Question

Three impedances \( 6 + j(10/3)\, \Omega \) each are connected in delta. The per-phase impedance of equivalent star circuit will be

• A. \( 18 + j(10/9)\, \Omega \)
• B. \( 2 + j(10/3)\, \Omega \)
• C. \( 2 + j(10/9)\, \Omega \)
• D. \( 18 + j10\, \Omega \)

Hint:

Use \( Z_{star} = \frac{Z_{delta}}{3} \) for balanced delta-to-star conversions directly.

Answer 1

The Correct Option is C

Delta to star conversion formula: \[ Z_{star} = \frac{Z_{delta}}{3} \] Given: \[ Z_{delta} = 6 + j(10/3)\, \Omega \] Then, \[ Z_{star} = \frac{6 + j(10/3)}{3} = 2 + j(10/9)\, \Omega \]