You are given `n` resistors each of resistance `r` .These are first connected to get minimum resistance .In the secoind case these are again connected differently to get maximum possible resistance. Compute the ratio between the minimum and maximum value of resistance so obtained.

Answer

`(1)/(n^(2))`

Explanation

To get minimum posibale resistance, the resistors are to be connected in parallel .Then the connected minimum resistance is given by
`(1)/(R_(min)) = (1)/(r ) + (1)/(r ) + … n "times" = (n)/(r ) or R_(min) = r//n`
For maximum position resistance, the resistors are to be conneted in series. Then the combined maximum resistance is given by
`R_(min) = r+ r + .... "times" =nr`
`:.(R_(min))/(R_(min)) = (r//n)/(nr) = (1)/(n^(2))`