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(A) Let the two resistances 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

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$4$

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(A) Let the two resistances 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 the condition $S = nP$,we substitute the expressions:$R_1 + R_2 = n \left( \frac{R_1 R_2}{R_1 + R_2} \right)$.Rearranging the terms,we get $(R_1 + R_2)^2 = n R_1 R_2$.Using the algebraic identity $(R_1 + R_2)^2 = (R_1 - R_2)^2 + 4 R_1 R_2$,we have:$(R_1 - R_2)^2 + 4 R_1 R_2 = n R_1 R_2$.Dividing by $R_1 R_2$,we get $n = 4 + \frac{(R_1 - R_2)^2}{R_1 R_2}$.Since the term $\frac{(R_1 - R_2)^2}{R_1 R_2}$ is always greater than or equal to $0$,the minimum value of $n$ occurs when $R_1 = R_2$,which gives $n = 4$.

The resistance of the series combination of two resistances is $S$. When they are joined in parallel,the total resistance is $P$. If $S = nP$,then the minimum possible value of $n$ is

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