Explanation

(a) If the cells are connected as shown in figure, they are said to be connected in mixed grouping. Let there be $n\ $cells in series in one row and $m\ $rows of cells in parallel. Suppose all the cells are identical. Let each cell be of emf E and internal resistance $r$.
In each row, there are n cells in series, therefore their total internal resistance=$nr$.
Their total $emf = nE$.
Since, there are $m$ rows of cells in parallel therefore, total internal resistance $\left( r_{p} \right)$ of all the cells is given by
$\frac{1}{r_{p}} = \frac{1}{nr} + \frac{1}{nr} + \ldots upto\ m\ terms$
$\ \ \ \ \ = \frac{m}{nr}$
$or\ \ \ r_{p} = \frac{nr}{m}$
Total resistance in their circuit $= R + nr/m$
Effective emf of all the cells$= \ nE$
The current in the external resistance is given by
$I = \frac{nE}{R + nr/m} = \frac{mnE}{mR + nr}$
The current I will be maximum, if $mR + nr$ is minimum.
Mathematically, it can be shown that $mR + nr$ is minimum, if
$\ \ \ \ mR = nr$
$or\ \ \ \ R = \frac{nr}{m}$
Here, $mn = 45$ (given )….(i)
and $m \times 2.5 = n \times 0.5$
Or $n = 5m\ \ $ ………..(ii)
From Eqs. (i) and (ii), we have
$\ \ \ 5m^{2} = 45\ $
$or\ \ \ \ m^{2} = 9\ $
$or\ \ \ \ \ \ m = 3$
∴ n=15