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

To solve the problem, we need to analyze the behavior of resistors when they are connected in parallel and in series.

### Step 1: Resistors in Parallel
When resistors \( R_1, R_2, \ldots, R_n \) are connected in parallel, the formula for the equivalent resistance \( R_p \) is given by:

\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \]
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