A battery of e.m.f. 10 V is connected to resistance as shown in the figure below. The potential difference $V_A-V_B$  between the points A and B is:

1. –2 V

2. 2 V

3. 5 V

4. $\frac{20}{11}V$

What is the equivalent resistance of the circuit​​​​​​?

1. 6 Ω

2. 7 Ω

3. 8 Ω

4. 9 Ω

If each resistance in the figure is 9 Ω, then the reading of the ammeter is:

1. 5 A

2. 8 A

3. 2 A

4. 9 A

Equivalent resistance across terminals A and B will be:

1. 1 Ω

2. 2 Ω

3. 3 Ω

4. 4 Ω

The total current supplied to the circuit by the battery is:

1. 1 A

2. 2 A

3. 4 A

4. 6 A

In circuit shown below, the resistances are given in ohms and the battery is assumed ideal with emf equal to 3 volt. The voltage across the resistance R₄ is:

1. 0.4 V

2. 0.6 V

3. 1.2 V

4. 1.5 V

A battery of e.m.f. E and internal resistance r is connected to a variable resistor R as shown below. Which one of the following is true​​​​​​?

The current in the arm CD of the circuit will be:
           
1. $i_1+i_2$

2. $i_2+i_3$

3. $i_1+i_3$

4. $i_1-i_2+i_3$

Consider the circuit shown in the figure below. The current I₃ is equal to:

1. 5 A

2. 3 A

3. –3 A

4. –5/6 A

For measurement of potential difference, the potentiometer is preferred in comparison to the voltmeter because:

1. the potentiometer is more sensitive than the voltmeter.

2. the resistance of the potentiometer is less than
    the voltmeter.

3. the potentiometer is cheaper than the voltmeter.

4. the potentiometer does not take current from the circuit.