In the following circuit,the switch S is clos | vedclass

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(A) The circuit consists of two capacitors $C_1$ and $C_2$ connected in series with a resistor $R$ and a battery of $EMF$ $E$. When the switch $S$ is

Vedclass · Accepted Answer

${C_{eq}}E\,[1 - \exp ( - t/R{C_{eq}})]$

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(A) The circuit consists of two capacitors $C_1$ and $C_2$ connected in series with a resistor $R$ and a battery of $EMF$ $E$. When the switch $S$ is closed at $t = 0$,the capacitors begin to charge. The equivalent capacitance of the series combination is given as $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$. The charging equation for a series $RC$ circuit is $Q(t) = Q_0(1 - e^{-t/	au})$,where $Q_0$ is the maximum charge and $	au = RC_{eq}$ is the time constant. The maximum charge on the equivalent capacitor is $Q_0 = C_{eq}E$. Since the capacitors are in series,the charge on each capacitor is the same and equal to the charge on the equivalent capacitor. Thus,the charge on $C_1$ as a function of time is $Q(t) = C_{eq}E[1 - \exp(-t/RC_{eq})]$.

In the following circuit,the switch $S$ is closed at $t = 0.$ The charge on the capacitor $C_1$ as a function of time will be given by $\left( {{C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).$

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