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

[Omnibus]

@LarryC,

Read very carefully the pages you have posted and try to understand what that standard text is trying to tell you. I can explain it to you but cannot understand it for you. Read carefully and think.

[broli]

Since the beginning I and others have tried to tell you that your basic formula's and methodology were wrong. But you had the mistaken belief that since no one complained about Stearn's use of these formula's, that you were right.

Stearn had a pulsed DC circuit which he insisted that the pulse had to be flat. Yes, it is okay to use those formula's for DC, pulsed DC, or AC resistive only circuits.

The page listed below is from electricity one-seven revised second edition page 4-35. It shows the formula's used to calculate the true power actually consumed in a series RL circuit, which is what you are testing. 

Now, I and others don't want to hear one of your usual misleading statements and don't even bother trying to trash the book. Show the Page and list the book, showing that your formula's can be used for any circuit other then DC or AC resistive only. 

Regards, Larry

No offense Larry but that's stupid. You know that formula only counts for PERFECT SINUSODIAL waves.  What that formula does in theory, that is integrate the wave over time, is done in practice with the actual time stepped data. You cannot even begin to argue against that.

If more power is indeed being sent back it doesn't matter what the shape of the wave is. If you can integrate over some million values and show a strong negative power average the case is closed.

[LarryC]

@LarryC,

Read very carefully the pages you have posted and try to understand what that standard text is trying to tell you. I can explain it to you but cannot understand it for you. Read carefully and think.

So you decided to use a lame attempt at misdirection and not show the text proof. Please enlighten us with your explaination. If I don't understand I'm sure someone else here will.

And don't try to use where they stated Ptrue = I^2R, where R = Z cos angle. Your R is incorrectly using the coil resistance.

Regards, Larry

[broli]

So you decided to use a lame attempt at misdirection and not show the text proof. Please enlighten us with your explaination. If I don't understand I'm sure someone else here will.

And don't try to use where they stated Ptrue = I^2R, where R = Z cos angle. Your R is incorrectly the coil resistance.

Regards, Larry

Power is voltage times current. The load is irrelevant now. Those two quantities however give you an indication of what the load is like. If the wave is mostly above the zero line it's resistive, if it's centered on the zero line it's pure inductive...however if it's mostly below it, it's acting as a source thus pumping more energy back. By mostly I mean literally taking the average by using the raw data from the scope and not some fantasy of ideal sine waves. By using the raw current and voltage data the shape of the wave becomes IRRELEVANT. Omnibus got this probe for doing just that. Eliminating information of the load, which might be some crazy beast due to all the parasitic and ohmic crap. With the current probe you don't need to know anything about the load before hand. You just measure voltage from source and current from source. Take the raw data and integrate it which will give you a REAL average. If this average is positive, nothing special. If it's 0, means no losses. If it's negative means energy is pumped back.

You can choose to deny this if your ego is at stake. But it won't change the truth.

[Omnibus]

So you decided to use a lame attempt at misdirection and not show the text proof. Please enlighten us with your explaination. If I don't understand I'm sure someone else here will.

And don't try to use where they stated Ptrue = I^2R, where R = Z cos angle. Your R is incorrectly using the  resistance.

Regards, Larry

No, I, of course, won't use "Ptrue = I^2R, where R = Z cos angle". Instead, I will use this and please read it carefully and try to understand it:

In resistive circuits all of the power is dissipated by the load but if you recall what you learned from Volume 3 in an RL circuit only a portion of the input power is dissipated. The part delivered to the inductance is returned to the source each time each time the magnetic field around the inductance collapsews

Did you read that carefully, Larry? I guess not. So, I'll repeat it: in an RL circuit the power which is dissipated is only that due to the dissipative (Ohmic, active) resistance. that is, I^2R where R is the dissipative (Ohmic, active) resistance having nothing to do with inductance L. The part of the power delivered to the inductnce is not dissipated. Again -- the part of the power delivered to the inductance is not dissipated when integrating over a full cycle. Can this come across to you or you need more explanation? Here in the discussion at hand we are only concerned with the dissipated power. Only the power dissipated is the ouput power (the power delivered to the inductance is not dissipated and therefore is not of interest to us in this discussion). Therefore, because the only dissipated power is I^2R where R is the dissipative (Ohmic, active) resistance having nothing to do with inductance or L, that's what we calculate as output power in our case. Get it?