CHAPTER 4
Problems and Solutions
不要用看的，要動手！
Problem 4
How long will it take the following waveform to pass through 10
cycles?
ω = 2,000 rad/s = 2π/T
T = 2 π ÷ 2000 rad/s = 3.142 ms
走10個cycles 需要 10× T = 31.42 ms
Problem 5
At what instant of time will the following waveform be 6 V?
Associate t = 0s with θ of sinθ equal to 0°.
v  12 sin( 200  30)
6  12 sin 
   30
t0
Problem 6
a. At what angle θ (closest to 0°) will the following waveform
reach 3 mA? i = 8.6 × 10-3 sin 500t
b. At what time t will it occur?
v  12 sin( 200  30)
3
i  8.6  10 sin 500 t
 3  10
3
3
 8.6  10 sin 500 t
sin 500 t  0.349
500t  sin
t
1
0.349  20.426  0.3565rad
0.3565rad
500rad / s
 0.713ms
Problem 7
What is the phase relationship between the following pairs of
waveforms?
相角-72°
相角-16°
a. 電流領先電壓 56°
b. 電流領先電壓140°
Problem 8
Write the sinusoidal expression for a current i that has a peak
value of 6 μA and leads the following voltage by 40°
i的 peak value 為6 μA ，且領先電壓40°
 i (t) = 6×10-6A sin(1000t+46°)
Problem 9
Write the sinusoidal expression for a voltage V that has a peak
value of 48 mV and lags the following current I by 60°
v的 peak value 為48 mV ，且落後電流60°
 v (t) = 48×10-3V sin(ωt-90°)
Problem 10
Determine the effective value of each of the following.
a.Veff 
b.I eff 
v peak
 45.25V
2
i peak
 4.243mA
2
c.Veff  Vdc  20V
Problem 11
Write the sinusoidal expression for each quantity using the
information provided
a. Ieff = 36 mA, f = 1 kHz, phase angle = 60°
b. Veff = 8V, f =60 Hz, phase angle = -10°
a.i( t ) 
2  36mA sin( 2f  60)
3
 50.9  10 A sin( 6,283.2 t  60)
b.v ( t ) 
2  8V sin( 2f  10)
 11.31V sin( 377 t  10)
Problem 12
Determine the average value of the following.
Problem 13
Determine the average value of the waveform in Fig. 4.80.
v av 
A
1
A BC D
8ms
( 2 ms )( 4 V )
B  (1ms )( 2 V )
2
C
1
(1ms )( 2 V )
2
v av  0.25V
D
1
2
(1ms )( 2 V )
Problem 14
Determine the average value of the waveform in Fig. 4.81.
求出交點
v av 
A  BC D E
8ms
v av  0.5mV
Problem 15
Determine the average value of i2 from θ= 0 to π if i = 6 sin θ
(integral calculus required).
i av



(6 sin ) d
2
0

 18A
2
Problem 16
a. Determine the sinusoidal expression for the voltage drop
across a 1.2-kΩ resistor if the current iR is 8×10-3 sin 200t.
b. Find the power delivered to the resistor.
c. What is the power factor of the load?
a.v peak  i peak  R  9.6V
b.I eff 
i peak
2
 5.657mA
v R ( t )  9.6V sin 200t
PI
2
eff
 R  38.4mW
Problem 17
a. Find the sinusoidal expression for the current through a 2.2-kΩ
resistor if the power delivered to the resistor is 3.6 W at a
frequency of 1000 Hz.
b. Find the sinusoidal expression for the voltage across the
resistor.
a.P  I
2
eff
 R  3.6 W
i peak 
I eff  40.45mA
2I eff  57.2mA
i( t )  57.2mA sin( 6.283 10 t )
3
b.v peak  i peak  R  125.84V
v( t )  125.84V sin( 6.283 10 t )
3
Problem 18
a. Find the sinusoidal expression for the voltage drop across a
20-mH coil if the current iL is 4 sin(500t + 60°)
b. Find the power delivered to the coil.
c. What is the power factor of the load?
a.X L  L  (500rad / s)  20mH  10
v p  i L X L  ( 4A )(10)  40V
v( t )  40V sin( 500 t  150) 電壓領先電流90º
b.P  0W
c.Power factor FP  0
Problem 19
Determine the sinusoidal expression for the current ic of a
10-μF capacitor if the vo1tage across the capacitor is Vc =
20× 10-3 sin(2000t + 30°)
XC 
ip 
1
C
v peak
XC


