Question

A wire of resistance \( 9 \, \Omega \) is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be:

Hint:

When calculating the equivalent resistance of a network of resistors, carefully identify series and parallel combinations. In this problem, the two resistors between the vertices are in parallel, with the other two sides forming a series combination.

Answer 1

The total resistance of the wire is \( 9 \, \Omega \). When this wire is bent into an equilateral triangle, the resistance of each side is calculated as: \[ R_{{side}} = \frac{9}{3} = 3 \, \Omega \] To find the equivalent resistance between two vertices, we consider a parallel circuit. One side, with \( 3 \, \Omega \), is in parallel with the other two sides, which are in series: \[ R_{{series}} = 3 + 3 = 6 \, \Omega \] The circuit then consists of a \( 3 \, \Omega \) resistance in parallel with a \( 6 \, \Omega \) resistance. The equivalent resistance \( R_{{eq}} \) is determined by: \[ \frac{1}{R_{{eq}}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} \] \[ R_{{eq}} = \frac{6}{3} = 2 \, \Omega \] Therefore, the equivalent resistance across any two vertices of the triangle is \( \boxed{2 \, \Omega} \).