1.

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°I had been asked this question in homework.This key question is from Advanced Problems on C.T. and P.T. in portion Extension of Instrument Ranges of Electrical Measurements

Answer»

Correct choice is (a) 3.1°

The EXPLANATION is: 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°.



Discussion

No Comment Found

Related InterviewSolutions