A wire has a resistance of 12 , Omega. It is | vedclass

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(D) The resistance of a wire is directly proportional to its length $(R \propto L)$.Since the wire is bent into an equilateral triangle,the total resi

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$8/3 \, \Omega$

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(D) The resistance of a wire is directly proportional to its length $(R \propto L)$.Since the wire is bent into an equilateral triangle,the total resistance of $12 \, \Omega$ is divided into three equal parts.Resistance of each side of the triangle = $12 \, \Omega / 3 = 4 \, \Omega$.When we calculate the effective resistance between any two corners,two sides are in series with each other,and this combination is in parallel with the third side.Resistance of the two sides in series = $4 \, \Omega + 4 \, \Omega = 8 \, \Omega$.Now,this $8 \, \Omega$ resistance is in parallel with the third side of $4 \, \Omega$.The effective resistance $R_{eq} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{8 	imes 4}{8 + 4} = \frac{32}{12} = \frac{8}{3} \, \Omega$.

$A$ wire has a resistance of $12 \, \Omega$. It is bent in the form of an equilateral triangle. The effective resistance between any two corners of the triangle is:

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