n resistances, each of `r Omega`, when connected in parallel give an equivalent resistance of R `Omega`. If these resistances were connected in series, the combination would have a resistance in homs equal to

Options

A. `n^(2)R`
B. `(R )/(n^(2))`
C. `(R )/(n)`
D. nR

Answer

A

Explanation

Resistance in parallel combination, `R=(r )/(n) rArr r=Rn`
Resistance in series combination, `R.=nr=n^(2)R`