A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is:

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Resistance, R = ρ l A R α l s a y R e f f = l = 18 a s l 1 = l 2 = l 3 = l / 3 ⇒ R 1 = R 2 = R 3 = R = 18 / 3 = 6 So from diagram 2 R = 12 a n d R = 6 a l s o w e k n o w R A B = R 2 R R R R 2 R + R R R A B = 6 × 12 6 + 12 = 4 Ω

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