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Chapter 9, Problem 1.
Given the sinusoidal voltage v(t) = 50 cos (30t + 10 o ) V, find: (a) the amplitude V m ,(b)
the period T, (c) the frequency f, and (d) v(t) at t = 10 ms.
Chapter 9, Solution 1.
(a) Vm = 50 V.
2π
= 0.2094s = 209.4ms
ω 30
(c ) Frequency f = ω/(2π) = 30/(2π) = 4.775 Hz.
(d) At t=1ms, v(0.01) = 50cos(30x0.01rad + 10˚)
= 50cos(1.72˚ + 10˚) = 44.48 V and ωt = 0.3 rad.
(b) Period T =
2π
=
Chapter 9, Problem 2.
A current source in a linear circuit has
i s = 8 cos (500 π t - 25 o ) A
(a) What is the amplitude of the current?
(b) What is the angular frequency?
(c) Find the frequency of the current.
(d) Calculate i s at t = 2ms.
Chapter 9, Solution 2.
(a)
amplitude = 8 A
(b)
ω = 500π = 1570.8 rad/s
(c)
f =
(d)
Is = 8∠-25° A
Is(2 ms) = 8 cos((500π )(2 × 10 -3 ) − 25°)
= 8 cos(π − 25°) = 8 cos(155°)
= -7.25 A
ω
= 250 Hz
2π
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
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Chapter 9, Problem 3.
Express the following functions in cosine form:
(a) 4 sin ( ω t - 30 o )
(b) -2 sin 6t
(c) -10sin( ω t + 20 o )
Chapter 9, Solution 3.
(a)
4 sin(ωt – 30°) = 4 cos(ωt – 30° – 90°) = 4 cos(ωt – 120°)
(b)
-2 sin(6t) = 2 cos(6t + 90°)
(c)
-10 sin(ωt + 20°) = 10 cos(ωt + 20° + 90°) = 10 cos(ωt + 110°)
Chapter 9, Problem 4.
(a) Express v = 8 cos(7t = 15 o ) in sine form.
(b) Convert i = -10 sin(3t - 85 o ) to cosine form.
Chapter 9, Solution 4.
(a)
v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°)
(b)
i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°)
Chapter 9, Problem 5.
Given v 1 = 20 sin( ω t + 60 o ) and v 2 = 60 cos( ω t - 10 o ) determine the phase angle
between the two sinusoids and which one lags the other.
Chapter 9, Solution 5.
v1 = 20 sin(ωt + 60°) = 20 cos(ωt + 60° − 90°) = 20 cos(ωt − 30°)
v2 = 60 cos(ωt − 10°)
This indicates that the phase angle between the two signals is 20° and that v1 lags
v2.
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Chapter 9, Problem 6.
For the following pairs of sinusoids, determine which one leads and by how much.
(a) v(t) = 10 cos(4t - 60 o ) and i(t) = 4 sin (4t + 50 o )
(b) v 1 (t) = 4 cos(377t + 10 o ) and v 2 (t) = -20 cos 377t
(c) x(t) = 13 cos 2t + 5 sin 2t and y(t) = 15 cos(2t -11.8 o )
Chapter 9, Solution 6.
(a)
v(t) = 10 cos(4t – 60°)
i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°)
Thus, i(t) leads v(t) by 20°.
(b)
v1(t) = 4 cos(377t + 10°)
v2(t) = -20 cos(377t) = 20 cos(377t + 180°)
Thus, v2(t) leads v1(t) by 170°.
(c)
x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90°)
X = 13∠0° + 5∠-90° = 13 – j5 = 13.928∠-21.04°
x(t) = 13.928 cos(2t – 21.04°)
y(t) = 15 cos(2t – 11.8°)
phase difference = -11.8° + 21.04° = 9.24°
Thus, y(t) leads x(t) by 9.24°.
Chapter 9, Problem 7.
If f( φ ) = cos φ + j sin φ , show that f( φ ) = e jφ .
Chapter 9, Solution 7.
If f(φ) = cosφ + j sinφ,
df
= -sinφ + j cos φ = j (cos φ + j sin φ) = j f (φ )
dφ
df
= j dφ
f
Integrating both sides
ln f = jφ + ln A
f = Aejφ = cosφ + j sinφ
f(0) = A = 1
i.e. f(φ) = ejφ = cosφ + j sinφ
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Chapter 9, Problem 8.
Calculate these complex numbers and express your results in rectangular form:
15∠45o
+ j2
(a)
3 − j4
8∠ − 20 o
10
+
(b)
(2 + j )(3 − j 4)
− 5 + j12
o
(c) 10 + (8 ∠ 50 ) (5 – j12)
Chapter 9, Solution 8.
(a)
(b)
15∠45°
15∠45°
+ j2 =
+ j2
5∠ - 53.13°
3 − j4
= 3∠98.13° + j2
= -0.4245 + j2.97 + j2
= -0.4243 + j4.97
(2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57°
8∠ - 20°
(-5 − j12)(10)
8∠ - 20°
10
+
=
+
11.18∠ - 26.57°
25 + 144
(2 + j)(3 - j4) - 5 + j12
= 0.7156∠6.57° − 0.2958 − j0.71
= 0.7109 + j0.08188 − 0.2958 − j0.71
= 0.4151 − j0.6281
(c)
10 + (8∠50°)(13∠-68.38°) = 10+104∠-17.38°
= 109.25 – j31.07
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Chapter 9, Problem 9.
Evaluate the following complex numbers and leave your results in polar form:
⎛
3 ∠60 o ⎞
⎟
(a) 5 ∠30 o ⎜⎜ 6 − j8 +
2 + j ⎟⎠
⎝
(b)
(10 ∠60 o ) (35 ∠ − 50 o )
(2 + j 6) − (5 + j )
Chapter 9, Solution 9.
(a)
(5∠30°)(6 − j8 + 1.1197 + j0.7392) = (5∠30°)(7.13 − j7.261)
= (5∠30°)(10.176∠ − 45.52°) =
50.88∠–15.52˚.
(b)
(10∠60°)(35∠ − 50°)
= 60.02∠–110.96˚.
(−3 + j5) = (5.83∠120.96°)
Chapter 9, Problem 10.
Given that z 1 = 6 – j8, z 2 = 10 ∠ -30 o , and z 3 = 8e − j120 , find:
o
(a) z 1 + z 2 + z 3
(b)
z1 z 2
z3
Chapter 9, Solution 10.
(a) z1 = 6 − j8, z 2 = 8.66 − j 5, and z 3 = −4 − j 6.9282
z1 + z 2 + z 3 = 10.66 − j19.93
(b)
z1 z 2
= 9.999 + j 7.499
z3
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Chapter 9, Problem 11.
Find the phasors corresponding to the following signals:
(a) v(t) = 21 cos(4t - 15 o ) V
(b) i(t) = -8 sin(10t + 70 o ) mA
(c) v(t) = 120 sin (10t – 50 o ) V
(d) i(t) = -60 cos(30t + 10 o ) mA
Chapter 9, Solution 11.
(a)
V = 21 < −15o V
(b) i(t ) = 8sin(10t + 70o + 180o ) = 8cos(10t + 70o + 180o − 90o ) = 8cos(10t + 160o )
I = 8 < 160o mA
(c ) v(t ) = 120sin(103 t − 50o ) = 120 cos(103 t − 50o − 90o )
V = 120 < −140o V
(d) i(t ) = −60 cos(30t + 10o ) = 60 cos(30t + 10o + 180o )
I = 60 < 190o mA
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Chapter 9, Problem 12.
Let X = 8 ∠40 o and and Y = 10 ∠ − 30 o Evaluate the following quantities and express
your results in polar form:
(a) (X + Y)X*
(b) (X -Y)*
(c) (X + Y)/X
Chapter 9, Solution 12.
Let X = 8∠40° and Y = 10∠-30°. Evaluate the following quantities and express
your results in polar form.
(X + Y)/X*
(X - Y)*
(X + Y)/X
X = 6.128+j5.142; Y = 8.66–j5
(14.788 + j0.142)(8∠ − 40°)
= (14.789∠0.55°)(8∠ − 40°) = 118.31∠ − 39.45°
= 91.36-j75.17
(a)
(X + Y)X* =
(b)
(X - Y)* = (–2.532+j10.142)* = (–2.532–j10.142) = 10.453∠–104.02˚
(c)
(X + Y)/X = (14.789∠0.55˚)/(8∠40˚) = 1.8486∠–39.45˚
= 1.4275–j1.1746
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Chapter 9, Problem 13.
Evaluate the following complex numbers:
2 + j3
7 − j8
(a)
+
1 − j6
−5 + j11
o
(5 ∠10 )(10∠ − 40o )
(b)
(4∠ − 80o )(−6∠50o )
2 + j3 − j 2
(c)
− j 2 8 − j5
Chapter 9, Solution 13.
(a) (−0.4324 + j 0.4054)+ (−0.8425 − j 0.2534) = − 1.2749 + j 0.1520
(b)
50∠ − 30 o
= − 2.0833 = –2.083
24∠150 o
(c) (2+j3)(8-j5) –(-4) = 35 +j14
Chapter 9, Problem 14.
