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 3em Ω. Numerical answer type. Work it out, then open the answer.

Answer: 1. Same emf: E = 0.25(5 + r) = 0.5(2 + r) 1.25 + 0.25r = 1 + 0.5r ⇒ 0.25r = 0.25 r = 1 Ω

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Cells: EMF, Internal Resistance & Combinations

When an external resistance of $5\ \Omega$ is connected across terminals of a cell, a current of 0.25 A flows through it. When the $5\ \Omega$ resistor is replaced by a $2\ \Omega$ resistor, a current of 0.5 A flows through it. The internal resistance of the cell is $\underline{\hspace{3em}}\ \Omega$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 1

Same emf: $E = 0.25(5 + r) = 0.5(2 + r)$
$1.25 + 0.25r = 1 + 0.5r \Rightarrow 0.25r = 0.25$
$r = 1\ \Omega$

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