Question 4.5: A slip-ring-type three-phase induction motor rotates at a sp...

We know the synchronous speed of the rotating magnetic field produced is expressed as

Here, f = 50 Hz but number of poles of the stator winding has not been mentioned. The number of poles can be 2, 4, 6, 8, etc.

for               p=2 ,  N_{s} =\frac{120×50}{2}= 3000 rpm

for               p=4 ,   N_{s} =\frac{120×50}{4}= 1500  rpm

for               p=6 ,  N_{s} =\frac{120×50}{6}= 1000  rpm

for              p=8 ,   N_{s} =\frac{120×50}{8}=750  rpm

We know that an induction motor runs at a speed slightly less than the synchronous speed. Here, N_{r} = 1440  rpm (given). Synchronous speed corresponding to this rotor speed must, therefore, be 1500 rpm Thus, N_{r} = 1440  rpm and N_{s} = 1500  rpm.

Slip,                S=\frac{N_{s} – N_{r} }{N_{s}}=\frac{1500-1400}{1500}=0.04

Rotor frequency,           f_{r} = S ×  f=0.04×50 =2 Hz