Simplify the following expressions:
(a)
(5 − j 6) − (2 + j8)
(−3 + j 4)(5 − j ) + (4 − j 6)
(b)
(240∠75o + 160∠ − 30o )(60 − j80)
(67 + j84)(20∠32o )
⎛ 10 + j 20 ⎞
(c) ⎜
⎟
⎝ 3 + j4 ⎠
2
(10 + j 5)(16 − j120)
Chapter 9, Solution 14.
(a)
3 − j14
14.318∠ − 77.91°
=
= 0.7788∠169.71° = − 0.7663 + j0.13912
− 7 + j17 18.385∠112.38°
(b)
(62.116 + j 231.82 + 138.56 − j80)(60 − j80)
24186 − 6944.9
=
= − 1.922 − j11.55
(67 + j84)(16.96 + j10.5983)
246.06 + j 2134.7
(c)
(− 2 + j 4)2
(260 − j120) = (20∠ − 126.86°)(16.923∠ − 12.38°) =
338.46∠ − 139.24° = − 256.4 − j 221
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Chapter 9, Problem 15.
Evaluate these determinants:
(a)
10 + j 6 2 − j 3
−5
−1+ j
(b)
20∠ − 30° − 4∠ − 10°
16∠0°
3∠45°
1− j − j
(c)
1
j
j
1
0
−j
1+ j
Chapter 9, Solution 15.
(a)
10 + j6 2 − j3
-5
-1 + j
= -10 – j6 + j10 – 6 + 10 – j15
= –6 – j11
(b)
(c)
20∠ − 30° - 4∠ - 10°
= 60∠15° + 64∠-10°
16∠0°
3∠45°
= 57.96 + j15.529 + 63.03 – j11.114
= 120.99 + j4.415
1− j − j 0
j
1 −j
1
j 1+ j
1− j − j 0
j
1 −j
= 1 + 1 + 0 − 1 − 0 + j2 (1 − j) + j2 (1 + j)
= 1 − 1 (1 − j + 1 + j)
= 1 – 2 = –1
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Chapter 9, Problem 16.
Transform the following sinusoids to phasors:
(a) -10 cos (4t + 75 o )
(b) 5 sin(20t - 10 o )
(c) 4 cos2t + 3 sin 2t
Chapter 9, Solution 16.
(a)
-10 cos(4t + 75°) = 10 cos(4t + 75° − 180°)
= 10 cos(4t − 105°)
The phasor form is 10∠-105°
(b)
5 sin(20t – 10°) = 5 cos(20t – 10° – 90°)
= 5 cos(20t – 100°)
The phasor form is 5∠-100°
(c)
4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°)
The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87°
Chapter 9, Problem 17.
Two voltages v1 and v2 appear in series so that their sum is v = v1 + v2. If v1 = 10
cos(50t - π )V and v2 = 12cos(50t + 30 o ) V, find v.
3
Chapter 9, Solution 17.
V = V1 + V2 = 10 < −60o + 12 < 30o = 5 − j8.66 + 10.392 + j 6 = 15.62 < −9.805o
v = 15.62 cos(50t − 9.805o ) V = 15.62cos(50t–9.8˚) V
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Chapter 9, Problem 18.
Obtain the sinusoids corresponding to each of the following phasors:
(a) V 1 = 60 ∠ 15 o V, ω = 1
(b) V 2 = 6 + j8 V, ω = 40
(c) I 1 = 2.8e − jπ 3 A, ω = 377
(d) I 2 = -0.5 – j1.2 A, ω = 10 3
Chapter 9, Solution 18.
(a)
v1 ( t ) = 60 cos(t + 15°)
(b)
V2 = 6 + j8 = 10∠53.13°
v 2 ( t ) = 10 cos(40t + 53.13°)
(c)
i1 ( t ) = 2.8 cos(377t – π/3)
(d)
I 2 = -0.5 – j1.2 = 1.3∠247.4°
i 2 ( t ) = 1.3 cos(103t + 247.4°)
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
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Chapter 9, Problem 19.
Using phasors, find:
(a) 3cos(20t + 10º) – 5 cos(20t- 30º)
(b) 40 sin 50t + 30 cos(50t - 45º)
(c) 20 sin 400t + 10 cos(400t + 60º) -5 sin(400t - 20º)
Chapter 9, Solution 19.
(a)
3∠10° − 5∠-30° = 2.954 + j0.5209 – 4.33 + j2.5
= -1.376 + j3.021
= 3.32∠114.49°
Therefore,
3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t + 114.49°)
(b)
40∠-90° + 30∠-45° = -j40 + 21.21 – j21.21
= 21.21 – j61.21
= 64.78∠-70.89°
Therefore,
40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89°)
(c)
Using sinα = cos(α − 90°),
20∠-90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4.699
= 6.7101 – j6.641
= 9.44∠-44.7°
Therefore,
20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°)
= 9.44 cos(400t – 44.7°)
Chapter 9, Problem 20.
A linear network has a current input 4cos( ω t + 20º)A and a voltage output 10
cos( ωt +110º) V. Determine the associated impedance.
Chapter 9, Solution 20.
I = 4 < 20o ,
Z=
V = 10 < 110o
V 10 < 110o
=
= 2.5 < 90o = j 2.5 Ω
o
4 < 20
I
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Chapter 9, Problem 21.
Simplify the following:
(a) f(t) = 5 cos(2t + 15(º) – 4sin(2t -30º)
(b) g(t) = 8 sint + 4 cos(t + 50º)
(c) h(t) =
t
∫ (10 cos 40t + 50 sin 40t )dt
0
Chapter 9, Solution 21.
(a) F = 5∠15 o − 4∠− 30 o − 90 o = 6.8296 + j 4.758 = 8.3236∠34.86 o
f (t ) = 8.324 cos(30t + 34.86 o )
(b) G = 8∠ − 90 o + 4∠50 o = 2.571 − j 4.9358 = 5.565∠ − 62.49 o
g (t ) = 5.565 cos(t − 62.49 o )
(c) H =
(
)
1
10∠0 o + 50∠ − 90 o ,
jω
ω = 40
i.e. H = 0.25∠ − 90 o + 1.25∠ − 180 o = − j0.25 − 1.25 = 1.2748∠ − 168.69 o
h(t) = 1.2748cos(40t – 168.69°)
Chapter 9, Problem 22.
An alternating voltage is given by v(t) = 20 cos(5t - 30 o ) V. Use phasors to find
t
dv
10v(t ) + 4 − 2 ∫ v(t )dt
dt
−∞
Assume that the value of the integral is zero at t = - ∞ .
Chapter 9, Solution 22.
t
dv
Let f(t) = 10v(t ) + 4 − 2 ∫ v(t )dt
dt
−∞
2V
F = 10V + jω 4V −
, ω = 5, V = 20∠ − 30 o
jω
F = 10V + j20V − j0.4V = (10 + j20.4)(17.32 − j10) = 454.4∠33.89 o
f ( t ) = 454.4 cos(5t + 33.89 o )
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
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Chapter 9, Problem 23.
Apply phasor analysis to evaluate the following.
(a) v = 50 cos( ω t + 30 o ) + 30 cos( ω t + 90 o )V
(b) i = 15 cos( ω t + 45 o ) - 10 sin( ω t + 45 o )A
Chapter 9, Solution 23.
(a) V = 50 < 30o + 30 < 90o = 43.3 + j 25 − j30 = 43.588 < −6.587 o
v = 43.588cos(ωt − 6.587 o ) V = 43.49cos(ωt–6.59˚) V
(b) I = 15 < 45o − 10 < 45o − 90o = (10.607 + j10.607) − (7.071 − j 7.071) = 18.028 < 78.69o
i = 18.028cos(ωt + 78.69o ) A = 18.028cos(ωt+78.69˚) A
Chapter 9, Problem 24.
Find v(t) in the following integrodifferential equations using the phasor approach:
(a) v(t) + ∫ v dt = 10 cos t
(b)
dv
+ 5v(t ) + 4∫ v dt = 20 sin(4t + 10 o )
dt
Chapter 9, Solution 24.
(a)
V
= 10∠0°, ω = 1
jω
V (1 − j) = 10
10
V=
= 5 + j5 = 7.071∠45°
1− j
Therefore,
v(t) = 7.071 cos(t + 45°)
V+
(b)
4V
= 20∠(10° − 90°), ω = 4
jω
⎛
4⎞
V ⎜ j4 + 5 + ⎟ = 20 ∠ - 80°
j4 ⎠
⎝
20∠ - 80°
V=
= 3.43∠ - 110.96°
5 + j3
Therefore,
v(t) = 3.43 cos(4t – 110.96°)
jωV + 5V +
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Chapter 9, Problem 25.
Using phasors, determine i(t) in the following equations:
di
(a) 2 + 3i (t ) = 4 cos(2t − 45 o )
dt
di
(b) 10 ∫ i dt + + 6i (t ) = 5 cos(5t + 22 o )
dt
Chapter 9, Solution 25.
(a)
2jωI + 3I = 4∠ - 45°, ω = 2
I (3 + j4) = 4∠ - 45°
4∠ - 45° 4∠ - 45°
I=
=
= 0.8∠ - 98.13°
3 + j4
5∠53.13°
Therefore,
i(t) = 0.8 cos(2t – 98.13°)
(b)
I
+ jωI + 6I = 5∠22°, ω = 5
jω
(- j2 + j5 + 6) I = 5∠22°
5∠22°
5∠22°
I=
=
= 0.745∠ - 4.56°
6 + j3 6.708∠26.56°
Therefore,
i(t) = 0.745 cos(5t – 4.56°)
10
Chapter 9, Problem 26.
