Three 2 Ω resistors are connected in a delta (triangle) network. What is the equivalent resistance between any two terminals?

When three $2\,\Omega$ resistors are connected in a delta (triangle) configuration across terminals A, B, and C: looking across any two terminals (e.g., A and B), one $2\,\Omega$ resistor ($R_{AB}$) is connected in parallel with the series combination of the other two resistors ($R_{BC} + R_{CA} = 2 + 2 = 4\,\Omega$).

The equivalent resistance is: $R_{eq} = \frac{2 \times (2 + 2)}{2 + (2 + 2)} = \frac{2 \times 4}{6} = \frac{8}{6} = \frac{4}{3}\,\Omega \approx 1.333\,\Omega$.

Option Analysis: