Two identical capacitors A and B,charged to t | vedclass

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Question

(A) In circuit $A$,the diode is reverse-biased. Therefore,no current flows through the circuit. The capacitor $A$ remains fully charged. Thus,$Q_{A} =

Vedclass · Accepted Answer

$Q_{A}=VC, Q_{B}=\frac{VC}{e}$

Vedclass · Answer

(A) In circuit $A$,the diode is reverse-biased. Therefore,no current flows through the circuit. The capacitor $A$ remains fully charged. Thus,$Q_{A} = CV$.In circuit $B$,the diode is forward-biased. The capacitor discharges through the resistor $R$. The charge on the capacitor at any time $t$ is given by $q(t) = Q_{0} e^{-\frac{t}{RC}}$,where $Q_{0} = CV$.At time $t = CR$,the charge $Q_{B}$ is:$Q_{B} = CV e^{-\frac{CR}{RC}} = CV e^{-1} = \frac{CV}{e}$.Therefore,$Q_{A} = CV$ and $Q_{B} = \frac{CV}{e}$.

Two identical capacitors $A$ and $B$,charged to the same potential $V$,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):

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