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(C) 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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(C) 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 $p = \frac{R_1 R_2}{R_1 + R_2}$.Given $s = np$,we have $R_1 + R_2 = n \left( \frac{R_1 R_2}{R_1 + R_2} \right)$.This simplifies to $n = \frac{(R_1 + R_2)^2}{R_1 R_2}$.Using the Arithmetic Mean-Geometric Mean inequality ($AM$-$GM$),we know that $(R_1 + R_2)^2 \ge 4 R_1 R_2$.Therefore,$n = \frac{(R_1 + R_2)^2}{R_1 R_2} \ge 4$.The minimum value of $n$ occurs when $R_1 = R_2$,which gives $n = \frac{(2R)^2}{R^2} = 4$.

The equivalent resistance of a series combination of two resistors is $s$. When they are connected in parallel,the equivalent resistance is $p$. If $s = np$,then the minimum value for $n$ is (Round off to the nearest integer).

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