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QJEEM21Pha5bb: JEE Main 2021 Physics - Class 12 JEEM21Pha5bb Solution: The equivalent resistance of series combination of two…

The equivalent resistance of 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)

Step-by-Step Verified Solution

s=nps = np

R1+R2=n[R1R2R1+R2]{R_1} + {R_2} = n\left[ {{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \right]

⇒\Rightarrow R12+R22+2R1R2=nR1R2R_1^2 + R_2^2 + 2{R_1}{R_2} = n{R_1}{R_2}

⇒\Rightarrow R12+(2−n)R1R2+R22=0R_1^2 + (2 - n){R_1}{R_2} + R_2^2 = 0

For real roots, b2 - 4ac ≥\ge 0

[(2−n)R2]2−4×1×R22{[(2 - n){R_2}]^2} - 4 \times 1 \times R_2^2 ≥\ge 0

⇒\Rightarrow (2−4)2R22≥4R22{(2 - 4)^2}R_2^2 \ge 4R_2^2

⇒\Rightarrow 2 −- n ≥\ge ±\pm 2

⇒\Rightarrow 2 −- n ≥\ge −- 2

⇒\Rightarrow n ≥\ge 4

So, minimum value for n = 4

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