The loop equation for a series RLC circuit gives
t
di
+ 2i + ∫ i dt = cos 2t
−∞
dt
Assuming that the value of the integral at t = - ∞ is zero, find i(t) using the phasor
method.
Chapter 9, Solution 26.
I
= 1∠0°, ω = 2
jω
⎛
1⎞
I ⎜ j2 + 2 + ⎟ = 1
j2 ⎠
⎝
1
I=
= 0.4∠ - 36.87°
2 + j1.5
Therefore,
i(t) = 0.4 cos(2t – 36.87°)
jωI + 2I +
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Chapter 9, Problem 27.
A parallel RLC circuit has the node equation
dv
= 50v + 100∫ v dt = 110 cos(377t − 10o )
dt
Determine v(t) using the phasor method. You may assume that the value of the integral at
t = - ∞ is zero.
Chapter 9, Solution 27.
V
= 110∠ - 10°, ω = 377
jω
⎛
j100 ⎞
⎟ = 110∠ - 10°
V ⎜ j377 + 50 −
⎝
377 ⎠
V (380.6∠82.45°) = 110∠ - 10°
V = 0.289 ∠ - 92.45°
jωV + 50V + 100
Therefore, v(t) = 0.289 cos(377t – 92.45°).
Chapter 9, Problem 28.
Determine the current that flows through an 8- Ω resistor connected to a voltage source
vs = 110 cos 377t V.
Chapter 9, Solution 28.
i( t ) =
v s ( t ) 110 cos(377 t )
=
= 13.75 cos(377t) A.
R
8
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Chapter 9, Problem 29.
What is the instantaneous voltage across a 2- µ F capacitor when the current through it is
i =4 sin(10 6 t +25 o ) A?
Chapter 9, Solution 29.
Z=
1
1
=
= - j 0.5
6
jωC j (10 )(2 × 10 -6 )
V = IZ = (4∠25°)(0.5∠ - 90°) = 2 ∠ - 65°
Therefore
v(t) = 2 sin(106t – 65°) V.
Chapter 9, Problem 30.
A voltage v(t) = 100 cos(60t + 20 o ) V is applied to a parallel combination of a 40-k Ω
resistor and a 50- µ F capacitor. Find the steady-state currents through the resistor and the
capacitor.
Chapter 9, Solution 30.
Since R and C are in parallel, they have the same voltage across them. For the resistor,
100 < 20o
⎯⎯
→ IR = V / R =
= 2.5 < 20o mA
V = IR R
40k
o
iR = 2.5cos(60t + 20 ) mA
For the capacitor,
iC = C
dv
= 50 x10−6 (−60) x100sin(60t + 20o ) = −300sin(60t + 20o ) mA
dt
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Chapter 9, Problem 31.
A series RLC circuit has R = 80 Ω , L = 240 mH, and C = 5 mF. If the input voltage is
v(t) = 10 cos 2t find the currrent flowing through the circuit.
Chapter 9, Solution 31.
jω L = j 2 x 240 x10−3 = j 0.48
1
1
C = 5mF
⎯⎯
→
=
= − j100
jωC j 2 x5 x10−3
Z = 80 + j 0.48 − j100 = 80 − j 99.52
L = 240mH
⎯⎯
→
V
10 < 00
I= =
= 0.0783 < 51.206o
Z 80 − j 99.52
i (t ) = 78.3cos(2t + 51.206o ) mA = 78.3cos(2t+51.26˚) mA
Chapter 9, Problem 32.
For the network in Fig. 9.40, find the load current I L .
Figure 9.40
For Prob. 9.32.
Chapter 9, Solution 32.
I=
V 100 < 0o
=
= 12.195 − 9.756 = 15.62 < −38.66o A
Z
5 + j4
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Chapter 9, Problem 33.
A series RL circuit is connected to a 110-V ac source. If the voltage across the resistor is
85 V, find the voltage across the inductor.
Chapter 9, Solution 33.
110 = v 2R + v 2L
v L = 110 2 − v 2R
v L = 110 2 − 85 2 = 69.82 V
Chapter 9, Problem 34.
What value of ω will cause the forced response v o in Fig. 9.41 to be zero?
Figure 9.41
For Prob. 9.34.
Chapter 9, Solution 34.
v o = 0 if ωL =
ω=
1
ωC
1
(5 × 10
−3
)(20 × 10
−3
⎯
⎯→ ω =
1
LC
= 100 rad/s
)
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Chapter 9, Problem 35.
Find current i in the circuit of Fig. 9.42, when v s (t) = 50 cos200t V.
Figure 9.42
For Prob. 9.35.
Chapter 9, Solution 35.
vs (t ) = 50 cos 200t
⎯⎯
→
Vs = 50 < 0o , ω = 200
1
1
=
=−j
jωC j 200 x5 x10−3
⎯⎯
→ jω L = j 20 x10−3 x 200 = j 4
⎯⎯
→
5mF
20mH
Z in = 10 − j + j 4 = 10 + j 3
Vs 50 < 0o
I=
=
= 4.789 < −16.7o
Z in 10 + j 3
i (t ) = 4.789 cos(200t − 16.7 o ) A
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Chapter 9, Problem 36.
In the circuit of Fig. 9.43, determine i. Let v s = 60 cos(200t - 10 o )V.
Figure 9.43
For Prob. 9.36.
Chapter 9, Solution 36.
Let Z be the input impedance at the source.
100 mH
10 µF
⎯
⎯→
⎯
⎯→
jωL = j 200 x100 x10 −3 = j 20
1
1
=
= − j 500
jωC j10 x10 −6 x 200
1000//-j500 = 200 –j400
1000//(j20 + 200 –j400) = 242.62 –j239.84
Z = 2242.62 − j 239.84 = 2255∠ − 6.104 o
I=
60∠ − 10 o
= 26.61∠ − 3.896 o mA
o
2255∠ − 6.104
i = 266.1 cos(200t − 3.896 o ) mA
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Chapter 9, Problem 37.
Determine the admittance Y for the circuit in Fig. 9.44.
Figure 9.44
For Prob. 9.37.
Chapter 9, Solution 37.
Y=
1 1
1
+ +
= 0.25 − j 0.025 S = 250–j25 mS
4 j8 − j10
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Chapter 9, Problem 38.
Find i(t) and v(t) in each of the circuits of Fig. 9.45.
Figure 9.45
For Prob. 9.38.
Chapter 9, Solution 38.
1
(a)
F ⎯
⎯→
6
1
1
=
= - j2
jωC j (3)(1 / 6)
- j2
(10 ∠45°) = 4.472∠ - 18.43°
4 − j2
Hence, i(t) = 4.472 cos(3t – 18.43°) A
I=
V = 4I = (4)(4.472∠ - 18.43°) = 17.89∠ - 18.43°
Hence, v(t) = 17.89 cos(3t – 18.43°) V
(b)
1
F ⎯
⎯→
12
⎯→
3H ⎯
1
1
=
= - j3
jωC j (4)(1 / 12)
jωL = j (4)(3) = j12
V 50∠0°
= 10∠36.87°
=
Z 4 − j3
Hence, i(t) = 10 cos(4t + 36.87°) A
I=
V=
j12
(50∠0°) = 41.6 ∠33.69°
8 + j12
Hence, v(t) = 41.6 cos(4t + 33.69°) V
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Chapter 9, Problem 39.
For the circuit shown in Fig. 9.46, find Z eg and use that to find current I. Let ω = 10
rad/s.
Figure 9.46
For Prob. 9.39.
Chapter 9, Solution 39.
Z eq = 4 + j 20 + 10 //(− j14 + j 25) = 9.135 + j 27.47 Ω
I=
V
12
=
= 0.4145 < −71.605o
Z eq 9.135 + j 27.47
i (t ) = 0.4145cos(10t − 71.605o ) A = 414.5cos(10t–71.6˚) mA
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Chapter 9, Problem 40.
In the circuit of Fig. 9.47, find i o when:
(a) ω = 1 rad/s
(c) ω = 10 rad/s
(b) ω = 5 rad/s
Figure 9.47
For Prob. 9.40.
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Chapter 9, Solution 40.
(a)
For ω = 1 ,
1H ⎯
⎯→
jωL = j (1)(1) = j
1
1
0.05 F ⎯
⎯→
=
= - j20
jωC j (1)(0.05)
- j40
Z = j + 2 || (- j20) = j +
= 1.98 + j0.802
2 − j20
4 ∠0°
4∠0°
V
=
=
= 1.872 ∠ - 22.05°
Z 1.98 + j0.802 2.136∠22.05°
Hence, i o ( t ) = 1.872 cos(t – 22.05°) A
Io =
(b)
For ω = 5 ,
⎯→
1H ⎯
jωL = j (5)(1) = j5
1
1
0.05 F ⎯
⎯→
=
= - j4
jωC j (5)(0.05)
- j4
Z = j5 + 2 || (- j4) = j5 +
= 1.6 + j4.2
1 − j2
4∠0°
4∠0°
V
=
= 0.89∠ - 69.14°
=
Z 1.6 + j4 4.494∠69.14°
Hence, i o ( t ) = 0.89 cos(5t – 69.14°) A
Io =
(c)
For ω = 10 ,
⎯→ jωL = j (10)(1) = j10
1H ⎯
1
1
0.05 F ⎯
⎯→
=
= - j2
jωC j (10)(0.05)
- j4
Z = j10 + 2 || (- j2) = j10 +
= 1 + j9
2 − j2
4 ∠0°
V 4∠0°
=
= 0.4417 ∠ - 83.66°
=
Z 1 + j9 9.055∠83.66°
Hence, i o ( t ) = 0.4417 cos(10t – 83.66°) A
Io =
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Chapter 9, Problem 41.
