Q. No.
| 1. | \(6~\text V\) | 2. | \(8~\text V\) |
| 3. | \(10~\text V\) | 4. | \(4~\text V\) |
| 1. | \(3.75~\text{V}\) | 2. | \(4.25~\text{V}\) |
| 3. | \(4~\text{V}\) | 4. | \(0.375~\text{V}\) |
A set of '\(n\)' equal resistors, of value '\(R\)' each, are connected in series to a battery of emf '\(E\)' and internal resistance '\(R\)'. The current drawn is \(I.\) Now, if '\(n\)' resistors are connected in parallel to the same battery, then the current drawn becomes \(10I.\) The value of '\(n\)' is:
| 1. | \(10\) | 2. | \(11\) |
| 3. | \(20\) | 4. | \(9\) |
1. \(0.5~\Omega\)
2. \(0.8~\Omega\)
3. \(1.0~\Omega\)
4. \(0.2~\Omega\)
A cell having an emf \(\varepsilon\) and internal resistance \(r\) is connected across a variable external resistance \(R\). As the resistance \(R\) is increased, the plot of potential difference \(V\) across \(R\) is given by:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
A current of \(2~\text{A}\) flows through a \(2~\Omega\) resistor when connected across a battery. The same battery supplies a current of \(0.5~\text{A}\) when connected across a \(9~\Omega\) resistor. The internal resistance of the battery is:
| 1. | \(\dfrac{1}{3}~\Omega\) | 2. | \(\dfrac{1}{4}~\Omega\) |
| 3. | \(1~\Omega\) | 4. | \(0.5~\Omega\) |
2. \(-r\) and \(E\)
1. 1.5 V
2. 1.0 V
3. 0.5 V
4. 3.2 V
A battery is charged at a potential of \(15\) V for \(8\) hours when the current flowing is \(10\) A. The battery on discharge supplies a current of \(5\) A for \(15\) hours. The mean terminal voltage during discharges is \(14\) V. The "Watt hour" efficiency of the battery is:
1. \(80\%\)
2. \(90\%\)
3. \(87.5\%\)
4. \(82.5\%\)
For a cell, the terminal potential difference is \(2.2\) V when the circuit is open and reduces to \(1.8\) V when the cell is connected to the resistance of \(R = 5~\Omega\). The internal resistance of cell (\(r\)) is:
| 1. | \(\dfrac{10}{9}~ \Omega\) | 2. | \(\dfrac{9}{10}~ \Omega\) |
| 3. | \(\dfrac{11}{9}~ \Omega\) | 4. | \(\dfrac{5}{9}~ \Omega\) |
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