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 VI 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

The generated e.m.f from 45-pole armature having 400 turns driven at 7

1.	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 VI 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» 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.

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