According to this diagram, the potential difference across the terminals is:

(internal resistance of cell=r)

1. $\mathrm{V}=\mathrm{E}-\mathrm{ir}$

2. $\mathrm{V}=\mathrm{E}+\mathrm{ir}$

3. $\mathrm{V}=\mathrm{E}$

4. Zero

The value of current through 2$\mathrm{\Omega }$ is:

1. 1.0 A

2. 1.5 A

3. 5.0 A

4. 2.1 A

The reading of the ammeter in the circuit below is:

1. 5 A

2. 15 A

3. 20 A

4. 25 A

A thousand cells of the same emf E and same internal resistance r are connected in series in the same order without external resistance. The potential drop across 399 cells is found to be:

1. Zero

2. 399 E

3. 601 E

4. 1000 E

Five cells each of e.m.f. E and internal resistance r are connected in series. If due to oversight, one cell is connected wrongly, then the equivalent e.m.f. of the combination is:

1. 5 E

2. 2 E

3. 3 E

4. 4 E

Two batteries of different e.m.f.'s and internal resistances are connected in series with each other and with an external load resistor. The current is 3.0 A. When the polarity of one battery is reversed, the current becomes 1.0 A. The ratio of the e.m.f.'s of the two batteries is:

1. 2.5

2. 2

3. 1.5

4. 1

In the figure, the e.m.f. of the cell is 2V and the internal resistance is negligible. The resistance of the voltmeter is 80 ohm. The reading of the voltmeter will be:

1. 2 volt

2. 1.33 volt

3. 1.60 volt

4. 0.80 volt

Calculate the current shown by ammeter A in the circuit diagram.

1. 0.1 A

2. 0.2 A

3. 0.3 A

4. 0.4 A

Ends A and B are connected to positive and negative terminals of a cell of emf 12 V respectively. The value of voltage shown by the voltmeter of negligible resistance is:

1. Zero

2. 3 V

3. 6 V

4. 12 V

In the circuit shown, R₁ is increased. What happens to the reading of the voltmeter (ideal)?

1. Increases

2. Decreases

3. First increases then decrease

4. Does not change