JEE Main 2023 · 1 February · evening · Single correct · +4 / −1
Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be :
- (a) ((n-1) R)/n²
- (b) (n² R)/(n-1)
- (c) ((n-1) R)/(2 n-1)
- (d) ((n-1) R)/n
Answer and solution
Answer (a) ((n-1) R)/n²
Solution
Physics sir · 4 steps
Two adjacent corners split the wire into one side and the other n − 1 sides, joined in parallel.
- Resistance per side. The whole wire has resistance R and is bent into n equal sides. So each side has resistance R/n.
- Two paths. Between two adjacent corners, current can go directly along one side, R/n, or the long way round through the other n-1 sides, (n-1)R/n. Both paths join the same two points, so they are in parallel.
R_(eq) = (R/n × (n-1)R/n)/(R/n + (n-1)R/n) = ((n-1)R²/n²)/R = (n-1)R/n²
- Quick check. For a triangle, n = 3: each side is R/3, and R/3 in parallel with 2R/3 gives 2R/9. The formula gives (3-1)R/9 = 2R/9 as well.
- So. The equivalent resistance is (n-1)R/n², which is option (a).
The idea to keep. For any closed loop of uniform wire, the two points cut it into two arcs. Share the total resistance by length, then put the two arcs in parallel.
Answer from the NTA final answer key. Solution by PrepWiser.