 − j0.5294 = 0542∠ - 77.47°
Therefore, i o = 542 cos(2t – 77.47°) mA
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Chapter 10, Problem 66.
At terminals a-b, obtain Thevenin and Norton equivalent circuits for the network
depicted in Fig. 10.109. Take ω = 10 rad/s.
Figure 10.109
For Prob. 10.66.
Chapter 10, Solution 66.
ω = 10
0.5 H ⎯
⎯→
jωL = j (10)(0.5) = j5
1
1
10 mF ⎯
⎯→
=
= - j10
jωC j (10)(10 × 10 -3 )
To find Z th , consider the circuit in Fig. (a).
-j10 Ω
Vx
+
10 Ω
Vo
j5 Ω
2 Vo
1A
−
(a)
Vx
Vx
+
,
j5 10 − j10
19 Vx
V
- 10 + j10
1+
= x ⎯
⎯→ Vx =
10 − j10
j5
21 + j2
1 + 2 Vo =
Z N = Z th =
where Vo =
10Vx
10 − j10
Vx
14.142 ∠135°
=
= 0.67∠129.56° Ω
1
21.095∠5.44°
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To find Vth and I N , consider the circuit in Fig. (b).
12∠0° V
-j10 Ω
− +
+
+
10 Ω
-j2 A
Vo
j5 Ω
I
2 Vo
Vth
−
−
(b)
where
Thus,
(10 − j10 + j5) I − (10)(- j2) + j5 (2 Vo ) − 12 = 0
Vo = (10)(- j2 − I )
(10 − j105) I = -188 − j20
188 + j20
I=
- 10 + j105
Vth = j5 (I + 2 Vo ) = j5 (−19I − j40) = − j95 I + 200
Vth =
− j95 (188 + j20)
+ 200 = 29.73 + j1.8723
- 10 + j105
Vth = 29.79∠3.6° V
IN =
Vth
29.79∠3.6°
=
= 44.46∠–125.96° A
Z th 0.67∠129.56°
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Chapter 10, Problem 67.
Find the Thevenin and Norton equivalent circuits at terminals a-b in the circuit of Fig.
10.110.
Figure 10.110
For Prob. 10.67.
Chapter 10, Solution 67.
Z N = Z Th = 10 //(13 − j5) + 12 //(8 + j6) =
Va =
10
(60∠45 o ) = 13.78 + j21.44,
23 − j5
10(13 − j5) 12(8 + j6)
+
= 11.243 + j1.079Ω
23 − j5
20 + j6
Vb =
(8 + j6)
(60∠45 o ) = 12.069 + j26.08Ω
20 + j6
VTh = Va − Vb = 1.711 − j4.64 = 4.945∠ − 69.76 o V,
V
4.945∠ − 69.76°
I N = Th =
= 0.4378∠ − 75.24 o A
Z Th
11.295∠5.48°
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Chapter 10, Problem 68.
Find the Thevenin equivalent at terminals a-b in the circuit of Fig. 10.111.
Figure 10.111
For Prob. 10.68.
Chapter 10, Solution 68.
1H
⎯
⎯→
jωL = j10x1 = j10
1
1
1
F
⎯
⎯→
=
= − j2
1
20
jω C
j10 x
20
We obtain VTh using the circuit below.
Io
4Ω
a
+
6<0
-
o
+
Vo/3
j10(− j2)
= − j2.5
j10 − j2
Vo = 4I o x (− j2.5) = − j10I o
1
− 6 + 4I o + Vo = 0
3
+
-
-j2 j10
Vo
4Io
b
j10 //(− j2) =
(1)
(2)
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Combining (1) and (2) gives
Io =
6
,
4 − j10 / 3
VTh = Vo = − j10I o =
− j60
= 11.52∠ − 50.19 o
4 − j10 / 3
v Th = 11.52 sin(10 t − 50.19 o )
To find RTh, we insert a 1-A source at terminals a-b, as shown below.
Io
4Ω
a
+
+
-
Vo/3
-j2 j10
Vo
4Io
-
1
4I o + Vo = 0
3
1 + 4I o =
⎯⎯→
1<0o
V
Io = − o
12
Vo Vo
+
− j2 j10
Combining the two equations leads to
Vo =
1
= 1.2293 − j1.4766
0.333 + j0.4
V
Z Th = o = 1.2293 − 1.477Ω
1
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Chapter 10, Problem 69.
For the differentiator shown in Fig. 10.112, obtain V o /V s . Find vo (t ) when v s (t) = V m
sin ωt and ω = 1/RC.
Figure 10.112
For Prob. 10.69.
