JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Current Electricity

A wire of total resistance $9\Omega$ bent into an equilateral triangle has each side of resistance $\frac{9}{3}=3\Omega$.

Between any two vertices, say A and B, there are two parallel paths: the direct side AB ($3\Omega$) and the path through the third vertex, A→C→B ($3+3=6\Omega$).

These two paths are in parallel:
$R_{eq}=\frac{3\times 6}{3+6}=\frac{18}{9}=2\Omega$

The equivalent resistance is 2 ohm.