Answer: Option B
:
B
Let m cells be connected in series and n such groups are connected in parallel
If the emf of each cell is E and internal resistance r then the total emf of m in series is mE and the total internal resistance is mr. When n such groups are in parallel the effective internal resistance is mr/n. Then the current through an external resistance R is
$l=\frac{mE}{R+\frac{mr}{n}}=\frac{mnE}{nR+mr}=\frac{mnE}{(\sqrt{nR}-{\sqrt{mr)}}^2+2\sqrt{mnRr}}$
Now i will be maximum if the denominator is the minimum i.e. if nR=mr
Using R= 3ohm and r=1ohm we have 3n = m
But mn = 48
therefore $\frac{m\times m}{3}=48$
which gives m= 12
thus n=4