A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 2 Ω each and leaves by the corner R. The currents i₁ in ampere is . [Figure: triangle with P at the top, Q bottom left, R bottom right, each side 2 Ω. The 6 A arrives at P through a lead containing anot…

Question

A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 2 Ω each and leaves by the corner R. The currents i₁ in ampere is . [Figure: triangle with P at the top, Q bottom left, R bottom right, each side 2 Ω. The 6 A arrives at P through a lead containing another 2 Ω and leaves from R. i₁ is marked in side PQ and i₂ in side PR.]

Accepted Answer

2. From P to R there are two paths: PR directly (2 Ω) and PQR (4 Ω). The lead resistor above P carries the full 6 A and does not affect the split. i₁=6×2/(2+4)=2 A (and i₂=4 A).

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