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

[broli]

My problem is not so much with your P=I^2 R. It your use of P = EI where you have negatives signs almost half the time causing the graph to show Omni OU incorrectly.

P = V^2 Z is the correct formula is this case. Z is for a LCR circuit. Where the inductor and capacitor is parallel and the R is in series.


Circuits dominated by inductance reactance has the voltage leading the current.
Circuits dominated by capacitive reactance has the current leading the voltage.
Omni showed the same in his test results.


The capacitive information that I recently posted, predicts that using your current formulas that Omni OU power will be shown using a capacitor only.

The larger the capacitor the lower the frequency to show Omni OU.

If Omni does make this simple test with the correct type capacitor(no polarised electrolytic), he should ramp up the frequency slowly as it will appear like a short to the function generator at some point. Should stop at the same amp's as produced by the 700HZ to compare the results.


Regards, Larry

I respond to this seriously.

1st point:
Yes power goes up and down that's also why we are AVERAGING these values. What do you think root mean square means? It's just the theoretical AVERAGE of a perfect sine wave. But we are not assuming we are dealing with a sine wave, let alone a perfect one, thus averaging the raw data... you can't get a more accurate result than that.

2nd point:
It seems like you are late to the parasitic capacitance party as this has been the argument all along, not the formulas or data. Omnibus already did a controlled experiment on this showing that a reasonable sized cap in parallel still shows OU. This is a good argument but to many people, including me, not inconclusive proof.
So this is why I and others hope omnibus can do the temperature measurements of the coil to prove that indeed current is going through resistance.

This is where we are "at". I hope you are now enlightened and either contribute in a constructive way or step aside and not waste the time of the people wanting to progress. That's fair enough, right?

[LarryC]

I respond to this seriously.

1st point:
Yes power goes up and down that's also why we are AVERAGING these values. What do you think root mean square means? It's just the theoretical AVERAGE of a perfect sine wave. But we are not assuming we are dealing with a sine wave, let alone a perfect one, thus averaging the raw data... you can't get a more accurate result than that.

Broli, thanks for getting serious, your honest response helps to clear up some of the confusion.

1st point:
Sorry, but RMS is not just for a perfect sine wave and it is not considered average.

RMS is computed as follows:
It is equal to the square root of the average value of the squares of all the instantaneous value of current or voltage during one-half cycle. The Oscilloscope shows this calculation. The one-half cycle eliminates the sign issue.

The effective value or RMS of an ac voltage or current is the value that will cause the same amount of heat to be produced in a circuit containing only resistance that would be caused by a dc voltage or current of the same value. That is why it is the accepted standard.

In a perfect sine wave RMS is equal to .707 of peak, Average is equal to .637 of peak. Note my early post scope shot that my RMS value was not .707 of peak, because it was not perfect.

2nd point:
It seems like you are late to the parasitic capacitance party as this has been the argument all along, not the formulas or data. Omnibus already did a controlled experiment on this showing that a reasonable sized cap in parallel still shows OU. This is a good argument but to many people, including me, not inconclusive proof.
So this is why I and others hope omnibus can do the temperature measurements of the coil to prove that indeed current is going through resistance.

Sorry, I totally missed the post showing that Omnibus got OU with a capacitor. I'm interested in the scope shot and data, so could this be shown.
It would be easy to calculate Z with a capacitive only circuit.

'going through resistance'. Current is not going thru the resistance. This is a complex issue and I admit I don't know the answer. What I do know is at 700K Hz the inductive reactance is massive so that only a miniscule current can flow thru the coil. The circuit is dominated by capacitance. So there is a current flowing into the coil from one side or the other, maybe halfway and flowing back out. Is R the same at this point? 

This is where we are "at". I hope you are now enlightened and either contribute in a constructive way or step aside and not waste the time of the people wanting to progress. That's fair enough, right?

Thanks, you have enlightened me into your mindset. You have helped me with some of my misconceptions as I hope I helped you with some of yours. Looking forward to the temperature measurements.


Regards, Larry

[trilson]

You realise the average of a sine wave is 0, right?

Putting that aside though, I think a deeper understanding of electromagnetism (and for some people, even basic electronics) is required here. Impedance and resistance shouldn't be thought of in absolute terms, and the difference between the two needs to be clearly understood.

Has anyone performed the necessary calculations using Maxwell's equations (and their relativistic forms, not to rule anything out)?

[broli]

Broli, thanks for getting serious, your honest response helps to clear up some of the confusion.

1st point:
Sorry, but RMS is not just for a perfect sine wave and it is not considered average.

RMS is computed as follows:
It is equal to the square root of the average value of the squares of all the instantaneous value of current or voltage during one-half cycle. The Oscilloscope shows this calculation. The one-half cycle eliminates the sign issue.

The effective value or RMS of an ac voltage or current is the value that will cause the same amount of heat to be produced in a circuit containing only resistance that would be caused by a dc voltage or current of the same value. That is why it is the accepted standard.

In a perfect sine wave RMS is equal to .707 of peak, Average is equal to .637 of peak. Note my early post scope shot that my RMS value was not .707 of peak, because it was not perfect.



Sorry, I totally missed the post showing that Omnibus got OU with a capacitor. I'm interested in the scope shot and data, so could this be shown.
It would be easy to calculate Z with a capacitive only circuit.

'going through resistance'. Current is not going thru the resistance. This is a complex issue and I admit I don't know the answer. What I do know is at 700K Hz the inductive reactance is massive so that only a miniscule current can flow thru the coil. The circuit is dominated by capacitance. So there is a current flowing into the coil from one side or the other, maybe halfway and flowing back out. Is R the same at this point? 


Thanks, you have enlightened me into your mindset. You have helped me with some of my misconceptions as I hope I helped you with some of yours. Looking forward to the temperature measurements.


Regards, Larry

Now we are getting to the real crooks of the problem. Let me quote some of wikipedia on rms power:

However, if the current is a time-varying function, I(t), this formula must be extended  to reflect the fact that the current (and thus the instantaneous power)  is varying over time. If the function is periodic (such as household AC  power), it is nonetheless still meaningful to talk about the average  power dissipated over time, which we calculate by taking the simple  average of the power at each instant in the waveform or, equivalently,  the squared current.

http://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power

So this is really like the chicken and the egg problem. But in our case the mean is much stronger than the root mean square. The latter is just a by product of the former if you follow some assumptions. The P = R*I^2 asks you take the mean of I and then square it. This is coincidently called rms. The problem with using rms values with non purely resistive circuits is that you could have 0 REAL power yet have substantially more apparent power calculated from these rms values:

http://en.wikipedia.org/wiki/AC_power#Basic_calculations_using_real_numbers

This is due to the inherit way root mean square is calculated by starting from a resistive energy point of view. That is why it has to be corrected by using impedance knowledge.

However if only instantaneous power and its average/mean was used you will always be safe and get only REAL power out of your average formula of power (P=U*I). Because you are not starting off by assuming you have resistive load and calculating the rms voltage and current. This is fundamentally calculus. By integrating over an arbitrary function you get the average area under that function, then by dividing by the length of the area you get the average height.

Has all doubt been cleared now about the calculation?

[Omnibus]

@broli,

Very well put. I hope @LarryC could finally get a grip of this difference between apparent and real power when reactive components are involved.