SSC JE Electrical Engineering 2014 Set 2 Paper

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Question : 185 of 200

Marks: +1, -0

A 3-phase 4 pole induction motor works on 3-phase 50 c/s supply. If the slip of the motor is 4%. The actual speed will be :

720 rpm
1550 rpm
1460 rpm
1440 rpm

Solution:

∵

N_s ={ 120\f} /P

=

{120×50}/4= 1500

r.p.m.
Now %

S ={ N_s-N_ r }/ N_s × 100

∴

4=({1500-N_r}/1500)×100

∴

60 = 1500 – N_r

∴

N_r = 1500 – 60 = 1440

r.p.m

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