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and VL � ILZR � (3.023 mA �0.011°)(1 k� �0°)
� 3.023 V �0.011°
c. The results clearly indicate that the capacitor had little effect on the
frequencies of interest. In addition, note that most of the supply current
reached the load for the typical parameters employed.
EXAMPLE 16.6 For the network of Fig. 16.12:
I 1
6 mA
∠ 20°
a. Determine the current I.
b. Find the voltage V.
I
R2 10 k�
I
4 mA
R1 2 k� 2 ∠ 0°
R3 6.8 k�
X C
20 k�
FIG. 16.12
Example 16.6.
Solutions:
a. The rules for parallel current sources are the same for dc and ac networks.
That is, the equivalent current source is their sum or difference
(as phasors). Therefore,
IT � 6 mA �20° � 4 mA �0°
� 5.638 mA � j 2.052 mA � 4 mA
� 1.638 mA � j 2.052 mA
� 2.626 mA �51.402°
Redrawing the network using block impedances will result in the
network of Fig. 16.13 where
Z1 � 2 k� �0° � 6.8 k� �0° � 1.545 k� �0°
and Z2 � 10 k� �j 20 k� �22.361 k� ��63.435°
Note that I and V are still defined in Fig. 16.13.
Current divider rule:
I � �
4.057 A �51.402°
� ����
23.093 � 10
� 0.176 mA �111.406°
3 4.057 A �51.402°
���
11.545 � 10 ��60.004°
3 � j 20 � 10 3
Z1IT (1.545 k� �0°)(2.626 mA �51.402°)
� ����
Z1 � Z2 1.545 k��10 k��j 20 k�
b. V � IZ 2
� (0.176 mA �111.406°)(22.36 k� ��63.435°)
� 3.936 V �47.971°
+
V
–
I T
ILLUSTRATIVE EXAMPLES ⏐⏐⏐ 715
2.626 mA ∠ 51.402°
Z 1
Z 2
FIG. 16.13
Network of Fig. 16.12 following the
assignment of the subscripted impedances.
I
+
V
–