Two identical capacitors A and B are charged | vedclass

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

Vedclass · Accepted Answer

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

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(B) The time $t = CR$ is known as the time constant of the $RC$ circuit. It is the time in which the charge on the capacitor decreases to $\frac{1}{e}$ times its initial charge $(Q_0 = CV)$.In figure $(i)$,the $p-n$ junction diode is in forward bias,allowing current to flow through the circuit. Consequently,the charge on the capacitor decays according to the equation $q(t) = Q_0 e^{-t/CR}$. At $t = CR$,the charge becomes $q = CV e^{-CR/CR} = \frac{CV}{e}$.In figure $(ii)$,the $p-n$ junction diode is in reverse bias,which acts as an open circuit. Therefore,no current flows through the circuit,and the charge on the capacitor does not decay. It remains at its initial value of $Q = CV$.

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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