Problem

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

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

#### Step-by-Step Solution

Solution 1

Refer to the Figure E7.1 in the text book.

For , the capacitor is fully charged and conducts no current since the capacitor acts like an open circuit to dc.

Draw the following circuit diagram for . 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 $$8 e^{-\frac{1}{0.6}} \mathrm{~V}$$.