A wire of resistance 9 ohm is bent in a form of equilateral triangle. the equivalent resistance between any two of its vertices will be
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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