n resistors each of resistance R first combine to give maximum effective resistgance and then combine to give minimum. The ratio of the maximum resistance is

Options

A. n
B. `n^(2)`
C. `n^(2)-1`
D. `n^(3)`

Answer

B

Explanation

To get maximum equivalent resistance all resistances must be connected in series
`therefore (R_("eq"))_("max")=R+R+R...n"times"=nR`
To get minimum equivalent resistance all resistances myst be connected in parallel.
`therefore (1)/(R_("eq"))_("min")=(1)/(R)+(1)/(R)+....n"time", (1)/(R_("eq"))_("min")=(n)/(R)`
`(R_("eq"))_("min")=(R)/(n)therefore (R_("eq"))_("max")/(R_("eq"))_("min")=(nR)/(R//n)=n^(2)`