You are given 48 cells, each of emf 2 V and internal resistance 1 Ω. How will you connect them so that the current through an extemal resistance of 3 Ω is the maximum?
Let m cells be connected in series and n such groups be connected in parallel. If the ernf of each cell is E and internal resistance r, then the total ernf of m cells 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 I =
mE
R+
mr
n
=
mnE
nR+mr
=
mnE
(√nR−√mr)2+2√mnRr
Now, I will be maximum if the denominator is minimum, i.e. if nR = mr. Given R = 3Ω and r = 1Ω. Using these values, we have 3n = m. But mn = 48 (given). Therefore,
m×n
3
= 48, which gives m = 12. Thus, n = 4. Hence, the correct choice is (2).