Page 298 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 24] ELECTRIC CURRENT 283

The efficiency of the generator is therefore

output 7200 W
Eff = = = 0.80 = 80%
input 8952 W

SOLVED PROBLEM 24.18
A 12-V storage battery is charged by a current of 20 A for 1 h. (a) How much power is required to charge
the battery at this rate? (b) How much energy has been provided during the process?
(a) P = IV = (20 A)(12 V) = 240 W
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(b) W = Pt = (240 W)(3600 s) = 8.64 × 10 J

SOLVED PROBLEM 24.19
The kilowatthour (kWh) is an energy unit equal to the energy delivered by a source whose power is 1 kW
in1hof operation. How much energy in kilowatthours does a 240-V clothes dryer that draws 15 A use
in 45 min of operation?
The power of the dryer is

P = IV = (15 A)(240 V) = 3600 W = 3.6kW

and the time interval is

45 min
t = = 0.75 h
60 min/h
Hence

W = Pt = (3.6kW)(0.75 h) = 2.7 kWh

SOLVED PROBLEM 24.20
The 12-V battery of a certain car has a capacity of 80 Ah, which means that it can furnish a current of
80 A for 1 h, a current of 40 A for 2 h, and so forth. (a) How much energy is stored in the battery? (b)If
the car’s lights require 60 W of Power, how long can the battery keep them lighted when the engine (and
hence its generator) is not running?
(a) The 80-Ah capacity of the battery is a way to express the amount of charge it can transfer from one of its
terminals to the other. Here the amount of charge is

5
5
q = (80 Ah)(3600 s/h) = 2.88 × 10 A·s = 2.88 × 10 C
and so the energy the battery can provide is

5
6
W = qV = (2.88 × 10 C)(12 V) = 3.46 × 10 J
(b) Since P = W/t,

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W 3.46 × 10 J 4
t = = = 5.8 × 10 s = 16 h
P 60 W