A series parallel combination battery consisting of a large number `N = 300` of identical cells, each will an inernal resistance `r = 0.3 Omega`, is loaded with an external resistances `R = 10 Omega`. Find the number `n` of parallel groups consisting of an equal number of cells connected in series, at which the external resistance generates the highest thermal power.

Explanation

Let, there are `x` number of cells, connected in series in each of the `n` parallel groups then, `n x = N` or `x = (N)/(n)` ....(1)
Now, for any one of the loop consisting of `x` celss and the resistor `R`, from loop rule
`i R + (i)/(n) xr - x xi = 0`
So, `i = (x xi)/(R + (xr)/(n)) = ((N)/(n) xi)/(R + (N r)/(n^(2)))`, using (1)
Heat generated in the resitor `R`,
`Q = i^(2) R = ((N n xi)/(n^(2) R + N R))^(2) R` ....(2)
and for `Q` to be maximum, `(dQ)/(dn) = 0`, which yields
`n = sqrt((Nr)/(R)) = 3`