SSC JE Electrical Engineering 22 Jan 2018 Evening Paper

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Question : 195 of 200

Marks: +1, -0

The percentage full load slip in a synchronous motor is

0.01
0.005
1
Zero

Solution:

Percent slip

{N_S-N_R}/N_r×100

At full load

N_r = N_S

∴ % S = 0.

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