1
6
( 2000rad / s)(10  10 F)
20mV
50
 50
 0.4mA
i C ( t )  0.4mA sin( 2000 t  120)
電流領先電壓90º
Problem 20
a. For the following pairs determine whether the element is a
resistor, inductor, or capacitor.
b. Determine the resistance, inductance, or capacitance.
1. v = 16 sin(200t + 80° )
i = 0.04 sin(200t－10° )
2. v = 0.12 sin(1000t + 10° )
i = 6×10-3cos(1000t + 10° )
v = 16 sin(200t + 80° )
i = 0.04 sin(200t－10° )
電流落後電壓90º，為電感。
XL 
L
v peak
i peak
XL



16V
 400  L
0.04A
400
200rad / s
2H
v = 0.12 sin(1000t + 10°)
i = 6×10-3cos(1000t + 10° ) = 6×10-3sin(1000t+100°)
電流領先電壓90º，為電容。
XC 
C
v peak

i peak
1
X C
0.12V
 20 
6mA

1
C
1
(1000rad / s)( 20)
 50F
Problem 21
a. For the following pairs determine the power delivered to the
load.
b. Find the power factor and indicate whether it is inductive or
capacitive.
1. v = 1600 sin(377t + 360° )
i = 0.8 sin(377t + 60° )
2. v = 100 sin(106t－ 10° )
i = 0.2 sin(106t － 40° )
a.P1 
(1600V )(0.8A )
P2 
cos 60  320 W
2
(100V )( 0.2A )
cos 30  8.66 W
2
b.FP1  cos 60  0.5
leading
FP 2  cos 30  0.866
lagging
Problem 22
Convert the following to the other domain.
Problem 23
Perform the following operations. State your answer in polar
form.
Problem 24
Using phasor notation, determine the vo1tage (in the time domain)
across a 2.2-kΩ resistor if the current through the resistor is i =
20× 10-3 sin (400t + 30°).
i( t )  I( j)
I( j)  (0.707)( 20mA )30  14.142mA 30
V ( j)  I( j)  Z R  (14.142mA 30)( 2.2k0)  31.11V30
 time domain
v( t ) 
2  31.11V sin( 400 t  30)  44V sin( 400 t  30)
I( j)  i( t )  (0.707)( 20mA )30  14.142mA 30
v( t )  V( j) 
2  31.11sin( 400t  30)  44V sin( 400t  30)
Problem 25
Using phasor notation, determine the current (in the time domain)
through a 20-mH coil if the voltage across the coil is vL = 4
sin(1000t + 10°).
I( j)  i( t )
v( t )  V ( j)
V ( j)  (0.707)( 4V )10  2.828V10
v( t )  V( j)
X L  L  (1000rad / s)( 20mH )  20
I( j) 
V( j)

ZL
2.828V10
2090
 141.4mA   80
 time domain
i( t ) 
2  141.4mA sin( 1000t  80)  0.2A sin( 1000 t  80)
Problem 26
Using phasor notation, determine the voltage (in the time domain)
across a 10-μF capacitor if the current ic = 40×10-3 sin(10t + 40°).
i( t )  I( j)
I( j)  (0.707)( 40mA )40  28.284mA 40
XC 
1
C
ZC 
1
C
  90  10k  90
V ( j)  I( j) Z C  282.84V  50
 time domain
v( t ) 
2  282.84V sin( 10 t  50)  400A sin( 10 t  50)
Problem 27
For the system in Fig. 4.82, determine the vo1tage v1 in the time
domain.
V1 ( j)  E ( j)  V2 ( j)  (0.707)(100V)0  (0.707)( 20V )60
 (70.71  j0)  (7.071  j12.247)  63.639  j12.247
 64.807  10.893
 time domain
v1 ( t ) 
2 (64.807 V) sin( 377 t  10.893)  91.64V sin( 377 t  10.893)
再度提醒
V1 ( j)  E( j)  V2 ( j)  (0.707)(100V)0  (0.707)( 20V)60
 (70.71  j0)  (7.071  j12.247)  63.639  j12.247
 64.807  10.893

