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 $\left( {{C_{eq}}\, = {\kern 1pt} \,\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).$
JEE MAIN 2018, Diffcult
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During charging charge on the capacitor increases with time. Charge on the capacitor
$\mathrm{C}_{1}$ as a function of time, $\mathrm{Q}=\mathrm{Q}_{0}\left(1-\mathrm{e}^{-t / \mathrm{R} C}\right)$
$\mathrm{Q}=\mathrm{C}_{\mathrm{eq}} \mathrm{E}\left[1-\mathrm{e}^{-t / \mathrm{RC}_{\mathrm{eq}}}\right]$
$\left(\because Q_{0}=C_{e q} E\right)$
Both capacitor will have charge as they are connected in series
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