Electrical Calculations - Basic Mathematics for Water and Wastewater Operators - page 135

+

R 1
6 ohms

-

FIGURE 10.2

Determining current in a simple circuit.

To observe the effect of source voltage on circuit current, in the next example we
use the circuit shown in Figure 10.2 but double the voltage to 6 volts.

■ Example 10.2
Problem: Given that E = 6 volts and R = 6 ohms, what is I ?

Solution:

E
R

6
6

I === 1ampere

Notice that as the source of voltage doubles, the circuit current also doubles.

Note: Circuit current is directly proportional to applied voltage and will change by
the same factor that the voltage changes.

To verify that current is inversely proportional to resistance, assume that the resis-
tor in Figure 10.2 has a value of 12 ohms.

■ Example 10.3
Problem: Given that E = 3 volts and R = 12 ohms, what is I ?

Solution:

E
R

3
12

I === 025 . mpere

Comparing the current of 0.25 amp for the 12-ohm resistor to the 0.5-amp cur-
rent obtained with the 6-ohm resistor shows that doubling the resistance will reduce
the current to one-half the original value. The point here is that circuit current is
inversely proportional to the circuit resistance .
Recall that, if we know any two quantities ( E , I , or R ), we can calculate the third.
In many circuit applications, current is known and either the voltage or the resistance
will be the unknown quantity. To solve a problem in which voltage ( E ) and resistance
( R ) are known, the basic formula for Ohm's law must be transposed to solve for I . The
Ohm's law equations can be memorized and practiced effectively by using an Ohm's
law circle (see Figure 10.3). To find the equation for E , I , or R when two quantities are
known, cover the unknown third quantity with your finger, ruler, or piece of paper
as shown in Figure 10.4.