MGVCL Exam Paper (30-07-2021 Shift 3)
A 10-pole induction motor is supplied by a 6-pole alternator, which is driven at 1400 rpm. If the motor runs with a slip of 5%, what is its speed?
Frequency of natural oscillation is given by, fn = {((dPe/dδ)at(δo))/M)} dPe/dδ = ((V1*V2)/X*(cosδ)) = (1.1/0.6)*cosδ = (1.1/0.6)*0.5 = 0.91 M = (H*s)/(πf) = 4/(50π)
EXAMIANS