`N` sources of current with different emf's are connected as shown in Fig. The emf of the sources are proportional to their internal resistancs, i.e., `E = alpha R`, where `alpha` is an assigned constatant. The lead wire resistance is neglible. Find:
(a) the current in the circuit ,
(b) the potential differences between points `A` and `B` dividing the circuit in `n` and `N- n` links.

Explanation

(a) Current `i = ( N epsilon)/( NR) = ( N alphaR )/( NR) = alpha`, as `epsilon= alpha R ` ( given )
(b) `phi_(A) -hi_(B) = n epsilon- ni R = n alpha R - n alpha R = 0 `