boylistad

406 ⏐⏐⏐ CAPACITORS
R 1
7 k�
E 120 V
R2 5 k� R3 + vC –
C
40 �F �
+ 40 V –
FIG. 10.56
Example 10.11.
18 k�
R 4
2 k�
b. At t � 9 ms,
vC � ETh(1 � e �t/t ) � 7(1 � e �(9�10�3 )/(6�10�3 )
)
� 7(1 � e �1.5 ) � 7(1 � 0.223)
vC � 7(0.777) � 5.44 V
and iC � � ETh
�t/t �3 �1.5
�e � (0.233 � 10 )e
R
� (0.233 � 10 �3 )(0.223)
iC � 0.052 � 10 �3 � 0.052 mA
Using Eq. (10.23) with Vf � 0 V and Vi � 5.44 V, we find that
vC � Vf � (Vi � Vf)e �t/t′
becomes vC � 0 V � (5.44 V � 0 V)e �t/t′
� 5.44e �t/t′
with t′ � R4C � (10 k�)(0.2 mF) � 2 ms
and vC � 5.44e �t/2ms
By Eq. (10.22),
Ii � �0.054 mA
and iC � Iie �t/t � �(0.54 � 10 �3 )e �t/2ms
5.44 V
�
10 k�
c. See Fig. 10.55.
E Th = 7
0
0.233
0.052
– 0.54
v C (V)
5
i C (mA)
10
10
5 �'
V i = 5.44 V
15
5�
20 25 30 35
5 �'
15
t (ms)
0 5
20 25 30 35 t (ms)
5�
FIG. 10.55
The resulting waveforms for the network of Fig. 10.52.
EXAMPLE 10.11 The capacitor of Fig. 10.56 is initially charged to
40 V. Find the mathematical expression for v C after the closing of the
switch.