Rectified Voltage Waveform

  1. 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
    [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. jcsd
  3. Delphi51

    Delphi51 3,410
    Homework Helper

    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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