A uniform metallic wire of length L is mounte | vedclass

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(B) Let $a$ be the side length of the equilateral triangle,$r$ be the radius of the circle,and $x$ be the resistance per unit length of the wire. Then

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$9 / 8$

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(B) Let $a$ be the side length of the equilateral triangle,$r$ be the radius of the circle,and $x$ be the resistance per unit length of the wire. Then,$L = 3a = 2 \pi r$,which gives $a = L / 3$ and $r = L / (2 \pi)$.In configuration $1$,the equivalent resistance across $AB$ is the parallel combination of one side (resistance $ax$) and two sides (resistance $2ax$):$R_{AB} = \frac{(ax)(2ax)}{ax + 2ax} = \frac{2a^2 x^2}{3ax} = \frac{2}{3} ax = \frac{2}{3} \left( \frac{L}{3} \right) x = \frac{2Lx}{9}$.The power dissipated is $P_1 = \frac{V^2}{R_{AB}} = \frac{9V^2}{2Lx}$.In configuration $2$,the equivalent resistance across $PQ$ is the parallel combination of two semicircles,each of length $\pi r$ (resistance $\pi rx$):$R_{PQ} = \frac{(\pi rx)(\pi rx)}{\pi rx + \pi rx} = \frac{\pi rx}{2} = \frac{\pi (L / 2\pi) x}{2} = \frac{Lx}{4}$.The power dissipated is $P_2 = \frac{V^2}{R_{PQ}} = \frac{4V^2}{Lx}$.The ratio of power dissipated is $\frac{P_1}{P_2} = \frac{9V^2 / 2Lx}{4V^2 / Lx} = \frac{9}{8}$.

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