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## 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 = ∞ ## 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

$v_{C}(t)=5\left(1-e^{\frac{-5}{1.5}}\right)=4.8216 V$

c. At t = ∞

$v_{C}(t)=5\left(1-e^{\frac{-\infty}{1.5}}\right)=5(1-0)=5 V$

This value is called steady-state voltage, a term that will be explained and discussed later in this chapter.