Question -
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?
Options:
A .  8 cells in series 6 such groups in parallel
B .  12 cells in series 4 such groups in parallel
C .  16 cells in series 3 such groups in parallel
D .  24 cells in series 2 such groups in parallel
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=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= 3ohm and r=1ohm we have 3n = m But mn = 48 therefore m×m3=48 which gives m= 12 thus n=4
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