JEE Main 2018 — Capacitance Question with Solution

In given circuit capacitors $C_1$ and $C_2$ are connected in series , so equivalent capacitance will be 

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{{\displaystyle 1}}{{\displaystyle C_1}} +\frac{{\displaystyle 1}}{{\displaystyle C_2}} \\ \Rightarrow C_{eq}=\frac{{\displaystyle C_1{C}_2}}{{\displaystyle C_1+C_2}} \end{gathered}$$

at $t = 0$, the switch is closed and capacitors start charged, at time $t$ 

charge on the capacitor is given by 

$q= q_0 \left[1-e^{-t/\tau }\right]$

where $q_0$ is the maximum charge on the capacitor at steady state and $\tau$ is the time constant.

and $q_0 =C_{eq}E$ and $\tau = R{C}_{eq}$

Then from the above equation

$q= C_{eq}E \left[1-e^{-t/R{C}_{eq}}\right]$

$\therefore q= C_{eq}E \left[1-exp\left(\frac{t}{R{C}_{eq}}\right)\right]$
Both capacitor will have same charge as they are connected in series