The generated e.m.f from 50-pole armature having 400 conductors driven at 20 rev/sec having flux per pole as 30 mWb, with lap winding is ___________(a) 230 V(b) 140 V(c) 240 V(d) 250 VThe question was posed to me in semester exam.This intriguing question originated from Dynamics in portion Dynamics of Electrical Drives of Electric Drives

The generated e.m.f from 50-pole armature having 400 conductors driven

1.	The generated e.m.f from 50-pole armature having 400 conductors driven at 20 rev/sec having flux per pole as 30 mWb, with lap winding is ___________(a) 230 V(b) 140 V(c) 240 V(d) 250 VThe question was posed to me in semester exam.This intriguing question originated from Dynamics in portion Dynamics of Electrical Drives of Electric Drives
Answer» Right CHOICE is (c) 240 V Best explanation: The generated 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. In lap winding number of parallel paths are EQUAL to the number of poles. Eb = .03×50×400×1200÷60×50= 240 V.

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