Let ABC be the equilateral triangle formed. AB = BC = CA = L (say)

The original length of the wire = 3L, whose resistance is 9 Ω.

Since, resistance is directly proportional to the length, so each side of the triangle have equal resistance.

That is, R _{ AB } = R _{ AC } = R _{ CA } = 3 Ω

Let us find the resistance across A and B.

R _{ AC } and R _{ BC } are in series, so, they have a combined resistance of 3+3 = 6 Ω

This 6 Ω is in parallel with R _{ AB } . So the equivalent resistance across A and B is, R _{ eq } = (6×3)/(6+3) = 2 Ω.

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