1
Chapter 23 Homework Solutions
1.
SSM REASONING The voltage across the capacitor reaches its maximum instantaneous value when
the generator voltage reaches its maximum instantaneous value. The maximum value of the capacitor
voltage first occurs one-fourth of the way, or one-quarter of a period, through a complete cycle (see the
voltage curve in Figure 23.4).
SOLUTION The period of the generator is T  1/ f  1/ (5.00 Hz) = 0.200 s . Therefore, the least amount
of time that passes before the instantaneous voltage across the capacitor reaches its maximum value is
1
4
4.
T  41 ( 0.200 s) = 5.00  10 –2 s .
REASONING As the frequency f of the generator increases, the capacitive reactance XC of the capacitor
1
decreases, according to X C 
(Equation 23.2), where C is the capacitance of the capacitor. The
2 f C
decreasing capacitive reactance leads to an increasing rms current Irms, as we see from
V
I rms  rms (Equation 23.1), where Vrms is the constant rms voltage across the capacitor. We know that
XC
Vrms is constant, because it is equal to the constant rms generator voltage. The fuse is connected in series
with the capacitor, so both have the same current Irms. We will use Equations 23.1 and 23.2 to determine
the frequency f at which the rms current is 15.0 A.
SOLUTION Solving Equation 23.2 for f, we obtain
f 
1
2 C X C
(1)
In terms of the rms current and voltage, Equation 23.1 gives the capacitive reactance as X C 
Vrms
I rms
.
Substituting this relation into Equation (1) yields
f 
1

2 C X C
I rms
1
15.0 A


 9470 Hz
 Vrms  2 CVrms 2 63.0 106 F  4.00 V 
2 C 

 I rms 


18. REASONING For a series RCL circuit the total impedance Z and the phase angle  are given by
Z  R2   X L  X C 
2
(23.7)
tan  
XL  XC
R
(23.8)
where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. In the present
case, there is no capacitance, so that X C  0  . Therefore, these equations simplify to the following:
Z  R 2  X L2
(1)
tan  
XL
R
(2)
2
We are given neither R nor XL. However, we do know the current and voltage when only the resistor is
V
connected and can determine R from these values using R  rms (Equation 20.14). In addition, we know
I rms
the current and voltage when only the inductor is connected and can determine XL from these values using
V
X L  rms (Equation 23.3). Since the generator frequency is fixed, this value for XL also applies for the
I rms
series combination of the resistor and the inductor.
SOLUTION With only the resistor connected, Equation 20.14 indicates that the resistance is
V
112 V
R  rms 
 224 
I rms 0.500 A
With only the inductor connected, Equation 23.3 indicates that the inductive reactance is
XL 
Vrms
I rms

112 V
 2.80 102 
0.400 A
a. Using these values for R and XL in Equations (1) and (2), we find that the impedance is
Z  R 2  X L2 
 224 2   2.80  102  
2
 359 
b. The phase angle between the current and the voltage of the generator is
tan  
XL
R
or
2
 XL 
1  2.80  10  

tan

  51.3

224

 R 


  tan 1 
19. SSM REASONING The voltage supplied by the generator can be found from Equation 23.6,
Vrms  I rms Z . The value of I rms is given in the problem statement, so we must obtain the impedance of the
circuit.
SOLUTION The impedance of the circuit is, according to Equation 23.7,
Z
R 2  ( X L – X C ) 2  ( 275  ) 2  (648  – 415  ) 2  3.60  10 2 
The rms voltage of the generator is
Vrms  I rms Z  ( 0.233 A )( 3.60  10 2  ) = 83.9 V
3
30. REASONING The inductance L and the capacitance C of a series RCL circuit determine the resonant
frequency f0 according to
1
f0 
(23.10)
2 LC
As we see from Equation 23.10, the smaller the inductance L, the larger the resonant frequency, and the
larger the inductance, the smaller the resonant frequency. Therefore, in part (a) we will use the largest
frequency to determine the minimum inductance, and in part (b) we will use the smallest frequency to find
the maximum inductance. We note that 1 MHz = 1×106 Hz.
SOLUTION
a. Squaring both sides of Equation 23.10 and solving for L, we obtain
f02 
1
4 LC
2
L
or
1
4 f02C
(1)
2
The minimum inductance is obtained when the resonant frequency is greatest, so Equation (1) gives
L
1

4 2 9.0 106 Hz
 1.8 1011 F
2
 1.7 105 H
b. Using Equation (1) once more, this time with the smallest resonant frequency, yields the maximum
inductance:
L
1

4 2 4.0 106 Hz
 1.8 1011 F
2
 8.8 105 H
33. SSM REASONING The current in an RCL circuit is given by Equation 23.6, I rms  Vrms / Z , where the
impedance Z of the circuit is given by Equation 23.7 as Z  R 2  ( X L  X C ) 2 . The current is a
maximum when the impedance is a minimum for a given generator voltage. The minimum impedance
occurs when the frequency is f 0 , corresponding to the condition that X L  X C , or 2 f 0 L  1 / ( 2 f 0 C ) .
Solving for the frequency f 0 , called the resonant frequency, we find that
f0 
1
2
LC
Note that the resonant frequency depends on the inductance and the capacitance, but does not depend on
the resistance.
4
SOLUTION
a. The frequency at which the current is a maximum is
f0 
1
2
LC

1
2  (17 .0  10
–3
H)(12.0  10 –6 F)
 352 Hz
b. The maximum value of the current occurs when f  f 0 . This occurs when X L  X C , so that Z  R .
Therefore, according to Equation 23.6, we have
I rms 
Vrms
Z

V rms
R

155 V
 15.5 A
10.0 
37. REASONING Since the resonant frequency f0 is known, we may use Equation 23.10, f 0 
1
to
2  LC
find the inductance L, provided the capacitance C can be determined. The capacitance can be found by
using the definitions of capacitive and inductive reactances.
SOLUTION
a. Solving Equation 23.10 for the inductance, we have
L 
1
4  f 02 C
(1)
2
where f0 is the resonant frequency. From Equations 23. 2 and 23.4, the capacitive and inductive reactances
are
1
XC 
and
X L  2 f L
2 f C
where f is any frequency. Solving the first of these equations for f, substituting the result into the second
L
equation, and solving for C yields C 
. Substituting this result into Equation (1) above and
XL XC
solving for L gives
L 
1
2 f 0
XL XC 
1
2  1500 Hz
b
g b 30.0  gb5.0  g 
1.3  10 3 H
b. The capacitance is
1.3  10 3 H
L
C 

 8.7  10 6 F
XL XC
30.0  5.0 
b
gb
g