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 maximum value for n is ______. (Round off to the nearest integer)

Answer: (4)

s = np

\(\begin{array}{l}R_{1}+R_{2}=n\left [ \frac{R_{1}R_{2}}{R_{1}+R_{2}} \right ]\end{array} \)

R12+ R22+ 2R1R2 = nR1R2

R12+(2–n)R1R2 +R12 = 0

(Considering this quadratic equation, for real roots)

b2 – 4ac≥0

[(2–n)R2]2 – 4×1×R22 = 0

(2–n)2R22 = 4R22

2-n = ±2

So n = 4

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