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NEET 2011 (AIPMT, Prelims) · Cells: EMF, Internal Resistance & Combinations

Cells: EMF, Internal Resistance & Combinations question from NEET 2011 (AIPMT, Prelims)

Q44NEET 2011 · AIPMT · PRELIMSEasy

Cells: EMF, Internal Resistance & Combinations

A current of 2 A flows through a 2 $\Omega$ resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 $\Omega$ resistor. The internal resistance of the battery is

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Answer: (b)

Same emf in both cases: $E = I(R + r)$
$2(2 + r) = 0.5(9 + r) \Rightarrow 4 + 2r = 4.5 + 0.5r$
$1.5r = 0.5 \Rightarrow r = \frac{1}{3}\ \Omega$ . So the answer is (b).

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