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 \(5-\)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. $\frac{18\mathrm{V}}{5}$

2. $\frac{5\mathrm{V}}{9}$

3. $\frac{9\mathrm{V}}{35}$

4. $\frac{5\mathrm{V}}{18}$

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:

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 Ω resistance is (in the figure below):

1.  0.69 A
2.  0.92 A
3.  1.30 A
4.  1.6 A