Fundamental Engineering Concepts - Mathematics and Statistics for the Environment - page 258

+

R 1
6 ohms

-

FIGURE 11.21 Determining current in a simple circuit.

To observe the effect of source voltage on circuit current, we use the circuit shown in Figure
11.21 but double the voltage to 6 volts. Notice that, as the source of voltage doubles, the circuit cur-
rent also doubles.

■ EXAMPLE 11.19
Problem: Given that E = 6 volts and R = 6 ohms, what is I ?

Solution:

E
R

6
6

I === 1amp

Key Point: 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 resistor in Figure
11.21 has a value of 12 ohms.

■ EXAMPLE 11.20
Problem: Given that E = 3 volts and R = 12 ohms, what is I ?

Solution:

E
R

3
12

I === 0.25 amps

Comparing the current of 0.25 amp for the 12-ohm resistor to the 0.5-amp current 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 current ( I ) and resistance ( R ) are known, for example, the basic formula
for Ohm's law must be transposed to solve for E . The Ohm's law equations can be memorized and
practiced effectively by using an Ohm's law circle (see Figure 11.22). 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 11.23.

■ EXAMPLE 11.21
Problem: An electric light bulb draws 0.5 A when operating on a 120-V DC circuit. What is the
resistance of the bulb?