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. When the switch is first closed, the voltage across the capacitor cannot
change instantaneously, and VR � E � Vi � 8 V � 2 V � 6 V.
The current therefore jumps to a level determined by Ohm’s law:
VR 6 V
IRmax � � ��� 0.06 mA
R 100 k�
The current will then decay to zero amperes with the same time constant
calculated in part (a), and
c. See Fig. 24.21.
i C � 0.06 mAe �t/t
EXAMPLE 24.9 Sketch v C for the step input shown in Fig. 24.22.
Assume that the �4 mV has been present for a period of time in excess
of five time constants of the network. Then determine when v C � 0 V
if the step changes levels at t � 0 s.
vi 10 mV
0 t
–4 mV
Solution:
Vi ��4 mV Vf � 10 mV
t � RC � (1 k�)(0.01 mF) � 10 ms
By Eq. (24.6),
vC � Vf � (Vi � Vf)e �t/RC
� 10 mV � (�4 mV � 10 mV)e �t/10ms
and v C � 10 mV � 14 mV e �t/10ms
The waveform appears in Fig. 24.23.
10
v C (mV)
+
v i
–
FIG. 24.22
Example 24.9.
t = 3.37 ms
R
1 k�
–
+
4 mV C 0.01 mF vC +
–
0 10
–4
20
5t
30 40 50 60 70 80 t (ms)
FIG. 24.23
v C for the network of Fig. 24.22.
TRANSIENT R-C NETWORKS ⏐⏐⏐ 1103
8
2
0
0
v C (V)
0.1
iC (mA)
0.1
0.06
0.2 0.3 0.4 0.5 0.6 0.7 0.8
5t
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
FIG. 24.21
v C and i C for the network of Fig. 24.20.
t (s)
t (s)