Find v(t) in the RLC circuit of Fig. 9.48.
Figure 9.48
For Prob. 9.41.
Chapter 9, Solution 41.
ω = 1,
1H ⎯
⎯→
jωL = j (1)(1) = j
1
1
1F ⎯
⎯→
=
= -j
jωC j (1)(1)
- j+1
Z = 1 + (1 + j) || (- j) = 1 +
= 2− j
1
I=
Vs
10
=
,
Z 2− j
I c = (1 + j) I
(1 − j)(10)
= 6.325∠ - 18.43°
2− j
v(t) = 6.325 cos(t – 18.43°) V
V = (- j)(1 + j) I = (1 − j) I =
Thus,
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Chapter 9, Problem 42.
Calculate v o (t) in the circuit of Fig. 9.49.
Figure 9.49
For Prob. 9.42.
Chapter 9, Solution 42.
ω = 200
50 µF ⎯
⎯→
1
1
=
= - j100
jωC j (200)(50 × 10 -6 )
0.1 H ⎯
⎯→
jωL = j (200)(0.1) = j20
50 || -j100 =
Vo =
(50)(-j100) - j100
=
= 40 − j20
50 − j100
1 - j2
j20
j20
(60∠0°) =
(60∠0°) = 17.14 ∠90°
j20 + 30 + 40 − j20
70
Thus, v o ( t ) = 17.14 sin(200t + 90°) V
or
v o ( t ) = 17.14 cos(200t) V
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Chapter 9, Problem 43.
Find current I o in the circuit shown in Fig. 9.50.
Figure 9.50
For Prob. 9.43.
Chapter 9, Solution 43.
Z in = 50 + j80 //(100 − j 40) = 50 +
Io =
j80(100 − j 40)
= 105.71 + j 57.93
100 + j 40
60 < 0o
= 0.4377 − 0.2411 = 0.4997 < −28.85o A = 499.7∠–28.85˚ mA
Z in
Chapter 9, Problem 44.
Calculate i(t) in the circuit of Fig. 9.51.
Figure 9.51
For prob. 9.44.
Chapter 9, Solution 44.
ω = 200
10 mH ⎯
⎯→ jωL = j (200)(10 × 10 -3 ) = j2
1
1
5 mF ⎯
⎯→
=
= -j
jωC j (200)(5 × 10 -3 )
1 1
1
3+ j
Y= + +
= 0.25 − j0.5 +
= 0.55 − j0.4
4 j2 3 − j
10
1
1
= 1.1892 + j0.865
Z= =
Y 0.55 − j0.4
6∠0°
6∠0°
=
= 0.96 ∠ - 7.956°
I=
5 + Z 6.1892 + j0.865
Thus, i(t) = 0.96 cos(200t – 7.956°) A
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Chapter 9, Problem 45.
Find current I o in the network of Fig. 9.52.
Figure 9.52
For Prob. 9.45.
Chapter 9, Solution 45.
We obtain I o by applying the principle of current division twice.
I
I2
Z1
I2
-j2 Ω
Z2
(a)
Z 1 = - j2 ,
Io
2Ω
(b)
Z 2 = j4 + (-j2) || 2 = j4 +
I2 =
Z1
- j2
- j10
(5∠0°) =
I=
- j2 + 1 + j3
1+ j
Z1 + Z 2
Io =
⎛ - j ⎞⎛ - j10 ⎞ - 10
- j2
⎟⎜
⎟=
I2 = ⎜
= –5 A
2 - j2
⎝1 - j ⎠⎝ 1 + j ⎠ 1 + 1
- j4
= 1 + j3
2 - j2
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Chapter 9, Problem 46.
If i s
=
5 cos(10t + 40 o ) A in the circuit of Fig. 9.53, find i o .
Figure 9.53
For Prob. 9.46.
Chapter 9, Solution 46.
i s = 5 cos(10 t + 40°) ⎯
⎯→ I s = 5∠40°
1
1
0.1 F ⎯
⎯→
=
= -j
jωC j (10)(0.1)
⎯→
0.2 H ⎯
Z1 = 4 || j2 =
Let
jωL = j (10)(0.2) = j2
j8
= 0.8 + j1.6 ,
4 + j2
Io =
Z1
0.8 + j1.6
Is =
(5∠40°)
3.8 + j0.6
Z1 + Z 2
Io =
(1.789∠63.43°)(5∠40°)
= 2.325∠94.46°
3.847 ∠8.97°
Z2 = 3 − j
Thus, i o ( t ) = 2.325 cos(10t + 94.46°) A
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Chapter 9, Problem 47.
In the circuit of Fig. 9.54, determine the value of i s (t).
Figure 9.54
For Prob. 9.47.
Chapter 9, Solution 47.
First, we convert the circuit into the frequency domain.
Ix
5∠0˚
Ix =
+
−
2Ω
j4
-j10
20 Ω
5
5
5
=
=
= 0.4607∠52.63°
− j10(20 + j4) 2 + 4.588 − j8.626 10.854∠ − 52.63°
2+
− j10 + 20 + j4
is(t) = 460.7cos(2000t +52.63˚) mA
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Chapter 9, Problem 48.
Given that v s (t) = 20 sin(100t - 40 o ) in Fig. 9.55, determine i x (t).
Figure 9.55
For Prob. 9.48.
Chapter 9, Solution 48.
Converting the circuit to the frequency domain, we get:
10 Ω
V1 30 Ω
Ix
20∠-40˚
+
−
j20
-j20
We can solve this using nodal analysis.
V1 − 20∠ − 40° V1 − 0
V −0
+
+ 1
=0
10
j20
30 − j20
V1(0.1 − j0.05 + 0.02307 + j0.01538) = 2∠ − 40°
2∠40°
= 15.643∠ − 24.29°
0.12307 − j0.03462
15.643∠ − 24.29°
=
= 0.4338∠9.4°
30 − j20
= 0.4338 sin(100 t + 9.4°) A
V1 =
Ix
ix
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Chapter 9, Problem 49.
Find v s (t) in the circuit of Fig. 9.56 if the current i x through the 1- Ω resistor is 0.5 sin
200t A.
Figure 9.56
For Prob. 9.49.
Chapter 9, Solution 49.
Z T = 2 + j2 || (1 − j) = 2 +
I
( j2)(1 − j)
=4
1+ j
Ix
1Ω
j2 Ω
j2
j2
I=
I,
j2 + 1 − j
1+ j
1+ j
1+ j
I=
Ix =
j2
j4
Ix =
Vs = I Z T =
-j Ω
where I x = 0.5∠0° =
1
2
1+ j
1+ j
(4) =
= 1 − j = 1.414∠ - 45°
j4
j
v s ( t ) = 1.414 sin(200t – 45°) V
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Chapter 9, Problem 50.
Determine v x in the circuit of Fig. 9.57. Let i s (t) = 5 cos(100t + 40 o )A.
Figure 9.57
For Prob. 9.50.
Chapter 9, Solution 50.
Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3)
= -j10Ω.
j10
5∠40˚
Ix
+
-j10
20 Ω
vx
−
Using the current dividing rule:
− j10
5∠40° = − j2.5∠40° = 2.5∠ − 50°
− j10 + 20 + j10
Vx = 20I x = 50∠ − 50°
Ix =
v x = 50 cos(100t − 50°) V
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Chapter 9, Problem 51.
If the voltage v o across the 2- Ω resistor in the circuit of Fig. 9.58 is 10 cos2t V, obtain
is.
Figure 9.58
For Prob. 9.51.
Chapter 9, Solution 51.
0.1 F ⎯
⎯→
1
1
=
= - j5
jωC j (2)(0.1)
0.5 H ⎯
⎯→
jωL = j (2)(0.5) = j
The current I through the 2-Ω resistor is
Is
1
I=
Is =
,
1 − j5 + j + 2
3 − j4
I s = (5)(3 − j4) = 25∠ - 53.13°
where I =
10
∠0° = 5
2
Therefore,
i s ( t ) = 25 cos(2t – 53.13°) A
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Chapter 9, Problem 52.
If V o = 8 ∠ 30 o V in the circuit of Fig. 9.59, find I s. .
Figure 9.59
For Prob. 9.52.
Chapter 9, Solution 52.
5 || j5 =
j25
j5
=
= 2.5 + j2.5
5 + j5 1 + j
Z1 = 10 ,
Z 2 = - j5 + 2.5 + j2.5 = 2.5 − j2.5
I2
IS
Z1
Z2
Z1
10
4
Is =
Is =
I
12.5 − j2.5
5− j s
Z1 + Z 2
Vo = I 2 (2.5 + j2.5)
I2 =
⎛ 4 ⎞
10 (1 + j)
⎟ I s (2.5)(1 + j) =
I
8∠30° = ⎜
5− j s
⎝ 5 − j⎠
(8∠30°)(5 − j)
Is =
= 2.884∠-26.31° A
10 (1 + j)
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Chapter 9, Problem 53.
