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How many accumulators with an e.m.f. of 2V and internal resistance of 0.2Ω should be connected in series to obtain a current of 5A in an external circuit with a 110V difference in potential across the battery terminals?

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Answer: n = 110.

Explanation:

If n accumulators are connected in series, the e.m.f. of the battery is ε = nε1 and the internal resistance is r = nr1. For a current I, the loss of voltage across the internal resistance of the battery is V' = Ir. According to Ohm's law,

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