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JEE Main 2021 (17 Mar, Shift 2) · Cells: EMF, Internal Resistance & Combinations

Cells: EMF, Internal Resistance & Combinations question from JEE Main 2021 (17 Mar, Shift 2)

Q28JEE Main 2021 · 17 MAR · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations

Two cells of emf 2E and E with internal resistance r1 and r2 respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is

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Answer: (b)

Both cells aid each other: $I = \dfrac{3E}{R + r_1 + r_2}$ .
Terminal p.d. of first cell is zero: $2E - Ir_1 = 0 \Rightarrow I = \dfrac{2E}{r_1}$ .
$\dfrac{3E}{R + r_1 + r_2} = \dfrac{2E}{r_1} \Rightarrow 3r_1 = 2R + 2r_1 + 2r_2$
So the answer is (b) $R = \dfrac{r_1}{2} - r_2$ .

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