A 50 Hz bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1 Ω. The phase angle between the primary and secondary current is
- A. 4.6°
- B. 85.4°
- C. 94.6°
- D. 175.4°
Answer: A. 4.6°
Explanation: Secondary AT = 500×5 = 2500; magnetising AT (200, in phase quadrature with the resistive-burden secondary current) gives δ = tan⁻¹(200/2500) = 4.6°.