Q. Two identical capacitors $A$ and $B$ are charged to the same potential $V$ and are connected in two circuits at $t = 0$, as shown in figure. The charge on the capacitors at time $t = CR$ are respectively

A) $VC,\ VC$
B) $\frac{VC}{e},\ VC$
C) $VC,\frac{VC}{e}$
D) $\frac{VC}{e},\frac{VC}{e}$
Explanation

(b) Time $t = CR$ is known as time constant. It is time in which charge on the capacitor decreases to $\frac{1}{e}$ times of it’s initial charge (steady state charge).
In figure (i) $PN$ junction diode is in forward bias, so current will flow the circuit $i.e.,$ charge on the capacitor decrease and in time $t$ it becomes $Q = \frac{1}{e}\left( Q_{o} \right);$ where
$Q_{o} = CV \Rightarrow Q = \frac{CV}{e}$
In figure (ii) $P$-$N$ junction diode is in reverse bias, so no current will flow through the circuit hence change on capacitor will not decay and it remains same, $i.e.,\ CV$ after time $t$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.