SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

Question : 124 of 200

Marks: +1, -0

A 220-V, 50-Hz, 6-pole, single-phase induction motor runs with 3% slip. Determine the rotor speed.

Solution:

f = 50 Hz
P = 6
s = 3% = 0.03

∴\N _{ \s }={120 \f }/{ \P }={120 × 50}/{6}=1000 \rpm

from above concept,

0.03={1000- \N _{ \r }}/{1000}

or,

1000 × 0.03=1000- \N _{ \r }

\N _{ \r }=970 \rpm

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