SSC JE Electrical Engineering 2014 Set 2 Paper

Question : 156 of 200

Marks: +1, -0

For a half wave rectified sine wave the ripple factor is :

Solution:

In Half wave Rectifier

V_{\rms} = V_{\max} /2

V_{\av}= V_{\max}/h

Ripple factor =

√{{ {V^2}_{\rms} -{V^2}_{\av}} /({{V}_{\av})^2

=√{( {V }_{\rms}/{V}_{\av})^2-1}

=√{(V_{\max}/2)^2/(V_{\max}/h)^2-1}

√{T^2_1/4-1}

= 1.21 Volt
So Ripple factor of Half wave Rectifier is 1.21 volt.

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