FIGURE 47.

CIRCUIT FOR EXAMPLE PROBLEM.

Solution:

Ia + Ib + ...

In = 0

The currents are placed into the equation with the proper signs.

I1 + I2 + I3 + I4 = 0

10 amps + (-3 amps) + I3 + (-5 amps) = 0

I3 + 2 amps = 0

I3 = -2 amps

I3 has a value of 2 amperes, and the negative sign shows it to be a current

leaving the junction.

Now, use figure 48 on the following page and solve for the magnitude and

direction of I3.

Given:

I1 = 6 amps

I2 = 3 amps

I4 = 5 amps

Solution:

I1 + I2 + I3 + I4 = 0

6 amps + (-3 amps) + I3 + (-5 amps) = 0

I3 + (-2 amps) = 0

I3 = +2 amps