The equivalent resistance of a series combination of two resistors is 's'. When they are connected in parallel, the equivalent resistance is 'p'.

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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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)