Chapter 10, Solution 69.
This is an inverting op amp so that
Vo - Z f
-R
=
=
= -jωRC
Vs
Zi
1 jωC
When Vs = Vm and ω = 1 RC ,
1
Vo = - j ⋅
⋅ RC ⋅ Vm = - j Vm = Vm ∠ - 90°
RC
Therefore,
v o ( t ) = Vm sin(ωt − 90°) = - Vm cos(ωt)
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Chapter 10, Problem 70.
The circuit in Fig. 10.113 is an integrator with a feedback resistor. Calculate vo (t ) if
v s = 2 cos 4 × 10 4 t V.
Figure 10.113
For Prob. 10.70.
Chapter 10, Solution 70.
This may also be regarded as an inverting amplifier.
2 cos(4 × 10 4 t ) ⎯
⎯→ 2 ∠0°, ω = 4 × 10 4
1
1
⎯→
=
= - j2.5 kΩ
10 nF ⎯
4
jωC j (4 × 10 )(10 × 10 -9 )
Vo - Z f
=
Vs
Zi
where Z i = 50 kΩ and Z f = 100k || (- j2.5k ) =
- j100
kΩ .
40 − j
Vo
j2
=
Vs 40 − j
Thus,
If Vs = 2 ∠0° ,
Vo =
Therefore,
j4
4∠90°
=
= 0.1∠91.43°
40 − j 40.01∠ - 1.43°
v o ( t ) = 0.1 cos(4x104 t + 91.43°) V
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Chapter 10, Problem 71.
Find vo in the op amp circuit of Fig. 10.114.
Figure 10.114
For Prob. 10.71.
Chapter 10, Solution 71.
8 cos(2t + 30 o )
0. 5µF
⎯
⎯→
⎯⎯→ 8∠30 o
1
1
=
= − j1MΩ
jωC j2x 0.5x10 − 6
At the inverting terminal,
Vo − 8∠30 o Vo − 8∠30 o 8∠30 o
+
=
− j1000k
10k
2k
⎯
⎯→
Vo (1 − j100) = 8∠30 + 800∠ − 60° + 4000 ∠ − 60°
Vo =
6.928 + j4 + 2400 − j4157 4800∠ − 59.9°
=
= 48∠29.53o
1 − j100
100∠ − 89.43°
vo(t) = 48cos(2t + 29.53o) V
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Chapter 10, Problem 72.
Compute io (t ) in the op amp circuit in Fig. 10.115 if v s = 4 cos10 4 t V.
Figure 10.115
For Prob. 10.72.
Chapter 10, Solution 72.
4 cos(10 4 t ) ⎯
⎯→ 4 ∠0°, ω = 10 4
1
1
⎯→
=
= - j100 kΩ
1 nF ⎯
4
jωC j (10 )(10 -9 )
Consider the circuit as shown below.
50 kΩ
Vo
+
−
4∠0° V
+
−
-j100 kΩ
At the noninverting node,
Vo
4 − Vo
=
50
- j100
Io =
Therefore,
⎯
⎯→ Vo =
Vo
Io
100 kΩ
4
1 + j0.5
Vo
4
=
mA = 35.78∠ - 26.56° µA
100k (100)(1 + j0.5)
i o ( t ) = 35.78 cos(104 t – 26.56°) µA
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Chapter 10, Problem 73.
If the input impedance is defined as Z in =V s /I s find the input impedance of the op amp
circuit in Fig. 10.116 when R1 = 10 k Ω, R2 = 20 k Ω, C1 = 10 nF, and ω = 5000 rad/s.
Figure 10.116
For Prob. 10.73.
Chapter 10, Solution 73.
As a voltage follower, V2 = Vo
1
1
=
= -j20 kΩ
3
jωC1 j (5 × 10 )(10 × 10 -9 )
1
1
⎯→
=
= -j10 kΩ
C 2 = 20 nF ⎯
3
jωC 2 j (5 × 10 )(20 × 10 -9 )
⎯→
C1 = 10 nF ⎯
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Consider the circuit in the frequency domain as shown below.
-j20 kΩ
Is 10 kΩ
20 kΩ
+
−
V1
VS
+
−
V2
Io
Vo
-j10 kΩ
Zin
At node 1,
Vs − V1 V1 − Vo V1 − Vo
=
+
10
- j20
20
2 Vs = (3 + j)V1 − (1 + j)Vo
(1)
At node 2,
V1 − Vo Vo − 0
=
20
- j10
V1 = (1 + j2)Vo
(2)
Substituting (2) into (1) gives
2 Vs = j6Vo
or
1
Vo = -j Vs
3
⎛2
1⎞
V1 = (1 + j2)Vo = ⎜ − j ⎟ Vs
⎝3
3⎠
Vs − V1 (1 3)(1 + j)
Vs
=
10k
10k
1+ j
=
30k
Is =
Is
Vs
Vs 30k
=
= 15 (1 − j) k
Is 1 + j
Z in = 21.21∠–45° kΩ
Z in =
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Chapter 10, Problem 74.
