Two cells of e.m.f.s E₁ and E₂ (E₁ > E₂) are connected as shown in… | MHT CET 2025 (23 Apr, Shift 2)

Q25MHT CET 2025 · 23 APR · SHIFT 2Medium

Potentiometer

Two cells of e.m.f.s $E_1$ and $E_2\ (E_1 > E_2)$ are connected as shown in figure.

When the potentiometer is connected between A and B, the balancing length of the potentiometer wire is 3.60 m. On connecting the potentiometer between A and C, the balancing length is 0.90 m. The ratio $E_1/E_2$ is

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Answer:

A–C gives a smaller length, so the cells oppose there.
A–B: $E_1 \propto 3.60$ ; A–C: $E_1 - E_2 \propto 0.90$ .
$E_2 \propto 3.60 - 0.90 = 2.70$ , so $\frac{E_1}{E_2} = \frac{3.60}{2.70} = \frac{4}{3}$ .
So the answer is (b) $4:3$ .

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