$A$ capacitor is connected to a cell of $emf$ $E$ having some internal resistance $r$. The potential difference across the

The circuit shown has been connected for a long time. The voltage across the capacitor is ............. $V$.

$A$ 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 $\,s$)

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.

The capacitor shown in the figure is in a steady state. The energy stored in the capacitor is

$A$ capacitor of capacitance $C = 10 \, \mu F$ is connected to a battery of emf $\varepsilon = 2 \, V$ through a resistor $R$. It is observed that it takes $t = 50 \, ms$ for the charge on the capacitor to reach $q = 12.6 \, \mu C$. Calculate the resistance $R$ of the circuit in $k \Omega$. (Take $1/e \approx 0.37$)