A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 2Ω each and leaves by the corner R. The currents i₁ in ampere is _________.

Solution

Physics sir · 4 steps

Current splits at P between two paths, and the shorter path gets the bigger share.

1. Ignore the feed resistor. The 2 Ω above P is in series with the whole triangle, so all 6 A passes through it. It does not change how the current divides at P.
2. Two paths from P. Current i₁ goes P to Q to R through 2 + 2 = 4 Ω. Current i₂ goes straight from P to R through 2 Ω. These two paths are in parallel between P and R.
3. Same voltage, Kirchhoff. Both paths have the same potential difference, so 4 i₁ = 2 i₂, giving i₂ = 2 i₁. At junction P, i₁ + i₂ = 6.

   i₁ = 6 × 2/(2 + 4) = 2 A

4. So. i₁ = 2 A and i₂ = 4 A, which together make 6 A. The answer is 2.

The common slip: 4. Giving the larger share to the longer path swaps the ratio and gives 4 A, which is actually i₂ along the direct 2 Ω side.

The idea to keep. In a parallel split, each branch carries current in proportion to the OTHER branch's resistance. A resistor in series with the whole network never changes the split.

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