Transient R-C Network

In Chapter 9 the general solution for the transient behavior of an R-C network with or without initial values was developed. The resulting equation for the voltage across a capacitor is repeated below for convenience. $$\bbox[10px,border:1px solid grey]{V_{C}=V_{f}+\left(V_{i}-V_{f}\right) e^{-t / R C}} \tag{1}$$ Recall that $ V_{i} $ is the initial voltage across the capacitor when the transient phase is initiated as shown in Fig. 1.
Fig. 1: Defining the parameters of Eq. (1).
Fig. 2: Example of the use of Eq. (1).
The voltage $ V_{f} $ is the steady-state (resting) value of the voltage across the capacitor when thetransient phase has ended. The transient period is approximated as $ 5 \tau $,where $ \tau $ is the time constant of the network and is equal to the product $ R C $.
For the case of Fig. 2, $ V_{i}=-2 \mathrm{~V}$, $V_{f}=+5 \mathrm{~V} $, and $$\begin{aligned}V_{C} &=V_{i}+\left(V_{f}-V_{i}\right)\left(1-e^{-f / R C}\right) \\ &=-2 \mathrm{~V}+[5 \mathrm{~V}-(-2 \mathrm{~V})]\left(1-e^{-t / R C}\right) \\ V_{C} &=-2 \mathrm{~V}+7 \mathrm{~V}\left(1-e^{-t R C}\right)\end{aligned}$$ For the case where $ t=\tau=R C $, $$\begin{aligned}v_{C} &=-2 \mathrm{~V}+7 \mathrm{~V}\left(1-e^{-t / 7}\right)=-2 \mathrm{~V}+7 \mathrm{~V}\left(1-e^{-1}\right) \\ &=-2 \mathrm{~V}+7 \mathrm{~V}(1-0.368)=-2 \mathrm{~V}+7 \mathrm{~V}(0.632) \\ v_{C} &=2.424 \mathrm{~V}\end{aligned}$$ as verified by Fig. 2.
Example 1: The capacitor of Fig. 3 is initially charged to $ 2 \mathrm{~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} $ and $ i_{C} $.
Fig. 3: Example 1.
Solution:
a. $$ V_{i}=2 \mathrm{~V} $$ $$ V_{f}( \text{ after } 5 \tau )=E=8 \mathrm{~V} $$ $$ \tau=R C=(100 \mathrm{k} \Omega)(1 \mu \mathrm{F})=100 \mathrm{~ms} $$ By Eq. (1), $$\begin{aligned}V_{C} &=V_{f}+\left(V_{i}-V_{j}\right) e^{-u / R C} \\ &=8 \mathrm{~V}+(2 \mathrm{~V}-8 \mathrm{~V}) e^{-u t} \end{aligned}$$ and $$v_{C}=8 \mathrm{~V}-6 \mathrm{~V} e^{-t / t}$$ b. When the switch is first closed, the voltage across the capacitor cannot change instantaneously, and $V_{R}=E-V_{i}=8 \mathrm{~V}-2 \mathrm{~V}=6 \mathrm{~V} $. The current therefore jumps to a level determined by Ohm's law: $$I_{R_{\max }}=\frac{V_{R}}{R}=\frac{6 \mathrm{~V}}{100 \mathrm{k} \Omega}=0.06 \mathrm{~mA}$$ The current will then decay to zero amperes with the same time constant calculated in part (a), and $$i_{C}=0.06 \mathrm{~mA} e^{-t / \tau}$$ c. See Fig. 4.
Fig. 4: $v_C$ and $i_C$ for the network of Fig. 3