In the given circuit 'a' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be √(x/2). The value of x is __________.

Solution

Physics sir · 4 steps

One term grows with m and the other shrinks with m, so set dR/dm = 0 to find the sweet spot.

1. Left group. Three equal resistors ma in parallel give ma/3.
2. Right group. Two equal resistors a/m in parallel give a/2m. The two groups are in series, so their resistances add.

   R(m) = ma/3 + a/2m

3. Minimise. dR/dm = a/3 - a/2m² = 0, so m² = 3/2 and m = √(3/2). This is a minimum because d²R/dm² = a/m³ > 0.
4. So. Matching √(3/2) with √(x/2) gives x = 3.

The common slip: 1.5. Reporting m² = 3/2 as x gives 1.5, but m = √(x/2) means x = 2m² = 3.

The idea to keep. When R has the form Am + B/m, the minimum comes at m = √(B/A). Reduce each parallel group of equal resistors first, then add the groups in series.

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