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

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(D) Since the resistance of a wire is directly proportional to its length $(R \propto L)$, when a wire of resistance $12 \, \Omega$ is bent into an eq

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$8/3 \, \Omega$

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(D) Since the resistance of a wire is directly proportional to its length $(R \propto L)$, when a wire of resistance $12 \, \Omega$ is bent into an equilateral triangle, the resistance of each of the three sides becomes $R_{side} = 12/3 = 4 \, \Omega$.When we calculate the effective resistance between any two corners, one side (of $4 \, \Omega$) is in parallel with the other two sides connected in series (which have a combined resistance of $4 \, \Omega + 4 \, \Omega = 8 \, \Omega$).Therefore, the effective resistance $R_{eq}$ is given by the parallel combination formula:$R_{eq} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{4 	imes 8}{4 + 8} = \frac{32}{12} = \frac{8}{3} \, \Omega$.

$A$ 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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