A fully charged $37 - \mu \mathrm { F }$ capacitor is discharged through a $1.0 - \mathrm { k } \Omega$ resistor. If the voltage across the capacitor is reduced to $7.6$ volts after just $20 \mathrm {~ms}$, what was the original potential across the capacitor?
A) $13 \mathrm {~V}$
B) $9.0 \mathrm {~V}$
C) $16 \mathrm {~V}$
D) $8.0 \mathrm {~V}$
E) $11 \mathrm {~V}$

Question

Quizplus · Accepted Answer

The voltage across a capacitor in a discharging $RC$ circuit is given by:$$V(t) = V_0e^{-t/\tau}$$where $V_0$ is the initial voltage, $\tau$ is the time constant, and $t$ is the time.The time constant is given by:$$\tau = RC$$where $R$ is the resistance and $C$ is the capacitance.In this case, $R = 1.0 \mathrm { k } \Omega$ and $C = 37 \mu \mathrm { F }$, so:$$\tau = (1.0 \mathrm { k } \Omega)(37 \mu \mathrm { F }) = 37 \mathrm { ms}$$After $20 \mathrm { ms}$, the voltage across the capacitor is:$$V(20 \mathrm { ms }) = V_0e^{-20 \mathrm { ms }/37 \mathrm { ms }} = 0.76V_0$$We are given that $V(20 \mathrm { ms }) = 7.6 \mathrm { V }$, so:$$0.76V_0 = 7.6 \mathrm { V }$$Solving for $V_0$, we get:$$V_0 = \frac{7.6 \mathrm { V }}{0.76} = 10 \mathrm { V }$$Therefore, the original potential across the capacitor was $10 \mathrm { V }$.

A Fully Charged 37−μF37 - \mu \mathrm { F }37−μF Capacitor Is Discharged Through A

A Fully Charged $37 - \mu \mathrm { F }$ Capacitor Is Discharged Through A