Question

The generated e.m.f from 2-pole armature having 2 conductors driven at 3000 rpm having flux per pole as 4000 mWb, with 91 parallel paths is ___________

(a) 8.64 V

(b) 8.56 V

(c) 8.12 V

(d) 8.79 V

I had been asked this question in a job interview.

I would like to ask this question from Energy Relations During Starting topic in chapter Starting of Electric Drives

Answer 1

The correct answer is (d) 8.79 V

To elaborate: The generated e.m.f can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. Eb = 4×2×3000×2÷60×91 = 8.79 V.