Evaluate the voltage gain A v = V o /V s in the op amp circuit of Fig. 10.117. Find A v at
ω = 0, ω → ∞, ω = 1 / R1C1 , and ω = 1 / R2 C 2 .
Figure 10.117
For Prob. 10.74.
Chapter 10, Solution 74.
Zi = R1 +
1
,
jωC1
Zf = R 2 +
1
jωC 2
1
⎛ C ⎞ ⎛ 1 + jω R 2 C 2 ⎞
V
- Zf
jωC 2
⎟⎟
Av = o =
=−
= − ⎜⎜ 1 ⎟⎟ ⎜⎜
1
Vs
Zi
C
1
j
R
C
+
ω
1 1 ⎠
⎝ 2⎠⎝
R1 +
jωC1
R2 +
Av = –
At ω = 0 ,
As ω → ∞ ,
Av = –
C1
C2
R2
R1
At ω =
1
,
R 1 C1
⎛ C ⎞ ⎛ 1 + j R 2 C 2 R 1C1 ⎞
⎟
A v = –⎜ 1 ⎟ ⎜
1+ j
⎠
⎝ C2 ⎠ ⎝
At ω =
1
,
R 2C2
⎛C ⎞⎛
⎞
1+ j
⎟
A v = –⎜ 1 ⎟ ⎜
⎝ C 2 ⎠ ⎝ 1 + j R 1C1 R 2 C 2 ⎠
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Chapter 10, Problem 75.
In the op amp circuit of Fig. 10.118, find the closed-loop gain and phase shift of the
output voltage with respect to the input voltage if C1 = C 2 = 1 nF, R1 = R2 = 100 k Ω ,
R3 = 20 k Ω , R4 = 40 k Ω , and ω = 2000 rad/s.
Figure 10.118
For Prob. 10.75.
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Chapter 10, Solution 75.
ω = 2 × 10 3
⎯→
C1 = C 2 = 1 nF ⎯
1
1
=
= -j500 kΩ
3
jωC1 j (2 × 10 )(1 × 10 -9 )
Consider the circuit shown below.
100 kΩ
-j500 kΩ
-j500 kΩ
V2
+
−
V1
40 kΩ
VS
+
−
100 kΩ
+
Vo
20 kΩ
−
Let Vs = 10V.
At node 1,
[(V1–10)/(–j500k)] + [(V1–Vo)/105] + [(V1–V2)/(–j500k)] = 0
or (1+j0.4)V1 – j0.2V2 – Vo = j2
(1)
[(V2–V1)/(–j5)] + (V2–0) = 0
or –j0.2V1 + (1+j0.2)V2 = 0 or V1 = (1–j5)V2
(2)
At node 2,
But
V2 =
V
R3
Vo = o
R3 + R4
3
From (2) and (3),
V1 = (0.3333–j1.6667)Vo
(3)
(4)
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Substituting (3) and (4) into (1),
(1+j0.4)(0.3333–j1.6667)Vo – j0.06667Vo – Vo = j2
(1.077∠21.8˚)(1.6997∠–78.69˚) = 1.8306∠–56.89˚ = 1 – j1.5334
Thus,
(1–j1.5334)Vo – j0.06667Vo – Vo = j2
and, Vo = j2/(–j1.6601) = –1.2499 = 1.2499∠180˚ V
Since Vs = 10,
Vo/Vs = 0.12499∠180˚.
Checking with MATLAB.
>> Y=[1+0.4i,-0.2i,-1;1,-1+5i,0;0,-3,1]
Y=
1.0000 + 0.4000i
0 - 0.2000i -1.0000
1.0000
-1.0000 + 5.0000i
0
0
-3.0000
1.0000
>> I=[2i;0;0]
I=
0 + 2.0000i
0
0
>> V=inv(Y)*I
V=
-0.4167 + 2.0833i
-0.4167
-1.2500 + 0.0000i (this last term is vo)
and, the answer checks.
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Chapter 10, Problem 76.
Determine V o and I o in the op amp circuit of Fig. 10.119.
Figure 10.119
For Prob. 10.76.
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Chapter 10, Solution 76.
Let the voltage between the -jk Ω capacitor and the 10k Ω resistor be V1.
