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Rectified Voltage Waveform 
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#1
Feb1309, 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 halfwave 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 [itex]P_{avg} = \frac{(V_{rms})^2}{R}[/itex] 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 [itex]\frac{2}{\pi}[/itex] 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 [itex]\frac{\sqrt2}{2} V_{peak}[/itex]. 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! 


#2
Feb1309, 08:39 PM

HW Helper
P: 3,394

I don't know of any formula for the halfwave 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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