You are given 48 cells each of emf 2 V and internal resistance 1 ohm. How will you connect them so that the current through an external resistance of 3 ohm is the maximum?
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=mER+mrn=mnEnR+mr=mnE(√nR−√mr)2+2√mnRr
Now i will be maximum if the denominator is the minimum i.e. if nR=mr
Using R= 3 ohm and r=1 ohm we have 3n = m
But mn = 48
therefore m×m3=48
which gives m= 12
thus n=4