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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 22C
RMS CURRENTS AND POTENTIAL DIFFERENCES
PROBLEM
In 1945, turbo-electric trains in the United States were capable of speeds
exceeding 160 km/h. Steam turbines powered the electric generators,
which in turn powered the driving wheels. Each generator produced
enough power to supply an rms potential difference of 6.0 × 103 V across
an 18 Ω resistor. What was the maximum potential difference across the
resistor? What was the maximum current in the resistor? What was the
rms current in the resistor? What was the generator’s power output?
SOLUTION
1. DEFINE
2. PLAN
Given:
∆Vrms = 6.0 × 103 V
Unknown:
∆Vmax = ?
R = 18 Ω
Imax = ?
Irms = ?
P =?
Choose the equation(s) or situation: Use the equation relating maximum and
rms potential differences to calculate the maximum potential difference. Use the
definition of resistance to calculate the maximum current, then use the equation
relating maximum and rms currents to calculate rms current. Power can be calculated from the product of rms current and rms potential difference.
∆Vmax = 2(∆Vrms)
∆V

Imax = max
R
Imax

Irms = 
2
P = ∆Vrms Irms
Substitute values into the equation(s) and solve:
∆Vmax = (1.41)(6.0 × 103 V)
= 8.5 × 103 V
(8.5 × 103 V)
Imax = 
(18 Ω)
= 4.7 × 102 A
Irms = (0.707)(4.7 × 102 A)
= 3.3 × 102 A
P = (6.0 × 103 V)(3.3 × 102 A)
= 2.0 × 106 W
4. EVALUATE
182
To determine whether severe rounding errors occurred through the various calculations, obtain the product of the maximum current and maximum potential
difference. The product of ∆Vmax and Imax should equal 2P, which for this problem equals 4.0 × 106 W.
Holt Physics Problem Workbook
Copyright © Holt, Rinehart and Winston. All rights reserved.
3. CALCULATE
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ADDITIONAL PRACTICE
1. In 1963, the longest single-span “rope way” for cable cars opened in California. The rope way stretched about 4 km from the Coachella Valley to
Mount San Jacinto. Suppose the rope way, which is actually a steel cable,
becomes icy. To de-ice the cable, you can connect its two ends to a
120 V (rms) generator. If the resistance of the cable is 6.0 × 10−2 Ω,
a. what will the rms current in the cable be?
b. what will the maximum current in the cable be?
c. what power will be dissipated by the cable, thus melting the ice?
2. In 1970, a powerful sound system was set up on the Ontario Motor
Speedway in California to make announcements to more than 200 000
people over the noise of 50 racing cars. The acoustic power of that system
was 30.8 kW. If the system was driven by a generator that provided an
rms potential difference of 120.0 V and only 10.0 percent of the supplied
power was transformed into acoustic power, what was the maximum
current in the sound system?
Copyright © Holt, Rinehart and Winston. All rights reserved.
3. Modern power plants typically have outputs of over 10 × 106 kW. But in
1905, the Ontario Power Station, built on the Niagara River, produced
only 1.325 × 105 kW. Consider a single generator producing this power
when it is connected to a single load (resistor). If the generated rms potential difference is 5.4 × 104 V, what is the maximum current and the
value of the resistor?
4. Stability of potential difference is a major concern for all high-emf
sources. In 1996, James Cross of the University of Waterloo in Canada,
constructed a compact power supply that produces a stable potential difference of 1.024 × 106 V. It can provide 2.9 × 10−2 A at this potential difference. If these values are the rms quantities for an alternating current
source, what are the maximum potential difference and current?
5. Certain species of catfish found in Africa have “power plants” similar to
those of electric eels. Though the electricity generated is not as powerful
as that of some eels, the electric catfish can discharge 0.80 A with a potential difference of 320 V. Consider an ac generator in a circuit with a
load. If its maximum values for potential difference and current are the
same as the potential difference and current for the catfish, what are the
rms values for the potential difference and current? What is the resistance of the load?
6. A wind generator installed on the island of Oahu, Hawaii, has a rotor
that is about 100 m in diameter. When the wind is strong enough, the
generator can produce a maximum current of 75 A in a 480 Ω load. Find
the rms potential difference across the load.
7. The world’s first commercial tidal power plant, built in France, has a power
output of only 6.2 × 104 kW, produced by 24 generators. Find the power
produced by each generator. If one of these generators is connected to a
120 kΩ resistor, find the rms and maximum currents in it.
Problem 22C
183
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Givens
Solutions
4. maximum emf =
8.00 × 103 V
maximum emf = NABw
maximum emf
B = 
NAw
N = 236
8.00 × 103 V
B = 
(236)(6.90 m)2(57.1 rad/s)
A = (6.90 m)2
w = 57.1 rad/s
B = 1.25 × 10−2 T
5. N = 1000 turns
−4
A = 8.0 × 10
−3
B = 2.4 × 10
maximum emf = NABw
2
m
T
maximum emf = 3.0 V
maximum emf
w = 
NAB
3.0 V
w = 
(1000)(8.0 × 10−4 m2)(2.4 × 10−3 T)
w = 1.6 × 103 rad/s
6. N = 640 turns
maximum emf = NABw
2
A = 0.127 m
maximum emf =
24.6 × 103 V
−2
B = 8.00 × 10
T
II
maximum emf
w = 
NAB
24.6 × 103 V
w = 
(640)(0.127 m2)(8.00 × 10−2 T)
w = 3.78 × 103 rad/s
7. f = 1.0 × 103 Hz
maximum emf = (250)(p)(12 × 10−2 m)2(0.22 T)(2p)(1.0 × 103 Hz)
B = 0.22 T
N = 250 turns
Copyright © Holt, Rinehart and Winston. All rights reserved.
r = 12 × 10
maximum emf = NABw = NAB(2pf ) = N(pr 3)Bw = N(pr 2)B(2pf )
−2
maximum emf = 1.6 × 104 V = 16 kV
m
Additional Practice 22C
1. ∆Vrms = 120 V
−2
R = 6.0 × 10
1
 = 0.707
2
Ω
∆Vrms
a. Irms = 
R
(120 V)
Irms = 
(6.0 × 10−2 Ω)
Irms = 2.0 × 103 A
b. Imax = (Irms) 2
(2.0 × 103 A)
Imax = 
(0.707)
Imax = 2.8 × 103 A
c. P = (Irms)(∆Vrms)
P = (2.0 × 103 A)(120 V)
P = 2.4 × 105 W
Section Two—Problem Workbook Solutions
II Ch. 22–3
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Givens
Solutions
2. P = 10.0(Acoustic power)
Acoustic power =
30.8 × 103 W
∆Vrms = 120.0 V
1
 = 0.707
2
P = ∆Vrms Irms
P
Irms = 
∆Vrms
Imax

