Question 127 of 190

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)

A${Q}_{A}=\dfrac{VC}{e},{Q}_{B}=\dfrac{CV}{2}$B${Q}_{A}=VC,{Q}_{B}=CV$C${Q}_{A}=VC,{Q}_{B}=\dfrac{VC}{e}$D${Q}_{A}=\dfrac{CV}{2},{Q}_{B}=\dfrac{VC}{e}$

Show answer and solution▾

Answer: (C) ${Q}_{A}=VC,{Q}_{B}=\dfrac{VC}{e}$

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 $t=CR$.
$∴{Q}_{A}=CV$
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
${Q}_{B}=Q\cdot {e}^{\dfrac{-t}{RC}}$
where, $t=RC$