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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 = \f

Vedclass · Accepted Answer

$4$

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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 implies $n = \frac{(R_1 + R_2)^2}{R_1 R_2}$. Using the Arithmetic Mean-Geometric Mean inequality,we know that $(R_1 + R_2) \geq 2 \sqrt{R_1 R_2}$. Squaring both sides,$(R_1 + R_2)^2 \geq 4 R_1 R_2$. Dividing by $R_1 R_2$,we get $\frac{(R_1 + R_2)^2}{R_1 R_2} \geq 4$. Thus,$n \geq 4$. The minimum value for $n$ is $4$.

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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