Two cells of e.m.f.s approximately 5 V and 10 V are to be compared using a potentiometer (sum and difference method) of length 400 cm, then, (a) first portion of 110 cm of wire should have a potential difference of 10 V. (b) a battery that runs the potentiometer should have a voltage of 5 V. (c) a battery that runs the potentiometer should have a voltage of more than 15 V. (d) a battery that runs the potentiometer should have a voltage of 10 V.

Answer: (c). In the sum method the wire must balance E₁ + E₂ ≈ 15 V. So the p.d. across the whole wire, and hence the driving battery emf, must exceed 15 V. So the answer is (c).

Two cells of e.m.f.s approximately 5 V and 10 V are to be compared… | MHT CET 2026 (12 May, Shift 1)

Q12MHT CET 2026 · 12 MAY · SHIFT 1Medium

Potentiometer

Two cells of e.m.f.s approximately $5$ V and $10$ V are to be compared using a potentiometer (sum and difference method) of length $400$ cm, then,

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Answer: (c)

In the sum method the wire must balance $E_1 + E_2 \approx 15\ \text{V}$ .
So the p.d. across the whole wire, and hence the driving battery emf, must exceed 15 V.
So the answer is (c).

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Potentiometer question from MHT CET 2026 (12 May, Shift 1)

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