Full wave rectifier with R load

Discussion in 'Homework Help' started by notoriusjt2, Oct 4, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    A single-phase full-wave bridge rectifier has a resistive load of 18ohms and an ac source of 120V rms. Determine the average power in the load.

    these average power questions are killing me....

    Vo=((2)(120)(1.414))/pi = 108v
    Io= Vm/R = 108/18 = 6A

    P=108*6=648w

    so what step am i forgetting to do here... i am never taking an electronics/math class online again.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I think that you need to find RMS voltage and current.
     
  3. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    so
    Vrms=108/1.414=76.37
    Irms=6/1.414=4.24

    P=76.37*4.24=324w

    would this be correct?
     
  4. Georacer

    Moderator

    Nov 25, 2009
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    P_{avg}=\left\langle \frac{V^2}{R} \right\rangle \\ <br />
=\frac{\langle V^2 \rangle}{R}\\<br />
=\frac{V_{rms}^2}{R} \\<br />
where\ \langle\ \rangle\ \text{is the time mean}

    Your answer is wrong, because Vrms is not connected to the average voltage with the relation you wrote. The Vrms of the rectified waveform is the same as the Vrms of the initial sine wave. Do the math and it will become apparent. Use the default RMS calculation formula.

    Is that clear?
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Hehe, No.
    Maybe you should read the question again.
    "A single-phase full-wave bridge rectifier has a resistive load of 18ohms and an ac source of 120V rms. Determine the average power in the load."
     
    Last edited: Oct 4, 2010
  6. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    yea that helps, but what do you mean when you say time mean?

    so if Pavg=(V^2rms)/R
    then
    Pavg=(120^2)/18
    Pavg=800w

    is that what you mean?
     
  7. Georacer

    Moderator

    Nov 25, 2009
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    Time mean is just the mean of a value over a time period, mean for short. Don't bother about it.

    Other than that, your answer seems correct.
     
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