`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

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 E and internal resistance r are connected in series. Find the expression for current in the circuit given below

n-identical cells each of emf E and internal resistance r are connected in series. Find the expression for current in the circuit given below.

Six cell each of emf E and internal resistance r are connected in series. If due to over sight, two cells are connected wrongly, then the equivalent emf and internal resistance of the combination is

A group of six identical cells each of emf E and internal resistance r are joined in series to form a loop. The terminal voltage across each cells is