If the capacitor connected across $AB$ is charged to $Q$ and then the switch is shifted to position $2$,calculate the heat released after that.

$A$ $2 \ \mu F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$ is (in $\%$)

$A$ capacitor of capacitance $C_{0}$ is charged to a potential $V_{0}$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S,$ the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by

$A$ capacitor of capacity $C_1$ is charged to potential $V_1$ and then disconnected. An uncharged capacitor of capacity $C_2$ is connected in parallel with $C_1$. The resultant potential $V_2$ is

Two capacitors of capacitances $3 \mu F$ and $6 \mu F$ are charged to a potential of $12 V$ each. They are now connected to each other,with the positive plate of each joined to the negative plate of the other. The potential difference across each will be $volt$.

Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using a battery of $emf$ $V$ volt. Now,disconnecting $a$ and $b$,the terminals $b$ and $c$ are connected. Due to this,what will be the percentage loss of energy?