An initially uncharged capacitor $C$ is being charged by a battery of emf $E$ through a resistance $R$. Up to the instant when the capacitor is charged to a potential $E/2$, the ratio of the work done by the battery to the heat dissipated by the resistor is given by

At time $t = 0$,the switch $K$ is closed. Which of the following statements is true for the current '$i$' flowing through the resistor $AB$ in the given circuit?

$A$ 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 ................

In the given figure,the resistance of the coil of the galvanometer $G$ is $2\,\Omega$. The emf of the cell is $4\,V$. The ratio of the potential difference across $C_1$ and $C_2$ is:

$A$ capacitor of $8 \mu\text{F}$ is connected as shown in the circuit. Find the charge on the plates of the capacitor. (in $\mu\text{C}$)