Two cells E₁ and E₂ having equal e.m.f ' E ' and internal resistances… | MHT CET 2025 (19 Apr, Shift 2)

Q8MHT CET 2025 · 19 APR · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations

Two cells $E_1$ and $E_2$ having equal e.m.f ' $E$ ' and internal resistances $r_1$ and $r_2\ (r_1 > r_2)$ respectively are connected in series. This combination is connected to an external resistance ' R '. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be

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Answer: (a)

$I = \frac{2E}{R + r_1 + r_2}$
Terminal p.d. of $E_1$ : $E - Ir_1 = 0 \Rightarrow I = \frac{E}{r_1}$
$\frac{2E}{R + r_1 + r_2} = \frac{E}{r_1} \Rightarrow R + r_1 + r_2 = 2r_1$
$R = r_1 - r_2$ . So the answer is (a).

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Cells: EMF, Internal Resistance & Combinations question from MHT CET 2025 (19 Apr, Shift 2)

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