2 (64.807 V) sin( 377 t  10.893)  91.64V sin( 377 t  10.893)
Problem 28
For the system in Fig. 4.83, determine the current i in the time
domain.
I( j)  I1 ( j)  I 2 ( j)  ...
 1.273A  60
 time domain
i( t )  ...  1.8A sin( 400 t  60)
Problem 29 For the series ac network in Fig. 4.84, determine:
a. The reactance of the capacitor.
b. The total impedance and the impedance diagram.
c. The current I
d. The voltages VR and VC using Ohm's law.
e. The voltages VR and VC using the voltage-divider rule.
f. The power to R
g. The power supplied by the voltage source e.
h. The phasor diagram. i. The Fp of the network.
j. The current and voltages in the time domain.
a. The reactance of the capacitor.
XC 
1
C

1
(1000rad / s)(0.1F)
 10k
b. The total impedance and the impedance diagram.
ZT  ZR  ZC  2k  j10k  10.198k  78.69
c. The current I
I( j) 
E( j)
ZT

1200
10.198k  78.69
 11.767 mA 78.69
d. The voltages VR and VC using Ohm's law.
e. The voltages VR and VC using the voltage-divider rule.
VC  I( j) ZC  (11.767mA 78.69)(10k  90)
 117.67V  11.31
VC  E ( j) 
ZC
ZT
f. The power to R
P  I R  (11.767mA ) (2k)  276.925mW
2
2
g. The power supplied by the voltage source e.
S  EI  (11.767mA )(120V)  1.412VA
i. The Fp of the network.
FP 
ZR
 0.196
leading
ZT
j.The current and voltages in the time domain.
Problem 30
Repeat Problem 29 for the network in Fig. 4.85 , after making the
appropriate changes in parts (a), (d) and (e).
Problem 31
Determine the voltage vL (in the time domain) for the network in
Fig. 4.86 using the voltage-divider rule.
Problem 32
For the series RLC network in Fig. 4.87, determine:
a. ZT b. I. c. VR, VL, VC using Ohm's law.
d. VL using the voltage-divider rule.
e. The power to R. f. The Fp.
g. The phasor and impedance diagrams.
Problem 33
Determine the voltage VC for the network in Fig. 4.88 using the
voltage-divider rule.
Problem 34 For the parallel RC network in Fig. 4.89,
determine:
a. The admittance diagram
b. YT, ZT
c. I.
d. IR, IC using Ohm's law.
e. The total power delivered to the network.
f. The power factor of the network.
g. The admittance diagram
Problem 35
Repeat Prob1em 34 for the network in Fig. 4.90, replacing IC
with IL in part (d).
Problem 36
Find the currents I1 and I2 in Fig. 4.91 using the current divider
rule. If necessary, review Section 2.10.
I1 ( j)  I 
I 2 ( j)  I 
ZL
ZL  ZR
ZR
ZL  ZR
 2A0
 2A0
0.8k90
0.8k90  1.2k0
1.2k0
0.8k90  1.2k0
 1.11A56.31
 1.66A  33.69
Problem 37 For the parallel RLC network in Fig. 4.92,
determine:
a. The admittance diagram.
b. YT, ZT
c. I, IR, IL, and IC.
d. The total delivered power.
e. The power factor of the network.
f. The sinusoidal format of I, IR, IL, and IC.
g. The phase relationship between e and iL.
Y T  Y R  Y L  Y C  1mS  0   1mS   90   0 . 5 mS  90 
 1mS  j0 . 5 mS  1 . 118 mS   26 . 565 
ZT 
1
YT
 0 . 894 k   26 . 565 
E
I
 ...  55 . 93 mA  3 . 435 
ZT
IR 
IC 
E
 50 mA  0 
ZR
E
 25 mA  120 
ZC
FP  cos 26 . 565   0 . 894
IL 
E
 50 mA   60 
ZL
2
P  I R  R  2 .5 W
lagging
Problem 38 For the network in Fig. 4.93, determine:
a. The short-circuit currents I1 and I2 ,
b. The voltages V1 and V2 .
c. The source current I.
a.I1  I 
b.V1  0
c.I 
E
R2
E
 5A
R2
V2  20
 5A
I2  0
Problem 39
Determine the current I and the voltage V for the network in Fig.
4.94
I
E
R1  R 2
V  VR 2 
 10A
E R2
R1  R 2
 100V