24 identical cells, each of internal resistance `0.5 Omega`, are arranged in a parallel combination of n rows, each row containing m cells in series. The combination is connected across a resistor of `3 Omega`. In order to send maximum current through the resistor, we should have

Options

A. m=12, n=2
B. m=8, n=3
C. m=2, n=2
D. m=3, n=8

Answer

A

Explanation

N=24=mn
For current to be maximum
`R_(Internal)=R_(external)`
`rArr(mr)/(n)=3rArr(m)/(n)(0.5)=3rArr(m)/(n)=6`
m=6n, substituting the values, we get
`24=6n^(2)`
`rArrn=2rArrm=12`