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For the particular case of two parallel resistors, as shown in Fig.
6.34,
R T �
R1R2 �
and I1 � I � — I
R RT 1 � R2 —
R
R2I and I1 �
R
(6.10)
1 � R2 Similarly for I 2,
I 2
� R1
R1I �
R1 � R2 R 1R 2
� R1 � R 2
Note difference in subscripts.
(6.11)
In words, for two parallel branches, the current through either branch is
equal to the product of the other parallel resistor and the input current
divided by the sum (not the total parallel resistance) of the two parallel
resistances.
EXAMPLE 6.17 Determine the current I2 for the network of Fig. 6.35
using the current divider rule.
Solution:
4
I2 � � � ��(6 A) � �
12
1
R1Is (4 k�)(6 A)
�� ��
�(6 A)
R1 � R2 4 k��8 k�
3
� 2 A
EXAMPLE 6.18 Find the current I 1 for the network of Fig. 6.36.
R T
R 1
I 1
1
I = 42 mA
6 � R2 24 � R3 48 �
FIG. 6.36
Example 6.18
Solution: There are two options for solving this problem. The first is
to use Eq. (6.9) as follows:
1 1 1 1
� � � �����0.1667 S � 0.0417 S � 0.0208 S
RT 6 � 24 � 48 �
� 0.2292 S
CURRENT DIVIDER RULE ⏐⏐⏐ 185
I s = 6 A
R 1
I s = 6 A
R 1
FIG. 6.34
Developing an equation for current division
between two parallel resistors.
I 1
I 2
4 k� R 8 k�
2
FIG. 6.35
Example 6.17.
I
R 2
I 2