Find I o in the circuit of Fig. 9.60.
Figure 9.60
For Prob. 9.53.
Chapter 9, Solution 53.
Convert the delta to wye subnetwork as shown below.
Z1
Io
Z2
2Ω
Z3
+
10 Ω
60∠ − 30 o V
8Ω
-
Z
− j2x 4
8∠ − 90°
j6x 4
=
= −1 − j1,
Z2 =
= 3 + j3,
4 + j4 5.6569∠45°
4 + j4
12
Z3 =
= 1.5 − j1.5
4 + j4
( Z 3 + 8) //( Z 2 + 10) = (9.5 − j1.5) //(13 + j3) = 5.691∠0.21° = 5.691 + j0.02086
Z = 2 + Z1 + 5.691 + j0.02086 = 6.691 − j0.9791
Z1 =
Io =
60∠ − 30 o
60∠ − 30 o
=
= 8.873∠ − 21.67 o A
o
Z
6.7623∠ − 8.33
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Chapter 9, Problem 54.
In the circuit of Fig. 9.61, find V s if I o = 2 ∠ 0 o A.
Figure 9.61
For Prob. 9.54.
Chapter 9, Solution 54.
Since the left portion of the circuit is twice as large as the right portion, the
equivalent circuit is shown below.
+ −
+
2Z
V2
−
Vs
−
V1
Z
+
V1 = I o (1 − j) = 2 (1 − j)
V2 = 2V1 = 4 (1 − j)
Vs = −V1 − V2 = −6 (1 − j)
Vs = 8.485∠–135° V
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Chapter 9, Problem 55.
* Find Z in the network of Fig. 9.62, given that V o = 4 ∠ 0 o V.
Figure 9.62
For Prob. 9.55.
* An asterisk indicates a challenging problem.
Chapter 9, Solution 55.
12 Ω
-j20 V
+
−
I
I1
I2
-j4 Ω
Z
+
Vo
j8 Ω
−
Vo
4
= = -j0.5
j 8 j8
I (Z + j8) (-j0.5)(Z + j8) Z
I2 = 1
=
= +j
- j4
- j4
8
Z
Z
I = I 1 + I 2 = -j0.5 + + j = + j0.5
8
8
- j20 = 12 I + I 1 (Z + j8)
⎛Z j⎞ - j
- j20 = 12 ⎜ + ⎟ + (Z + j8)
⎝ 8 2⎠ 2
⎛3
1⎞
- 4 - j26 = Z ⎜ − j ⎟
⎝2
2⎠
- 4 - j26 26.31∠261.25°
Z=
=
= 16.64∠279.68°
3
1 1.5811∠ - 18.43°
−j
2
2
Z = 2.798 – j16.403 Ω
I1 =
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Chapter 9, Problem 56.
At ω = 377 rad/s, find the input impedance of the circuit shown in Fig. 9.63.
Figure 9.63
For Prob. 9.56.
Chapter 9, Solution 56.
1
1
=
= − j 53.05
jωC j 377 x50 x10−6
60mH
⎯⎯
→ jω L = j 377 x60 x10−3 = j 22.62
Z in = 12 − j 53.05 + j 22.62 // 40 = 21.692 − j 35.91 Ω
50µ F
⎯⎯
→
Chapter 9, Problem 57.
At ω = 1 rad/s, obtain the input admittance in the circuit of Fig. 9.64.
Figure 9.64
For Prob. 9.57.
Chapter 9, Solution 57.
2H
⎯
⎯→
jωL = j 2
1
=−j
jω C
j2(2 − j)
Z = 1 + j2 //( 2 − j) = 1 +
= 2.6 + j1.2
j2 + 2 − j
1F
⎯
⎯→
Y = 1 = 0.3171 − j0.1463 S
Z
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Chapter 9, Problem 58.
Find the equivalent impedance in Fig. 9.65 at ω = 10 krad/s.
Figure 9.65
For Prob. 9.58.
Chapter 9, Solution 58.
2µ F
100mH
1
1
=
= − j 50
4
jωC j10 x 2 x10−6
⎯⎯
→ jω L = j104 x100 x10−3 = j1000
⎯⎯
→
Z in = (400 − j 50) //(1000 + j1000) =
(400 − j 50)(1000 + j1000)
= 336.24 + j 21.83 Ω
1400 + j 950
Chapter 9, Problem 59.
For the network in Fig. 9.66, find Z in . Let ω = 10 rad/s.
Figure 9.66
For Prob. 9.59.
Chapter 9, Solution 59.
0.25F
0.5H
⎯⎯
→
⎯⎯
→
1
1
=
= − j 0.4
jωC j10 x0.25
jω L = j10 x0.5 = j 5
Z in = j5 (5 − j0.4) =
(5∠90°)(5.016∠ − 4.57°)
= 3.691∠42.82°
6.794∠42.61°
= 2.707+j2.509 Ω.
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Chapter 9, Problem 60.
Obtain Z in for the circuit in Fig. 9.67.
Figure 9.67
For Prob. 9.60.
Chapter 9, Solution 60.
Z = (25 + j15) + (20 − j 50) //(30 + j10) = 25 + j15 + 26.097 − j 5.122 = 51.1 + j 9.878Ω
Chapter 9, Problem 61.
Find Z eq in the circuit of Fig. 9.68.
Figure 9.68
For Prob. 9.61.
Chapter 9, Solution 61.
All of the impedances are in parallel.
1
1
1
1
1
=
+
+ +
Z eq 1 − j 1 + j2 j5 1 + j3
1
= (0.5 + j0.5) + (0.2 − j0.4) + (- j0.2) + (0.1 − j0.3) = 0.8 − j0.4
Z eq
1
Z eq =
= 1 + j0.5 Ω
0.8 − j0.4
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Chapter 9, Problem 62.
For the circuit in Fig. 9.69, find the input impedance Z in at 10 krad/s.
Figure 9.69
For Prob. 9.62.
Chapter 9, Solution 62.
2 mH ⎯
⎯→
jωL = j (10 × 10 3 )(2 × 10 -3 ) = j20
1
1
1 µF ⎯
⎯→
=
= - j100
3
jωC j (10 × 10 )(1 × 10 -6 )
50 Ω
+
1∠0° A
+
V
j20 Ω
−
Vin
+
2V
−
-j100 Ω
V = (1∠0°)(50) = 50
Vin = (1∠0°)(50 + j20 − j100) + (2)(50)
Vin = 50 − j80 + 100 = 150 − j80
Z in =
Vin
= 150 – j80 Ω
1∠0°
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Chapter 9, Problem 63.
For the circuit in Fig. 9.70, find the value of Z T ⋅ .
Figure 9.70
For Prob. 9.63.
Chapter 9, Solution 63.
First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with
the corresponding delta.
200 + j150 + j300
= 20 + j45
10
200 + j450
200 + j450
z2 =
= 30 − j13.333, z3 =
= 10 + j22.5
j15
20
z1 =
8Ω
–j12 Ω
–j16 Ω
z2
10 Ω
z1
ZT
z3
–j16 Ω
10 Ω
Now all we need to do is to combine impedances.
z 2 (10 − j16) =
(30 − j13.333)(10 − j16)
= 8.721 − j8.938
40 − j29.33
z3 (10 − j16) = 21.70 − j3.821
ZT = 8 − j12 + z1 (8.721 − j8.938 + 21.7 − j3.821) = 34.69 − j6.93Ω
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Chapter 9, Problem 64.
Find Z T and I in the circuit of Fig. 9.71.
Figure 9.71
For Prob. 9.64.
Chapter 9, Solution 64.
− j10(6 + j8)
= 19 − j5Ω
6 − j2
30∠90°
I=
= −0.3866 + j1.4767 = 1.527∠104.7° A
ZT
ZT = 4 +
Chapter 9, Problem 65.
Determine Z T and I for the circuit in Fig. 9.72.
Figure 9.72
For Prob. 9.65.
Chapter 9, Solution 65.
Z T = 2 + (4 − j6) || (3 + j4)
(4 − j6)(3 + j4)
ZT = 2 +
7 − j2
Z T = 6.83 + j1.094 Ω = 6.917∠9.1° Ω
I=
V
120 ∠10°
=
= 17.35∠0.9° A
Z T 6.917 ∠9.1°
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Chapter 9, Problem 66.
For the circuit in Fig. 9.73, calculate Z T and V ab⋅ .
Figure 9.73
For Prob. 9.66.
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Chapter 9, Solution 66.
(20 − j5)(40 + j10) 170
=
(12 − j)
60 + j5
145
Z T = 14.069 – j1.172 Ω = 14.118∠-4.76°
Z T = (20 − j5) || (40 + j10) =
I=
V
60∠90°
=
= 4.25∠94.76°
Z T 14.118∠ - 4.76°
I
I1
I2
20 Ω
j10 Ω
+
Vab
−
40 + j10
8 + j2
I=
I
60 + j5
12 + j
20 − j5
4− j
I2 =
I=
I
60 + j5
12 + j
I1 =
Vab = -20 I 1 + j10 I 2
- (160 + j40)
10 + j40
Vab =
I+
I
12 + j
12 + j
- 150
(-12 + j)(150)
Vab =
I=
I
12 + j
145
Vab = (12.457 ∠175.24°)(4.25∠97.76°)
Vab = 52.94∠273° V
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Chapter 9, Problem 67.
