Condensers of capacity $4 \mathrm{\mu F}\text{,} 5 \mathrm{\mu F}\text{,} 6 \mathrm{\mu F}$ are connected first in series. The effective capacitance is $C_1$. When they are connected in parallel, the effective capacitance is $C_2$. Then the ratio $\frac{C_2}{C_1}$ will be

The capacitance between the points $P$ and $Q$ in the following circuit is 

Six capacitors, each of capacitance of $2 \mu F$ are connected as shown in the figure. The effective capacitance between $A \text{and} B$ is

Two condensers, one of capacity C and the other of capacity C / 2, are connected to a V-volt battery as shown. 

The work done in fully charging both the condensers is

A capacitor of $30 \mu \mathrm{F}$ charged up to $500$ volt is connected in parallel with another capacitor of $15$ $\mu \mathrm{F}$ which is charged up to $300 \mathrm{V}$. The common potential is

Three capacitors $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are connected to a battery of e.m.f. $120 \mathrm{V}$ as shown in the figure. The potential difference across $C_{1}$ is

A $8 \mu \mathrm{F}$ capacitor is fully charged across a $12 \mathrm{V}$ battery. It is then disconnected from the battery and connected to an uncharged capacitor. If the voltage across the capacitor becomes $3 \mathrm{V}$, then the capacitance of the uncharged capacitor will be

A condenser of capacity $2 \mathrm{\mu F}$ is charged to a potential of $100 \mathrm{V}$. It is now connected to an uncharged condenser of capacity $3 \mathrm{\mu F}$. The common potential will be

Three capacitors, each of capacitance $2 \mu \mathrm{F}$ are connected as shown in the figure. The capacitance between $X$ and $Y$ will be