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a. Find the mathematical expression for the voltage across the capacitor
once the switch is closed.
b. Find the mathematical expression for the current during the transient
period.
c. Sketch the waveform for each from initial value to final value.
Solutions:
a. Substituting the open-circuit equivalent for the capacitor will result
in a final or steady-state voltage vC of 24 V.
The time constant is determined by
t � (R1 � R2)C � (2.2 k��1.2 k�)(3.3 mF)
� 11.22 ms
with 5t � 56.1 ms
Applying Eq. (10.23):
vC � Vf � (Vi � Vf)e �t/t
� 24 V � (4 V � 24 V)e �t/11.22ms
and vC � 24 V � 20 Ve �t/11.22ms
b. Since the voltage across the capacitor is constant at 4 V prior to the
closing of the switch, the current (whose level is sensitive only to
changes in voltage across the capacitor) must have an initial value of
0 mA. At the instant the switch is closed, the voltage across the
capacitor cannot change instantaneously, so the voltage across the
resistive elements at this instant is the applied voltage less the initial
voltage across the capacitor. The resulting peak current is
Im � � � �5.88 mA
The current will then decay (with the same time constant as the
voltage vC) to zero because the capacitor is approaching its opencircuit
equivalence.
The equation for iC is therefore:
iC � 5.88 mAe �t/11.22ms
E � VC 24 V � 4 V 20 V
� �� �
R1 � R2 2.2 k��1.2 k� 3.4 k�
c. See Fig. 10.50.
24 V
4 V
0
0
v C
i C
5.88 mA
5�
56.1 ms
56.1 ms
FIG. 10.50
v C and i C for the network of Fig. 10.49.
t
t
INITIAL VALUES ⏐⏐⏐ 401