A wire of resistance \( 9 \, \Omega \) is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be:
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Solution and Explanation
The total resistance of the wire is \( 9 \, \Omega \), and it is bent to form an equilateral triangle. The resistance of each side of the triangle is: \[ R_{{side}} = \frac{9}{3} = 3 \, \Omega \] When two vertices are connected, the equivalent resistance across the two vertices is found by considering the resistances in parallel. 
One side between the vertices is \( 3 \, \Omega \), and the other two sides are in series: \[ R_{{series}} = 3 + 3 = 6 \, \Omega \] Now, the two resistances \( 3 \, \Omega \) and \( 6 \, \Omega \) are in parallel. The equivalent resistance \( R_{{eq}} \) is: \[ \frac{1}{R_{{eq}}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} \] \[ R_{{eq}} = \frac{6}{3} = 2 \, \Omega \] Thus, the equivalent resistance across any two vertices is \( \boxed{2 \, \Omega} \).
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