A simple RC circuit consists of a 1-$\mu$F capacitor in series with a 2800-$\varOmega$resistor, a 6-V battery, and an open switch. Initially, the capacitor is uncharged. How long after the switch is closed will the voltage across the capacitor be 4.1 V? A)3 × 10^-⁴ s B)3 s C)3 × 10^-⁸ s D)3 × 10^-⁶ s E)3 × 10^-³ s

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Quizplus · Accepted Answer

The time constant $\tau$ for an RC circuit is given by $\tau = RC$. Here, $R = 2800 \, \Omega$ and $C = 1 \, \mu F = 1 \times 10^{-6} \, F$. Thus, $\tau = 2800 \times 10^{-6} = 2.8 \times 10^{-3} \, s$. The voltage across the capacitor as a function of time is $V(t) = V_0(1 - e^{-t/\tau})$. Solving $4.1 = 6(1 - e^{-t/2.8 \times 10^{-3}})$ gives $t \approx 3 \times 10^{-3} \, s$.

A Simple RC Circuit Consists of a 1- μ\muμ F Capacitor in Series with a 2800-

A Simple RC Circuit Consists of a 1- $\mu$ F Capacitor in Series with a 2800-