`N` identical cells, each 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 the required for series connection `(n lt (N)/(2))`. Let `E_(0)` be the emf of resulting battery and `r_(0)` be its internal resistance. ThenA. `epsilon_0 = (N - n) epsilon, r_0 = (N - n) r`B. `epsilon_0 = (N - 2n) epsilon, r_0 = (N - 2n) r`C. `epsilon_0 = (N - 2n) epsilon, r_0 = Nr`D. `epsilon_0 = (N - n) epsilon, r_0 = Nr`]

`N` identical cells, each emf `E` and internal resistance `r` are join

1.	`N` identical cells, each 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 the required for series connection `(n lt (N)/(2))`. Let `E_(0)` be the emf of resulting battery and `r_(0)` be its internal resistance. ThenA. `epsilon_0 = (N - n) epsilon, r_0 = (N - n) r`B. `epsilon_0 = (N - 2n) epsilon, r_0 = (N - 2n) r`C. `epsilon_0 = (N - 2n) epsilon, r_0 = Nr`D. `epsilon_0 = (N - n) epsilon, r_0 = Nr`]
Answer» Correct Answer - C `i=(nE)/(nr) = ( E)/( r)` `p.d` across each cell, `V = E - ir = E - (E )/( r) r = 0`.

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