JEE Main 2025 Jan 24 Shift 1, Physics Q23: Introduction And Ohm’s Law

The original 9 $\Omega$ wire is split equally among the three sides of the triangle, so each side has resistance 3 $\Omega$.

Looking between two vertices (say $B$ and $C$), one side $BC$ of resistance 3 $\Omega$ is in parallel with the series combination of the other two sides ($BA$ and $AC$), which totals $3+3=6\Omega$.

$R_{eq}=\frac{3\cdot 6}{3+6}=\frac{18}{9}=2\Omega$

$\therefore$ Answer: $2$.