Engproguides Question 12 Exam - Electrical - Engineer Boards
Jump to content
Engineer Boards
Sign in to follow this  

Engproguides Question 12 Exam

Recommended Posts

A 480VRMS AC single phase source is sent through a single phase half wave rectifier.  What is the average voltage at the output of the rectifier?


Vmax = sqrt(2)*Vrms= 679

Vavg = (1/2pi) * integral of 679sin(theta) from 0 to pi = 216 V


To me, when I see the solution, Vmax was chosen as if it was a single phase full wave rectifier.  Since it's half wave, I thought Vmax = 2*Vrms = 960 instead of 679, which gives me a Vavg = 306 instead.


If the solution is correct as written, when do I use Vmax = 2*Vrms for single phase half wave rectifiers.  Is it the voltage at the load and I should use Vmax=sqrt(2)*Vrms for Vavg?


Share this post

Link to post
Share on other sites

In case anyone else needs the solution...

For a half-wave rectified sine wave, the peak voltage is given by:

Vp = 2 * Vrms

But this is only for a half-wave rectified sine wave. The peak values of a sine wave does not change when rectified, but the RMS value does change. So you must first find the peak value of the un-rectified sine wave:

Vp = sqrt(2)*Vrms = 1.414 * 480 = 679

Once the voltage is rectified, you can use the formulas for a half-wave rectified sine wave:

Vavg = Vp / pi = 679 / 3.14 = 216 V

Vrms = Vp / 2 = 679 / 2 = 339.5


In summary, a sine wave and half-wave rectified sine wave will have the same peak value, but different RMS value.

Edited by Chattaneer

Share this post

Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Sign in to follow this  

  • Create New...