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EXAMPLE 6.13 Determine the currents I3 and I4 of Fig. 6.27 using
Kirchhoff’s current law.
Solution: We must first work with junction a since the only unknown
is I3. At junction b there are two unknowns, and both cannot be
determined from one application of the law.
At a:
At b:
I = 5 A
I 2 = 3 A
I 1 = 2 A
a
I 1
I 2 = 4 A
R 1
R 2
a
I 3
b
FIG. 6.27
Example 6.13.
Σ I entering � Σ I leaving
I 1 � I 2 � I 3
2 A � 3 A � I 3
I 3 � 5 A
Σ I entering � Σ I leaving
I 3 � I 5 � I 4
5 A � 1 A � I 4
I 4 � 6 A
b
c
R 3
R 4
I 3
I 4
d
FIG. 6.28
Example 6.14.
I 4
I 5 = 1 A
EXAMPLE 6.14 Determine I 1, I 3, I 4, and I 5 for the network of Fig.
6.28.
Solution: At a:
Σ Ientering � Σ Ileaving I � I1 � I2 5 A � I1 � 4 A
Subtracting 4 A from both sides gives
5 A � 4 A � I 1 � 4 A � 4 A
I 1 � 5 A � 4 A � 1 A
I 5
R 5
KIRCHHOFF’S CURRENT LAW ⏐⏐⏐ 181