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

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 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:

Three capacitors of capacitances \(3~\mu\text{F}\), \(9~\mu\text{F}\) and \(18~\mu\text{F}\) are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases \(\frac{C_s}{C_p}\) 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. \( 900~\mu\text{C}\)

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. \(3C, 3V\)

The equivalent capacitance of the following arrangement is:
               
1. \(18~\mu \text{F}\)
2. \(9~\mu \text{F}\)
3. \(6~\mu \text{F}\)
4. \(12~\mu \text{F}\)

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

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

The equivalent capacitance between points \(a\) and \(b\) in the network shown below is:

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

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. \(\left(1 - \frac{C_{2}}{C_{1}} \right) V\)