Alternating Current Practice
1.
An ac current has an rms value of 3.54 A. Determine the peak value of the current.
(a) 1.25 A
(c) 3.75 A
(e) 7.08 A
(b) 2.50 A
(d) 5.00 A
 2. A 220- resistor is connected across an ac voltage source V = (150 V) sin [2(60 Hz)t].
What is the average power delivered to this circuit?
(a) 51 W
(c) 280 W
(e) 550 W
(b) 110 W
(d) 320 W
 3.
A lamp uses an average power of 55 W when it is connected to an rms voltage of 120 V.
Which entry in the following table is correct for this circuit?
lamp resistance R (
Irms (A)
(a)
(b)
(c)
(d)
260
22
130
170
0.46
3.8
0.65
0.57
(e)
38
1.2
 4. When a 1500-W hair dryer is in use, the current passing through the dryer may be
represented as
I = (17.7 A) sin (120t). What is the rms current for this circuit?
(a) 17.7 A
(c) 85.7 A
(e) 8.85 A
(b) 12.5 A
(d) 25.0 A
Questions 5 through 9 pertain to the situation described below:
+ 10
The figure shows variation of
the current through the heating
element with time in an iron
when it is plugged into a
standard 120 V, 60 Hz outlet.
I (A)
t
– 10
 5. What is the peak voltage?
(a) 10 V
(b) 60 V
t1
(c) 120 V
(d) 170 V
 6. What is the rms value of the current in this circuit?
(a) 1.4 A
(c) 11 A
(b) 7.1 A
(d) 14 A
 7.
 8.
What is the resistance of the iron?
(a) 24 
(c) 17 
(b) 7.1 
(d) 12 
t2
(e) 240 V
(e) 18 A
(e) 1.8 
If t1 = 0.050 s, what is the value of t2? Note: The origin for the graph is not necessarily at t = 0 s.
(a) 0.067 s
(c) 0.10 s
(e) 61 s
(b) 0.079 s
(d) 0.60 s
 9. What is the approximate average power dissipated in the iron?
(a) 450 W
(c) 850 W
(b) 600 W
(d) 1200 W
(e) 1700 W
Answers to AC Multiple Choice Questions
1. D
2. A
3. A
4. B
5. D
6. B
7. C
8. A
9. C
Chap 20 Problems - Solutions
30. REASONING Substituting V  12 V0 into Equation 20.7 gives a result that can be
solved directly for the desired time.
SOLUTION From Equation 20.7 we have
V  12 V0  V0 sin 2 f t
or
1
2
 sin 2 f t
Using the inverse trigonometric sine function, we find
2 f t  sin 1
 12   0.524
In this result, the value of 0.524 is in radians and corresponds to an angle of 30.0º.
Thus we find that the smallest value of t is
t
0.524
0.524

 1.39 103 s
2 f
2  60.0 Hz 
31. REASONING AND SOLUTION The expression relating the peak current, I0, to the
rms-current, Irms, is
I
2.50 A
I rms  0 
 1.77 A
2
2
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32. REASONING
a. The average power P delivered to the copy machine is equal to the square of the
2
rms-current Irms times the resistance R, or P  I rms
R (Equation 20.15b). Both Irms
and R are known.
b. According to the discussion in Section 20.5, the peak power Ppeak is twice the
average power, or Ppeak  2 P .
SOLUTION
a. The average power is
2
2
P  I rms
R   6.50 A  18.6   786 W
(20.15b)
b. The peak power is twice the average power, so
Ppeak  2P  2  786 W   1572 W
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33.
SSM REASONING The average power is given by Equation 20.15c as
P
/ R . In this expression the rms voltage Vrms appears. However, we seek the
peak voltage V0. The relation between the two types of voltage is given by Equation
2
Vrms
20.13 as Vrms  V0 / 2 , so we can obtain the peak voltage by using Equation 20.13 to
substitute into Equation 20.15c.
SOLUTION Substituting Vrms from Equation 20.13 into Equation 20.15c gives
P
2
Vrms
R
V / 2 

0
R
2

V02
2R
Solving for the peak voltage V0 gives
V0  2RP  2  4.0  55 W   21 V
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36.
REASONING
a. We can obtain the frequency of the alternating current by comparing this specific
expression for the current with the more general one in Equation 20.8.
b. The resistance of the light bulb is, according to Equation 20.14, equal to the rmsvoltage divided by the rms-current. The rms-voltage is given, and we can obtain the rmscurrent by dividing the peak current by 2 , as expressed by Equation 20.12.
c. The average power is given by Equation 20.15a as the product of the rms-current and
the rms-voltage.
SOLUTION
a. By comparing I   0.707 A  sin  314 Hz  t  with the general expression (see
Equation 20.8) for the current in an ac circuit, I  I 0 sin 2  f t , we see that
2 f t   314 Hz  t
or
f =
314 Hz
 50.0 Hz
2
b. The resistance is equal to Vrms/Irms, where the rms-current is related to the peak
current I0 by I rms  I 0 / 2 . Thus, the resistance of the light bulb is
R
Vrms
I rms

Vrms
2 120.0 V 

 2.40  102 
I0
0.707 A
2
(20.14)
c. The average power is the product of the rms-current and rms-voltage:
 I 
 0.707 A 
P  I rmsVrms   0 Vrms  
(20.15a)
 120.0 V   60.0 W
2 

 2
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