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

Question

Aicharya · Accepted Answer

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

$R_{\text {eq }}=\frac{\left(\frac{2 R}{3}\right)\left(\frac{R}{3}\right)}{\frac{2 R}{3}+\frac{R}{3}}=\frac{2 R}{9}=R_1$ (lets say)
For the wire bent into a square, each side has a resistance $\frac{R}{4}$.

$\begin{aligned} & R_{\text {eq }}=\frac{\left(\frac{3 R}{4}\right)\left(\frac{R}{4}\right)}{\frac{3 R}{4}+\frac{R}{4}}=\frac{3 R}{16}=R_3 \text { (lets say) } \\ & \Rightarrow \frac{R_1}{R_3}=\frac{\frac{2 R}{9}}{\frac{3 R}{16}}=\frac{32}{27}\end{aligned}$

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into $a$ square.…

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