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(A) Let the two resistors be $R_1$ and $R_2$.In series,the equivalent resistance is $S = R_1 + R_2$.In parallel,the equivalent resistance is $P = \fra

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$4$

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(A) Let the two resistors be $R_1$ and $R_2$.In series,the equivalent resistance is $S = R_1 + R_2$.In parallel,the equivalent resistance is $P = \frac{R_1 R_2}{R_1 + R_2} = \frac{R_1 R_2}{S}$.Given $S = nP$,we substitute $P$: $S = n \left( \frac{R_1 R_2}{S} \right) \Rightarrow n = \frac{S^2}{R_1 R_2} = \frac{(R_1 + R_2)^2}{R_1 R_2}$.Expanding this,$n = \frac{R_1^2 + R_2^2 + 2R_1 R_2}{R_1 R_2} = \frac{R_1}{R_2} + \frac{R_2}{R_1} + 2$.Using the Arithmetic Mean-Geometric Mean inequality ($AM$ $\ge$ $GM$),for any positive $x$,$x + \frac{1}{x} \ge 2$.Here,let $x = \frac{R_1}{R_2}$. Then $n = x + \frac{1}{x} + 2$.The minimum value of $x + \frac{1}{x}$ is $2$ (when $x = 1$,i.e.,$R_1 = R_2$).Therefore,the minimum value of $n = 2 + 2 = 4$.

The equivalent resistance of two resistors in series is $S$. When they are connected in parallel,the equivalent resistance is $P$. If $S = nP$,then the minimum possible value of $n$ is:

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