Q. No.
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Two cells of emf \(E\) and internal resistance \(r_1\) and \(r_2\) are connected in series through an external resistance \(R\). The value of \(R\) for which the potential difference across one of the cells becomes zero will be:
| 1. | \(\dfrac{r_{1} r_{2}}{r_{1} + r_{2}}\) | 2. | \(r_{1} + r_{2}\) |
| 3. | \(|r_{1} - r_{2}|\) | 4. | \(\dfrac{r_{1}}{r_{2}}\) |
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In the given circuit diagram, \(3\) identical bulbs are connected. If bulb \(B_3\) get fused suddenly, how will the brightness of bulbs \(B_1\) and \(B_2\) change?
| 1. | Brightness of bulb \(B_1\) will increase whereas brightness of bulb \(B_2\) will decrease |
| 2. | Brightness of bulb \(B_2\) will increase whereas brightness of bulb \(B_1\) will decrease |
| 3. | Brightness of both bulbs \(B_1\) and \(B_2\) will increase |
| 4. | Brightness of bulb \(B_1\) will increase whereas brightness of bulb \(B_2\) will remain the same |
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What is the reading of the voltmeter of resistance \(1200~\Omega\) connected in the following circuit diagram?
| 1. | \(2.5\) V | 2. | \(5.0\) V |
| 3. | \(7.5\) V | 4. | \(40\) V |
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The dependence of resistivity \((\rho)\) on the temperature \((T)\) of a semiconductor is, roughly, represented by:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
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Current through the \(2~\Omega\) resistance in the electrical network shown is:
| 1. | zero | 2. | \(1~\text A\) |
| 3. | \(3~\text A\) | 4. | \(5~\text A\) |
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Two batteries, one of emf \(18~\text{V}\) and internal resistance \(2~\Omega\) and the other of emf \(12~\text V\) and internal resistance \(1~\Omega,\) are connected as shown. Reading of the voltmeter is:
(if a voltmeter is ideal)
1. \(14~\text V\)
2. \(15~\text V\)
3. \(18~\text V\)
4. \(30~\text V\)
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A current of \(2\) A is to be sent through a resistor of \(5 ~\Omega.\) Number of cells required in series, if each has emf \(2\) V and internal resistance \(0.5~\Omega,\) are:
1. \(40\)
2. \(30\)
3. \(20\)
4. \(10\)
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A \(50\) kW dc generator produces a potential difference of \(250\) V. If the resistance of the transmission line is \(1~\Omega\), what percentage of the original power is lost during transmission?
1. \(80\%\)
2. \(40\%\)
3. \(20\%\)
4. \(10\%\)
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A potential divider is used to give outputs of \(2~\text{V}\) and \(3~\text{V}\) from a \(5~\text{V}\) source, as shown in the figure.
| 1. | \({R}_1=1~\text{k} \Omega, {R}_2=1 ~\text{k} \Omega, {R}_3=2 ~\text{k} \Omega\) |
| 2. | \({R}_1=2 ~\text{k} \Omega, {R}_2=1~\text{k} \Omega, {R}_3=2~\text{k} \Omega\) |
| 3. | \({R}_1=1 ~\text{k} \Omega, {R}_2=2~ \text{k} \Omega, {R}_3=2~ \text{k} \Omega\) |
| 4. | \({R}_1=3~\text{k} \Omega, {R}_2=2~\text{k} \Omega, {R}_3=2~ \text{k} \Omega\) |
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