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This is a somewhat theoretical question, but one that has some impact on power theory.

The circuit in the figure is composed of a real DC source (with non-negligible internal resistance) and feeds a non-linear load, composed of an ideal switch that switches every T/2. This problem is proposed in https://doi.org/10.1109/EPE.2019.8777983 by L. Czarnecki (a renowned power system engineer).

enter image description here

It seems clear that the instantaneous power is null, since the voltage is 0 when the current flows and the current is also 0 when the switch is open. It seems quite intuitive. Mathematically, the instantaneous power is defined by \$p(t)=u(t)i(t)=0\$ and the active power \$P\$ is the average value, so it should be also 0. The problem arises if you use the Fourier series for a square wave. In this case (\$\omega=\frac{2\pi}{T}\$),

$$u(t)=\frac{100}{2}\left(1+\frac{4}{\pi}\sum_{n=1,3,5...}^{\infty}\frac{1}{n}\sin{n\omega t}\right)$$ $$i(t)=\frac{100}{2}\left(1-\frac{4}{\pi}\sum_{n=1,3,5...}^{\infty}\frac{1}{n}\sin{n\omega t}\right)$$

In this case, it can be seen that the power is no longer equal to 0, as there is at least one DC term

$$p(t)=\frac{100}{2}\frac{100}{2}+...=2500+... \quad \text{[W]}$$

What do you think about this? For me, it makes more sense physically the first proposal, but I think there's something that I miss...

Update: I update the summation index to reflect odd sequence.

SOLVED: Indeed, all the terms sum up to zero. The key is that the cross products between 1 and the summations cancel out and the product of summations is

$$\left(\frac{100}{2}\frac{4}{\pi}\sum_{n=1,3,5...}^{\infty}\frac{1}{n}\sin{n\omega t}\right)\left(-\frac{100}{2}\frac{4}{\pi}\sum_{n=1,3,5...}^{\infty}\frac{1}{n}\sin{n\omega t}\right)=-2500 $$

so \$p(t)=0\$ in both time and frequency domain

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It seems clear that the instantaneous power is null, since the voltage is 0 when the current flows and the current is also 0 when the switch is open. It seems quite intuitive.

No, when 100 amps taken by the resistor (switch closed), one side is shorted to 0 volts and the other is at 100 volts hence, the power dissipated during that closure is 10 kW.

What you seem to be considering is the power dissipated by the switch and, of course, for an ideal switch, that will be zero.

In this case, it can be seen that the power is no longer equal to 0, as there is at least one DC term

Just because there is at least one term that is DC positive does not mean that there are not other terms; such as sin(nwt)*-sin(nwt), that produce DC negative terms that cancel the positive term. Do the math.

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  • \$\begingroup\$ Of course, there is some power dissipated at the internal resistor of the source, but this is not the question. My load is an ideal switch, not the internal resistor. In other words, \$u(t)\$ and \$i(t)\$ are the voltage and current feeding my load. So, my concern is still unanswered... why the fourier decomposition shows that there is active power? \$\endgroup\$ – paketecuento Jan 8 at 9:54
  • \$\begingroup\$ You write "instantaneous power" but, without a definition of what this power applies to, the reader MUST assume the whole circuit. \$\endgroup\$ – Andy aka Jan 8 at 9:59
  • \$\begingroup\$ thank you for your comment, but maybe you didn't read it well. I stated that instantaneous power is defined by \$p(t)=u(t)i(t)\$ and the load is composed by an ideal switch that switches every T/2. \$\endgroup\$ – paketecuento Jan 8 at 10:43
  • \$\begingroup\$ so, you were right Andy... the product of sines, cancel the pure DC. Thank you for your hint. \$\endgroup\$ – paketecuento Jan 8 at 11:35
  • \$\begingroup\$ @paketecuento LOL we got there in the end! \$\endgroup\$ – Andy aka Jan 8 at 11:38
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I will illustrate this phenomenon with a funny puzzle.

A "black box" with two terminals is inserted in an electrical circuit... and you have connected an ammeter (in series), a voltmeter (in parallel) and a wattmeter (mixed) to the element. They show that there is voltage, there is current... but there is no power.

The question is, "What is there inside the box?"

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  • \$\begingroup\$ are you kidding me? :P \$\endgroup\$ – paketecuento Jan 8 at 9:56
  • \$\begingroup\$ No... but my professor from 80s was joking with us like that:-) \$\endgroup\$ – Circuit fantasist Jan 8 at 10:42
  • \$\begingroup\$ AC or DC circuit? \$\endgroup\$ – Transistor Jan 8 at 11:40
  • \$\begingroup\$ I think he meant the simple DC case... no frequency, no phase shift... \$\endgroup\$ – Circuit fantasist Jan 8 at 12:03
  • \$\begingroup\$ I already provided the waveform... a square with offset. \$\endgroup\$ – paketecuento Jan 8 at 12:04

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