boylistad

Solution: The network is redrawn in Fig. 10.57.
E Th:
R Th:
+
40 V
–
C
Thévenin
40 mF
R 2
5 k�
R 3
R 1
7 k�
R4 18 k�
FIG. 10.57
Network of Fig. 10.56 redrawn.
2 k�
E 120 V
R 18 k�(120 V)
ETh ��� 3E
����
R3 � R1 � R4 18 k��7 k��2 k�
� 80 V
RTh � 5 k��18 k� � (7 k��2 k�)
� 5 k��6 k�
� 11 k�
Therefore, Vi � 40 V and Vf � 80 V
and t � RThC � (11 k�)(40 mF) � 0.44 s
Eq. (10.23): vC � Vf � (Vi � Vf)e �t/t
� 80 V � (40 V � 80 V)e �t/0.44s
and v C � 80 V � 40 Ve �t/0.44s
EXAMPLE 10.12 Forthe network of Fig. 10.58, find the mathematical
expression for the voltage v C after the closing of the switch (at t � 0).
R 2
10 �
I 20 mA R1 = 6 �
FIG. 10.58
Example 10.12.
+
C 500 mF vC –
Solution:
RTh � R1 � R2 � 6 ��10 ��16 �
ETh � V1 � V2 � IR1 � 0
� (20 � 10 �3 A)(6 �) � 120 � 10 �3 V � 0.12 V
and t � RThC � (16 �)(500 � 10 �6 F) � 8 ms
so that vC � 0.12(1 � e �t/8ms )
THÉVENIN EQUIVALENT: t � R ThC ⏐⏐⏐ 407