In the adjoining figure,three capacitors $C_1$,$C_2$,and $C_3$ are connected to a battery of potential $V$. The correct condition is (Symbols have their usual meanings):

$A$ 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:

The potential difference between the points $P$ and $Q$ in the adjoining circuit will be :-

In the circuit shown,$q_2$ and $q_3$ are respectively (Initially all capacitors are uncharged).

Three identical capacitors (each of capacitance $C$,initial charge zero) are connected in series and charged by a battery of emf $E$. After removing the battery,two resistors are connected to these capacitors as shown in the figure. The heat dissipated in each of the resistors is:

Four capacitors are connected in a circuit as shown in the figure. The effective capacitance in $\mu F$ between points $A$ and $B$ will be