The effective capacity of the network between terminals \(\mathrm{A}\) and \(\mathrm{B}\) is:

A capacitor of capacity \(C_1\) is charged up to \(V\) volt and then connected to an uncharged capacitor \(C_2\). Then final P.D. across each will be:
1. \(\frac{C_{2} V}{C_{1} + C_{2}}\)
2. \(\frac{C_{1} V}{C_{1} + C_{2}}\)
3. \(\left(1 + \frac{C_{2}}{C_{1}}\right)\)
4. \(\left(1 - \frac{C_{2}}{C_{1}} \right) V\)

Three capacitors each of capacity \(4\) µF are to be connected in such a way that the effective capacitance is \(6\) µF. This can be done by:

A network of four capacitors of capacity equal to ${\mathrm{C}}_1=\mathrm{C},{\mathrm{ C}}_2=2\mathrm{C},{\mathrm{ C}}_3=3\mathrm{C}\text{, and }{\mathrm{C}}_4=4\mathrm{C}$ is conducted to a battery as shown in the figure. The ratio of the charges on ${\mathrm{C}}_2\text{ and }{\mathrm{C}}_4$ is:

1. $\frac{7}{4}$

2. $\frac{22}{3}$

3. $\frac{3}{22}$

4. $\frac{4}{7}$