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Th
6 �
R 1
R 2
12 �
b
R3 a
R4 3 � 4 �
RTh (a)
c
c′
6 �
FIG. 9.46
Solving for R Th for the network of Fig. 9.45.
R 3
R 1
b a
RTh 3 � 4 �
(b)
R 4
R 2
c,c′
12 �
Step 4: The circuit is redrawn in Fig. 9.47. The absence of a direct connection
between a and b results in a network with three parallel
branches. The voltages V 1 and V 2 can therefore be determined using the
voltage divider rule:
R1E (6 �)(72 V) 432 V
V1 ����� ���48 V
R1 � R3 6 ��3 � 9
V2 �
R 2E (12 �)(72 V) 864 V
��� ���54 V
12 ��4 � 16
� R2 � R 4
+
E 72 V
–
E
+
V1 R1 –
6 �
+
KVL
ETh R2 –
b a
R 3
3 �
R 4
12 �
Assuming the polarity shown for ETh and applying Kirchhoff’s voltage
law to the top loop in the clockwise direction will result in
� V ��ETh � V1 � V2 � 0
and ETh � V2 � V1 � 54 V � 48 V � 6 V
Step 5 is shown in Fig. 9.48.
FIG. 9.47
Determining E Th for the network of Fig. 9.45.
Thévenin’s theorem is not restricted to a single passive element, as
shown in the preceding examples, but can be applied across sources,
whole branches, portions of networks, or any circuit configuration, as
shown in the following example. It is also possible that one of the methods
previously described, such as mesh analysis or superposition, may
have to be used to find the Thévenin equivalent circuit.
EXAMPLE 9.10 (Two sources) Find the Thévenin circuit for the network
within the shaded area of Fig. 9.49.
4 �
+
V 2
–
THÉVENIN’S THEOREM ⏐⏐⏐ 335
–
ETh = 6 V
+
R 1
E 1
0.8 k�
– 6 V
R Th = 5 �
E 2
R 2
+ 10 V
4 k�
R 4
1.4 k�
R L
FIG. 9.48
Substituting the Thévenin equivalent circuit
for the network external to the resistor R L of
Fig. 9.44.
R 3 6 k� R L
FIG. 9.49
Example 9.10.
a
b