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## Q. 5.3

Capacitor Voltage

The switch in the circuit shown in Figure 5.11 has been closed for a long time period and opens at t = 0.

a. Find an expression for $v_C (t)$

b. Find the capacitor voltage after 60 ms

c. Find $v_C (t)$ when t tends to infinity

## Verified Solution

Because the switch was closed for a long time, the initial condition of the capacitor voltage will be zero, that is, $v_{C}(0)=0$. Note that for simplicity the minus or plus sign for time zero has been removed because the capacitor voltage is the same for both cases.

a. This problem can be solved using two methods:

Method 1: Applying KCL (note that there is only one non-reference node):

$-10 \times 10^{-3}+\frac{v_{C}(t)}{5 \times 10^3}+10 \times 10^{-6} \frac{dv_{C}(t)}{dt}=0$

$\frac{dv_{C}(t)}{dt}=\frac{-1}{5 \times 10^3 \times 10 \times 10^{-6}}\left(v_{C}(t)-5 \times 10^3 \times 10 \times 10^{-3}\right)$

or:

$\frac{dv_{C}(t)}{v_{C}(t)-50}=\frac{-dt}{0.05}$

Integrating both sides of this equation yields:

$\ln \frac{v_{C}(t)-50}{0-50}=\frac{-t}{0.05} \rightarrow 1-\frac{v_{C}(t)}{50}=e^{-20 t} \rightarrow v_{C}(t)=50\left(1-e^{-20 t}\right)$

Method 2: Another approach is to find the Thévenin equivalent circuit for the given circuit. Figure 5.12(a) shows the desired circuit.

The circuit time constant corresponds to:

$\tau=R C=5 \times 10^3 \times 10 \times 10^{-6}=0.05 s$

Substituting $V_{S}$ and τ into Equation (5.12) yields:

$v_{C}(t)=V_S(1-e^{-{\frac{t}{\tau}}})$     (5.12)

$v_{C}(t)=50\left(1-e^{\frac{-t}{0.05}}\right)=50\left(1-e^{-20 t}\right)$

b. When t = 60 ms:

$v_{C}(0.06)=50\left(1-e^{-20 \times 0.06}\right)=50(1-0.3012)=34.9403$

c. When t → ∞ , the capacitor is fully charged. In this case, the capacitor will be equivalent to an open circuit as shown in Figure 5.12(b).

$v_{C}(\infty)=10 mA \times 5 k\Omega=50 V$