SSC JE Electrical Engineering 2017 Set 2 Paper

Question : 162 of 200

Marks: +1, -0

The full load slip of a 4 - pole, 50Hz , phase Iduction motor rotating at 1410 r.p.m. is ___:

Solution:

Slip =

{N_S-N_r}/N_S

N_S={120×f}/P={120×50}/4=1500 \rpm

Slip =

{1500- 1410}/1500=690

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