1. | \(9~{\mu \text{F}}\) | 2. | \(2~{\mu \text{F}}\) |
3. | \(3~{\mu \text{F}}\) | 4. | \(6~{\mu \text{F}}\) |
1. | \(30~\mu \text{F}\) | 2. | \(15~\mu \text{F}\) |
3. | \(25~\mu\text{F}\) | 4. | \(20~\mu\text{F}\) |
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:
1. | \(0.9~\mu\text{F}\) | 2. | \(0.09~\mu\text{F}\) |
3. | \(0.1~\mu\text{F}\) | 4. | \(0.01~\mu\text{F}\) |
1. | \(10 ~\mu \text{F}, ~6~\mu \text{F}\) | 2. | \(8 ~\mu \text{F}, ~8~\mu \text{F}\) |
3. | \(12~\mu \text{F},~ 4~\mu \text{F}\) | 4. | \(1.2~\mu \text{F},~1.8~\mu \text{F}\) |
The equivalent capacitance of the combination shown in the figure is:
1. | \(\frac{C}{2}\) | 2. | \(\frac{3C}{2}\) |
3. | \(3C\) | 4. | \(2C\) |
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.
2.
3.
4. \(3C,~3V\)
A network of four capacitors of capacity equal to is conducted to a battery as shown in the figure. The ratio of the charges on is:
1.
2.
3.
4.
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:
1. | connecting all of them in a series. |
2. | connecting them in parallel. |
3. | connecting two in series and one in parallel. |
4. | connecting two in parallel and one in series. |
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.
2.
3.
4.
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. \(10~\mu\text{F}\)