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Finding Irms for rectified half wave

  1. Jun 25, 2009 #1
    I'm given a rectified half wave, first half of sine wave (lasts 4.0ms) has peak of 10A, followed by a flat line for another 4.0ms

    how am i supposed to calculate the Irms? I can't apply Irms= Ipeak/ .707 so what should i do?

    could someone explain why it turns out as 5.0A? 10A/2? if so, why?
  2. jcsd
  3. Jun 25, 2009 #2

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    Welcome to PF.

    The rms of a periodic function is given by:

    f_{rms} = \sqrt {{\textstyle{1 \over T}}\int\limits_0^T {(f(t))^2 } dt}
    where T is the period. Here, [itex]
    f(t) = (\sin \omega t + 0)

    Your T is 8 ms. Try it now.
  4. Jun 25, 2009 #3


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    If you had a sinewave with a peak current of 10 A the RMS value would be 0.707 times 10 A or 7.07 amps.
    You could replace that with a DC current of 7.07 amps and it would work the same at heating something.

    Now, look at that DC current and every 8 mS, turn it off for 4 mS and then back on again for 4 mS. 50% on 50% off. What would be the RMS current then?
    Last edited: Jun 25, 2009
  5. Jun 25, 2009 #4
    thx a lot vk6kro! so it would be like [(10/.707)^2*(4ms) + (0)]^ 1/2 which works out to be 5A! nice simple explanation, couldn't understand the complex equation given in the previous reply :( didn't noe what was omega ( 2 pi / 8ms ? ) but thx guys
  6. Jun 25, 2009 #5


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    The answer isn't 5 amps, sorry. That is wrong.

    Turning a 7.07 amp current off for 50 % of the time gives you half the 7.07 amps heating ability.
    So, 3.535 amps.

    You multiply the peak value by .707 to get RMS but you were showing a divide, I think. You can divide by 1.414 instead if you like. Gives the same result.
    Last edited: Jun 25, 2009
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