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

Let some minimum value of resistance be `R_(" min")`

Let `R_("max") and R_("min")` be maximum and minimum resistance.
For parallel connection,
`(1)/(R_(p)) = (1)/(R_(1)) + (1)/(R_(2)) + .... + (1)/(R_("min")) + ... + (1)/(R_(n))`
By multiplying by `R_("min") ` on both side,
`(R_("min"))/(R_(P)) = (R_("min"))/(R_(1)) + (R_("min"))/(R_(2)) + ..... + (R_("min"))/(R_(n))`
`(R_("min"))/(R_(P)) gt ` 1 (Right side value is more than 1)
`therefore (R_(p))/(R_("min")) lt 1 `
`therefore R_(p) lt R_("min")`

Series connection :
From `R_(1), R_(2), ...., R_(n) R_("max") ` be maximum value,
For series connection,
`R_(S) = R_(1) + R_(2) + .... + R_("max") + .... + R_(n)`
`R_(s) = R_("max") + (R_(1) + .... + R_(n))`
`R_(s) gt R_("max")`
( Value of summation of resistance other than `R_("max") ` is positive.

For parallel connection :
From figure (a) `R_("min")` resistance is minimum resistance for loop given in figure (b). But in figure (b) there are (n- 1) loop so additional resistor (n - 1) . Hence. current obtained in figure (b) is larger that current obtained in figure (a).
`therefore ` Equivalent resistance of figure (b) will be less than `R_("min")`
`therefore R_(p) lt R_("min")`
For series connection :
In figure (c) provide maximum resistance `R_("max")` Thus, current in figure (d) will be less than current in figure (c).
thus, equivalent resistance of figure (d) `lt R_("max") therefore R_(S) gt R_("max")`