Question

N identical cells, each of emf E and internal resistance r, are joined in series. Out of these n cells are incorrectly connected, i.e., these terminals are connected in reverse of that required for series connection (n < N/3). E₀ and r₀ represent the resulting emf and internal resistance of the combination.

Then E₀ and r₀ are given by:

(1) (N - 2n)E, r₀ = Nr

(2) \((\frac{N-n}{2})E,\,r_0 =(\frac{N-n}{2})r\)

(3) \((n-n)E, r_0 = (\frac{N}{n})r\)

(4) \((2N-n)E, r_0=(N-2n)r\)

Answer 1

(1) (N - 2n)E, r₀ = Nr

F₀ = (N– n) E – nE = (N–2n)E 

r_n = r + r + .......... n times = Nr