Question

A uniform wire of resistance $9\Omega$ is cut into 3 equal parts. They are connected in the form of equilateral triangle ABC. A cell of emf 2V and negligible internal resistance is connected across B and C. Potential difference across AB is

• A. 1V
• B. 2V
• C. 3V
• D. 0.5V

Hint:

Use current division rule to find current through each branch.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Each part has resistance $3\Omega$. The triangle has resistances $3\Omega$ on each side.
Step 2: Detailed Explanation:
When cell is connected across B and C, the resistances AB and AC are in series (total $6\Omega$) and this combination is in parallel with BC ($3\Omega$). Equivalent resistance = $\frac{6 \times 3}{6 + 3} = 2\Omega$. Current from cell = $\frac{2}{2} = 1$ A. Current through AB = current through AB-AC branch = $\frac{1}{3}$ A (since current divides in ratio of resistances). Potential difference across AB = $\frac{1}{3} \times 3 = 1$ V.
Step 3: Final Answer:
The potential difference across AB is 1V.