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Current Electricity question

2004 · Shift 0 · Q131
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Current Electricity question

2004 · Shift 0 · Q131

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
The resistance of the series combination of two resistances is S.S.S. When they are jointed in parallel the total resistance is P.P.P. If S=nPS = nPS=nP then the Minimum possible value of nnn is
  1. A
    222
  2. B
    333
  3. C
    444
  4. D
    111
View written solutionFree

Correct answer: C

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

  2. Their series combination is S=R1+R2.S = R_1 + R_2.S=R1​+R2​.

  3. Their parallel combination 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. Substitute the expressions for SSS and PPP: 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. Multiply both sides by (R1+R2)(R_1+R_2)(R1​+R2​): (R1+R2)2=nR1R2.(R_1+R_2)^2 = nR_1R_2.(R1​+R2​)2=nR1​R2​. Hence, 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 R1R2+R2R1≥2,\frac{R_1}{R_2} + \frac{R_2}{R_1} \ge 2,R2​R1​​+R1​R2​​≥2, with equality when R1=R2R_1 = R_2R1​=R2​.

    Therefore, n≥2+2=4.n \ge 2 + 2 = 4.n≥2+2=4.

  8. So the minimum possible value of nnn is 4.\boxed{4}.4​.

  9. Checking options:

    • A: 222 ❌
    • B: 333 ❌
    • C: 444 ✅
    • D: 111 ❌

Therefore, the correct option is C.

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