Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rₚ=1 Ω as shown in the figure. An external resistance of Rₑ=2 Ω is connected via the sliding contact. The electric current in the circuit is :

Solution

Physics sir · 3 steps

The slider splits the wire in two, and the 2 Ω resistor sits in parallel with only the left half.

1. Split the wire. The sliding contact is at the middle, so the 1 Ω wire becomes two halves of 0.5 Ω each. The external resistor Rₑ = 2 Ω joins the left end of the wire to the slider, so it is in parallel with the left half only.

   R_(left) = (0.5 × 2)/(0.5 + 2) = 1/2.5 = 0.4 Ω

2. Add the right half. This parallel pair is in series with the right half of the wire. Total resistance is 0.4 + 0.5 = 0.9 Ω.
3. So. The battery has no internal resistance shown, so the current from the cell is I = (0.9 V)/(0.9 Ω) = 1.0 A, option (c).

The idea to keep. With a sliding contact, first mark which part of the wire lies between the two ends of the external resistor. Only that part is in parallel with it; the rest stays in series.

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