Irms = 
2
Imax
P
 = 

2 ∆Vrms
P 2
Imax = 
Imax
∆Vrms
(10.0)(30.8 × 103 W)
= 
(120.0 V)(0.707)
Imax = 3.63 × 103 A
3. P = 1.325 × 108 W
∆Vrms = 5.4 × 104 V
II
1
 = 0.707
2
(∆V )2

P = ∆Vrms I rms = (Irms )2R = rms
R
Imax

Irms = 
2
2 P
Imax = 2 Irms = 
∆Vrms
1.325 × 108 W
Imax = 
(5.4 × 104 V)(0.707)
Imax = 3.5 × 103 A
(∆V )2

R = rms
P
(5.4 × 104 V)2
R = 
(1.325 × 108 W)
4. ∆Vrms = 1.024 × 106 V
Irms = 2.9 × 10−2 A
1
 = 0.707
2
∆Vmax = ∆Vrms
2
(1.024 × 106 V)
∆Vmax = 
(0.707)
∆Vmax = 1.45 × 106 V = 1.45 MV
Imax = Irms 2
(2.9 × 10−2 A)
Imax = 
(0.707)
Imax = 4.1 × 10−2 A
II Ch. 22–4
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
R = 22 Ω
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Givens
5. ∆Vmax = 320 V
Imax = 0.80 A
1
 = 0.707
2
Solutions
∆Vmax

∆Vrms = 
2
∆Vrms = (320 V)(0.707)
∆Vrms = 2.3 × 102 V
Imax

Irms = 
2
Irms = (0.80 A)(0.707)
Irms = 0.57 A
∆V ax ∆Vrms
 = 
R = m
Imax
Irms
(320 V)
(230 V)
R =  = 
(0.80 A) (0.57 A)
R = 4.0 × 102 Ω
6. Imax = 75 A
R = 480 Ω
1
 = 0.707
2
II
∆Vmax

∆Vrms = 
2
∆Vmax = (Imax )(R)
ImaxR

∆Vrms = 
2
∆Vrms = (75 A)(480 Ω)(0.707)
∆Vrms = 2.5 × 104 V = 25 kV
7. Ptot = 6.2 × 107 W
Ptot = 24 P
Copyright © Holt, Rinehart and Winston. All rights reserved.
R = 1.2 × 105 Ω
1
 = 0.707
2
P ot
P = (Irms )2R = t
24
6.2 × 107 W
P = 
24
P = 2.6 × 106 W = 2.6 MW
RP
(2.6 × 10 W)
= 
(1.2×1
0Ω)
Irms =
6
Irms
5
Irms = 4.7 A
Imax = 2 Irms
4.7 A
Imax = 
0.707
Imax = 6.6 A
Section Two—Problem Workbook Solutions
II Ch. 22–5