Two capacitors each of $1\,\mu F$ capacitance are connected in parallel and are then charged by a $200\,V$ $d.c.$ supply. The total energy of their charges (in $joules$) is

In the given circuit, what are the potential differences between points $A$ and $B$ $(V_{AB})$ and between points $B$ and $C$ $(V_{BC})$?

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.

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. $(i)$ The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then,the value of $E_1 / E_2$ is:

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

Calculate the charge on the second capacitor before and after the switch in the circuit is closed.