Find the charge on the capacitor $1 \text{ s}$ after opening the switch,given that the switch was closed for a long time $(t = \infty)$.

As shown in the figure,if a capacitor $C$ is charged by connecting it with a battery of voltage $V$ through a resistance $R$,then the energy supplied by the battery will be:

In the circuit shown in the figure,the switch $S$ is initially open and the capacitor is initially uncharged. $I_1, I_2$ and $I_3$ represent the current in the resistance $2\,\Omega, 8\,\Omega$ and $4\,\Omega$ respectively.

$A$ capacitor of capacitance $C$ is discharging through a resistor $R$. Let $t_1$ be the time taken for the energy stored in the capacitor to reduce to half of its initial value,and $t_2$ be the time taken for the charge to reduce to one-fourth of its initial value. Then the ratio $t_1 / t_2$ is:

$A$ capacitor is discharging through a resistor $R$. Consider in time $t_{1}$,the energy stored in the capacitor reduces to half of its initial value and in time $t_{2}$,the charge stored reduces to one-eighth of its initial value. The ratio $t_{1} / t_{2}$ will be ................

The capacitance of $C_1$ and $C_2$ shown in the diagram is $C$. $C_1$ is initially uncharged and $C_2$ is given a charge $2C\varepsilon$ as shown. Which of the following graph represents charge on plate $B$ of capacitor $C_2$ as a function of time?