At ω = 10 3 rad/s find the input admittance of each of the circuits in Fig. 9.74.
Figure 9.74
For Prob. 9.67.
Chapter 9, Solution 67.
(a)
20 mH ⎯
⎯→
jωL = j (10 3 )(20 × 10 -3 ) = j20
1
1
12.5 µF ⎯
⎯→
=
= - j80
3
jωC j (10 )(12.5 × 10 -6 )
Z in = 60 + j20 || (60 − j80)
( j20)(60 − j80)
Z in = 60 +
60 − j60
Z in = 63.33 + j23.33 = 67.494 ∠20.22°
Yin =
(b)
1
= 14.8∠-20.22° mS
Z in
10 mH ⎯
⎯→
20 µF ⎯
⎯→
jωL = j (10 3 )(10 × 10 -3 ) = j10
1
1
=
= - j50
3
jωC j (10 )(20 × 10 -6 )
30 || 60 = 20
Z in = - j50 + 20 || (40 + j10)
(20)(40 + j10)
Z in = - j50 +
60 + j10
Z in = 13.5 − j48.92 = 50.75∠ - 74.56°
Yin =
1
= 19.7∠74.56° mS = 5.24 + j18.99 mS
Z in
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Chapter 9, Problem 68.
Determine Y eq for the circuit in Fig. 9.75.
Figure 9.75
For Prob. 9.68.
Chapter 9, Solution 68.
Yeq =
1
1
1
+
+
5 − j2 3 + j - j4
Yeq = (0.1724 + j0.069) + (0.3 − j0.1) + ( j0.25)
Yeq = 0.4724 + j0.219 S
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Chapter 9, Problem 69.
Find the equivalent admittance Y eq of the circuit in Fig. 9.76.
Figure 9.76
For Prob. 9.69.
Chapter 9, Solution 69.
1
1
1
1
= +
= (1 + j2)
Yo 4 - j2 4
Yo =
4
(4)(1 − j2)
=
= 0.8 − j1.6
1 + j2
5
Yo + j = 0.8 − j0.6
1
1 1
1
=
+
+
= (1) + ( j0.333) + (0.8 + j0.6)
Yo ′ 1 - j3 0.8 − j0.6
1
= 1.8 + j0.933 = 2.028∠27.41°
Y′
o
Yo ′ = 0.4932∠ - 27.41° = 0.4378 − j0.2271
Yo ′ + j5 = 0.4378 + j4.773
1
1
1
0.4378 − j4.773
= +
= 0 .5 +
Yeq 2 0.4378 + j4.773
22.97
1
= 0.5191 − j0.2078
Yeq
Yeq =
0.5191 − j0.2078
= 1.661 + j0.6647 S
0.3126
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Chapter 9, Problem 70.
Find the equivalent impedance of the circuit in Fig. 9.77.
Figure 9.77
For Prob. 9.70.
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Chapter 9, Solution 70.
Make a delta-to-wye transformation as shown in the figure below.
a
Zan
Zbn
Zeq
n
Zcn
b
c
8Ω
2Ω
-j5 Ω
(- j10)(10 + j15)
(10)(15 − j10)
=
= 7 − j9
5 − j10 + 10 + j15
15 + j5
(5)(10 + j15)
= 4.5 + j3.5
=
15 + j5
(5)(- j10)
=
= -1 − j3
15 + j5
Z an =
Z bn
Z cn
Z eq = Z an + (Z bn + 2) || (Z cn + 8 − j5)
Z eq = 7 − j9 + (6.5 + j3.5) || (7 − j8)
(6.5 + j3.5)(7 − j8)
13.5 − j4.5
= 7 − j9 + 5.511 − j0.2
Z eq = 7 − j9 +
Z eq
Z eq = 12.51 − j9.2 = 15.53∠-36.33° Ω
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Chapter 9, Problem 71.
Obtain the equivalent impedance of the circuit in Fig. 9.78.
Figure 9.78
For Prob. 9.71.
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Chapter 9, Solution 71.
We apply a wye-to-delta transformation.
j4 Ω
Zab
b
a
Zbc
Zac
Zeq
-j2 Ω
1Ω
c
2 − j2 + j4 2 + j2
=
= 1− j
j2
j2
2 + j2
Z ac =
= 1+ j
2
2 + j2
Z bc =
= -2 + j2
-j
Z ab =
( j4)(1 − j)
= 1.6 − j0.8
1 + j3
(1)(1 + j)
= 0.6 + j0.2
1 || Z ac = 1 || (1 + j) =
2+ j
j4 || Z ab + 1 || Z ac = 2.2 − j0.6
j4 || Z ab = j4 || (1 − j) =
1
1
1
1
=
+
+
Z eq - j2 - 2 + j2 2.2 − j0.6
= j0.5 − 0.25 − j0.25 + 0.4231 + j0.1154
= 0.173 + j0.3654 = 0.4043∠64.66°
Z eq = 2.473∠-64.66° Ω = 1.058 – j2.235 Ω
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Chapter 9, Problem 72.
Calculate the value of Z ab in the network of Fig. 9.79.
Figure 9.79
For Prob. 9.72.
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Chapter 9, Solution 72.
Transform the delta connections to wye connections as shown below.
a
j2 Ω
j2 Ω
-j18 Ω
-j9 Ω
j2 Ω
R1
R2
R3
b
- j9 || - j18 = - j6 ,
R1 =
(20)(20)
= 8 Ω,
20 + 20 + 10
(20)(10)
R3 =
= 4Ω
50
R2 =
(20)(10)
= 4Ω,
50
Z ab = j2 + ( j2 + 8) || (j2 − j6 + 4) + 4
Z ab = 4 + j2 + (8 + j2) || (4 − j4)
(8 + j2)(4 − j4)
Z ab = 4 + j2 +
12 - j2
Z ab = 4 + j2 + 3.567 − j1.4054
Z ab = 7.567 + j0.5946 Ω
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Chapter 9, Problem 73.
Determine the equivalent impedance of the circuit in Fig. 9.80.
Figure 9.80
For Prob. 9.73.
Chapter 9, Solution 73.
Transform the delta connection to a wye connection as in Fig. (a) and then
transform the wye connection to a delta connection as in Fig. (b).
a
j2 Ω
j2 Ω
-j18 Ω
-j9 Ω
j2 Ω
R1
R2
R3
b
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( j8)(- j6)
48
=
= - j4.8
j8 + j8 − j6 j10
Z 2 = Z1 = -j4.8
( j8)( j8) - 64
Z3 =
=
= j6.4
j10
j10
Z1 =
(2 + Z1 )(4 + Z 2 ) + (4 + Z 2 )(Z 3 ) + (2 + Z1 )(Z 3 ) =
(2 − j4.8)(4 − j4.8) + (4 − j4.8)( j6.4) + (2 − j4.8)( j6.4) = 46.4 + j9.6
46.4 + j9.6
= 1.5 − j7.25
j6.4
46.4 + j9.6
Zb =
= 3.574 + j6.688
4 − j4.8
46.4 + j9.6
Zc =
= 1.727 + j8.945
2 − j4.8
Za =
(6∠90°)(7.583∠61.88°)
= 07407 + j3.3716
3.574 + j12.688
(-j4)(1.5 − j7.25)
= 0.186 − j2.602
- j4 || Z a =
1.5 − j11.25
(12∠90°)(9.11∠79.07°)
= 0.5634 + j5.1693
j12 || Z c =
1.727 + j20.945
j6 || Z b =
Z eq = ( j6 || Z b ) || (- j4 || Z a + j12 || Z c )
Z eq = (0.7407 + j3.3716) || (0.7494 + j2.5673)
Z eq = 1.508∠75.42° Ω = 0.3796 + j1.46 Ω
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Chapter 9, Problem 74.
Design an RL circuit to provide a 90 o
leading
phase shift.
Chapter 9, Solution 74.
One such RL circuit is shown below.
20 Ω
V
20 Ω
+
+
j20 Ω
Vi = 1∠0°
j20 Ω
Vo
−
Z
We now want to show that this circuit will produce a 90° phase shift.
Z = j20 || (20 + j20) =
V=
( j20)(20 + j20) - 20 + j20
=
= 4 (1 + j3)
20 + j40
1 + j2
Z
1 + j3 1
4 + j12
Vi =
= (1 + j)
(1∠0°) =
Z + 20
6 + j3 3
24 + j12
Vo =
⎛ j ⎞⎛ 1
⎞ j
j20
⎟⎜ (1 + j) ⎟ = = 0.3333∠90°
V =⎜
⎠ 3
20 + j20
⎝1 + j ⎠⎝ 3
This shows that the output leads the input by 90°.
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Chapter 9, Problem 75.
Design a circuit that will transform a sinusoidal voltage input to a cosinusoidal voltage
output.
Chapter 9, Solution 75.
Since cos(ωt ) = sin(ωt + 90°) , we need a phase shift circuit that will cause the
output to lead the input by 90°. This is achieved by the RL circuit shown
below, as explained in the previous problem.
10 Ω
10 Ω
+
Vi
+
j10 Ω
j10 Ω
−
Vo
−
This can also be obtained by an RC circuit.
