N identical cells, each of emf epsilon and internal resistances r are joined in series. Out of these, n cells are wrongly connected, i.e. their terminals are connected in reverse order as required for series connection. If epsilon_(o) be the emf of the resulting battery and r_(o) be its internal resistance then for n lt N//2

N identical cells, each of emf epsilon and internal resistances r are

1.	N identical cells, each of emf epsilon and internal resistances r are joined in series. Out of these, n cells are wrongly connected, i.e. their terminals are connected in reverse order as required for series connection. If epsilon_(o) be the emf of the resulting battery and r_(o) be its internal resistance then for n lt N//2
Answer» `epsilon_(0) = (N-n)EPSILON,r_(o) = (N-n)R` `epsilon_(0 )=(N-n)epsilon, r_(o) = (N-2n) r` v `epsilon_(o)= (N-2n) epsilon, r_(o) = NR` `epsilon_(o) = (N-n)epsilon, r_(o) = Nr` ANSWER :C

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