Q 12-02-195JEE MainJEE Main 2018 (16 Apr, Shift 1)Easy
In a circuit, two uncharged capacitors $C_1$ and $C_2$ are connected in series with a cell of emf $E$, a resistor $R$ and a switch $S$. The switch $S$ is closed at $t = 0$. The charge on the capacitor $C_1$ as a function of time will be given by $\left(C_{eq} = \dfrac{C_1C_2}{C_1 + C_2}\right)$
Answer: (C) $C_{eq}E\left[1 - \exp\left(-\dfrac{t}{RC_{eq}}\right)\right]$
The two series capacitors act as one capacitor $C_{eq}$ charging through $R$:
$$q(t) = C_{eq}E\left(1 - e^{-t/RC_{eq}}\right)$$
Capacitors in series carry the same charge, so this is also the charge on $C_1$.
Solution by Sreeraj P, M.Sc Physics