Introductory Physics Volume Two

3.9 More Examples 71
15Ω
5Ω
10Ω
15V
+
-
10V
+
-
15Ω
Label the currents, including best guesses for the directions:
I 1
I 2
I 3
15Ω
5Ω
10Ω
+
-
15V
+
-
10V
I 1
A
15Ω
B
Apply Kirchoff’s loop rules. First to the lefthand loop, beginning at
point A and moving clockwise:
15V − (15Ω)I 1 − (5Ω)I 2 + 10V − (15Ω)I 1 = 0,
then to the righthand loop, beginning at point B and moving counterclockwise:
(10Ω)I 3 − (5Ω)I 2 + 10V = 0.
Note that for a resistor, the conventional positive current flows from
high to low potential, which determines the sign to use when moving
from one side of a resistor to the other.
Now apply Kirchoff’s junction rule at the point labeled B:
I 2 + I 3 − I 1 = 0
Simplifying, we have three equations with three unknowns:
25 − 30I 1 − 5I 2 = 0
10 + 10I 3 − 5I 2 = 0
I 2 + I 3 − I 1 = 0
To solve, let first eliminate I 3 :
10 + 10I 3 − 5I 2 = 10 + 10(I 1 − I 2 ) − 5I 2 = 0