2∠30 o − V1 V1 − Vo V1 − Vo
=
+
− j4k
10k
20k
⎯
⎯→
(1)
2∠30 o = (1 − j0.6)V1 + j0.6Vo
= 1.7321+j1
Also,
V1 − Vo
V
= o
10k
− j2k
⎯⎯→
V1 = (1 + j5)Vo
(2)
Solving (2) into (1) yields
2∠30° = (1 − j0.6)(1 + j5)Vo + j0.6Vo = (1 + 3 − j0.6 + j5 + j6)Vo
= (4+j5)Vo
2∠30°
Vo =
= 0.3124∠ − 21.34 o V
6.403∠51.34°
>> Y=[1-0.6i,0.6i;1,-1-0.5i]
Y=
1.0000 - 0.6000i
0 + 0.6000i
1.0000
-1.0000 - 5.0000i
>> I=[1.7321+1i;0]
I=
1.7321 + 1.0000i
0
>> V=inv(Y)*I
V=
0.8593 + 1.3410i
0.2909 - 0.1137i = Vo = 0.3123∠–21.35˚V. Answer checks.
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Chapter 10, Problem 77.
Compute the closed-loop gain V o /V s for the op amp circuit of Fig. 10.120.
Figure 10.120
For Prob. 10.77.
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Chapter 10, Solution 77.
Consider the circuit below.
R3
2
R1
1
+
−
VS
At node 1,
V1
V1
C2
R2
−
+
+
Vo
C1
−
Vs − V1
= jωC V1
R1
Vs = (1 + jωR 1C1 ) V1
(1)
At node 2,
0 − V1 V1 − Vo
=
+ jωC 2 (V1 − Vo )
R3
R2
⎛ R3
⎞
V1 = (Vo − V1 ) ⎜
+ jωC 2 R 3 ⎟
⎝R2
⎠
⎛
⎞
1
⎟ V1
Vo = ⎜1 +
⎝ (R 3 R 2 ) + jωC 2 R 3 ⎠
(2)
From (1) and (2),
Vo =
⎛
⎞
Vs
R2
⎜1 +
⎟
1 + jωR 1C1 ⎝ R 3 + jωC 2 R 2 R 3 ⎠
Vo
R 2 + R 3 + jωC 2 R 2 R 3
=
Vs (1 + jωR 1C 1 ) ( R 3 + jωC 2 R 2 R 3 )
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Chapter 10, Problem 78.
Determine vo (t ) in the op amp circuit in Fig. 10.121 below.
Figure 10.121
For Prob. 10.78.
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Chapter 10, Solution 78.
2 sin(400t ) ⎯
⎯→ 2∠0°, ω = 400
1
1
⎯→
=
= - j5 kΩ
0.5 µF ⎯
jωC j (400)(0.5 × 10 -6 )
1
1
⎯→
=
= - j10 kΩ
0.25 µF ⎯
jωC j (400)(0.25 × 10 -6 )
Consider the circuit as shown below.
20 kΩ
10 kΩ V
1
2∠0° V
+
−
-j5 kΩ
V
+
−
V
40 kΩ
-j10 kΩ
10 kΩ
20 kΩ
At node 1,
V
V − V2 V1 − Vo
2 − V1
= 1 + 1
+
10
- j10
- j5
20
4 = (3 + j6) V1 − j4 V2 − Vo
(1)
V1 − V2 V2
=
10
− j5
V1 = (1 − j0.5) V2
(2)
At node 2,
But
20
1
Vo = Vo
20 + 40
3
From (2) and (3),
1
V1 = ⋅ (1 − j0.5) Vo
3
Substituting (3) and (4) into (1) gives
1
4
1⎞
⎛
4 = (3 + j6) ⋅ ⋅ (1 − j0.5) Vo − j Vo − Vo = ⎜1 + j ⎟ Vo
3
3
6⎠
⎝
24
Vo =
= 3.945∠ − 9.46°
6+ j
Therefore,
v o ( t ) = 3.945 sin(400t – 9.46°) V
V2 =
(3)
(4)
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Chapter 10, Problem 79.
For the op amp circuit in Fig. 10.122, obtain vo (t ) .
Figure 10.122
For Prob. 10.79.
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Chapter 10, Solution 79.
5 cos(1000 t ) ⎯
⎯→ 5∠0°, ω = 1000
1
1
⎯→
=
= - j10 kΩ
0.1 µF ⎯
jωC j (1000)(0.1 × 10 -6 )
1
1
⎯→
=
= - j5 kΩ
0.2 µF ⎯
jωC j (1000)(0.2 × 10 -6 )
Consider the circuit shown below.
