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

Physics · Current Electricity

JEE Main 2026 — 5 April, Evening Shift — Question 23

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

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

Using $i = E/(R+r)$ : $0.25 = E/(5+r)$ and $0.5 = E/(2+r)$ . Dividing gives $2 = (5+r)/(2+r) \Rightarrow 4+2r = 5+r \Rightarrow r=1\Omega$ .

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge