JEE Physics 2025 Question with Solution

Step-by-step solution

Standard Method

Given: The total resistance of the wire is $9 \, \Omega$ and it is bent to form an equilateral triangle.

Find: The equivalent resistance across any two vertices.

Since the wire is bent into an equilateral triangle, each side has equal resistance.

$$R_{\text{side}} = \frac{9}{3} = 3 \, \Omega$$

Between any two vertices, one path has resistance $3 \, \Omega$ and the other path has two sides in series.

$$R_{\text{series}} = 3 + 3 = 6 \, \Omega$$

Now these two paths are in parallel.

$$\frac{1}{R_{\text{eq}}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$

Therefore,

$$R_{\text{eq}} = 2 \, \Omega$$

Thus, the equivalent resistance across any two vertices is $2 \, \Omega$.

Common mistakes

• Assuming the full $9 \, \Omega$ lies directly between the two chosen vertices is incorrect because the wire is uniformly distributed over three equal sides. First divide the total resistance equally, so each side is $3 \, \Omega$.

• Treating all three sides as if they are in series is wrong because current can flow through two different paths between the selected vertices. One path is $3 \, \Omega$, while the other is $6 \, \Omega$, and these two paths are in parallel.

• Using the parallel formula incorrectly after finding $3 \, \Omega$ and $6 \, \Omega$ gives a wrong result. For parallel combination, add reciprocals first, then invert to get the equivalent resistance.