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(D) The total resistance of the wire is $R_{total} = 18 \Omega$. When the wire is bent into an equilateral triangle,the wire is divided into three equ

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$4$

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(D) The total resistance of the wire is $R_{total} = 18 \Omega$. When the wire is bent into an equilateral triangle,the wire is divided into three equal segments,each forming one side of the triangle. The resistance of each side is $R_{side} = \frac{18 \Omega}{3} = 6 \Omega$. When we calculate the effective resistance between any two vertices,one side of the triangle acts as a resistor in parallel with the other two sides connected in series. The resistance of the two sides in series is $R_s = 6 \Omega + 6 \Omega = 12 \Omega$. Now,this $12 \Omega$ resistor is in parallel with the third side of $6 \Omega$. The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6} = \frac{1+2}{12} = \frac{3}{12} = \frac{1}{4}$. Therefore,$R_{eq} = 4 \Omega$.

$A$ straight wire of resistance $18 \Omega$ is bent in the form of an equilateral triangular loop. The effective resistance between any two vertices of the triangle is (in $\Omega$)

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