Chapter 9, Problem 76.
For the following pairs of signals, determine if v 1 leads or lags v 2 and by how much.
(a) v 1 = 10 cos(5t - 20 o ),
v2
=
(b) v 1 = 19 cos(2t - 90 o ),
v2
=6
(c) v 1
=-
4 cos10t ,
v2
= 15
8 sin5t
sin2t
sin10t
Chapter 9, Solution 76.
(a) v2 = 8sin 5t = 8cos(5t − 90o )
v1 leads v2 by 70o.
(b) v2 = 6sin 2t = 6 cos(2t − 90o )
v1 leads v2 by 180o.
(c ) v1 = −4 cos10t = 4 cos(10t + 180o )
v2 = 15sin10t = 15cos(10t − 90o )
v1 leads v2 by 270o.
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Chapter 9, Problem 77.
Refer to the RC circuit in Fig. 9.81.
(a) Calculate the phase shift at 2 MHz.
(b) Find the frequency where the phase shift is 45 o .
Figure 9.81
For Prob. 9.77.
Chapter 9, Solution 77.
(a)
- jX c
V
R − jX c i
1
1
=
= 3.979
where X c =
ωC (2π)(2 × 10 6 )(20 × 10 -9 )
Vo =
Vo
- j3.979
=
=
Vi 5 - j3.979
Vo
=
Vi
3.979
25 + 15.83
3.979
5 + 3.979
2
2
∠(-90° + tan -1 (3.979 5))
∠(-90° − 38.51°)
Vo
= 0.6227 ∠ - 51.49°
Vi
Therefore, the phase shift is 51.49° lagging
(b)
θ = -45° = -90° + tan -1 (X c R )
45° = tan -1 (X c R ) ⎯
⎯→ R = X c =
ω = 2πf =
f=
1
ωC
1
RC
1
1
=
= 1.5915 MHz
2πRC (2π )(5)(20 × 10 -9 )
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Chapter 9, Problem 78.
A coil with impedance 8 + j6 Ω is connected in series with a capacitive reactance X. The
series combination is connected in parallel with a resistor R. Given that the equivalent
impedance of the resulting circuit is 5 ∠ 0 o Ω find the value of R and X.
Chapter 9, Solution 78.
8+j6
R
Z
-jX
Z = R //[8 + j (6 − X )] =
R[8 + j (6 − X )]
=5
R + 8 + j (6 − X )
i.e 8R + j6R – jXR = 5R + 40 + j30 –j5X
Equating real and imaginary parts:
8R = 5R + 40 which leads to R=13.333Ω
6R-XR =30-5X which leads to X= 6 Ω.
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Chapter 9, Problem 79.
(a) Calculate the phase shift of the circuit in Fig. 9.82.
(b) State whether the phase shift is leading or lagging (output with respect to input).
(c) Determine the magnitude of the output when the input is 120 V.
Figure 9.82
For Prob. 9.79.
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Chapter 9, Solution 79.
(a)
Consider the circuit as shown.
20 Ω
V2
40 Ω
V1
30 Ω
+
Vi
+
j10 Ω
j30 Ω
j60 Ω
−
Vo
−
Z2
Z1
( j30)(30 + j60)
Z1 = j30 || (30 + j60) =
= 3 + j21
30 + j90
( j10)(43 + j21)
Z 2 = j10 || (40 + Z1 ) =
= 1.535 + j8.896 = 9.028∠80.21°
43 + j31
Let Vi = 1∠0° .
Z2
(9.028∠80.21°)(1∠0°)
Vi =
Z 2 + 20
21.535 + j8.896
V2 = 0.3875∠57.77°
V2 =
Z1
3 + j21
(21.213∠81.87°)(0.3875∠57.77°)
V2 =
V2 =
Z1 + 40
43 + j21
47.85∠26.03°
V1 = 0.1718∠113.61°
V1 =
j60
j2
2
V1 =
V1 = (2 + j)V1
30 + j60
1 + j2
5
Vo = (0.8944∠26.56°)(0.1718∠113.6°)
Vo = 0.1536 ∠140.2°
Vo =
Therefore, the phase shift is 140.2°
(b)
The phase shift is leading.
(c)
If Vi = 120 V , then
Vo = (120)(0.1536∠140.2°) = 18.43∠140.2° V
and the magnitude is 18.43 V.
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Chapter 9, Problem 80.
Consider the phase-shifting circuit in Fig. 9.83. Let V i = 120 V operating at 60 Hz. Find:
(a) V o when R is maximum
(b) V o
when
R is minimum
(c) the value of R that will produce a phase shift of 45 o
Figure 9.83
For Prob. 9.80.
Chapter 9, Solution 80.
200 mH ⎯
⎯→
Vo =
jωL = j (2π )(60)(200 × 10 -3 ) = j75.4 Ω
j75.4
j75.4
Vi =
(120∠0°)
R + 50 + j75.4
R + 50 + j75.4
(a)
When R = 100 Ω ,
(75.4∠90°)(120∠0°)
j75.4
Vo =
(120 ∠0°) =
167.88∠26.69°
150 + j75.4
Vo = 53.89∠63.31° V
(b)
When R = 0 Ω ,
(75.4∠90°)(120 ∠0°)
j75.4
Vo =
(120∠0°) =
90.47 ∠56.45°
50 + j75.4
Vo = 100∠33.55° V
(c)
To produce a phase shift of 45°, the phase of Vo = 90° + 0° − α = 45°.
Hence, α = phase of (R + 50 + j75.4) = 45°.
For α to be 45°,
R + 50 = 75.4
Therefore,
R = 25.4 Ω
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Chapter 9, Problem 81.
The ac bridge in Fig. 9.37 is balanced when R 1 = 400 Ω , R 2 = 600 Ω , R 3 = 1.2k Ω , and
C 2 = 0.3 µF . Find R x and C x . Assume R 2 and C 2 are in series.
Chapter 9, Solution 81.
Let
Z1 = R 1 ,
Z2 = R 2 +
1
,
jωC 2
Zx =
Z3
Z
Z1 2
Rx +
R3 ⎛
1
1 ⎞
⎜R 2 +
⎟
=
jωC x R 1 ⎝
jωC 2 ⎠
Rx =
R3
1200
(600) = 1.8 kΩ
R2 =
400
R1
Z 3 = R 3 , and Z x = R x +
1
.
jωC x
R1
⎛ 400 ⎞
1 ⎛R3 ⎞ ⎛ 1 ⎞
⎟(0.3 × 10 -6 ) = 0.1 µF
=⎜ ⎟⎜ ⎟ ⎯
C2 = ⎜
⎯→ C x =
⎝ 1200 ⎠
Cx ⎝ R1 ⎠ ⎝ C2 ⎠
R3
Chapter 9, Problem 82.
A capacitance bridge balances when R 1 = 100 Ω , and R 2 = 2k Ω and C s = 40 µF . What
is C x the capacitance of the capacitor under test?
Chapter 9, Solution 82.
Cx =
R1
⎛ 100 ⎞
⎟(40 × 10 -6 ) = 2 µF
Cs = ⎜
⎝ 2000 ⎠
R2
Chapter 9, Problem 83.
An inductive bridge balances when R 1 = 1.2k Ω , R 2 = 500 Ω , and L s = 250 mH. What
is the value of L x , the inductance of the inductor under test?
Chapter 9, Solution 83.
Lx =
R2
⎛ 500 ⎞
⎟(250 × 10 -3 ) = 104.17 mH
Ls = ⎜
⎝
1200 ⎠
R1
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Chapter 9, Problem 84.
The ac bridge shown in Fig. 9.84 is known as a Maxwell bridge and is used for
accurate measurement of inductance and resistance of a coil in terms of a standard
capacitance C s⋅ Show that when the bridge is balanced,
R
L x = R2 R3Cs
and
R x = 2 R3
R1
Find L x and R x for R 1 = 40k Ω , R 2 = 1.6k Ω , R 3 = 4k Ω , and C s = 0.45 µ F.
Figure 9.84
Maxwell bridge; For Prob. 9.84.
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Chapter 9, Solution 84.
Let
1
Z2 = R 2 ,
,
jωC s
R1
jωC s
R1
Z1 =
=
1
jωR 1C s + 1
R1 +
jωC s
Z1 = R 1 ||
Since Z x =
Z 3 = R 3 , and Z x = R x + jωL x .
Z3
Z ,
Z1 2
R x + jωL x = R 2 R 3
jωR 1C s + 1 R 2 R 3
=
(1 + jωR 1C s )
R1
R1
Equating the real and imaginary components,
R 2R 3
Rx =
R1
ωL x =
R 2R 3
(ωR 1C s ) implies that
R1
L x = R 2 R 3Cs
Given that R 1 = 40 kΩ , R 2 = 1.6 kΩ , R 3 = 4 kΩ , and C s = 0.45 µF
R 2 R 3 (1.6)(4)
=
kΩ = 0.16 kΩ = 160 Ω
R1
40
L x = R 2 R 3 C s = (1.6)(4)(0.45) = 2.88 H
Rx =
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Chapter 9, Problem 85.
The ac bridge circuit of Fig. 9.85 is called a Wien bridge. It is used for measuring the
frequency of a source. Show that when the bridge is balanced,
f =
1
2π R2 R4C2C4
Figure 9.85
Wein bridge; For Prob. 9.85.
