A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into $a$ square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is
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TestHub · Accepted Answer

For the wire bent into an equilateral triangle, each side has a resistance $\frac{R}{3}$ .

$R_{eq} = \frac{( \frac{2 R}{3} ) ( \frac{R}{3} )}{\frac{2 R}{3} + \frac{R}{3}} = \frac{2 R}{9} = R_{1}$ (lets say)
For the wire bent into a square, each side has a resistance $\frac{R}{4}$ .

$R_{eq} = \frac{( \frac{3 R}{4} ) ( \frac{R}{4} )}{\frac{3 R}{4} + \frac{R}{4}} = \frac{3 R}{16} = R_{3} (lets say) \Rightarrow \frac{R _{1}}{R _{3}} = \frac{\frac{2 R}{9}}{\frac{3 R}{16}} = \frac{32}{27}$

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