Two parallel plate capacitors of capacitances $20 \mu F$ and $30 \mu F$ are charged to potentials of $30 \, V$ and $20 \, V$ respectively. If the similarly charged plates are connected together,then the common potential difference will be: (in $, V$)

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging batteries are then removed,and the capacitors are connected in parallel such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is:

Circuits $(a)$ and $(b)$ have charged capacitors with capacitances $C, 2C$ and $3C$ as shown,with an open switch $S$. When the switch is closed,determine the direction of charge flow for each circuit.

Two identical capacitors each of capacitance $5 \mu F$ are charged to potentials $2 kV$ and $1 kV$ respectively. Their negative ends are connected together. When the positive ends are also connected together,the loss of energy of the system is

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Consider the situation shown in the figure. The switch $S$ is kept open for a long time and then closed. Find the charge that flows through the battery when $S$ is closed.