JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Current Electricity

Bending the $18\Omega$ wire into an equilateral triangle splits it into three equal $6\Omega$ sides. Between two vertices the current has two routes in parallel: the single side at $6\Omega$ and the long way through the other two sides in series at $12\Omega$. In parallel these give $\frac{6\times 12}{6+12}=\frac{72}{18}=4\Omega$, and by symmetry every pair of vertices gives the same $4\Omega$.