A wire of resistance 20 Ω is divided into 10 equal parts, resulting pairs. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _________ Ω.

Solution

Physics sir · 3 steps

Cut first, then pair: each piece is 2 Ω, each parallel pair is 1 Ω, and five pairs in series add up.

1. Resistance of a piece. Resistance is proportional to length, R = ρ L / A. Cutting the wire into 10 equal parts gives each part 20/10 = 2 Ω.
2. One parallel pair. Two equal resistors in parallel give half of one: (2 × 2)/(2 + 2) = 1 Ω. The 10 parts make 5 such pairs.

   R_(eq) = 5 × 1 Ω = 5 Ω

3. So. The 5 pairs of 1 Ω in series give an equivalent resistance of 5 Ω.

The common slip: 50. Using 20 Ω for each part instead of 2 Ω makes each pair 10 Ω and the total 50, not 5.

The idea to keep. When a wire is cut into n equal parts, each part has R/n. Find the new piece resistance first, then combine the pieces step by step.

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