`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. Then

Options

A. `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

C

Explanation

`i=(nE)/(nr) = ( E)/( r)`
`p.d` across each cell, `V = E - ir = E - (E )/( r) r = 0`.