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N identical cells, each of emf E and internal resistance r are joined in series. Out of N cells, n cells are wrongly connected i.e., their terminals are connected in reverse of that required for series connection $(n < \frac{N}{2})$ . Let E₀ be the emf of resulting battery and r₀ be its internal resistance. Then

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E0 =N−n E, r0 = N−nr

E0 =N−2n E, r0 = N−2nr

E0 =N−2n E, r0 = Nr

E0 =N−n E, r0 = Nr

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detailed solution

Correct option is C

For every cell that is wrongly connected, the emf decreases by 2E. However, internal resistance does not depend on the direction and therefore remains same for all cells.

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N identical cells, each of emf E and internal resistance r are joined in series. Out of N cells, n cells are wrongly connected i.e., their terminals are connected in reverse of that required for series connection n < N2. Let E0 be the emf of resulting battery and r0 be its internal resistance. Then

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