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8/2/2019 Chapter9 Assignments
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Copyright ©2004 The McGraw-Hill Companies Inc.
1
Chapter 9, Problem 32 (26).
Two elements are connected in series as shown in Fig. 9.40.
If i = 12 cos (2t - 30°) A, find the element values.
Figure 9.40
Chapter 9, Solution 32 (26).
V = 180∠10°, I = 12∠-30°, w = 2
Ω+=°∠=°∠°∠
== 642.9 j49.11041530-12
01180
I
VZ
One element is a resistor with R = 11.49Ω.The other element is an inductor with wL = 9.642 or L = 4.821 H.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Copyright ©2004 The McGraw-Hill Companies Inc.
2
Chapter 9, Problem 46 (39).
If i s = 5 cos (10t + 40°) A in the circuit in Fig. 9.53, find io.
Figure 9.53
Chapter 9, Solution 46 (39).
°∠= → °+= 405)40t10cos(5i ss I
j-)1.0)(10( j
1
C j
1F1.0 ==
ω →
2 j)2.0)(10( jL jH2.0 ==ω →
Let 6.1 j8.02 j4
8 j2 j||41 +=+
==Z , j32 −=Z
)405(6.0 j8.3
j1.60.8s
21
1
o °∠++
=+
= IZZ
ZI
°∠=°∠
°∠°∠= 46.94325.2
97.8847.3
)405)(43.63789.1(oI
Thus, =)t(io 2.325 cos(10t + 94.46°) A
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Copyright ©2004 The McGraw-Hill Companies Inc.
3
Chapter 9, Problem 55 (44).
Find Z in the network of Fig. 9.62, given that Vo = 4∠0° V
Figure 9.62
Chapter 9, Solution 55 (44).
-j0.5
8 j
4
8 j
o
1 ===V
I
j8 j4-
)8 j((-j0.5)
j4-
)8 j(1
2 +=+
=+
=ZZZI
I
5.0 j8
j8
-j0.521 +=++=+=ZZ
III
)8 j(12 j20- 1 ++= ZII
)8 j(2 j-
2 j
812 j20- ++
+= ZZ
−=2
1 j
2
3 j26-4- Z °∠=
°∠°∠
=−
= 279.6864.1618.43-5811.1
25.26131.26
2
1 j
2
3
j26-4-Z
Z = 2.798 – j16.403 Ω
-j4 Ω
I I1
8 Ω
12 Ω Z
+
−
I2
-j20 V
+
Vo
−
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Copyright ©2004 The McGraw-Hill Companies Inc.
4
Chapter 9, Problem 66 (50).
For the circuit in Fig. 9.73, calculate ZT and Vab.
) j12(145
170
5 j60
)10 j40)(5 j20()10 j40(||)5 j20(T −=
++−
=+−=Z
=TZ 14.069 – j1.172 Ω = 14.118∠-4.76°
°∠=°∠
°∠== 76.9425.4
76.4-118.14
9060
TZ
VI
III j12
2 j8
5 j60
10 j401 +
+=
++
=
III j12
j4
5 j60
5 j202 +
−
=+
−
=
21ab 10 j20- IIV += IIV j12
40 j10
j12
)40 j(160-ab +
++
++
=
IIV145
j)(150)-12(
j12
150-ab
+=
+= )76.9725.4)(24.175457.12(ab °∠°∠=V
=abV 52.94∠273° V
+Vab
10 Ω20 ΩI2I1
I
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Copyright ©2004 The McGraw-Hill Companies Inc.
5
Chapter 9, Problem 79 (61).
(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
Chapter 9, Solution 79 (61).
Consider the circuit as shown.
21 j390 j30
)60 j30)(30 j()60 j30(||30 j1 +=
++
=+=Z
°∠=+=+
+=+= 21.80028.9896.8 j535.1
31 j43
)21 j43)(10 j()40(||10 j 12 ZZ
Let °∠= 01iV .
896.8 j535.21
)01)(21.80028.9(
20 i
2
2
2 +°∠°∠
=+
= VZ
ZV
°∠= 77.573875.02V
20 Ω 30 Ω
10 Ω 60 Ω+
Vi
−
+
Vo
−
40 Ω
30 Ω
V2
Z2
V1
Z1
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Copyright ©2004 The McGraw-Hill Companies Inc.
6
°∠°∠°∠
=+
+=
+=
03.2685.47
)77.573875.0)(87.81213.21(
21 j43
21 j3
40 22
1
1
1 VVZ
ZV
°∠= 61.1131718.01V
111o ) j2(5
2
2 j1
2 j
60 j30
60 jVVVV +=
+=
+=
)6.1131718.0)(56.268944.0(o °∠°∠=V
°∠= 2.1401536.0oV
Therefore, the phase shift is 140.2° The phase shift is leading.
If V120i =V , then
°∠=°∠= 2.14043.18)2.1401536.0)(120(oV V
and the magnitude is 18.43 V.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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7
Chapter 9, Problem 84 (66).
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,
L x = R2 R3 C s and 23
1
x
R R R
R=
Find L x and R x for R1 = 40 k Ω, R2 = 1.6 k Ω, R3 = 4 k Ω, and C s = 0.45 µ F.
Figure 9.84
Chapter 9, Solution 84 (66).
Lets
11 C j
1||R
ω=Z , 22 R =Z , 33 R =Z , and xxx L jR ω+=Z .
1CR j
R
C j
1R
C j
R
s1
1
s
1
s
1
1 +ω=
ω+
ω=Z
Since 2
1
3
x ZZ
ZZ = ,
)CR j1(R
R R
R
1CR jR R L jR s1
1
32
1
s1
32xx ω+=+ω
=ω+
Equating the real and imaginary components,
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Copyright ©2004 The McGraw-Hill Companies Inc.
8
1
32
x R
R R R =
)CR (R
R R L
s11
32
xω=ω implies that
s32x CR R L =
Given that Ω= k 40R 1 , Ω= k 6.1R 2 , Ω= k 4R 3 , and F45.0Cs µ=
=Ω=Ω== k 16.0k 40
)4)(6.1(
R
R R R
1
32
x 160 Ω
=== )45.0)(4)(6.1(CR R L s32x 2.88 H
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Copyright ©2004 The McGraw-Hill Companies Inc.
9
Chapter 9, Problem 90 (72).
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:
1V 145V, V 50V, V 110V s o= = =
Use these measurements to determine the values of L and R.
Figure 9.90
Chapter 9, Solution 90 (72).
Let °∠= 0145sV , L377 jL)60)(2( jL jX =π=ω=
jXR 80
0145
jXR 80
s
++°∠
=++
=V
I
jXR 80
)145)(80(801 ++
== IV
jXR 80
)145)(80(50
++=
(1)
jXR 80
)0145)( jXR () jXR (o ++
°∠+=+= IV
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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10
jXR 80
)145)( jXR (110
+++
=
(2)
From (1) and (2),
jXR
80
110
50
+=
=+5
11)80( jXR
30976XR 22 =+
(3)
From (1),
23250
)145)(80(
jXR 80 ==++ 53824XR R 1606400 22 =+++
47424XR R 160 22 =++
(4)
Subtracting (3) from (4),
= → = R 16448R 160 102.8 Ω
From (3),
204081056830976X2 =−=
= → == LL37786.142X 0.3789 H