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Power in fullwave rectified circuit

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A sinusoidal alternating current is fullwave rectified. The rectified current will produce in the same load

    a. the same power
    b. 0.71 times the power
    c. 1.41 times the power
    d. half the power
    e. twice the power


    2. Relevant equations
    Pmax=Imax x Vmax

    Paverage=Irms x Vrms


    3. The attempt at a solution
    I guess the question is asking about the average power. For full wave rectified :

    [tex]I__rms =\frac{I_max}{\sqrt{2}}[/tex]

    [tex]V__rms =\frac{V_max}{\sqrt{2}}[/tex]


    Paverage=Irms x Vrms


    = [tex]\frac{I_max}{\sqrt{2}}\times \frac{V_max}{\sqrt{2}}[/tex]

    = [tex]\frac{P_max}{2}[/tex]

    So, the answer is (d) ?

    If the question asking about the max. power, then the answer will be the same ?

    And how to determine whether the question is asking about max. or average power?

    Thanks
     
  2. jcsd
  3. Nov 2, 2009 #2

    Delphi51

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    Homework Helper

    Go back to the basics. Power is the integral over a cycle of I*V or I²R dt.
    Sketch the waveforms and shade in the area of the integral in each case.
    How do the areas compare?
     
  4. Nov 2, 2009 #3
    Hi Delphi51
    I tried to sketch the waveform of sine wave and full wave rectified, then compared the areas. The areas will be the same so the answer should be (a) ?
     
  5. Nov 2, 2009 #4

    Delphi51

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    Homework Helper

  6. Nov 2, 2009 #5
    Hi Delphi51

    Now I see that what I've done on the first post didn't answer the question. It's just a little work to show that the average power is half of the max. power.

    Thanks a lot !!
     
  7. Nov 2, 2009 #6

    Delphi51

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    Homework Helper

    Yes, nothing to do with average power. Just a comparison between sine wave and rectified sine wave.
     
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