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.

In the circuit shown in the figure below, if the potential at point \(\mathrm{A}\) is taken to be zero, the potential at point \(\mathrm{B}\) will be:

1. \(+1\) V
2. \(-1\) V
3. \(+2\) V
4. \(-2\) V

The net resistance of the circuit between \(A\) and \(B\) is:

Two batteries, one of emf \(18\) volts and internal resistance \(2~\Omega\) and the other of emf \(12\) V and internal resistance \(1~\Omega,\) are connected as shown. The voltmeter \(\mathrm{V}\) will record a reading of: 

1. \(18\) V
2. \(30\) V
3. \(14\) V
4. \(15\) V

A \(5\text-\)ampere fuse wire can withstand a maximum power of \(1\) watt in a circuit. The resistance of the fuse wire is:

For the network shown in the figure below, the value of the current \(i\) is:

   
1. \(\frac{18V}{5}\)
2. \(\frac{5V}{9}\)
3. \(\frac{9V}{35}\)
4. \(\frac{5V}{18}\)

A car battery of emf \(12~\text{V}\) and internal resistance \(5\times 10^{-2}~\Omega\) receives a current of \(60~\text{A}\) from an external source. The terminal voltage of the battery is:

If there are two bulbs of (\(40~\text{W},200~\text{V}\)), and (\(100~\text{W},200~\text{V}\)), then the correct relation for their resistance is:
1. \(R_{40}<R_{100}\)
2. \(R_{40}>R_{100}\)
3. \(R_{40}=R_{100}\)
4. no relation can be predicted

When three identical bulbs are connected in series, the consumed power is \(10\) W. If they are now connected in parallel then the consumed power will be:
1. \(30\) W
2. \(90\) W
3. \(\frac{10}{3}\) W
4. \(270\) W

The current in \(8~\Omega\) resistance is (in the figure below):

1. \(0.69\) A
2. \(0.92\) A
3. \(1.30\) A
4. \(1.6\) A