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 given circuit, the potential difference across the plates of the capacitor $C$ in steady state is (in $V$)

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.

Assertion : $A$ capacitor blocks direct current in the steady state.
Reason : The capacitive reactance of the capacitor is inversely proportional to frequency $f$ of the source of $emf$.

The switch $S$ shown in the figure is kept closed for a long time and then opened at $t = 0$. The current in the middle $20\, \Omega$ resistor at $t = 0.25\, ms$ is:

$A$ 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: