SSC JE Electrical Engineering 2013 Paper

Question : 108 of 200

Marks: +1, -0

A 4-pole generator is running at 1200 rpm. The frequency and time period of the emf generated is its coils are respectively

Solution:

\N_δ={120×f}/p

f={\N_δ×p}/120={1200×4}/120=40\Hz

f=1/T⇒T=1/40=0.025\sec

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