A $2.0 - \mu \mathrm { F }$ capacitor is charged to $12 \mathrm {~V}$ and then discharged through a $4.0 - \mathrm { M } \Omega$ resistor. How long will it take for the voltage across the capacitor to drop to $3.0 \mathrm {~V}$ ?
A) $11 \mathrm {~s}$
B) $8.0 \mathrm {~s}$
C) $22 \mathrm {~s}$
D) $24 \mathrm {~s}$

Question

Quizplus · Accepted Answer

The time constant $\tau$ is given by $\tau = RC = (4.0 \times 10^6 \, \Omega)(2.0 \times 10^{-6} \, \mathrm{F}) = 8 \, \mathrm{s}$. The voltage across the capacitor as a function of time is $V(t) = V_0 e^{-t/\tau}$. Solving $3 = 12 e^{-t/8}$ gives $t = 8 \ln(4) \approx 11 \, \mathrm{s}$.

A 2.0−μF2.0 - \mu \mathrm { F }2.0−μF Capacitor Is Charged To

A $2.0 - \mu \mathrm { F }$ Capacitor Is Charged To