A wire of resistance 9 Ω is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be _________ ohm.

Solution

Physics sir · 3 steps

Each side gets 3 ohm; between two vertices you see one side in parallel with the other two in series.

1. Split the wire. The wire is uniform, so its resistance divides equally among the three equal sides. Each side has 9/3 = 3 Ω.
2. Two paths. Between any two vertices, one path is the single side joining them, 3 Ω. The other path goes through the third vertex, two sides in series, 3 + 3 = 6 Ω.

   R_(eq) = (3 × 6)/(3 + 6) = 18/9 = 2 Ω

3. So. The two paths are in parallel, so the equivalent resistance across any two vertices is 2 ohm.

The common slip: 4.5. Treating the two paths as 9 Ω and 9 Ω in parallel, without first splitting the wire into sides, gives 4.5 instead of 2.

The idea to keep. When a wire is bent into a shape, first share its resistance by length. Then find the paths between the two terminals and combine them in parallel.

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