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

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0.5 Ω

1/3 Ω

1/4 Ω

1 Ω

answer is B.

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Detailed Solution

Let internal resistance of battery be r then according to questionE2+r=2 and E9+r=0.5 ⇒r=13Ω

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