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

When all resistances are connected in parallel, the resultant resistance `R_(p)` is given by
`(1)/(R_(p))=(1)/(R_(1))+. . . . .+(1)/(R_(n))`
On multiplying both sides by `R_("min")` we have
`(R_("min"))/(R_(p))=(R_("min"))/(R_(1))+(R_("min"))/(R_(@))+. . . .+(R_("min"))/(R_(n))`
Here, in RHSlt there exist one term `(R_("min"))/(R_("min"))=1` and other terms are positive, so we have
`(R_("min"))/(R_(p))=(R_("min"))/(R_(1))+(R_("min"))/(R_(2))+ . . . .+(R_("min"))/(R_(n)) gt 1`
This shows that the resultant resistance `R_(p) lt R_("min")`
thus, in parallel combination, the equivalent resistance of resistors is less than the minimum resistance available in combination of resistors. Now, in series combination, the equivalent resistance is given by
`R_(S)=R_(1)+ . . . +R_(n)`
Here, in RHS, there exist one term having resistance `R_(max)`
So, we have
or `R_(S)=R_(1)+ . . .+R_(max)+ . . .+. . .+R_(n)`
`R_(S)=R_(1)+. . .+R_(max). . .+R_(n)=R_(max)+ . . .(R_(1)+. . .+)R_(n)`
or `R_(S)geR_(max)`
or `R_(S)=R_(max)(R_(1)+. . . +R_(n))`
Thus, in series combination, the equivalent resistance of resistors of resistors is greater than the maximum resistance available in combination resistors. Physical interpretation.
In Fig (b), `R_("min")` provides an equivalent route as in Fig. (a) for current. But in addition there are (n-1) routes by the remaining (n-1) resistors. Current in fig. (b) is greater than current in Fig. (a). Effective resistance in Fig (b) `lt R_("min")` Second circuit evidently affords a greater resistance.
In figure. (d), `R_(max)` provides an equivalent route as in fig. (c) for current. Current in figure. `(d) lt ` current in figure (c) Effective resistance in fig. (b) `lt R_(max)` second circuit evidently affords a greater resistance.

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