20 kΩ
-j10 kΩ
10 kΩ
Vs = 5∠0° V
−
+
+
−
40 kΩ
V1
−
+
-j5 kΩ
Since each stage is an inverter, we apply Vo =
+
Vo
−
- Zf
V to each stage.
Zi i
Vo =
- 40
V1
- j5
(1)
V1 =
- 20 || (- j10)
Vs
10
(2)
and
From (1) and (2),
⎛ - j8 ⎞⎛ - (20)(-j10) ⎞
⎟ 5∠0°
⎟⎜
Vo = ⎜
⎝ 10 ⎠⎝ 20 − j10 ⎠
Vo = 16 (2 + j) = 35.78∠26.56°
Therefore,
v o ( t ) = 35.78 cos(1000t + 26.56°) V
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Chapter 10, Problem 80.
Obtain vo (t ) for the op amp circuit in Fig. 10.123 if v s = 4 cos(1000t − 60°) V.
Figure 10.123
For Prob. 10.80.
Chapter 10, Solution 80.
4 cos(1000t − 60°) ⎯
⎯→ 4∠ - 60°, ω = 1000
1
1
⎯→
=
= - j10 kΩ
0.1 µF ⎯
jωC j (1000)(0.1 × 10 -6 )
1
1
⎯→
=
= - j5 kΩ
0.2 µF ⎯
jωC j (1000)(0.2 × 10 -6 )
The two stages are inverters so that
⎛ 20
20 ⎞⎛ - j5 ⎞
⎟
Vo = ⎜
V ⎟⎜
⋅ (4∠ - 60°) +
50 o ⎠⎝ 10 ⎠
⎝ - j10
=
-j 2
-j
⋅ ( j2) ⋅ (4∠ - 60°) + ⋅ Vo
2
2 5
(1 + j 5) Vo = 4∠ - 60°
Vo =
Therefore,
4∠ - 60°
= 3.922 ∠ - 71.31°
1+ j 5
v o ( t ) = 3.922 cos(1000t – 71.31°) V
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Chapter 10, Problem 81.
Use PSpice to determine V o in the circuit of Fig. 10.124. Assume ω = 1 rad/s.
Figure 10.124
For Prob. 10.81.
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Chapter 10, Solution 81.
We need to get the capacitance and inductance corresponding to –j2 Ω and j4 Ω.
1
1
− j2
⎯⎯
→ C=
=
= 0.5F
ω X c 1x 2
X
j4
⎯⎯
→ L = L = 4H
ω
The schematic is shown below.
When the circuit is simulated, we obtain the following from the output file.
FREQ
VM(5)
VP(5)
1.592E-01 1.127E+01 -1.281E+02
From this, we obtain
Vo = 11.27∠128.1o V.
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Chapter 10, Problem 82.
Solve Prob. 10.19 using PSpice.
Chapter 10, Solution 82.
The schematic is shown below. We insert PRINT to print Vo in the output file. For AC
Sweep, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we print out the output file which includes:
FREQ
1.592 E-01
which means that
VM($N_0001)
7.684 E+00
VP($N_0001)
5.019 E+01
Vo = 7.684∠50.19o V
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Chapter 10, Problem 83.
Use PSpice to find vo (t ) in the circuit of Fig. 10.125. Let i s = 2 cos(10 3 t ) A.
Figure 10.125
For Prob. 10.83.
Chapter 10, Solution 83.
The schematic is shown below. The frequency is f = ω / 2π =
1000
= 159.15
2π
When the circuit is saved and simulated, we obtain from the output file
FREQ
1.592E+02
Thus,
VM(1)
6.611E+00
VP(1)
-1.592E+02
vo = 6.611cos(1000t – 159.2o) V
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Chapter 10, Problem 84.
Obtain V o in the circuit of Fig. 10.126 using PSpice.
Figure 10.126
For Prob. 10.84.
Chapter 10, Solution 84.
The schematic is shown below. We set PRINT to print Vo in the output file. In AC
Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we obtain the output file which includes:
FREQ
VM($N_0003)
1.592 E-01
1.664 E+00
VP($N_0003)
-1.646
E+02
Namely,
Vo = 1.664∠-146.4o V
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Chapter 10, Problem 85.
Use PSpice to find V o in the circuit of Fig. 10.127.
Figure 10.127
For Prob. 10.85.
Chapter 10, Solution 85.
The schematic is shown below. We let ω = 1 rad/s so that L=1H and C=1F.
