Let there be n resistors `R_(1) . . . R_(n)` with `R_(max)=max(R_(1) . . . . R_(n))` and `R_("min")=min{R_(1) . . . R_(n)}`. Show that when they are connected in parallel the resultant resistance `R_(p)=R_("min")` and when they are connected in series, the resultant resistance `R_(S) gt R_(max)`. Interpret the result physically.

Explanation

The net resistance of n resistors `(R_(1), R_(2)….R_(n))` in series is given by : `R_(a) = R_(1) + R_(2) + R_(3) + ….+ R_(n)`
`rArr R_(S) gt R_("max")` = max `(R_(1) ….R_(n))`, which means that in a series combination, the net resistance will be greater than the value of maximum resistance in the combination. The net resistance of n resistors `(R_(1), R_(2) .....R_(n))` in parallel is given by:
`(1)/(R_(P)) = (1)/(R_(1)) + (1)/(R_(2)) + ...+ (1)/(R_(n))`
`R_("min") = "min" (R_(1), R_(2) ...R_(n))`
For the least value of resistance in the combination, its reciprocal will have the maximum value, which means that reciprocal of sum of all resistances i.e., `(1)/(R_(P))` will be greater than that value. This means that `R_(P)` will be minimum. `rArr R_(P) lt R_("min") = "min" (R_(1), R_(2) ....R_(n))`. Which means that in a parallel combination, the net resistance will be smaller than the value of minimum resistance in the combination.