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 |
The dependence of resistivity \((\rho)\) on the temperature \((T)\) of a semiconductor is, roughly, represented by:
1. | 2. | ||
3. | 4. |
Current through the \(2~\Omega\) resistance in the electrical network shown is:
1. | zero | 2. | \(1\) A |
3. | \(3\) A | 4. | \(5\) A |
Two batteries, one of emf \(18~\text{V}\) and internal resistance \(2~\Omega\) and the other of emf \(12\) V and internal resistance \(1~\Omega\), are connected as shown. Reading of the voltmeter is:
(if voltmeter is ideal)
1. \(14\) V
2. \(15\) V
3. \(18\) V
4. \(30\) V
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. | \({R}_1=3~\text{k} \Omega, {R}_2=2~\text{k} \Omega, {R}_3=2~ \text{k} \Omega\) |
1. | \(10^{\circ}\text{C}\) | 2. | \(5^{\circ}\text{C}\) |
3. | \(20^{\circ}\text{C}\) | 4. | \(15^{\circ}\text{C}\) |
Power consumed in the given circuit is \(P_1\). On interchanging the position of \(3~\Omega\) and \(12~\Omega\) resistances, the new power consumption is \(P_2\). The ratio of \(\frac{P_2}{P_1}\) is:
1. | \(2\) | 2. | \(1 \over 2\) |
3. | \(3 \over 5\) | 4. | \(2 \over 5\) |
For the given circuit, the value of the resistance in which the maximum heat is produced is:
1. \(2~\Omega\)
2. \(6~\Omega\)
3. \(4~\Omega\)
4. \(12~\Omega\)
A charged particle having drift velocity of \(7.5\times10^{-4}~\text{ms}^{-1}\) in an electric field of \(3\times10^{-10}~\text{Vm}^{-1}\), has mobility of:
1. \(2.5\times 10^{6}~\text{m}^2\text{V}^{-1}\text{s}^{-1}\)
2. \(2.5\times 10^{-6}~\text{m}^2\text{V}^{-1}\text{s}^{-1}\)
3. \(2.25\times 10^{-15}~\text{m}^2\text{V}^{-1}\text{s}^{-1}\)
4. \(2.25\times 10^{15}~\text{m}^2\text{V}^{-1}\text{s}^{-1}\)
The figure below shows a network of currents. The current \(i\) will be:
1. \(3~\text{A}\)
2. \(13~\text{A}\)
3. \(23~\text{A}\)
4. \(-3~\text{A}\)