Two identical capacitors A and B are charged | vedclass

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Question

(B) The time $t = CR$ is known as the time constant. It is the time in which the charge on a discharging capacitor decreases to $\frac{1}{e}$ times it

Vedclass · Accepted Answer

$\frac{VC}{e}, VC$

Vedclass · Answer

(B) The time $t = CR$ is known as the time constant. It is the time in which the charge on a discharging capacitor decreases to $\frac{1}{e}$ times its initial charge $(Q_0 = CV)$.In figure $(i)$,the $PN$ junction diode is in forward bias,allowing current to flow through the circuit. Thus,the capacitor discharges,and at time $t = CR$,the charge becomes $Q = Q_0 e^{-t/CR} = CV e^{-CR/CR} = \frac{CV}{e}$.In figure $(ii)$,the $PN$ junction diode is in reverse bias,so no current flows through the circuit. Therefore,the charge on the capacitor does not decay and remains at its initial value,$CV$.

$X=x$	$P(X=x)$
$0$	$0.4$
$1$	$0.3$
$2$	$0.1$
$3$	$0.1$
$4$	$0.1$

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 the figure. The charge on the capacitors at time $t = CR$ are respectively:

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