Find vC(t) for t > 0 in the circuit shown in Fig. E7.1.Figure E7.1

Determine the initial voltage across the capacitor using voltage division.

$$ \begin{aligned} v_{C}\left(0^{-}\right) &=12\left(\frac{4 \mathrm{k} \Omega+2 \mathrm{k} \Omega}{3 \mathrm{k} \Omega+4 \mathrm{k} \Omega+2 \mathrm{k} \Omega}\right) \\ &=12\left(\frac{6 \mathrm{k} \Omega}{9 \mathrm{k} \Omega}\right) \\ &=12\left(\frac{2}{3}\right) \\ &=8 \mathrm{~V} \end{aligned} $$

Therefore, the initial voltage, \(v_{C}(0-)\) across the capacitor is \(8 \mathrm{~V}\).

Since the voltage across the capacitor cannot change instantaneously, the voltage across the capacitor \(t>0\) is \(8 \mathrm{~V}\).

Therefore, \(v_{C}(0+)=8 \mathrm{~V}\).

At the switch is opens. For , the circuit is shown in Figure 2.

Apply Kirchhoff's current law to the circuit shown in Figure \(2 .\)

\((100 \mu) \frac{d v_{C}(t)}{d t}+\frac{v_{C}(t)}{6 \mathrm{k}}=0\)

\(\frac{d v_{C}(t)}{d t}+\frac{v_{C}(t)}{(6 \mathrm{k})(100 \mu)}=0\)

\(\frac{d v_{c}(t)}{d t}+\frac{v_{c}(t)}{0.6}=0 \ldots \ldots .(1)\)

The equation (1) is in the form of \(\frac{d x_{c}(t)}{d t}+a x_{c}(t)=0\).

Where,

\(x_{C}(t)=v_{C}(t)\)

\(a=\frac{1}{0.6}\)

The form of the solution to this homogeneous equation is,

\(v_{C}(t)=K_{2} e^{-a t} \mathrm{~V}\)

\(v_{C}(t)=K_{2} e^{-\frac{1}{0.6} \ldots \ldots(2)}\)

Substitute \(t=0\) in equation (2). \(v_{C}(0)=K_{2} e^{-\frac{0}{0.6}}\)

\(8=K_{2}(1)\)

\(K_{2}=8\)

Substitute \(K_{2}=8\) in equation (2).

\(v_{C}(t)=8 e^{\frac{1}{0.6}} \mathrm{~V}\)

Therefore, the voltage across the capacitor, \(v_{C}(t)\) for \(t>0\) is \(8 e^{-\frac{1}{0.6}} \mathrm{~V}\).