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When the circuit is saved and simulated, we obtain from the output file
FREQ
VM($N_0001) VP($N_0001)
1.592E-01 4.471E-01 1.437E+01
From this, we conclude that
Vo = 447.1∠14.37˚ mV
Checking using MATLAB and nodal analysis we get,
>> Y=[1.5,-0.25,-0.25,0;0,1.25,-1.25,1i;-0.5,-1,1.5,0;0,1i,0,0.5-1i]
Y=
1.5000
0
-0.5000
0
-0.2500
-0.2500
1.2500
-1.2500
-1.0000
1.5000
0 + 1.0000i
0
0
0 + 1.0000i
0
0.5000 - 1.0000i
>> I=[0;0;2;-2]
I=
0
0
2
-2
>> V=inv(Y)*I
V=
0.4331 + 0.1110i = Vo = 0.4471∠14.38˚, answer checks.
0.6724 + 0.3775i
1.9260 + 0.2887i
-0.1110 - 1.5669i
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Chapter 10, Problem 86.
Use PSpice to find V 1 , V 2 , and V 3 in the network of Fig. 10.128.
Figure 10.128
For Prob. 10.86.
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Chapter 10, Solution 86.
The schematic is shown below. We insert three pseudocomponent PRINTs at nodes 1, 2,
and 3 to print V1, V2, and V3, into the output file. Assume that w = 1, we set Total Pts =
1, Start Freq = 0.1592, and End Freq = 0.1592. After saving and simulating the circuit,
we obtain the output file which includes:
FREQ
VM($N_0002)
1.592 E-01
6.000 E+01
FREQ
VM($N_0003)
1.592 E-01
2.367 E+02
FREQ
VM($N_0001)
1.592 E-01
1.082 E+02
VP($N_0002)
3.000
E+01
VP($N_0003)
-8.483
E+01
VP($N_0001)
1.254
E+02
Therefore,
V1 = 60∠30o V V2 = 236.7∠-84.83o V V3 = 108.2∠125.4o V
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Chapter 10, Problem 87.
Determine V 1 , V 2 , and V 3 in the circuit of Fig. 10.129 using PSpice.
Figure 10.129
For Prob. 10.87.
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Chapter 10, Solution 87.
The schematic is shown below. We insert three PRINTs at nodes 1, 2, and 3. We set
Total Pts = 1, Start Freq = 0.1592, End Freq = 0.1592 in the AC Sweep box. After
simulation, the output file includes:
FREQ
VM($N_0004)
1.592 E-01
1.591 E+01
FREQ
VM($N_0001)
1.592 E-01
5.172 E+00
FREQ
VM($N_0003)
1.592 E-01
2.270 E+00
VP($N_0004)
1.696
E+02
VP($N_0001)
-1.386
E+02
VP($N_0003)
-1.524
E+02
Therefore,
V1 = 15.91∠169.6o V V2 = 5.172∠-138.6o V V3 = 2.27∠-152.4o V
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Chapter 10, Problem 88.
Use PSpice to find vo and io in the circuit of Fig. 10.130 below.
Figure 10.130
For Prob. 10.88.
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Chapter 10, Solution 88.
The schematic is shown below. We insert IPRINT and PRINT to print Io and Vo in the
output file. Since w = 4, f = w/2π = 0.6366, we set Total Pts = 1, Start Freq = 0.6366,
and End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes:
FREQ
VM($N_0002)
6.366 E-01
3.496 E+01
1.261
FREQ
IM(V_PRINT2)
IP
6.366 E-01
8.912 E-01
VP($N_0002)
E+01
(V_PRINT2)
-8.870 E+01
Therefore,
Vo = 34.96∠12.6o V, Io = 0.8912∠-88.7o A
vo = 34.96 cos(4t + 12.6o)V,
io = 0.8912cos(4t - 88.7o )A
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Chapter 10, Problem 89.
The op amp circuit in Fig. 10.131 is called an inductance simulator. Show that the input
impedance is given by
Ζin =
Vin
= jωLeq
Ι in
where
Leq =
R1R3 R4
C
R2
Figure 10.131
For Prob. 10.89.
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Chapter 10, Solution 89.
Consider the circuit below.
R1
R2
Vin
2
R3
1
Vin
C
4
R4
3
−
+
Iin
+
−
−
+
Vin
At node 1,
0 − Vin Vin − V2
=
R1
R2
R2
- Vin + V2 =
V
R 1 in
(1)
At node 3,
V2 − Vin Vin − V4
=
R3
1 jωC
Vin − V2
- Vin + V4 =
jωCR 3
(2)
From (1) and (2),
- Vin + V4 =
- R2
V
jωCR 3 R 1 in
Thus,
I in =
Vin − V4
R2
=
V
R4
jωCR 3 R 1 R 4 in
Z in =
Vin jωCR 1R 3 R 4
=
= jωL eq
I in
R2
where
L eq =
R 1R 3 R 4C
R2
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Chapter 10, Problem 90.
