A galvanometer together with an unknown resistance in series is connected across two identical batteries each of 1.5 V. When the batteries are connected in series, the galvanometer records a current of 1A, and when the batteries are in parallel, the current is 0.6 A. What is the internal resistance of the battery ?

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1Ω

12Ω

23Ω

13Ω

answer is D.

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Detailed Solution

Is =2E2r+R=1A=32r+R=1Ip=Er2+R=0.6A=3r+2R=0.6ASolving r=1/3Ω

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