Question

The generated e.m.f from 45-pole armature having 400 turns driven at 70 rev/sec having flux per pole as 90 mWb, with 17 parallel paths is ___________

(a) 13341.17 V

(b) 12370.14 V

(c) 14700.89 V

(d) 15690.54 V

I had been asked this question by my college director while I was bunking the class.

This intriguing question comes from Modified Speed Torque Characteristics of DC Series Motors topic in division Characteristics of DC & AC Motors of Electric Drives

Answer 1

The correct choice is (a) 13341.17 V

The explanation is: 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. One turn is equal to two conductors. Eb = .09×45×400×2×4200÷60×17 = 13341.17 V.