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):

$A$ source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key $K$ is closed,the total energy stored across the combination is $E_{1}$. Now key $K$ is opened and a dielectric of dielectric constant $K_{d} = 5$ is introduced between the plates of both capacitors. The total energy stored across the combination is now $E_{2}$. The ratio $E_{1} / E_{2}$ will be:

In the circuit shown in the figure,four capacitors are connected to a battery. The charge on the $5 \ \mu\text{F}$ capacitor is ....... $\mu\text{C}$.

In the given circuit, if point $C$ is connected to the earth and a potential of $+2000 \, V$ is applied to point $A$, the potential at point $B$ is.....$V$.

Four capacitors with capacitances $C_1 = 1 \ \mu F, C_2 = 1.5 \ \mu F, C_3 = 2.5 \ \mu F$ and $C_4 = 0.5 \ \mu F$ are connected as shown and are connected to a $30 \ V$ source. The potential difference between points $a$ and $b$ is: (in $V$)

Consider the circuit shown in the figure below. All the resistors are identical. The charge stored in the capacitor,once it is fully charged is