The metre bridge shown is in a balanced position with \(\frac{P}{Q} = \frac{l_1}{l_2}\). If we now interchange the position of the galvanometer and the cell, will the bridge work? If yes, what will be the balanced condition?
1. Yes, \(\frac{P}{Q}=\frac{l_1-l_2}{l_1+l_2}\)
2. No, no null point
3. Yes, \(\frac{P}{Q}= \frac{l_2}{l_1}\)
4. Yes, \(\frac{P}{Q}= \frac{l_1}{l_2}\)
1. | \(2~\Omega\) | 2 | \(1~\Omega\) |
3. | \(0.5~\Omega\) | 4. | zero |
In the circuit shown in the figure below, the current supplied by the battery is:
1. \(2\) A
2. \(1\) A
3. \(0.5\) A
4. \(0.4\) A
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. \(-3~\text{A}\)
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. | \(2 \over 5\) |
1. | \(10^{\circ}\text{C}\) | 2. | \(5^{\circ}\text{C}\) |
3. | \(20^{\circ}\text{C}\) | 4. | \(15^{\circ}\text{C}\) |
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