A $1.0 - \mu \mathrm { F }$ 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 $500.0 - k \Omega$ resistance, what is the voltage drop across the capacitor $9.0 \mathrm {~ms}$ after beginning the discharge?
A) $- 16 \mathrm {~V}$
B) $16 \mathrm {~V}$
C) $880 \mathrm {~V}$
D) $920 \mathrm {~V}$

Question

Quizplus · Accepted Answer

The voltage across the capacitor during discharge can be calculated using the formula $V = V_0 e^{-\frac{t}{RC}}$, where $V_0$ is the initial voltage (900 V), $t$ is the time (9 ms or 0.009 s), $R$ is the resistance (500 kΩ or $500 \times 10^3$ Ω), and $C$ is the capacitance (1 μF or $1 \times 10^{-6}$ F). Plugging in the values, we get $V = 900 e^{-\frac{0.009}{500 \times 10^3 \times 1 \times 10^{-6}}}$ which calculates to approximately 880 V.

A 1.0−μF1.0 - \mu \mathrm { F }1.0−μF Capacitor Is Charged Until It Acquires a Potential Difference Of Charged

A $1.0 - \mu \mathrm { F }$ Capacitor Is Charged Until It Acquires a Potential Difference Of Charged