A series-parallel combination battery consists of 300 identical cells, each with an internal resistance `0.3 Omega`. It is connected to the external resistance `10 Omega`. Find the number of parallel goups cosisting of equal number of cells connected in series, at which the external resistance generated the higher thermal power.

Explanation

Let there be m rows of cells and n cells in each row of a battery.
Total number of cells,
`N = nm or n = N//m`
Let `epslion`,r be the mef and internal resistance of each cell and R be the external resistance. Then total internal resistance of all the m rows of cells ` = nr//m`.
Total resistance of the whole circuit `= (R + nr//m)`
Total emf of all the cells `=n epsilon`
Current in the external resistance R will be
`I = (n epsilon)/(R + nr//m) = ((N//m)epsilon)/([R + (N//m) r//m]) = (m N epsilon)/(m^(2)R + N r)`
Heat generated in resistance R is
`H = I^(2)R= ((m N epsilon)/(m^(2)r + Nr))^(2) R`
For H to be maximum, `(dH)/(dm) = 0`
Differentiating (ii), w.r.t. m and equation to zero, we get
`m =sqrt(Nr//R) = sqrt(300 xx 0.3//10) = 3`