In the given electrical circuit,if the switch $S$ is closed,then the maximum energy stored in the inductor is: (in $J$)

In the following figure,the charge on each capacitor in the steady state will be.....$\mu C$.

$A$ part of a circuit is shown in the figure. If $V_A - V_B = 4 \, V$,then find the charge on the $4 \, \mu F$ capacitor connected to point $A$. (in $, \mu C$)

In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is

Two parallel plate capacitors with capacitances $C$ and $2C$ are connected in parallel. They are charged to a potential difference $V$. Now,if the battery is disconnected and a dielectric of dielectric constant $K$ is inserted between the plates of the capacitor with capacitance $C$,find the potential difference across each capacitor.

Find the potential difference $V_a - V_b$ between the points $a$ and $b$ shown in the figure.