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:

$A$ capacitor of capacity $0.1 \mu F$ connected in series to a resistor of $10 M \Omega$ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given, $\log _{10} 2=0.3010$ ) (in $\,s$)

$A$ $1 \mu F$ capacitor $C$ is connected to a battery of $10 V$ through a resistance of $1 \text{ M}\Omega$. The voltage across $C$ after $1 \text{ s}$ is approximately: (in $V$)

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

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

$A$ $4 \mu F$ capacitor and a resistance of $2.5 \, M\Omega$ are in series with a $12 \, V$ battery. Find the time after which the potential difference across the capacitor is $3$ times the potential difference across the resistor. (Given $\ln(2) = 0.693$)