Question 24.8: The capacitor in Fig. 24.20 is initially charged to 2 V befo...
The capacitor in Fig. 24.20 is initially charged to 2 V before the switch is closed. The switch is then closed.
a. Determine the mathematical expression for v_{C}.
b. Determine the mathematical expression for i_{C}.
c. Sketch the waveforms of v_{C} \text { and } i_{C} .
\text { a. } V_{i}=2 V.
V_{f}(\text { after } 5 \tau)=E=8 V.
\tau=R C=(100 k \Omega)(1 \mu F )=100 ms.
By Eq. (24.6),
v_{C}=V_{f}+\left(V_{i}-V_{f}\right) e^{-t / R C} (24.6)
v_{C}=V_{f}+\left(V_{i}-V_{f}\right) e^{-t / R C}.
=8 V +(2 V -8 V ) e^{-t / \tau}.
and v c=8 V -6 V e^{-t / \tau}.
b. When the switch is first closed, the voltage across the capacitor cannot change instantaneously, and V_{R}=E-V_{i}=8 V -2 V =6 V.
The current therefore jumps to a level determined by Ohm’s law:
I_{R_{\max }}=\frac{V_{R}}{R}=\frac{6 V }{100 k \Omega}=0.06 mA.
The current then decays to zero amperes with the same time constant calculated in part (a), and
i_{C}=0.06 mA e^{-t / \tau}.
c. See Fig. 24.21.
