Fundamental Engineering Concepts - Mathematics and Statistics for the Environment - page 277

I 1 I 2 I 3

I T

120 volts R 1 15 ohms R 2 15 ohms R 3 12 ohms

FIGURE 11.44 Illustration for Example 11.36.

Now find total current using Equation 11.34:

I T = I 1 + I 2 + I 3

I T = 8 + 8 + 10 = 26 amps

11.7.7.4 Parallel Circuits and Kirchhoff's Current Law
The division of current in a parallel network follows a definite pattern. This pattern is described by
Kirchhoff's current law , which is stated as follows:

The algebraic sum of the currents entering and leaving any junction of conductors is equal to zero.

This can be stated mathematically as

I a + I b + … + I n = 0 (11.35)

where I a , I b , …, I n are the currents entering and leaving the junction. Currents entering the junction
are assumed to be positive, and currents leaving the junction are considered negative. When solving
a problem using Equation 11.35, the currents must be placed in the equation with the proper polarity.

■ EXAMPLE 11.37
Problem: Solve for the value of I 3 in Figure 11.45.
Solution: First, give the currents the appropriate signs:

I 1 = +10 amps
I 2 = -3 amps
I 3 = ? amps
I 4 = -5 amps

Then, place these currents into Equation 11.35:

I a + I b + … + I n = 0

with the proper signs as follows:

I 1 + I 2 + I 3 + I 4 = 0

(+10) + (-3) + ( I 3 ) + (-5) = 0

I 2 = 3 A

I 1 = 10 A I 3 = ?

I 4 = 5 A

FIGURE 11.45 Illustration for Example 11.37.