SSC JE Electrical Engineering 2012 Paper

Question : 146 of 200

Marks: +1, -0

A pure sinusoidal current is being rectified. For the given maximum value of half wave rectified current is 50 A, then the rms value of full wave rectification will be

50/ π A
100/ π A
100 A
70.7 A

Validate

Solution:

(I_\rms)_\HWR=I_m/{√2π}=50A

(I_\rms)_\FWR={2I_m/{√2π}}

=100/√2=70.7A

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