A 8.0- $\mu \mathrm { F }$ uncharged capacitor is connected in series with a 6.0- $\mathrm { k } \Omega$ resistor, an ideal $20 - \mathrm { V } \mathrm { dc }$ source, and an open switch. If the switch is closed at time $t = 0.0 \mathrm {~s}$, what is the charge on the capacitor at $t = 9.0 \mathrm {~ms}$ ?
A) $0 \mathrm { C }$
B) $37 \%$ of the minimum charge
C) $96 \%$ of the maximum charge
D) $17 \%$ of the maximum charge
E) $68 \%$ of the minimum charge

Question

Quizplus · Accepted Answer

The charge on the capacitor as a function of time is given by $ Q(t) = Q_{\text{max}}(1 - e^{-t/RC}) $. Here, $ Q_{\text{max}} = C \times V = 8.0 \times 10^{-6} \times 20 = 1.6 \times 10^{-4} \, \text{C} $. The time constant $ \tau = RC = 6.0 \times 10^3 \times 8.0 \times 10^{-6} = 48 \times 10^{-3} \, \text{s} $. At $ t = 9.0 \, \text{ms} $, $ Q(9 \, \text{ms}) = 1.6 \times 10^{-4} (1 - e^{-9/48}) \approx 0.17 \times Q_{\text{max}} $.

A 80- μF\mu \mathrm { F }μF Uncharged Capacitor Is Connected in Series with a 6

A 80- $\mu \mathrm { F }$ Uncharged Capacitor Is Connected in Series with a 6