Q. No.
1. \(1~\mu\text F\)
2. \(0.5~\mu\text F\)
3. \(4~\mu\text F\)
4. \(2~\mu\text F\)
| 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. | \(20~\mu\text{F}\) |
| 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. \(\dfrac{C}{2}\)
2. \(\dfrac{3C}{2}\)
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 \(C_1\) 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. \(\left(1 - \frac{C_{2}}{C_{1}} \right) V\)
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