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(D) The total resistance of the wire is $12 \, \Omega$. Since it is bent into an equilateral triangle,each side will have a resistance of $R = \frac{1

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$\frac{8}{3} \, \Omega$

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(D) The total resistance of the wire is $12 \, \Omega$. Since it is bent into an equilateral triangle,each side will have a resistance of $R = \frac{12 \, \Omega}{3} = 4 \, \Omega$.When we calculate the equivalent resistance between any two corners,one side of the triangle is in parallel with the other two sides which are in series.The resistance of the two sides in series is $R_s = 4 \, \Omega + 4 \, \Omega = 8 \, \Omega$.Now,this $8 \, \Omega$ resistance is in parallel with the third side of $4 \, \Omega$.The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{8} + \frac{1}{4} = \frac{1+2}{8} = \frac{3}{8}$.Therefore,$R_{eq} = \frac{8}{3} \, \Omega$.

$A$ wire of resistance $12 \, \Omega$ is bent to form an equilateral triangle. What is the equivalent resistance between any two corners of the triangle?

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