A wire of resistance $9 \Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be _________ ohm.

<p>$$ \text{Each side of the triangle has a resistance of } R_{\text{side}} = \frac{9\,\Omega}{3} = 3\,\Omega. $$</p> <p>When measuring the equivalent resistance between any two vertices, there are two paths:</p> <p><p>A direct path along one side with resistance:</p> <p>$R_1 = 3\,\Omega.$</p></p> <p><p>An indirect path passing through the other two sides in series:</p> <p>$R_2 = 3\,\Omega + 3\,\Omega = 6\,\Omega.$</p></p> <p>These two paths are in parallel, so the equivalent resistance is calculated by:</p> <p>$$ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3\,\Omega} + \frac{1}{6\,\Omega} = \frac{2}{6\,\Omega} + \frac{1}{6\,\Omega} = \frac{3}{6\,\Omega}. $$</p> <p>Thus,</p> <p>$R_{\text{eq}} = \frac{6\,\Omega}{3} = 2\,\Omega.$</p> <p>The equivalent resistance across any two vertices of the equilateral triangle is therefore:</p> <p>$2\,\Omega.$</p>