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 $V$)

$27$ identical drops are charged at $22 \ V$ each. They combine to form a bigger drop. The potential of the bigger drop will be............ $V$.

$A$ parallel plate capacitor of capacity $100 \mu F$ is charged by a battery of $50 \text{ V}$. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance,the additional energy given by the battery to the capacitor in joules is

The diagram shows four capacitors with capacitances and breakdown voltages as mentioned. What should be the maximum value of the external emf source such that no capacitor breaks down? (in $kV$)

The numerical value of the charge on either plate of the capacitor $C$ shown in the figure is

$A$ capacitor $C_1 = 1\,\mu F$ can withstand a maximum voltage of $V_1 = 6\,kV$,and a capacitor $C_2 = 3\,\mu F$ can withstand a maximum voltage of $V_2 = 4\,kV$. If these two capacitors are connected in series,what is the maximum voltage (in $kV$) that can be applied to the combination?