Two capacitors are shown in the diagram. Each is charged to potential $V_0$. If capacitor $C$ is filled with a dielectric of constant $2$ and capacitor $2C$ is filled with a dielectric of constant $3$,then the key is closed. The final voltage on the capacitor $C$ will be:

In the circuit shown below, $C_1 = 10 \, \mu F$, $C_2 = C_3 = 20 \, \mu F$, and $C_4 = 40 \, \mu F$. If the charge on $C_1$ is $20 \, \mu C$, then the potential difference between $X$ and $Y$ is ......... $V$.

$A$ 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 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:

$A$ capacitor of capacitance $C$ is initially charged to a potential difference of $V$ $volt$. Now it is connected to a battery of $2V$ with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be

Four identical capacitors are connected as shown in the diagram. When a battery of $6 \, V$ is connected between $A$ and $B$,the total charge stored is found to be $1.5 \, \mu C$. The value of $C_1$ is ........ $\mu F$.

Find the equivalent capacitance across $AB$ in $\mu F$ (all capacitances are in $\mu F$).