To solve this problem, we first need to understand the configuration of the circuit.
Given: A uniform wire of resistance \(9\Omega\) is cut into 3 equal parts, meaning each part will have a resistance of \(3\Omega\).
These three wires are connected to form an equilateral triangle ABC, so the resistance of each side (AB, BC, CA) is \(3\Omega\).
An emf source of 2V is connected across B and C. We are asked to find the potential difference across AB.
Since the resistance between B and C is \(3\Omega\) and the same for other sides, let's analyze the current flow.
1. The resistance between points B and C is \(3\Omega\).
2. Using Ohm's Law (\(V = IR\)), the current \(I\) flowing through BC is calculated as:
\[I = \frac{V}{R} = \frac{2V}{3\Omega} = \frac{2}{3} A\]3. Since this is an equilateral triangle circuit, the symmetry implies the same current will split equally at points B and C to pass through sides AB and CA. Thus, the current through AB is
\[\frac{1}{3} A\].
4. Now, apply Ohm's Law to find the potential difference across AB:
\[V_{AB} = I \cdot R = \frac{1}{3} \times 3 = 1V\]Therefore, the potential difference across AB is 1V.
Hence, the correct answer is 1.