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Chapter 9, Solution 85.
Let
1
,
jωC 2
R4
- jR 4
Z4 =
=
jωR 4 C 4 + 1 ωR 4 C 4 − j
Z1 = R 1 ,
Since Z 4 =
Z2 = R 2 +
Z3
Z
Z1 2
Z 3 = R 3 , and Z 4 = R 4 ||
1
.
jωC 4
⎯
⎯→ Z1 Z 4 = Z 2 Z 3 ,
⎛
- jR 4 R 1
j ⎞
⎟
= R 3 ⎜R 2 −
ωR 4 C 4 − j
ωC 2 ⎠
⎝
jR 3
- jR 4 R 1 (ωR 4 C 4 + j)
= R 3R 2 −
2
2 2
ω R 4C4 + 1
ωC 2
Equating the real and imaginary components,
R 1R 4
= R 2R 3
2
ω R 24 C 24 + 1
(1)
R3
ωR 1 R 24 C 4
=
ω2 R 24 C 24 + 1 ωC 2
(2)
Dividing (1) by (2),
1
= ωR 2 C 2
ωR 4 C 4
1
ω2 =
R 2C2R 4C4
1
ω = 2πf =
R 2C2 R 4C4
f=
1
2π R 2 R 4 C 2 C 4
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Chapter 9, Problem 86.
The circuit shown in Fig. 9.86 is used in a television receiver. What is the total
impedance of this circuit?
Figure 9.86
For Prob. 9.86.
Chapter 9, Solution 86.
1
1
1
+
+
240 j95 - j84
Y = 4.1667 × 10 -3 − j0.01053 + j0.0119
Y=
1
1000
1000
=
=
Y 4.1667 + j1.37 4.3861∠18.2°
Z = 228∠-18.2° Ω
Z=
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Chapter 9, Problem 87.
The network in Fig. 9.87 is part of the schematic describing an industrial electronic
sensing device. What is the total impedance of the circuit at 2 kHz?
Figure 9.87
For Prob. 9.87.
Chapter 9, Solution 87.
1
-j
= 50 +
jωC
(2π)(2 × 10 3 )(2 × 10 -6 )
Z1 = 50 − j39.79
Z1 = 50 +
Z 2 = 80 + jωL = 80 + j (2π)(2 × 10 3 )(10 × 10 -3 )
Z 2 = 80 + j125.66
Z 3 = 100
1
1
1
1
=
+
+
Z Z1 Z 2 Z 3
1
1
1
1
=
+
+
Z 100 50 − j39.79 80 + j125.66
1
= 10 -3 (10 + 12.24 + j9.745 + 3.605 − j5.663)
Z
= (25.85 + j4.082) × 10 -3
= 26.17 × 10 -3 ∠8.97°
Z = 38.21∠-8.97° Ω
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Chapter 9, Problem 88.
A series audio circuit is shown in Fig. 9.88.
(a) What is the impedance of the circuit?
(b) If the frequency were halved, what would be the impedance of the circuit?
Figure 9.88
For Prob. 9.88.
Chapter 9, Solution 88.
(a)
(b)
Z = - j20 + j30 + 120 − j20
Z = 120 – j10 Ω
1
1
=
would cause the capacitive
ωC 2πf C
impedance to double, while ωL = 2πf L would cause the inductive
impedance to halve. Thus,
Z = - j40 + j15 + 120 − j40
Z = 120 – j65 Ω
If the frequency were halved,
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Chapter 9, Problem 89.
An industrial load is modeled as a series combination of a capacitance and a resistance as
shown in Fig. 9.89. Calculate the value of an inductance L across the series combination
so that the net impedance is resistive at a frequency of 50 kHz.
Figure 9.89
For Prob. 9.89.
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Chapter 9, Solution 89.
⎛
1 ⎞
⎟
Z in = jωL || ⎜ R +
jωC ⎠
⎝
⎛
1 ⎞
L
⎟
jωL ⎜ R +
+ jωL R
jωC ⎠
⎝
C
=
Z in =
1
⎛
1 ⎞
R + jωL +
⎟
R + j⎜ωL −
jωC
⎝
ωC ⎠
Z in =
⎞⎛
⎛
⎛L
1 ⎞⎞
⎟⎟
⎜ + jωL R ⎟⎜ R − j⎜ωL −
⎠⎝
⎝
⎝C
ωC ⎠⎠
⎛
1 ⎞
⎟
R + ⎜ωL −
⎝
ωC ⎠
2
2
To have a resistive impedance, Im(Z in ) = 0 . Hence,
⎛ L ⎞⎛
1 ⎞
⎟=0
ωL R 2 − ⎜ ⎟⎜ωL −
⎝ C ⎠⎝
ωC ⎠
1
ωR 2 C = ωL −
ωC
ω2 R 2 C 2 = ω2 LC − 1
L=
ω2 R 2 C 2 + 1
ω2 C
Now we can solve for L.
L = R 2 C + 1 /(ω 2 C)
= (2002)(50x10–9) + 1/((2πx50,000)2(50x10–9)
= 2x10–3 + 0.2026x10–3 = 2.203 mH.
Checking, converting the series resistor and capacitor into a parallel combination, gives
220.3Ω in parallel with -j691.9Ω. The value of the parallel inductance is ωL =
2πx50,000x2.203x10–3 = 692.1Ω which we need to have if we are to cancel the effect of
the capacitance. The answer checks.
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Chapter 9, Problem 90.
An industrial coil is modeled as a series combination of an inductance L and resistance R,
as shown in Fig. 9.90. Since an ac voltmeter measures only the magnitude of a sinusoid,
the following measurements are taken at 60 Hz when the circuit operates in the steady
state:
Vs = 145 V, V1 = 50 V,
Vo = 110 V
Use these measurements to determine the values of L and R.
Figure 9.90
For Prob. 9.90.
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Chapter 9, Solution 90.
Vs = 145∠0° ,
X = ωL = (2π)(60) L = 377 L
Let
Vs
145∠0°
I=
=
80 + R + jX 80 + R + jX
V1 = 80 I =
50 =
(80)(145)
80 + R + jX
(80)(145)
80 + R + jX
(1)
Vo = (R + jX) I =
110 =
(R + jX)(145∠0°)
80 + R + jX
(R + jX)(145)
80 + R + jX
(2)
From (1) and (2),
50
80
=
110
R + jX
⎛11 ⎞
R + jX = (80) ⎜ ⎟
⎝5⎠
R 2 + X 2 = 30976
(3)
From (1),
(80)(145)
= 232
50
6400 + 160R + R 2 + X 2 = 53824
160R + R 2 + X 2 = 47424
80 + R + jX =
(4)
Subtracting (3) from (4),
160R = 16448 ⎯
⎯→ R = 102.8 Ω
From (3),
X 2 = 30976 − 10568 = 20408
X = 142.86 = 377 L ⎯
⎯→ L = 0.3789 H
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Chapter 9, Problem 91.
Figure 9.91 shows a parallel combination of an inductance and a resistance. If it is
desired to connect a capacitor in series with the parallel combination such that the net
impedance is resistive at 10 MHz, what is the required value of C?
Figure 9.91
For Prob. 9.91.
Chapter 9, Solution 91.
1
+ R || jωL
jωC
-j
jωLR
Z in =
+
ωC R + jωL
Z in =
- j ω 2 L2 R + jωLR 2
=
+
ωC
R 2 + ω 2 L2
To have a resistive impedance, Im(Z in ) = 0 .
Hence,
-1
ωLR 2
+ 2
=0
ωC R + ω2 L2
1
ωLR 2
= 2
ωC R + ω2 L2
R 2 + ω 2 L2
C=
ω2 LR 2
where ω = 2π f = 2π × 10 7
9 × 10 4 + (4π 2 × 1014 )(400 × 10 −12 )
(4π 2 × 1014 )(20 × 10 − 6 )(9 × 10 4 )
9 + 16π 2
C=
nF
72π 2
C = 235 pF
C=
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,
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Chapter 9, Problem 92.
A transmission line has a series impedance of Z = 100 ∠ 75 Ω and a shunt admittance
o
of Y = 450 ∠ 48 µS . Find: (a) the characteristic impedance Z o = Z Y
o
(b) the propagation constant γ = ZY .
Chapter 9, Solution 92.
Z
=
(a) Z o =
Y
100∠75 o
= 471.4∠13.5 o Ω
o
−6
450∠48 x10
(b) γ = ZY = 100∠75 o x 450∠48 o x10 −6 = 0.2121∠61.5 o
Chapter 9, Problem 93.
A power transmission system is modeled as shown in Fig. 9.92. Given the following;
Source voltage
o
V s = 115 ∠ 0 V,
Source impedance
Line impedance
Load impedance
find the load current
Z s = 1 + j0.5 Ω ,
Z l = 0.4 + j0.3 Ω ,
Z L = 23.2 + j18.9 Ω ,
I L⋅
Figure 9.92
For Prob. 9.93.
Chapter 9, Solution 93.
Z = Zs + 2 Zl + ZL
Z = (1 + 0.8 + 23.2) + j(0.5 + 0.6 + 18.9)
Z = 25 + j20
VS
115∠0°
=
Z 32.02 ∠38.66°
I L = 3.592∠-38.66° A
IL =
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.
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