Rectified Voltage Waveform

by Machodog
Tags: rectified, voltage, waveform
Machodog is offline
Feb13-09, 06:52 PM
P: 5
1. The problem statement, all variables and given/known data
A sinusoidal voltage waveform is rectified so that it becomes a half-wave waveform (i.e. the bottom half has been "chopped off"). Determine:

i) The average p.d.
ii) The RMS p.d.
iii) The heat dissipated in a 200ohm resistor in 100s

2. Relevant equations
[latex]P_{avg} = \frac{(V_{rms})^2}{R}[/latex]

3. The attempt at a solution
I'm not sure, but I would guess that the average p.d. is the line that cuts the rectified waveform in such a way that the area above and below is the same. I have a vague recollection about it being [latex]\frac{2}{\pi}[/latex] times the peak voltage.
I am also quite uncertain about the RMS, how would the voltage being zero half the time affect this value? I'm guessing it will no longer be [latex]\frac{\sqrt2}{2} V_{peak}[/latex].

Please could anyone give me some clues on how to work out the RMS and average values?

EDIT:Oops sorry I did not mean to post the same thread twice...just ignore the other one!
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Delphi51 is offline
Feb13-09, 08:39 PM
HW Helper
P: 3,394
I don't know of any formula for the half-wave signal. I rather think you will have to find out what "root mean square" means and work it out from the definition - no doubt you will be integrating over a full wavelength (the V = zero part will be easy!).

The average potential will be the same sort of thing, the area divided by the time.

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