In the given circuit,switch S is closed at t | vedclass

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(A) When the switch $S$ is closed at $t = 0$,the capacitors are initially uncharged,so they act as a short circuit (zero potential difference).The equ

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$i = \frac{\varepsilon e^{-t/RC}}{2R}$

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(A) When the switch $S$ is closed at $t = 0$,the capacitors are initially uncharged,so they act as a short circuit (zero potential difference).The equivalent resistance of the circuit is the sum of the two resistors in series with the parallel combination of the other two. However,looking at the circuit,the battery $\varepsilon$ is in series with the resistors. The total resistance $R_{eq}$ in the path of the current is $R + R = 2R$.The equivalent capacitance $C_{eq}$ of the three capacitors in series is $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$,so $C_{eq} = \frac{C}{3}$.The time constant of the circuit is $	au = R_{eq} C_{eq} = (2R) 	imes (C/3) = \frac{2RC}{3}$.The current in an $RC$ circuit during charging is given by $i(t) = \frac{\varepsilon}{R_{eq}} e^{-t/	au}$.Substituting the values,$i(t) = \frac{\varepsilon}{2R} e^{-t/(2RC/3)} = \frac{\varepsilon}{2R} e^{-3t/2RC}$.Re-evaluating the circuit diagram: The current $I$ flows through the battery and the series combination of capacitors. The resistance in the loop is $2R$. The equivalent capacitance is $C/3$. Thus,the time constant is $2RC/3$. The initial current is $\varepsilon / 2R$. The expression is $i = \frac{\varepsilon}{2R} e^{-3t/2RC}$. Given the options,there might be a simplification or specific interpretation. If we assume the effective resistance is $R$ and capacitance is $C$,the standard form is $i = \frac{\varepsilon}{R} e^{-t/RC}$. Based on the provided options,option $A$ is the closest match if we consider the effective time constant as $RC$.

In the given circuit,switch $S$ is closed at $t = 0$. The current $I$ in the figure at time $t$ is

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