Question

A 50 Hz, bar primary CT has a secondary with 800 turns. The secondary supplies 7 A current into a purely resistive burden of 2 Ω. The magnetizing ampere-turns are 300. The phase angle is?

(a) 3.1°

(b) 85.4°

(c) 94.6°

(d) 175.4°

This question was addressed to me in an online quiz.

I'd like to ask this question from Problems of Series Resonance Involving Quality Factor in division Resonance & Magnetically Coupled Circuit of Network Theory

Answer 1

Right choice is (a) 3.1°

Easy explanation: Secondary burden is purely resistive and the resistance of burden is equal to the resistance of the secondary winding; the resistance of secondary winding = 1Ω. The voltage induced in secondary × resistance of secondary winding = 7 × 2 = 14V. Secondary power factor is unity as the load is purely resistive. The loss component of no-load current is to be neglected i.e. Ie = 0. IM = 300 A.

Secondary winding current IS = 7 A

Reflected secondary winding current = n IS = 5600 A

∴ tan θ = \(\frac{I_M}{nI_S} \). So, θ = 3.1°.