Two identical capacitors $A$ and $B$, charged to the same potential $5V$ are connected in two different circuits as shown below at time $t = 0$. If the charge on capacitors $A$ and $B$ at time $t = CR$ is $Q_A$ and $Q_B$ respectively, then (Here $e$ is the base of natural logarithm)
Answer: (C) $Q_A = VC,\ Q_B = \dfrac{VC}{e}$
Discharge current leaves the positive (left) plate, goes down through $R$, along the bottom wire, and must go up through the diode to reach the negative plate.
In circuit $A$ the diode points down, so it is reverse biased: no current flows and $Q_A = CV$.
In circuit $B$ the diode points up and conducts, so the capacitor discharges through $R$: $Q = CVe^{-t/RC}$. At $t = CR$, $Q_B = \dfrac{CV}{e}$.
Solution by Sreeraj P, M.Sc Physics