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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Equivalent DC-current of an AC-Current

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