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Equivalent DC-current of an AC-Current

  1. Jul 2, 2015 #1
    One has to calculate the equivalent DC-current in order to provide the same power on a load like a sine, triangular and square wave current do in average.

    I know the general formula for the average power of an AC-current is:
    [itex]p_{average} = \frac{R}{T} \int_0^T i(t)^2 \, dx [/itex]

    For the sine current, the integral goes from 0 to 2pi/w (as expected).
    Now, I don't understand why in the solution of this tasks the integral referring to the triangular function goes from 0 to T/4, and for the square wave function from 0 to T/2? Why simply to T?
    Last edited: Jul 2, 2015
  2. jcsd
  3. Jul 2, 2015 #2
    I'm not sure that I understand your question here...

    Average AC power is... [itex]P_{average} = V*I*PF[/itex] (forgive this poor quality...it's my first time using latex)
  4. Jul 2, 2015 #3
    I think you are saying that when you look up the standard solution for the problem, they integrate from 0 to T/4 and 0 to T/2 -- and you are thinking they should be going 0 to T?

    If that is the point of your question - it is because the waveforms are discontinuous and symmetrical. - i.e. you only have to do 1/2 of a complete square wave or 1/4 of a triangle wave to get the correct result.
  5. Jul 2, 2015 #4
    Looking to your title, are you wondering how to acquire the DC equivalence of an AC circuit so you don't have to use Diff Eq. to solve circuits? If so... [itex]L_{DC equivalence} = jwL[/itex] and [itex]C_{DC equivalence} = jwC^{-1}[/itex]

    Your source will change to use only the magnitude (in rms) of the value given and the phase angle of said source.
    [itex]V_{source} = V_{rms}sin(\omega t- \varphi)[/itex] will turn to [itex]V_{source} = V_{rms}\angle -\varphi[/itex]

    Once you transfer everything over to the frequency domain, you can once again use DC analysis methods to find whatever it is you're trying to find.
    Last edited: Jul 2, 2015
  6. Jul 2, 2015 #5


    Staff: Mentor

    I think you want RMS root mean square. Look it up on Wikipedia.

    RMS works for any wave shape. Based on ypthe original question, that is what you need.
  7. Jul 2, 2015 #6
    You have understood what is my problem. I don't get why I "only have to do 1/2 of a complete square wave or 1/4 of a triangle wave to get the correct result". Can you make this point clear to me?
  8. Jul 2, 2015 #7


    User Avatar
    Gold Member

    I think the formula OP uses IS to calculate the mean power by finding the mean of the square of the current.
    Because it is mean power being calculated, there is no root - P = R( I^2 ) or (V^2)/R
    RMS voltage and current (not squared) are calculated from the root of the mean power.

    With your AC wave, the instantaneous power in the resistance is varying as the voltage and current are varying - peak power when voltage or current are peak and zero power when they are zero. To find the DC equivalent - which is a constant voltage and current - we need to average over the whole cycle. Each complete cycle will be the same, because the variations in voltage and current are the same in each cycle.

    So the general rule is to integrate the square of the voltage or current over one complete cycle. This will work for any wave shape. The squaring ensures that the positive and negative parts of the cycle do not cancel each other. Since power is proportional to the square of the current or voltage, we just divide by the length of the cycle to get the average power. And if you want the equivalent DC current or voltage, you need to take the square root of that average (mean) - what people call the root mean square or RMS value.

    Now for waves with symmetry you MAY integrate over less than the whole cycle and average that.
    ##(\int_0^{360} \! sin^2(x) \, \mathrm{d}x)/1 = ## ##(\int_0^{180} \! sin^2(x) \, \mathrm{d}x)/2 = ## ##(\int_0^{90} \! sin^2(x) \, \mathrm{d}x)/4 ##

    In fact you can do 1/4 cycle of sine, square or triangle wave and the average of the squared values is the same as over a complete cycle. So for these 3 waves it does not matter whether you use 1/4 , 1/2 or a complete cycle *provided* you start at the beginning of a cycle (or any 1/4 cycle) and the wave is symmetric with no DC component. This would not be true for less symmetric waves, such as a sawtooth or a mixture of harmonics.
    Last edited: Jul 2, 2015
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