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

The terminal potential difference of a cell is greater than its emf when:

A battery is charged at a potential of \(15\) V for \(8\) hours when the current flowing is \(10\) A. The battery on discharge supplies a current of \(5\) A for \(15\) hours. The mean terminal voltage during discharges is \(14\) V. The "Watt hour" efficiency of the battery is:
1. \(80\%\)
2. \(90\%\)
3. \(87.5\%\)
4. \(82.5\%\)

Five equal resistances each of resistance \(R\) are connected as shown in the figure below. A battery of \(V\) volts is connected between \(A\) and \(B\). The current flowing in \(AFCEB\) will be:

   
1. \(\frac{V}{R}\)
2. \(\frac{V}{2R}\)
3. \(\frac{2V}{R}\)
4. \(\frac{3V}{R}\)