Page 297 - Schaum's Outline of Theory and Problems of Applied Physics

282 ELECTRIC CURRENT [CHAP. 24

Table 24-1

Known Quantities
Unknown
V, I I, R V, R P, I P, V P, R
Quantity
√
V = IR P/I PR
√
I = V/R P/V P/R
2
R = V/I P/I 2 V /P
2
2
P = VI I R V /R
ELECTRIC POWER
The rate at which work is done to maintain an electric current is given by the product of the current I and the
potential difference V :
P = IV
Power = (current)(potential difference)

When I is in amperes and V is in volts, P will be in watts.
If the conductor or device through which a current passes obeys Ohm’s law, the power consumed may be
expressed in the alternative forms
V 2
2
P = IV = I R =
R
Table 24.1 is a summary of the various formulas for potential difference V, current I, resistance R, and
power P that follow from Ohm’s law I = V/R and from the power formula P = VI.

SOLVED PROBLEM 24.15
The current through a 50- resistance is 2 A. How much power is dissipated as heat?

2
2
P = I R = (2A) (50 ) = 200 W
SOLVED PROBLEM 24.16
A 2-kW water heater is to be connected to a 240-V power line whose circuit breaker is rated at 10 A.
Will the breaker open when the heater is switched on?

The heater draws a current of
P 2000 W 1
I = = = 8 A
V 240 V 3
Since this current is less than 10 A, the breaker will not open.

SOLVED PROBLEM 24.17

A generator driven by a diesel engine that develops 12 hp delivers 30 A at 240 V. What is the efficiency
of the generator?
Since 1 hp = 746 W and P = IV ,

Input power = (12 hp)(746 W/hp) = 8952 W
Output power = (30 A)(240 V) = 7200 W