Fundamental Engineering Concepts - Mathematics and Statistics for the Environment - page 260

P = E × I (11.22)

where
P = Power (watts, W).
E = Voltage (volts, V).
I = Current (amps, A).

If we know the current ( I ) and the resistance ( R ) but not the voltage, we can find the power ( P ) by
using Ohm's law for voltage, so by substituting Equation 11.21:

E = I × R = IR

into Equation 11.22, we obtain:

P = IR × I = I 2 R (11.23)

In the same manner, if we know the voltage and the resistance but not the current, we can find the
P by using Ohm's law for current, so by substituting Equation 11.19:

E
R

I =

into Equation 11.22, we obtain:

2

PE E
R

E
R

=×= (11.2 4)

Note: If we know any two quantities, we can calculate the third.

■ EXAMPLE 11.22
Problem: The current through a 200-Ω resistor to be used in a circuit is 0.25 A. Find the power
rating of the resistor.

Solution: Because the current ( I ) and resistance ( R ) are known, use Equation 11.23 to find P :

P = I 2 × R = (0.25) 2 × 200 = 0.0625 × 200 = 12.5 W

Note: The power rating of any resistor used in a circuit should be twice the wattage calculated
by the power equation to prevent the resistor from burning out; thus, the resistor used in
Example 11.22 should have a power rating of 25 W.

■ EXAMPLE 11.23
Problem: How many kilowatts of power are delivered to a circuit by a 220-V generator that supplies
30 A to the circuit?

Solution: Because the voltage ( E ) and current ( I ) are given, use Equation 11.22 to find P :

P = E × I = 220 × 30 = 6600 W = 6.6 kW

■ EXAMPLE 11.24
Problem: If the voltage across a 30,000-Ω resistor is 450 V, what is the power dissipated in the resistor?