Capacitor Voltage In the circuit shown in Figure 5.7, the switch is closed at t = 0, and the initial voltage across the capacitor [latex]v_C (0^-) = 0[/latex]. a. Find an expression for the capacitor voltage, [latex]v_C (t)[/latex] b. Find the voltage across the capacitor at t = 5 s c. Find the capacitor voltage at t = ∞

a. The source voltage is: [latex]V_{S}=5 V[/latex] The time constant is: [latex]\tau=RC=15 \times 10^3 \times 100 \times 10^{-6}=1.5 s[/latex] Substituting these values in Equation (5.12): [latex]v_{C}(t)=V_S(1-e^{-{\frac{t}{\tau}}})[/latex] (5.12) [latex]v_{C}(t)=5\left(1-e^{\frac{-t}{1.5}}\right)[/latex] b. At t = 5 s [latex]v_{C}(t)=5\left(1-e^{\frac{-5}{1.5}}\right)=4.8216 V[/latex] c. At t = ∞ [latex]v_{C}(t)=5\left(1-e^{\frac{-\infty}{1.5}}\right)=5(1-0)=5 V[/latex] This value is called steady-state voltage, a term that will be explained and discussed later in this chapter.

Capacitor Voltage In the circuit shown in Figure 5.7, the switch is closed at t = 0, and the initial voltage across the capacitor vC (0^-) = 0. a. Find an expression for the capacitor voltage, vC (t) b. Find the voltage across the capacitor at t = 5 s c. Find the capacitor voltage at t = ∞

Q. 5.1

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 = ∞

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