Engproguides Question 12 Exam - Electrical - Engineer Boards
Engineer Boards

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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?

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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

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