Description : A 50 Hz, 3—phase induction motor has a full load speed of 1440 rpm. The number of poles of the motor is: (A)4 (8)6 (0)12 (D)8
Last Answer : Ans: A N= Ns (1-S) = NS –NS x S 1440 = Ns (1-S) Ns = 1440 / (1-S) Ns = (120 f/ p) = 120 x 50/p = 6000 p Ns will be closer to N i.e 1440 When P=2 ; Ns = 3000 rpm , not close to N When P=4 ; Ns = 1500 rpm , it is closer to N Therefore P =4 for N=1440