A wire of resistance $$9\,\Omega$$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be $$\underline{\hspace{2cm}}\,\Omega.$$

Correct Answer: 2

Wire of 9Ω bent into equilateral triangle. Equivalent resistance across two vertices.

Each side = 3Ω. Across two vertices: one 3Ω in parallel with two 3Ω in series (6Ω).

$$R_{eq} = \frac{3 \times 6}{3+6} = 2$$ Ω

The answer is 2.

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