P21 Homework Set #7 Solutions Chapter 26
P26.1. Prepare: Please refer to Figure 26.3 for an AC resistor circuit.
Solve: (a) For a circuit with a single resistor, the peak current is
IR 
E0
10 V

 0.050 A
R 200 
(b) The peak current is the same as in part (a) because the current is independent of frequency.
Assess: Ohm’s law applies to the peak current.
P26.3. Prepare: We have an AC one-resistor circuit.
Solve:
From Equation 26.8,
R
Vr2ms (10.0 V) 2

 50.0 
Pavg
2.0 W
Using the equation again,
Vr ms  Pavg R  (10.0 W)(50.0 )  22.4 V
Assess: As long as we work with rms voltages and currents, all the expressions for DC power carry over to AC power.
P26.5. Prepare: We are given PR  1600 W and Vrms  120 V.
Solve: (a) Equation 26.11 relates Vr ms , PR , and R .
PR 
(Vrms )2
R
Solve for R .
R
(Vrms )2 (120 V) 2

 9.0 
PR
1600 W
(b) First find I r ms from Equation 26.11, and then the peak current I R .
I r ms 
PR 1600 W

 13 A
Vr ms
120 V
I R  2 I r ms  2 (13 A)  19 A
(c) In Question 26.21 we learned that the peak power is twice the rms power.
ppeak  2 PR  2(1600 W)  3200 W
Assess: See Figure 26.5b to verify the last part.
P26.7. Prepare: Equation 26.11 relates the rms current, the rms voltage (75 kV), and the average power
(40 MW): PR  I rmsVrms.
Solve: Solve for I rms.
I r ms 
PR
40 MW

 530 A
Vr ms
75 kV
Assess: This is a large current indeed, and it would require large transmission lines; we could make the current
smaller by sending the power down the transmission lines at an even higher voltage.
P26.9. Prepare: The secondary of the transformer delivers a power of 60 W. In the development of the
transformer theory, we assumed the transformer was 100 percent efficient so the power at the primary must also
be 60 W. Knowing that the power is related to the rms voltage and current by P  VrmsI rms and that the rms voltage
at the primary is 120 V, we can determine the rms current at the primary.
Solve:
I1rms  P1 /V1rms  P2 /V1rms  60 W/(120 V)  0.50 A
Assess: We expect that a small bulb (60 W) on the secondary of a transformer connected to 120 V/(60 Hz)
would require a small current at the primary.
P26.11. Prepare: We have been given the rms voltage and current for the secondary (output) of the transformer
(power pack). We have also been given the rms voltage at the primary and we want to determine the rms current at the
primary. The key to this problem is the knowledge that in the development of the transformer theory, we assumed the
transformer was 100 percent efficient so the power at the secondary is the same as the power at the primary.
Solve: The power at the primary is equal to the power at the secondary: P1  P2 .
The power is related to the rms voltage and current by P  VrmsI rms , which allows us to write I1rmsV1rms  I 2rmsV2rms.
Solving for I1rms , obtain
I1rms  I 2rmsV2rms /V1rms  (0.40 A)(5.2 V)/(120 V)  0.017 A
Assess: Knowing that the power at the primary and secondary is the same, for the values given we should
expect a small primary current.
P26.15. Prepare: We will first convert 1000 kWh a month into kW. The average power consumed by a
typical American family is
Pavg  1000
kWh
kWh
1000
 1000

kW  1.389 kW
month
30  24 h 720
Solve: (a) Because P  (V)I with V as the voltage of the power line to the house,
Iavg 
Pavg
V

1.389 kW 1389 W

 12 A
120 V
120 V
(b) Becasue P  (V )2 /R,
Ravg 
( V )2 (120 V)2

 10 
Pavg
1389 W
Assess: Households use a number of devices, so an average current of 12 A and an average resistance of 10 
are reasonable.
P26.27. Prepare: Figure 26.15 shows a simple AC capacitor circuit. For f  100 Hz, 2f  2 (100 Hz) 
628.3 rad/s and for f  100 kHz, 2f 2(100 kHz)  628,300 rad/s. We will use Equations 26.20 and 26.21.
Solve: (a) The capacitive reactance at 100 Hz is
XC 
1
1

 5305   5300 
2 fC (628.3 rad/s)(0.30 10 6 F)
 IC 
VC
10.0 V

 1.88 103 A  1.9 mA
X C 5.305 103 
(b) The capacitive reactance at 100 kHz is
XC 
1
1

 5.305   5.3 
2 fC (6.283 105 rad/s)(0.30 10 6 F)
 IC 
VC
10.0 V

 1.88 A  1.9 A
X C 5.305 
Assess: Using reactance is just like using resistance in Ohm’s law. Because X C  f 1 , XC decreases with an
increase in f, as we observed.
P26.37. Prepare: Figure 26.19 shows an AC one-inductor circuit. We will use Equation 26.25. We also note
that VL  2 Vrms.
Solve: For a simple one-inductor circuit,
IL 
VL
V
VL
2Vr ms
2(6.0 V)
 L L


 1.39  103 H  1.4 mH
X L 2 fL
2 f I L 2 f I L 2 (15 103 Hz)(65 103 A)
Assess: Using reactance is just like using resistance in Ohm’s law. This, however, does not apply to instantaneous
values of current and voltage.
P26.43. Prepare: The relationship between the inductance, capacitance, and the resonance frequency in an
RLC circuit is given Equation 26.28: f0  1/(2 LC ).
We need to make the resonance frequency equal to the given desired tuning frequency, f0  740 kHz. We are
also given L  350 mH.
Solve: Solve the equation for C.
C
1
1

 1 3  1013 F  0 13 pF
2
2
3
4 f 0 L 4 (740  10 Hz) 2 (350 10 3 H)
2
Assess: A variable capacitor can be made from sets of parallel plates whose region of
overlap can be adjusted by turning a knob..