UPSEE Solved Model Paper 2

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).