Let there be $n$ resistors $R_1, \dots, R_n$ with $R_{\max} = \max \{R_1, \dots, R_n\}$ and $R_{\min} = \min \{R_1, \dots, R_n\}$. 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 $R_s > R_{\max}$. Interpret the result physically.

(N/A) $1$. Parallel Connection:
The equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$.
Since each $R_i \ge R_{\min}$,it follows that $\frac{1}{R_i} \le \frac{1}{R_{\min}}$.
Thus,$\frac{1}{R_p} = \sum_{i=1}^n \frac{1}{R_i} > \frac{1}{R_{\min}}$ (as there are $n$ terms).
Therefore,$R_p < R_{\min}$.
Physical Interpretation: In parallel,the current has multiple paths to flow,effectively reducing the total opposition to current flow below that of the smallest individual resistor.
$2$. Series Connection:
The equivalent resistance $R_s$ is given by $R_s = R_1 + R_2 + \dots + R_n$.
Since each $R_i > 0$,and $R_{\max}$ is one of the terms,$R_s = R_{\max} + \sum_{i \neq \max} R_i$.
Since $\sum_{i \neq \max} R_i > 0$,it follows that $R_s > R_{\max}$.
Physical Interpretation: In series,the current must pass through every resistor sequentially,so the total resistance is the sum of all individual resistances,which is necessarily greater than any single resistor.