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The total impedance is defined by
ZT � Z1 � Z2 with
Z1 � R �0° � 1 � �0°
(XC ��90°)(XL �90°) (2 � ��90°)(3 � �90°)
Z2 � ZC � ZL ���� � ���
�j XC � j XL �j 2 ��j 3 �
6 � �0° 6 � �0°
�����6 � ��90°
j 1 1 �90°
and
ZT � Z1 � Z2 � 1 ��j 6 ��6.08 � ��80.54°
E 120 V �0°
b. Is ������ � 19.74 A �80.54°
ZT 6.08 � ��80.54°
c. Referring to Fig. 16.2, we find that V R and V C can be found by a
direct application of Ohm’s law:
VR � IsZ1 � (19.74 A �80.54°)(1 � �0°) � 19.74 V �80.54°
VC � IsZ2 � (19.74 A �80.54°)(6 � ��90°)
� 118.44 V ��9.46°
d. Now that V C is known, the current I C can also be found using Ohm’s
law.
IC � � �59.22 A �80.54°
e. Pdel � I 2 sR � (19.74 A) 2 VC 118.44 V ��9.46°
� ���
ZC 2 � ��90°
(1 �) � 389.67 W
f. Fp � cos v � cos 80.54° � 0.164 leading
The fact that the total impedance has a negative phase angle (revealing
that Is leads E) is a clear indication that the network is capacitive in
nature and therefore has a leading power factor. The fact that the network
is capacitive can be determined from the original network by first
realizing that, for the parallel L-C elements, the smaller impedance predominates
and results in an R-C network.
EXAMPLE 16.2 For the network of Fig. 16.3:
a. If I is 50 A �30°, calculate I 1 using the current divider rule.
b. Repeat part (a) for I 2.
c. Verify Kirchhoff’s current law at one node.
Solutions:
a. Redrawing the circuit as in Fig. 16.4, we have
Z1 � R � j XL � 3 ��j 4 ��5 � �53.13°
Z2 ��j XC ��j 8 ��8 � ��90°
Using the current divider rule yields
Z2I (8 � ��90°)(50 A �30°) 400 ��60°
I1 ������ ���
Z2 � Z1 (�j 8 �) � (3 ��j 4 �) 3 � j 4
400 ��60°
��� �80 A ��6.87°
5 ��53.13°
I
ILLUSTRATIVE EXAMPLES ⏐⏐⏐ 711
I
I1 R 3 �
X L
4 �
FIG. 16.3
Example 16.2.
I 1
Z 1
I 2
X C
I 2
Z 2
8 �
FIG. 16.4
Network of Fig. 16.3 after assigning the block
impedances.