QJEEM25Ph5a7e: JEE Main 2025 Physics - Class 12 JEEM25Ph5a7e Solution: A wire of resistance is bent to form an equilateral triangle.…

$$\text{Each side of the triangle has a resistance of }{R}_{\text{side}}=\frac{9\mathrm{\Omega }}{3}=3\mathrm{\Omega }.$$

When measuring the equivalent resistance between any two vertices, there are two paths:

A direct path along one side with resistance:

$$R_1=3\mathrm{\Omega }.$$

An indirect path passing through the other two sides in series:

$$R_2=3\mathrm{\Omega }+3\mathrm{\Omega }=6\mathrm{\Omega }.$$

These two paths are in parallel, so the equivalent resistance is calculated by:

$$\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{3\mathrm{\Omega }}+\frac{1}{6\mathrm{\Omega }}=\frac{2}{6\mathrm{\Omega }}+\frac{1}{6\mathrm{\Omega }}=\frac{3}{6\mathrm{\Omega }}.$$

Thus,

$$R_{\text{eq}}=\frac{6\mathrm{\Omega }}{3}=2\mathrm{\Omega }.$$

The equivalent resistance across any two vertices of the equilateral triangle is therefore:

$$2\mathrm{\Omega }.$$