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

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Resistance, R = A ρl ​ R α l s a y R e ff ​ = 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 an d R = 6 a l s o w e kn o w R A B ​ = R 2 R ​ + R R ​ R 2 R ​ R R ​ ​ R A B ​ = 6 + 12 6 × 12 ​ = 4Ω

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