Three capacitors, each of capacitance \(0.3~\mu \text{F}\) are connected in parallel. This combination is connected with another capacitor of capacitance \(0.1~\mu \text{F}\) in series. Then the equivalent capacitance of the combination is:

The equivalent capacitance of the combination shown in the figure is: 

Three capacitors each of capacitance \(C\) and of breakdown voltage \(V\) are joined in series. The capacitance and breakdown voltage of the combination will be:
1. $\frac{C}{3},$ $\frac{V}{3}$

2. $3C,$ $\frac{V}{3}$

3. $\frac{C}{3},$ $3V$

4. \(3C,~3V\)

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. $\frac{4}{7}$

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 capacitor of capacity C₁ is charged up to V volt and then connected to an uncharged capacitor C₂. Then final P.D. across each will be:

1. $\frac{{\mathrm{C}}_2\mathrm{V}}{{\mathrm{C}}_1+{\mathrm{C}}_2}$

2. $\frac{{\mathrm{C}}_1\mathrm{V}}{{\mathrm{C}}_1+{\mathrm{C}}_2}$

3. $\left(1+\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}\right)$

4. $(1-\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1})\mathrm{V}$

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

1. \(6~\mu\text{F}~\)
2. \(20~\mu\text{F} ~\)
3. \(3~\mu\text{F}~\)
4. \(10~\mu\text{F}\)