SSC JE Electrical Engineering 2010 Paper

Question : 41 of 50

Marks: +1, -0

A 4-pole, 3-phase induction motor is running at 4% slip at full load. If the speed of the motor is 750 rpm, the supply frequency is

Solution:

No. of revolutions (

N_r

) = 750 rpm.
Slip at full load (s) = 0.04
No. of poles (P) = 4

N_S={120f}/P

...(i)

∵N_r=N_s(1-s)

∴N_s=N_r/{1-s}

=750/{1-0.04}

= 781.25 rpm
Putting the value of

N_s

in equation (i), we get

781.25={120f}/4

∴f={781.25×4}/120

= 26.04 Hz = 25 Hz

Go to Question:

Prev Question

Next Question