JEE Main 2020 — Alternating Current Question with Solution

Charge on capacitor after charging with a battery of potential $V$, $Q=CV$

Now, capacitor $A$ is connected as shown in the figure.

As diode is in reverse bias, so capacitor charge remains same after time $t=CR$.

$\therefore Q_A=CV$

In case $B$, capacitor is connected as shown in the figure.

Here, diode is in forward bias, so capacitor discharges and its charge decays as 

$Q_B=Q·e^{\frac{-t}{RC}}$

where, $t=RC$