Electrical Calculations - Basic Mathematics for Water and Wastewater Operators - page 137

E

220

I

10

FIGURE 10.6 Ohm's law illustration for Example 10.5.

I R 2.5 25

FIGURE 10.7 Ohm's law illustration for Example 10.6.

■ Example 10.6
Problem: Find E when I = 2.5 A and R = 25 W.

Solution: Place finger on E as shown in Figure 10.7. Use Equation 10.3 to find the
unknown E :

E = I × R = 2.5 × 25 = 62.5 V

Note: In the previous examples, we have demonstrated how the Ohm's law circle
can help solve simple voltage, current, and amperage problems. Beginning stu-
dents are cautioned, however, not to rely entirely on the use of this circle when
transposing simple formulas but rather to use it to supplement their knowledge of
the algebraic method. Algebra is a basic tool in the solution of electrical problems.
The importance of knowing how to use it should not be underemphasized, and its
use should not be bypassed after the operator has learned a shortcut method such
as the one indicated in this circle.

■ Example 10.7
Problem: An electric light bulb draws 0.5 A when operating on a 120-V DC circuit.
What is the resistance of the bulb?

Solution: The first step in solving a circuit problem is to sketch a schematic diagram
of the circuit itself, labeling each of the parts and showing the known values (see
Figure 10.8). Because I and E are known, we can use Equation 10.2 to solve for R :

E
I

120
05

R == = 240 ohms

.