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Current Electricity question

2021 · 17 Mar · Shift 1 · Q68
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Current Electricity question

2021 · 17 Mar · Shift 1 · Q68

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
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 ‾\underline{\hspace{2cm}}​. (Round off to the Nearest Integer)
Numerical answer
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Correct answer: 4

  1. Let the two resistances be R1R_1R1​ and R2R_2R2​.

  2. Their equivalent resistance in series is s=R1+R2.s = R_1 + R_2.s=R1​+R2​.

  3. Their equivalent resistance in parallel is p=R1R2R1+R2.p = \frac{R_1R_2}{R_1+R_2}.p=R1​+R2​R1​R2​​.

  4. Given that s=np,s = np,s=np, so R1+R2=n(R1R2R1+R2).R_1 + R_2 = n\left(\frac{R_1R_2}{R_1+R_2}\right).R1​+R2​=n(R1​+R2​R1​R2​​).

  5. Rearranging, n=(R1+R2)2R1R2.n = \frac{(R_1+R_2)^2}{R_1R_2}.n=R1​R2​(R1​+R2​)2​.

  6. Expand:

= \frac{R_1}{R_2} + 2 + \frac{R_2}{R_1}.$$ 7. Now use the standard inequality $$\frac{R_1}{R_2} + \frac{R_2}{R_1} \ge 2,$$ with equality when $R_1 = R_2$. Therefore, $$n \ge 2 + 2 = 4.$$ 8. Hence the minimum value of $n$ is $$\boxed{4}.$$ Since this is already an integer, rounding to the nearest integer still gives $4$.
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