In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E₁ and E₂ as well. For key K₁ (closed), the jockey is adjusted to touch the wire at point J₁ so that there is no deflection in the galvanometer. Now the first battery (E₁) is replaced by second battery (E₂) for working by making K₁ open and K₂ closed. The galvanometer gives then null deflection at J₂. The value of E₁/E₂ is a/b, where a = _________.

Solution

Physics sir · 5 steps

Follow the zigzag wire from A metre by metre. The balancing lengths, not the marked gaps, give the ratio.

1. Potentiometer principle. The wire is uniform, so the potential drop per unit length is the same everywhere. At null deflection each cell's emf equals the drop from A to the jockey point, so E ∝ balancing length measured from A.
2. Trace the wire. The 10 m wire is ten 1 m rows. Starting at A it runs right along row 1, left along row 2, right along row 3, and so on, reversing direction each time. Odd rows run left to right and even rows run right to left.
3. Length to J1. J₁ is on row 4, which runs right to left. It sits 20 cm from the left end, so it is 80 cm into that row. So AJ₁ = 3 m + 0.8 m = 3.8 m.
4. Length to J2. J₂ is on row 8, which also runs right to left. It sits 60 cm from the right end, where that row starts. So AJ₂ = 7 m + 0.6 m = 7.6 m.

   E₁/E₂ = AJ₁/AJ₂ = 3.8/7.6 = 1/2

5. So. E₁/E₂ = a/b = 1/2, so a = 1.

The common slip: 3. Comparing only the marked gaps, 20 cm and 60 cm, gives 1/3. That uses the wrong lengths, and the answer is still 1 only by accident. The correct balancing lengths are 3.8 m and 7.6 m.

The idea to keep. In a folded potentiometer wire, always measure the balancing length along the wire from the common end A. Note which way each row runs before you read off a marked distance.

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