Figure 10.132 shows a Wien-bridge network. Show that the frequency at which the phase
1
shift between the input and output signals is zero is f = π RC , and that the necessary
2
gain is A v =V o /V i = 3 at that frequency.
Figure 10.132
For Prob. 10.90.
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Chapter 10, Solution 90.
1
R
=
jωC 1 + jωRC
1
1 + jωRC
Z3 = R +
=
jωC
jωC
Consider the circuit shown below.
Let
Z 4 = R ||
Z3
Vi
+
−
R1
+ Vo
Z4
Vo =
R2
R2
Z4
Vi −
V
R1 + R 2 i
Z3 + Z 4
R
Vo
R2
1 + jωC
−
=
R
1 + jωRC R 1 + R 2
Vi
+
1 + jωC
jωC
=
jωRC
R2
−
2
jωRC + (1 + jωRC)
R1 + R 2
Vo
R2
jωRC
=
−
2
2 2
Vi 1 − ω R C + j3ωRC R 1 + R 2
For Vo and Vi to be in phase,
Vo
must be purely real. This happens when
Vi
1 − ω2 R 2 C 2 = 0
1
ω=
= 2πf
RC
or
At this frequency,
1
2πRC
Vo 1
R2
Av =
= −
Vi 3 R 1 + R 2
f=
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Chapter 10, Problem 91.
Consider the oscillator in Fig. 10.133.
(a) Determine the oscillation frequency.
(b) Obtain the minimum value of R for which oscillation takes place.
Figure 10.133
For Prob. 10.91.
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Chapter 10, Solution 91.
(a)
Let
V2 = voltage at the noninverting terminal of the op amp
Vo = output voltage of the op amp
Z p = 10 kΩ = R o
1
jωC
Z s = R + jωL +
As in Section 10.9,
Zp
V2
=
=
Vo Z s + Z p
Ro
R + R o + jωL −
j
ωC
ωCR o
V2
=
Vo ωC (R + R o ) + j (ω2 LC − 1)
For this to be purely real,
1
ωo2 LC − 1 = 0 ⎯
⎯→ ωo =
fo =
1
2π LC
=
LC
1
2π (0.4 × 10 -3 )(2 × 10 -9 )
f o = 180 kHz
(b)
At oscillation,
ωo CR o
Ro
V2
=
=
Vo ωo C (R + R o ) R + R o
This must be compensated for by
Vo
80
= 1+
=5
Av =
V2
20
Ro
1
=
R + Ro 5
⎯
⎯→ R = 4R o = 40 kΩ
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Chapter 10, Problem 92.
The oscillator circuit in Fig. 10.134 uses an ideal op amp.
(a) Calculate the minimum value of Ro that will cause oscillation to occur.
(b) Find the frequency of oscillation.
Figure 10.134
For Prob. 10.92.
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Chapter 10, Solution 92.
Let
V2 = voltage at the noninverting terminal of the op amp
Vo = output voltage of the op amp
Zs = R o
Z p = jωL ||
ωRL
1
1
=
|| R =
1
1
ωL + jR (ω2 LC − 1)
jωC
+ jωC +
jωL
R
As in Section 10.9,
ωRL
V2
ωL + jR (ω2 LC − 1)
=
=
ωRL
Vo Z s + Z p
Ro +
ωL + jR (ω2 LC − 1)
V2
ωRL
=
Vo ωRL + ωR o L + jR o R (ω2 LC − 1)
Zp
For this to be purely real,
ωo2 LC = 1 ⎯
⎯→ f o =
(a)
1
2π LC
At ω = ωo ,
ωo RL
V2
R
=
=
Vo ωo RL + ωo R o L R + R o
This must be compensated for by
Vo
Rf
1000k
Av =
= 1+
= 1+
= 11
V2
Ro
100k
Hence,
R
1
=
⎯
⎯→ R o = 10R = 100 kΩ
R + R o 11
(b)
fo =
1
2π (10 × 10 -6 )(2 × 10 -9 )
f o = 1.125 MHz
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Chapter 10, Problem 93.
Figure 10.135 shows a Colpitts oscillator. Show that the oscillation frequency is
1
fo =
2π LCT
where CT = C1C 2 / (C1 + C 2 ) . Assume Ri >> X C 2
Figure 10.135
A Colpitts oscillator; for Prob. 10.93.
