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10.11 THÉVENIN EQUIVALENT: t � R ThC
Occasions will arise in which the network does not have the simple
series form of Fig. 10.24. It will then be necessary first to find the
Thévenin equivalent circuit for the network external to the capacitive
element. E Th will then be the source voltage E of Eqs. (10.15) through
(10.20), and R Th will be the resistance R. The time constant is then t �
R ThC.
EXAMPLE 10.10 For the network of Fig. 10.52:
+
E 21 V
–
60 k�
R1 30 k�
R 2
10 k�
R 1
3
C = 0.2 mF
iC 2
+
vC R4 10 k�
FIG. 10.52
Example 10.10.
a. Find the mathematical expression for the transient behavior of the
voltage vC and the current iC following the closing of the switch
(position 1 at t � 0 s).
b. Find the mathematical expression for the voltage vC and current iC as a function of time if the switch is thrown into position 2 at t �
9 ms.
c. Draw the resultant waveforms of parts (a) and (b) on the same time
axis.
Solutions:
a. Applying Thévenin’s theorem to the 0.2-mF capacitor, we obtain
Fig. 10.53:
RTh � R1 � R2 � R3 � �10 k�
� 20 k��10 k�
RTh � 30 k�
ETh � � � (21 V) � 7 V
The resultant Thévenin equivalent circuit with the capacitor
replaced is shown in Fig. 10.54. Using Eq. (10.23) with Vf � ETh and Vi � 0 V, we find that
vC � Vf � (Vi � Vf)e �t/t
(60 k�)(30 k�)
��
90 k�
R2E (30 k�)(21 V) 1
� �� �
R2 � R1 30 k��60 k� 3
becomes v C � E Th � (0 V � E Th)e �t/t
or vC � ETh(1 � e �t/t )
with t � RC � (30 k�)(0.2 mF) � 6 ms
so that vC � 7(1 � e �t/6ms )
For the current: iC � � ETh
�t/RC
�e
R
� e �t/6ms 7 V
�
30 k�
i C � (0.233 � 10 �3 )e �t/6ms
–
THÉVENIN EQUIVALENT: t � R ThC ⏐⏐⏐ 405
R Th :
E Th :
+
E –
60 k�
R 1
60 k�
R1 21 V
R 2
30 k�
10 k�
10 k�
FIG. 10.53
Applying Thévenin’s theorem to the network of
Fig. 10.52.
R Th = 30 k�
E Th = 7 V C = 0.2 mF
i C
+
v C
FIG. 10.54
Substituting the Thévenin equivalent for the
network of Fig. 10.52.
R 3
R 3
R Th
R 2 30 k� E Th
–