JEE Main 2026 Apr 5 Shift 2, Physics Q24: Introduction And Ohm’s Law

JEE Main2026Apr 5, Shift 2Physics

When an external resistance of 5 $Ω$ is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 $Ω$ resistor is replaced by a 2 $Ω$ resistor, a current of 0.5 A flows through it. The internal resistance of the cell is _____ $Ω$ .

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Solution

Using $i = \frac{ε}{R + r}$ in both cases:

$0.25 (5 + r) = ε$ and $0.5 (2 + r) = ε$ .

Equating: $0.25 (5 + r) = 0.5 (2 + r) \Rightarrow 1.25 + 0.25 r = 1 + 0.5 r \Rightarrow 0.25 = 0.25 r \Rightarrow r = 1 Ω$ .

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JEE Main 2026 Apr 5 Shift 2, Physics Q24: Introduction And Ohm’s Law

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