Okay I already read this post which answers the question mathematically here,

at the end of the top answer the poster said:

"Thus, the only way to keep current from going to infinity is the condition I(T)=0 " So does this means that the indcutor does not have some property that it magically makes any periodic waveform have an average voltage of zero, but WE are setting that integral equal to ZERO and solving for that case?

So this thing about "average inductor voltage always equals 0" is more of a rule we must impose on it, rather than something it does on its own?

If i apply a voltage to an inductor for 50mS it will rise linearly and i remove the voltage for only 1ms it will return to the initial voltage? thats a periodic waveform if i keep doing it but seems to me it would rise for ever. Which makes me wonder how then is the average zero?

and same would apply to a capacitor?

This highlighted integral.
Basically , will it always equal zero because of some physical trait of the inductor that makes it always be in a steady state, or is this telling me that we must SET it equal to zero to operate in steady state? enter image description here

  • \$\begingroup\$ This is kind of like capacitors, where the average current must tend towards zero. Otherwise, if not, then charge must build up, one way or another, on the capacitor over time. And after enough charge, eventually, it will fail. Similarly, it's the volt-seconds (Webers) that need to stay close to zero for inductors. Webers are the inductor's equivalent to charge on a capacitor. Dividing accumulated charge on a capacitor by some finite time gives you average current. Dividing accumulated Webers on an inductor by some finite time gives you average voltage. \$\endgroup\$ – jonk Nov 14 '18 at 0:42
  • \$\begingroup\$ To DC, an ideal inductor is a short circuit and an ideal capacitor is an open circuit. \$\endgroup\$ – George White Nov 14 '18 at 0:56
  • \$\begingroup\$ You only get a voltage across an inductor when the current through it varies. So applying a sinusoidal current results in a pure sinusoidal voltage (no constant voltage component - no voltage bias), and the average value of a pure sinusoid over a complete cycle is zero. \$\endgroup\$ – Chu Nov 14 '18 at 1:36

So, that's a rule that applies to perfect inductors with the implied property that Bad Things happen if the current goes to infinity.

This is a pretty good model for the real world of power supply design, where long before you get to infinite current you get to enough current so that the magic smoke leaks out of something on your board (usually a chip, but sometimes the inductor or the board itself).

So it's only a condition imposed by the inductor in the sense that your circuit is damaged (or at least malfunctions badly) if you don't make sure it happens.

  • \$\begingroup\$ okay so it is something the designer has to make sure results to be true, and not just a given fact that hey if i throw this inductor here the average voltage and power loss will be zero. \$\endgroup\$ – Edwin Fairchild Nov 13 '18 at 23:19
  • \$\begingroup\$ There are circuit arrangements you can make around the inductor to insure that -- basically if the inductor is the circuit element in charge of its current, then that's as far as you need to go. But if you're driving an inductor with a low-impedance source all the time, then it's up to you. Usually in the latter case you rearrange your circuit, or you do something to measure the inductor current and servo it somehow. \$\endgroup\$ – TimWescott Nov 13 '18 at 23:54

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