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806 ⏐⏐⏐ NETWORK THEOREMS (ac)
R 1
R 2
X C I g
+
Eg –
Z Th
FIG. 18.44
Determining the Thévenin impedance for the
network of Fig. 18.40 using the approach
Z Th � E g / I g.
hI R 1 R 2
R 1
X C
FIG. 18.46
Example 18.11.
R 2
X C
Thévenin
Z Th = R 1 � R 2 – jX C
FIG. 18.47
Determining the Thévenin impedance for the
network of Fig. 18.46.
and
�mR2V
�� R1
� R2
E 1
oc ––––
ZTh � � � �mR
� ––––
2V
Isc �� 1
R
� �
1 � R2
(R1 � R2) � j XC � R 1 � R 2 � j X C
Method 3: See Fig. 18.44.
(R 1 � R 2) � j X C
I g �
E g
��
(R1 � R 2) � j X C
Eg and ZTh � �R1�R2� jXC � Ig
Th
In each case, the Thévenin impedance is the same. The resulting
Thévenin equivalent circuit is shown in Fig. 18.45.
Z Th = R 1 � R 2 – jX C
+
–
�R � 2V ETh = Thévenin
R1 + R2 +
FIG. 18.45
The Thévenin equivalent circuit for the network of Fig. 18.40.
EXAMPLE 18.11 Repeat Example 18.10 for the network of Fig.
18.46.
Solution: From Fig. 18.46, ETh is
hR1R2I ETh � Eoc ��hI(R1 � R2) � �
Method 1: See Fig. 18.47.
ZTh � R1 � R2 � j XC Note the similarity between this solution and that obtained for the previous
example.
Method 2: See Fig. 18.48.
I sc �
�hI(R 1 � R 2)
Eoc and ZTh � � � —— � R1 �
�(
R
R2 � j XC Isc
1 � R2)
hI
� �
(R � R ) � j X
1
�(R 1 � R 2)hI
��
(R 1 � R 2) � j X C
2
C
–
� R1 � R 2