STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[Omega_0]

You know, the thing is, the power levels are quite insignificant in this example,

I noticed that. I think the reason is that you have not considered the absolute values for V*I. So at certain bias the average of VI will be zero, as all negative values will cancel the positive ones. I guess this is what you are calling as "savings".

I think that the sign here implies only a polarity change (AC source), the energy is always flowing out of the source, so we should take absolute values of VI. If you substitute F3=ABS(D3*E3) and so on in the Excel sheet, everything returns to "normal".

What do you say about this? Am I thinking correctly?

[Omnibus]

I noticed that. I think the reason is that you have not considered the absolute values for V*I. So at certain bias the average of VI will be zero, as all negative values will cancel the positive ones. I guess this is what you are calling as "savings".

I think that the sign here implies only a polarity change (AC source), the energy is always flowing out of the source, so we should take absolute values of VI. If you substitute F3=ABS(D3*E3) and so on in the Excel sheet, everything returns to "normal".

What do you say about this? Am I thinking correctly?

Recall, we discussed that already. First, I think @teslaalset brought it up and then after a while @LarryC was adamant about it. However, think about a reactive component and a symmetric sine wave with no offset. If we're to consider only the absolute values then there will be dissipation in the reactive component when the cycle completes which isn't the case. A reactive component under such conditions returns to the source what has gone out and when the cycle completes there's no energy dissipated. The only component responsible for dissipation is the active (dissipative, Ohmic) resistance. So, we have to consider the signs of the current and voltage.

[Omega_0]

IMO, we need to see the complete picture in order to understand the results and to see their correctness. It looks like that the RC circuit is sending energy back into the sine wave source. But if you look deeper, the energy is not going back into any reusable form, such as increasing any potential etc.

In fact, there are no natural sine wave sources, it is normally a wave shaping arrangement, which simply shapes the DC(static) potential. I'm attaching a rough diagram of such a system. The output of such system is a varying voltage from 0-Vm, the Load resistor RL sees it as a potential of changing sign due to the coupling capacitor Cc, which blocks the actual potential and passes on only the variation. The controller device is usually a transistor which is continually modulating the value of Ro from 0-Ro, result is that the potential across Ro is also modulated.

You may argue that an AC generator (Dynamo) supplies an AC voltage, so its a natural AC source and energy can be fed back into it in usable form. But that's not the truth, there is a static energy source behind it, driving it, such the water in a river dam or the heat of burning coal. The question of whether energy can be given back to such sources is tricky. Things are not so simple here, and the equations may show something else.

So the bottom line is, the energy is always supplied by a DC source. Thus, the analysis (theoretical or experimental anything) should take into account all the involved components placed in an isolated environment.

In the sine wave energy source shown below, the ultimate energy source is the battery which never changes sign and is always connected to the circuit. Note that the energy is always flowing downhill here and there is nothing to store energy in that yellow box, which everyone calls an AC source.

It is little bit doubtful , if the analytical results are true, if your consider the full natural system.

[Omega_0]

So the question, whether we can take the sign of V*I as it is, or must take absolute values only, depends on the question, whether we can, in practice, re-charge the ultimate DC source powering the system back to its original charge or higher than that, every cycle.

If you can, then you can take the signs as it is.

If you consider an AC dynamo instead, then the question becomes, whether we can increase the gravitational potential of the water falling on it. Or in other words, will it put the water back in the dam, or say, will restore the coal back from CO2 and H2O. I'm saying this in broad terms but you get the point, which is, whether the energy flow can be reversed and entropy be restored in an thermodynamically isolated system. The result is a violation of law of thermodynamics, if it can be done.

I know that proving this analytically is difficult, but can be tried experimentally.

[Omnibus]

I understand your point but here we analyze only the workings of the device at hand. The question is whether or not there's an energy balance in this particular device, not outside of it. It's like when considering what the gas consumption is when a car travels from town A to town B. It's only the gas the car spent we need to know and not what the wear and tear is or what was the energy spent to make the car etc. Consider a gravity perpetuum mobile. We don't put into the energy balance equation the energy needed for the observer to subsist himself so that he be alive and be able to see the wheel or any other energy needed for this gravity wheel to exist at all. All such energies are external to the energy balance we're concerned about.