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186 ⏐⏐⏐ PARALLEL CIRCUITS
I = 27 mA
I 1 = 21 mA
R 1
R 2
7 �
FIG. 6.38
Example 6.20.
1
and RT ���4.363 �
0.2292 S
R T
4.363 �
with I1 � � I � � (42 mA) � 30.54 mA
R1 6 �
The second option is to apply Eq. (6.10) once after combining R2 and R3 as follows:
(24 �)(48 �)
24 � � 48 ���� �16 �
24 ��48 �
16 �(42 mA)
and I1 ��� �30.54 mA
16 ��6 �
Both options generated the same answer, leaving you with a choice
for future calculations involving more than two parallel resistors.
EXAMPLE 6.19 Determine the magnitude of the currents I 1, I 2, and I 3
for the network of Fig. 6.37.
I = 12 A
I 1
I 2
Solution: By Eq. (6.10), the current divider rule,
R
I1 � �
2I (4 �)(12 A)
� ��� �8 A
R1 � R2 2 ��4 �
Applying Kirchhoff’s current law,
I � I1 � I2 R 1
2 �
R 2
4 �
FIG. 6.37
Example 6.19.
and I2 � I � I1 � 12 A � 8 A � 4 A
or, using the current divider rule again,
R
I2 � �
1I (2 �)(12 A)
� ��� �4 A
R1 � R2 2 ��4 �
The total current entering the parallel branches must equal that leaving.
Therefore,
I3 � I � 12 A
or I3 � I1 � I2 � 8 A � 4 A � 12 A
EXAMPLE 6.20 Determine the resistance R1 to effect the division of
current in Fig. 6.38.
Solution: Applying the current divider rule,
R
I1 � �
2I
�
R1 �
R 2
I 3
P