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[Omega_0]

The situation here is very different than the example you stated. The battery, oscillator are very much a part of the system.
You can leave out the minor things which don't interfere in energy transfer, such as the drive circuit, probes etc.

Partial analysis is ok only if the part is reasonably isolated from the energy transfer path, else you may get an incorrect result.

[Omnibus]

Yeah, I know, but remember, we're interested in the power balance of this particular circuit (the RC circuit). You realize, of course, that other circuits powered by the same system won't demonstrate OU. In this sense the powering unit isn't part of the studied circuit.

[Omega_0]

I suggest you take an expert's opinion on whether to consider the signs or absolute values. Because if abs values make the ratio unity, it can be significant. Had the abs value lead to some other ratio, we could have dumped the issue.

Ultimately the whole energy cost must be calculated to have an usable OU. Analysis of a part may show OU, but the whole system may be simply consuming energy instead of generating it.

Its like a roller coaster, it goes up sometimes, if you see a small part of the track, but overall scene shows that it is always going down. It will be good if you invite comments from experts in this field.

[Omnibus]

Yes, I've done that. And, also that's common sense. Recall that example I gave you with the reactive elements. If you take the absolute values you'll see energy dissipation in the reactive elements (we're talking a full sine cycle without offset, not somewhere in between the cycle as in your roller coaster example). You know very well that that's not the case -- dissipation occurs only in the active element.

[Omega_0]

There is no doubt about that, a reactive element will never consume any energy, offset or no offset.

It however consumes reactive power, which is V*I, which is needed to "drive" it. So its ok to consider its sign here. The reactive energy drives the element and comes back. Net reactive energy consumed is zero.

The "real" power equivalent of V*I is the equivalent thermodynamic heat generated per second by this voltage and current in a resistance of value V/I. The sign here is meaningless, as heat is a scaler thing.

So when comparing the power in load resistor and "real" power of the source, engineers discard the sign. What you are doing here is comparing reactive power with real power. IMO, this may not be correct, but I lack the expertise to confirm this, so I told you to consult a more knowledgeable person.

I will be very interested in knowing your question and the answer.