An RC circuit consisting of a 150 µF capacitor and a 1 M$\varOmega$ resistor is connected to a 100-V battery for a long time,and then the battery is shorted.After 30 s,a second identical resistor is placed in parallel with the original.What is the voltage across the capacitor 30 s after this?
A)0 V
B)54.9 V
C)75.6 V
D)84.3 V
E)100 V

Question

Quizplus · Accepted Answer

Initially, when the capacitor is charged by the 100-V battery and then the battery is shorted, the capacitor starts discharging through the resistor. The time constant for an RC circuit is given by $\tau = RC$, where $R$ is the resistance and $C$ is the capacitance. With $R = 1 \, \text{M}\Omega$ and $C = 150 \, \mu\text{F}$, the time constant is $\tau = 1 \times 10^6 \, \Omega \times 150 \times 10^{-6} \, \text{F} = 150 \, \text{s}$.The voltage across the capacitor after a time $t$ during discharging can be found using the formula $V(t) = V_0 e^{-t/\tau}$, where $V_0$ is the initial voltage. After 30 s, the voltage across the capacitor before adding the second resistor is $V(30) = 100 \, \text{V} \times e^{-30/150} = 100 \, \text{V} \times e^{-1/5}$.When the second identical resistor is added in parallel, the total resistance is halved, so the new time constant is $\tau' = \frac{R}{2}C = \frac{1 \times 10^6}{2} \times 150 \times 10^{-6} = 75 \, \text{s}$. The voltage across the capacitor 30 s after adding the second resistor is calculated using the new time constant and the voltage just before adding the resistor as the initial voltage. Thus, $V'(30) = V(30) \times e^{-30/\tau'}$.Substituting $V(30)$ and calculating gives $V'(30) = 100 \, \text{V} \times e^{-1/5} \times e^{-30/75}$. Simplifying this with the values of $e^{-1/5} \approx 0.819$ and $e^{-30/75} \approx 0.670$, we get $V'(30) \approx 100 \times 0.819 \times 0.670 \approx 54.9 \, \text{V}$.

An RC Circuit Consisting of a 150 µF Capacitor and a 1