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. \(-3~\text{A}\)

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

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:
     

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:

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

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, \(\frac{P}{Q}= \frac{l_1}{l_2}\)