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

4 Answers

Answer :

4

Answer :

A

Answer :

d

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

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