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)

Solution

Physics sir · 4 steps

Write n as s divided by p, and you get a sum squared over a product. AM-GM finishes it.

1. Write s and p. For resistors R₁ and R₂, the series value is s = R₁ + R₂ and the parallel value is p = (R₁ R₂)/(R₁ + R₂).

   n = s/p = (R₁ + R₂)²/(R₁ R₂)

2. Use AM-GM. For positive numbers, (R₁ + R₂)/2 ≥ √(R₁ R₂). Squaring gives (R₁ + R₂)² ≥ 4R₁ R₂, so n ≥ 4.
3. Check equality. Equality holds when R₁ = R₂ = R. Then s = 2R and p = R/2, so n = 2R/(R/2) = 4.
4. So. The smallest possible value of n is 4, reached when the two resistors are equal.

The idea to keep. A ratio of series to parallel resistance is always (R₁+R₂)²/R₁R₂. AM-GM tells you it is smallest when the two resistors are equal.

Answer from the NTA final answer key. Solution by PrepWiser.