(Hint: Set the imaginary part of the impedance in the feedback circuit equal to zero.)
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Chapter 10, Solution 93.
As shown below, the impedance of the feedback is
jωL
1
jωC2
ZT =
1
jωC1
ZT
⎛
1
1 ⎞
⎟
|| ⎜ jωL +
jωC1 ⎝
jωC 2 ⎠
-j ⎛
-j ⎞
1
⎜ jωL +
⎟
− ωLC 2
ωC1 ⎝
ωC 2 ⎠
ω
=
ZT =
-j
-j
j (C1 + C 2 − ω2 LC1C 2 )
+ jωL +
ωC1
ωC 2
In order for Z T to be real, the imaginary term must be zero; i.e.
C1 + C 2 − ωo2 LC1C 2 = 0
ωo2 =
fo =
C1 + C 2
1
=
LC1C 2
LC T
1
2π LC T
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Chapter 10, Problem 94.
Design a Colpitts oscillator that will operate at 50 kHz.
Chapter 10, Solution 94.
If we select C1 = C 2 = 20 nF
C1 C 2
C1
CT =
=
= 10 nF
C1 + C 2
2
Since f o =
1
2π LC T
L=
,
1
1
=
= 10.13 mH
2
2
(2πf ) C T (4π )(2500 × 10 6 )(10 × 10 -9 )
Xc =
1
1
=
= 159 Ω
ωC 2 (2π )(50 × 10 3 )(20 × 10 -9 )
We may select R i = 20 kΩ and R f ≥ R i , say R f = 20 kΩ .
Thus,
C1 = C 2 = 20 nF,
L = 10.13 mH
R f = R i = 20 kΩ
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Chapter 10, Problem 95.
Figure 10.136 shows a Hartley oscillator. Show that the frequency of oscillation is
1
fo =
2π C (L1 + L2 )
Figure 10.136
A Hartley oscillator; For Prob. 10.95.
Chapter 10, Solution 95.
First, we find the feedback impedance.
C
ZT
L2
L1
⎛
1 ⎞
⎟
Z T = jωL1 || ⎜ jωL 2 +
jωC ⎠
⎝
⎛
j ⎞
⎟
jωL1 ⎜ jωL 2 −
⎝
ω2 L1C (1 − ωL 2 )
ωC ⎠
ZT =
=
j
j (ω2 C (L1 + L 2 ) − 1)
jωL1 + jωL 2 −
ωC
In order for Z T to be real, the imaginary term must be zero; i.e.
ωo2 C (L1 + L 2 ) − 1 = 0
1
ω o = 2π f o =
C ( L1 + L 2 )
fo =
1
2π C (L 1 + L 2 )
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Chapter 10, Problem 96.
Refer to the oscillator in Fig. 10.137.
(a) Show that
V2
1
=
Vo 3 + j (ωL / R − R / ωL )
(b) Determine the oscillation frequency f o .
(c) Obtain the relationship between R1 and R2 in order for oscillation to occur.
Figure 10.137
For Prob. 10.96.
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Chapter 10, Solution 96.
(a)
Consider the feedback portion of the circuit, as shown below.
jωL
Vo
V2 =
+
−
V1
R
V2
R
jωL
V
R + jωL 1
⎯
⎯→ V1 =
R + jωL
V2
jωL
(1)
Applying KCL at node 1,
Vo − V1 V1
V1
=
+
jωL
R R + jωL
⎛1
⎞
1
⎟
Vo − V1 = jωL V1 ⎜ +
⎝ R R + jωL ⎠
⎛ j2ωRL − ω2 L2 ⎞
⎟
Vo = V1 ⎜1 +
R (R + jωL) ⎠
⎝
(2)
From (1) and (2),
⎛ R + jωL ⎞⎛ j2ωRL − ω2 L2 ⎞
⎟V
⎟⎜1 +
Vo = ⎜
R (R + jωL) ⎠ 2
⎝ jωL ⎠⎝
Vo R 2 + jωRL + j2ωRL − ω2 L2
=
V2
jωRL
V2
=
Vo
1
R − ω2 L2
3+
jωRL
2
V2
1
=
Vo 3 + j (ωL R − R ωL )
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill 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 McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
(b)
Since the ratio
V2
must be real,
Vo
ωo L
R
−
=0
R
ωo L
R2
ωo L =
ωo L
ωo = 2πf o =
fo =
(c)
R
L
R
2π L
When ω = ωo
V2 1
=
Vo 3
This must be compensated for by A v = 3 . But
R2
Av = 1+
=3
R1
R 2 = 2 R1
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill 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 McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.