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a c
Using Eq. (15.32), we obtain
G 0.3 S
Fp � cos vT �����0.6 lagging
YT 0.5 S
Impedance approach: The input current I can also be determined
by first finding the total impedance in the following manner:
Z
ZT �
RZLZC ��� � 2 � �53.13°
ZRZL � ZLZC � ZRZC and, applying Ohm’s law, we obtain
E 100 V �53.13°
I � �� ��� �50 A �0°
Z 2 � �53.13°
T
15.9 CURRENT DIVIDER RULE
The basic format for the current divider rule in ac circuits is exactly
the same as that for dc circuits; that is, for two parallel branches with
impedances Z 1 and Z 2 as shown in Fig. 15.76,
Z
I1 � �
2IT Z
� or I2 � �
1IT �
Z1 �
Z1 �
I T = 5 A 30°
Z 2
R
1 �
X C
X L
8 �
2 �
FIG. 15.78
Example 15.16.
(15.33)
EXAMPLE 15.15 Using the current divider rule, find the current
through each impedance of Fig. 15.77.
Solution:
Z
IR �
LIT (4 � �90°)(20 A �0°) 80 A�90°
� ���� ���
ZR � ZL 3 � �0° � 4 � �90° 5 �53.13°
� 16 A �36.87°
Z (3 � �0°)(20 A �0°) 60 A�0°
IL �
RIT � ���� ���
ZR � ZL 5 � �53.13° 5 �53.13°
� 12 A ��53.13°
EXAMPLE 15.16 Using the current divider rule, find the current
through each parallel branch of Fig. 15.78.
Solution:
ZC IT (2 � ��90°)(5 A �30°) 10 A��60°
IR-L ��� ���� ���
ZC � ZR-L �j 2 ��1 ��j 8 � 1 � j 6
10 A��60°
��� � 1.644 A ��140.54°
6.083 �80.54°
Z 2
I T
CURRENT DIVIDER RULE ⏐⏐⏐ 667
I = 20 A
0°
I 1
I 2
Z 1
Z 2
I R
R 3 �
X L
I T
FIG. 15.76
Applying the current divider rule.
FIG. 15.77
Example 15.15.
I L
4 