## Chapter 5

## Q. 5.1

**Capacitor Voltage**

In the circuit shown in Figure 5.7, the switch is closed at t = 0, and the initial voltage across the capacitor v_C (0^-) = 0.

a. Find an expression for the capacitor voltage, v_C (t)

b. Find the voltage across the capacitor at t = 5 s

c. Find the capacitor voltage at t = ∞

## Step-by-Step

## Verified Solution

a. The source voltage is: V_{S}=5 V

The time constant is: \tau=RC=15 \times 10^3 \times 100 \times 10^{-6}=1.5 s Substituting these values in Equation (5.12):

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

v_{C}(t)=5\left(1-e^{\frac{-t}{1.5}}\right)b. At t = 5 s

v_{C}(t)=5\left(1-e^{\frac{-5}{1.5}}\right)=4.8216 Vc. At t = ∞

v_{C}(t)=5\left(1-e^{\frac{-\infty}{1.5}}\right)=5(1-0)=5 VThis value is called **steady-state** voltage, a term that will be explained and discussed later in this chapter.