Electrical Calculations - Basic Mathematics for Water and Wastewater Operators - page 139

into Equation 10.4, we obtain:

2

PE E
R

E
R

=×= (10.6)

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

■ Example 10.8
Problem: The current through a 200-W 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, we can use Equation
10.5 to find P :

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

■ Example 10.9
Problem: How many kilowatts of power are delivered to a circuit by a 220-V genera-
tor that supplies 30 A to the circuit?

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

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

■ Example 10.10
Problem: If the voltage across a 30,000-W resistor is 450 V, what is the power dis-
sipated in the resistor?

Solution: Because the resistance ( R ) and voltage ( E ) are known, we can use Equation
10.6 to find P :

2 2

E
R

450
30 000

( ) 202 500

30 000

,

P == = = 6.75 watts

, ,

In this section, P was expressed in terms of various pairs of the other three basic
quantities E , I , and R . In practice, we should be able to express any one of the three
basic quantities, as well as P , in terms of any two of the others. Figure 10.9 is a sum-
mary of the 12 basic formulas we should know. The four quantities E , I , R , and P are
at the center of the figure. Adjacent to each quantity are three segments. Note that in
each segment the basic quantity is expressed in terms of two other basic quantities,
and no two segments are alike.