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. \(10~\mu\text{F}\)

Three uncharged capacitors of capacities \(C_1, C_2~\text{and}~C_3~~\) are connected to one another as shown in the figure.

If points \(\mathrm{A}\), \(\mathrm{B}\), and \(\mathrm{D}\), are at potential \(V_1, V_2 ~\text{and}~V_3\) then the potential at \(\mathrm{O}\) will be:
1. \(\frac{V_1C_1+V_2C_2+V_3C_3}{C_1+C_2+C_3}\)
2. \(\frac{V_1+V_2+V_3}{C_1+C_2+C_3}\)
3. \(\frac{V_1(V_2+V_3)}{C_1(C_2+C_3)}\)
4. \(\frac{V_1V_2V_3}{C_1C_2C_3}\)

Three capacitors of capacitances 3 μF, 9 μF and 18 μF are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases $\left(\frac{{\displaystyle C_s}}{{\displaystyle C_p}}\right)$ will be:

1. 1 : 15

2. 15 : 1

3. 1 : 1

4. 1 : 3

Two capacitors of capacitance \(6~\mu\text{F}\) and \(3~\mu\text{F}\) are connected in series with battery of \(30~\text{V}\). The charge on \(3~\mu\text{F}\) capacitor at a steady state is:
        
1. \( 3 ~\mu\text{C}\)
2. \( 1.5 ~\mu\text{C}\)
3. \( 60~\mu\text{C}\)
4. \( 900~\mu\text{C}\)

The equivalent capacitance across \(A\) and \(B\) in the given figure is:

  
1. \( \frac{3}{2}C\)
2. \(C\)
3. \( \frac{2}{3}C\)
4. \( \frac{5}{3}C\)

The equivalent capacitance of the following arrangement is:
               
1. $18$ $\mu F$
2. $9$ $\mu F$
3. $6$ $\mu F$
4. $12$ $\mu F$

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

Two capacitors of capacity $2$ $\mu F$ and $3$ $\mu F$ are charged to the same potential difference of 6 V. Now they are connected with opposite polarity as shown. After closing switches $S_1$ $and$ $S_2$, their final potential difference becomes: