BASIC ELECTRONICS - OD1633 - LESSON 1/TASK 2
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
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