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?
The ratio of full load current to short circuit current = 1/6 Xsc = j/(1/6) External reactance required = j*((1/6) - 0.06)) = j*0.106 pu Full load current = (40*1000)/(√3*15) = 1539.6 A Per unit reactance = j*0.106 = (I*Xr)/V j*0.15 = (1539.6*Xb)/((15/√3)*1000)) = 0.60 ohm
EXAMIANS