A square-shaped wire with a resistance of 3, | vedclass

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(C) The total resistance of the square wire is $R_{total} = 3\, \Omega + 3\, \Omega + 3\, \Omega + 3\, \Omega = 12\, \Omega$.When this wire is bent in

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$3$

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(C) The total resistance of the square wire is $R_{total} = 3\, \Omega + 3\, \Omega + 3\, \Omega + 3\, \Omega = 12\, \Omega$.When this wire is bent into a circle,the total circumference corresponds to a resistance of $12\, \Omega$.For two diametrically opposite points,the circle is divided into two equal semicircular arcs.The resistance of each semicircular arc is $R' = \frac{12\, \Omega}{2} = 6\, \Omega$.These two $6\, \Omega$ resistances are connected in parallel between the diametrically opposite points.The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.Therefore,$R_{eq} = 3\, \Omega$.

$A$ square-shaped wire with a resistance of $3\, \Omega$ for each side is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in units of $\Omega$ will be:

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