To get maximum current through a resistance of `2.5Omega,` one can use m rows of cells, each row having n cells. The internal resistance of each cell is `0.5 Omega` what are the values of n and m, if the total number of cells is 45.

Options

A. 3,15
B. 5,6
C. 9,5
D. 15,3

Answer

D

Explanation

Since total number of cell is 45.
mm = 45 (i)
Now n cells are in series, then resistance in `nxx0.5Omega`
Now there are m rows with such series, then
`(1)/(R_(eq)) = (1)/(nxx0.5)+(1)/(nxx0.5)+….(m" times")`
`R_(eq) = (nxx0.5)/(m) = 2.5Omega [ :. I = (nE)/(R +r_(eq))]`
n = 5m (ii)
Using Eqs.(i) and (ii), m =3 and n = 15