Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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°,
Z=
I = 12∠-30°,
w = 2
V 180∠10°
=
= 15∠40° = 11.49 + j 9.642 Ω
I 12∠ - 30°
One element is a resistor with R = 11.49 Ω.
The other element is an inductor with wL = 9.642 or
L = 4.821 H.
Copyright ©2004 The McGraw-Hill Companies Inc.
1
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 9, Problem 46 (39).
If is = 5 cos (10t + 40°) A in the circuit in Fig. 9.53, find io.
Figure 9.53
Chapter 9, Solution 46 (39).
i s = 5 cos(10 t + 40°) 
→ I s = 5∠40°
Let
0.1 F 
→
1
1
=
= -j
jωC j (10)(0.1)
0.2 H 
→
jωL = j (10)(0.2) = j2
Z1 = 4 || j2 =
j8
= 0.8 + j1.6 ,
4 + j2
Z2 = 3 − j
Io =
Z1
0.8 + j1.6
(5∠40°)
Is =
3.8 + j0.6
Z1 + Z 2
Io =
(1.789∠63.43°)(5∠40°)
= 2.325∠94.46°
3.847 ∠8.97°
Thus, i o ( t ) = 2.325 cos(10t + 94.46°) A
Copyright ©2004 The McGraw-Hill Companies Inc.
2
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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).
12 Ω
-j20 V
I
I1
Z
I2
+
−
-j4 Ω
+
Vo
j8 Ω
−
I1 =
Vo
4
= = -j0.5
j 8 j8
I2 =
I 1 (Z + j8) (-j0.5)(Z + j8) Z
=
= +j
- j4
- j4
8
I = I 1 + I 2 = -j0.5 +
Z
Z
+ 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
Z = 2.798 – j16.403 Ω
Z=
- 4 - j26 26.31∠261.25°
=
= 16.64∠279.68°
3
1 1.5811∠ - 18.43°
−j
2
2
Copyright ©2004 The McGraw-Hill Companies Inc.
3
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 9, Problem 66 (50).
For the circuit in Fig. 9.73, calculate ZT and Vab.
Z T = (20 − j5) || (40 + j10) =
(20 − j5)(40 + j10) 170
=
(12 − j)
60 + j5
145
Z T = 14.069 – j1.172 Ω = 14.118∠-4.76°
I=
60∠90°
V
=
= 4.25∠94.76°
Z T 14.118∠ - 4.76°
I
I1
I2
20 Ω
j10 Ω
+
Vab
I1 =
40 + j10
8 + j2
I=
I
60 + j5
12 + j
I2 =
20 − j5
4− j
I=
I
60 + j5
12 + j
Vab = -20 I 1 + j10 I 2
- 150
(-12 + j)(150)
I=
I
12 + j
145
Vab = 52.94∠273° V
Vab =
Vab =
- (160 + j40)
10 + j40
I+
I
12 + j
12 + j
Vab = (12.457 ∠175.24°)(4.25∠97.76°)
Copyright ©2004 The McGraw-Hill Companies Inc.
4
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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.
20 Ω
V2
40 Ω
V1
30 Ω
+
Vi
+
j10 Ω
j30 Ω
−
j60 Ω
Vo
−
Z2
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
Z1 = j30 || (30 + j60) =
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 =
Copyright ©2004 The McGraw-Hill Companies Inc.
5
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Z1
3 + j21
(21.213∠81.87°)(0.3875∠57.77°)
V2 =
V2 =
43 + j21
47.85∠26.03°
Z1 + 40
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°
The phase shift is leading.
If Vi = 120 V , then
Vo = (120)(0.1536∠140.2°) = 18.43∠140.2° V
and the magnitude is 18.43 V.
Copyright ©2004 The McGraw-Hill Companies Inc.
6
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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 Cs. Show that when the bridge is balanced,
Lx = R2 R3 Cs
and
Rx =
R2
R3
R1
Find Lx and Rx for R1 = 40 kΩ, R2 = 1.6 kΩ, R3 = 4 kΩ, and Cs = 0.45 µF.
Figure 9.84
Chapter 9, Solution 84 (66).
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,
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Rx =
ωL x =
R 2R 3
R1
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 =
Copyright ©2004 The McGraw-Hill Companies Inc.
8
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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:
Vs = 145V,
V1 = 50V,
Vo = 110V
Use these measurements to determine the values of L and R.
Figure 9.90
Chapter 9, Solution 90 (72).
Let
Vs = 145∠0° ,
X = jωL = j (2π )(60) L = j377 L
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 =
(R + jX)(145∠0°)
80 + R + jX
Copyright ©2004 The McGraw-Hill Companies Inc.
9
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
110 =
(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
Copyright ©2004 The McGraw-Hill Companies Inc.
10