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796 ⏐⏐⏐ NETWORK THEOREMS (ac)
(0.5,0)�((0.995��5.71)*(3.61�33.69))/((0.995��5.71)�(3.61�33.69)) Enter
(1.311E0,35.373E�3)
Ans � Pol
(1.312E0�1.545E0)
and
Z 3 � R 3 � j X L � 3 k��j 2 k��3.61 k� �33.69°
Z T � Z 1 � Z 2 � Z 3
� 0.5 k��(0.995 k� ��5.71°) � (3.61 k� �33.69°)
� 1.312 k� �1.57°
Calculator Performing the above on the TI-86 calculator gives the
following result:
CALC. 18.1
E2 4 V �0°
Is � � ��� � 3.05 mA ��1.57°
ZT 1.312 k� �1.57°
Current divider rule:
Z
I3 �
2Is (0.995 k� ��5.71°)(3.05 mA ��1.57°)
� ������
Z2 � Z3 0.995 k� ��5.71° � 3.61 k� �33.69°
� 0.686 mA ��32.74°
with
V3 � (I3 �v)(R3 �0°)
� (0.686 mA ��32.74°)(3 k� �0°)
� 2.06 V ��32.74°
The total solution:
v3 � v3 (dc) � v3 (ac)
� 3.6 V � 2.06 V ��32.74°
v3 � 3.6 � 2.91 sin(qt � 32.74°)
The result is a sinusoidal voltage having a peak value of 2.91 V riding
on an average value of 3.6 V, as shown in Fig. 18.16.
6.51 V
3.6 V
0.69 V
0
v 3 32.74°
FIG. 18.16
The resultant voltage v 3 for the network of Fig. 18.12.
Dependent Sources
For dependent sources in which the controlling variable is not determined
by the network to which the superposition theorem is to be
applied, the application of the theorem is basically the same as for inde-
qt
Th