A 1.0-?F capacitor is charged until it acquires a potential difference of $900.0 \mathrm {~V}$ across its plates, and then the emf source is removed. If the capacitor is then discharged through a $500.0 - \mathrm { k } \Omega$ resistance, what is the voltage drop across the capacitor $9.0 \mathrm {~ms}$ after beginning the discharge?
A)880 V
B)920 V
C)16 V
D)-16 V

Question

Quizplus · Accepted Answer

The voltage across a capacitor during discharge is given by:$$V(t) = V_0e^{-t/\tau}$$where:* $V(t)$ is the voltage at time $t$* $V_0$ is the initial voltage* $\tau$ is the time constantThe time constant is given by:$$\tau = RC$$where:* $R$ is the resistance* $C$ is the capacitanceIn this problem, we have:* $V_0 = 900.0 \mathrm {~V}$* $R = 500.0 \mathrm {~k\Omega} = 500.0 \times 10^3 \mathrm {~\Omega}$* $C = 1.0 \mathrm {~\mu F} = 1.0 \times 10^{-6} \mathrm {~F}$So, the time constant is:$$\tau = RC = (500.0 \times 10^3 \mathrm {~\Omega})(1.0 \times 10^{-6} \mathrm {~F}) = 0.500 \mathrm {~s}$$After $9.0 \mathrm {~ms}$, the voltage across the capacitor is:$$V(t) = V_0e^{-t/\tau} = (900.0 \mathrm {~V})e^{-(9.0 \times 10^{-3} \mathrm {~s})/0.500 \mathrm {~s}} = 882.5 \mathrm {~V}$$So, the voltage drop across the capacitor is:$$\Delta V = V_0 - V(t) = 900.0 \mathrm {~V} - 882.5 \mathrm {~V} = 17.5 \mathrm {~V}$$Therefore, the correct answer is A.

A 10-?F Capacitor Is Charged Until It Acquires a Potential Difference