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In units, you can see that multiplying the current A = C/s by the
potential (V = J/C) yields
joules
joules
_ _ _
coulombs
=

×

second coulombs second
A joule/second is a unit of power called the watt (W). Th erefore,
electrical power is measured in units of watts, and
W = A·V
power (in watts) = current (in amps) × potential (in volts)
watts = amps × volts

This relationship is
P = IV A
equation 6.6
Household electrical devices are designed to operate on a
particular voltage, usually 120 or 240 volts (Figure 6.17). Th ey
therefore draw a certain current to produce the designed power.
Information about these requirements is usually found some-
where on the device. A lightbulb, for example, is usually stamped
with the designed power, such as 100 W. Other electrical devices
may be stamped with amp and volt requirements. You can deter-
mine the power produced in these devices by using equation 6.6,
that is, amps × volts = watts. Another handy conversion factor
to remember is that 746 watts is equivalent to 1.00 horsepower.

EXAMPLE 6.5
A 1,100 W hair dryer is designed to operate on 120 V. How much cur-
rent does the dryer require?

B
SOLUTION
The power (P) produced is given in watts with a potential diff erence

of 120 V across the dryer. The relationship between the units of amps,
volts, and watts is found in equation 6.6, P = IV.
P _
P = 1,100 W P = IV ∴ I =
V
V = 120 V
joules
_
I = ? A 1,100

second

= __
joules
_

120
coulomb
1,100 J
_ _ _
C

= ×

120 s J
J·C
_
= 9.2

s·J
C _

= 9.2
s
= 9.2 A C
FIGURE 6.17 What do you suppose it would cost to run each
EXAMPLE 6.6 of these appliances for one hour? (A) This lightbulb is designed to
operate on a potential difference of 120 volts and will do work at
An electric fan is designed to draw 0.5 A in a 120 V circuit. What is the
the rate of 100 W. (B) The finishing sander does work at the rate
power rating of the fan? (Answer: 60 W)
of 1.6 amp × 120 volts, or 192 W. (C) The garden shredder does
work at the rate of 8 amps × 120 volts, or 960 W.
152 CHAPTER 6 Electricity 6-14