Two identical circuits have a capacitor in series with a resistor and a switch. In circuit $A$, the resistor has resistance $\mathrm{R}$, while in circuit $\mathrm{B}$, its resistance is $4 \mathrm{R}$. The capacitors are identical. Each capacitor begins fully charged, and each circuit is open. When each circuit is completed (i.e., the switch in them is closed), the capacitor will begin to discharge. If a time $t$ is required for circuit A's capacitor to fully discharge (say, by $99 \%$ ), what time will be required for circuit B's capacitor to do the same?
A) $2 \mathrm{t}$
B) $0.25 \mathrm{t}$
C) $\mathrm{t}$
D) $4 \mathrm{t}$
E) $0.5 \mathrm{t}$

Question

Quizplus · Accepted Answer

The time required for a capacitor to discharge through a resistor is determined by the time constant $\tau = RC$. For circuit B, with a resistor of $4R$, the time constant is $4RC$, which is four times the time constant of circuit A ($RC$). Therefore, the time for circuit B's capacitor to discharge by $99\%$ is $4$ times that of circuit A.

Two Identical Circuits Have a Capacitor in Series with a Resistor