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PhysicsJEEM-2019-0112-S1-P17·January · 12 Jan · Shift 1

Current Electricity · Potentiometer

An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 Ω. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is:

  1. (1)480 Ω
  2. (2)495 Ω
  3. (3)395 Ω
  4. (4)490 Ω

Answer: 395 Ω

Worked solution

  1. 1

    Set up

    Potential drops uniformly along the wire: drop ∝ length.

    10 cm = 1/10 of the 1 m wire.

  2. 2

    Drop across the wire

    Drop across 10 cm = (1/10) × (drop across whole wire).

    So whole-wire drop = 10 × 5 mV = 50 mV = 0.05 V.

  3. 3

    Current (Ohm's law)

    I = 0.05 V ÷ 5 Ω = 0.01 A.

  4. 4

    Solve for R

    Series circuit: I = 4 / (R + 5) = 0.01 A.

    R + 5 = 400 → R = 395 Ω.

  5. 5

    Trap

    Dropping the 1/10 length fraction gives the wrong option.

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