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820 ⏐⏐⏐ NETWORK THEOREMS (ac)
1
Z Th
2
Z� 1
Z� 1 Z�1
FIG. 18.85
Substituting the Y equivalent for the upper D configuration of Fig. 18.84.
+
E Th
–
Z� 1
I 1 = 0
Z� 1
Z 2
Z� 1
FIG. 18.87
Finding the Thévenin voltage for the network
of Fig. 18.83.
+
–
E
Z 2
1
Z Th
The redrawn circuit (Fig. 18.86) shows
Z′ 1(Z′ 1 � Z 2)
ZTh � Z′ 1 ���
Z′1 � (Z′ 1 � Z 2)
3 � �90°( j 3 �� 8 �)
� j 3 �����
j 6 �� 8 �
(3 �90°)(8.54 �20.56°)
� j 3 ����
10 �36.87°
25.62 �110.56°
� j 3 ��� �j 3 � 2.56 �73.69°
10 �36.87°
� j 3 � 0.72 � j 2.46
ZTh � 0.72 ��j 5.46 �
and ZL � 0.72 ��j 5.46 �
Z 1
FIG. 18.84
Defining the subscripted impedances for the network of Fig. 18.83.
3
Th
For E Th, use the modified circuit of Fig. 18.87 with the voltage
source replaced in its original position. Since I 1 � 0, E Th is the voltage
across the series impedance of Z′ 1 and Z 2. Using the voltage divider
rule gives us
Z Th
(Z′ 1 � Z2)E ( j 3 ��8 �)(10 V �0°)
ETh ��� ����
Z′1 � Z2 � Z′ 1 8 ��j 6 �
(8.54 �20.56°)(10 V �0°)
����
10 �36.87°
ETh � 8.54 V ��16.31°
2
Z 1
Z 1
Z 2
Z� 1
3
Z� 1
Z 2
Z� 1
FIG. 18.86
Determining Z Th for the network of Fig. 18.83.