The given potentiometer has its wire of resistance 10Ω. When the sliding contact is in the middle of the potentiometer wire, the potential drop across 2Ω resistor is :

Solution

Physics sir · 4 steps

The 2 Ω sits across half the wire, so that half gets loaded. Combine it in parallel first, then divide 20 V.

1. Read the circuit. The 20 V cell drives current through the whole 10 Ω wire AB. The 2 Ω resistor connects A to the slider at the middle, so it is in parallel with the half of the wire from A to the slider, which is 5 Ω.
2. Combine resistances. The parallel part is (2 × 5)/(2 + 5) = 10/7 Ω. It is in series with the other 5 Ω half, so the total is 10/7 + 5 = 45/7 Ω.

   V_(2 Ω) = 20 × (10/7)/(10/7 + 5) = 20 × 10/45 = 40/9 V

3. Check with current. The current from the cell is I = 20/(45/7) = 28/9 A. The drop across the parallel part is 28/9 × 10/7 = 40/9 V, and the 2 Ω resistor carries this full drop.
4. So. The potential drop across the 2 Ω resistor is 40/9 V ≈ 4.44 V, option (c).

The idea to keep. A resistor connected to a potentiometer's slider draws current, so that section of wire is no longer a plain divider. Put the load in parallel with its section first, then use the voltage divider.

Answer from the NTA final answer key. Solution by PrepWiser.