A wire of resistance R is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points A and B is R / n. The value of n is :

Solution

Physics sir · 4 steps

Cut the wire into six equal edges, then use symmetry: the edge not touching A or B carries no current.

1. Each edge. A triangular pyramid has 6 edges, all of the same length. So each edge gets an equal share of the wire: r = R/6.
2. Use symmetry. Call the top vertex C and the centre vertex D. Each of them joins A through one edge r and B through one edge r, so C and D sit at the same potential. The edge CD therefore carries no current, and we can remove it.
3. Three parallel paths. What is left between A and B is the direct edge AB of resistance r, the path A–C–B of 2r, and the path A–D–B of 2r. All three are in parallel.

   1/R_(AB) = 1/r + 1/2r + 1/2r = 2/r ⇒ R_(AB) = r/2 = R/12

4. So. R_(AB) = R/12, so n = 12, which is option (c).

The idea to keep. In a symmetric network, look for two points at the same potential. A resistor joining them carries no current, so remove it and the network breaks into simple series and parallel paths.

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