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JEE Main 2026 (21 Jan, Shift 2) · Cells: EMF, Internal Resistance & Combinations

Cells: EMF, Internal Resistance & Combinations question from JEE Main 2026 (21 Jan, Shift 2)

Q3JEE Main 2026 · 21 JAN · SHIFT 2Easy

Cells: EMF, Internal Resistance & Combinations

A battery with EMF $E$ and internal resistance $r$ is connected across a resistance $R$ . The power consumption in $R$ will be maximum when:

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Answer: (d)

$P = \frac{E^2 R}{(R + r)^2}$ ; setting $\frac{dP}{dR} = 0$ gives $(R + r) - 2R = 0$ .
So $R = r$ , and the answer is (d).

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