Not the way I see it.
Each religion has their own set of beliefs, not facts, even the stuff they agree on.

Hi again,

Well then you have not talked to any fervent believers. Their doctrine is fact TO THEM or else they would be called agnostic.
That's just one example.

I totally understand your point of view, now i think it is time you tried to understand mine
If you still dont understand my point of view then i guess we will just have to leave it at that and move on. This kind of discussion can go on forever.

BTW, i dont make the rules, i just think them up and write them down (Cartman)
But lets get back on track here... now what was it we were talking about originally ?

 

Dear Guys,

We do have to give it to DrAI here. He is seems the best grounded in knowledge/experience, and most balanced in establishing logic amongst us all in this fundamental but critical little discussion we are fleshing out here. Below, I simulated the experiment he suggested on page 2 of this thread.




Please do note that the 2 AC sources are not in opposite polarization, or 180d out of phase. Its the oscilloscope setup that makes it seem that way at first glance. Since the ground is in the middle of 2 in-synch AC sources, their measurements are in reverse directions. If we were to connect AC sources serially, but put earth ground to only to lower source, scopes would show same phase overlapping AC outputs. If there was a differential probe facility in this simulator, I could show it. Perhaps someone could do this experiment with real equipment, and share pictures? Even a PC connected multimeter can show the voltage wave between outer pins of AC sources in this setup.

Now my conclusions are:

1- Confusions stem from where to take the reference from and the direction of measurement. That simple.
2- Midpoint of combined capacitors show 0 Volts relative to earth ground, as I humbly suspected and DrAI kindly confirmed.
3- Yes, even planet Earth should have a non-zero voltage potential relative to some other point in universe. But we just don't have to give a damn in this little case here.
4- If what you refer as "ground" in your circuit is not earth, and have a voltage difference relative to earth, then your capacitor midpoint will not be at 0V relative to earth. But it will be relative to your circuit's reference ground. Lets not mix things up.

This discussion would be a lot more fruitful for us all if we present our inputs as actual physical experiments instead of personal views from this point on. Especially ones with good lab equipment. Actual physical reality is the only thing we can ever gravitate to agree upon. As Crutschow's cool quote above nicely points. I will use that one, thanks.

Looking forward to see a few little experiments, if you want to continue on this thread please.

 

Looking forward to see a few little experiments, if you want to continue on this thread please.

Neat but this demo is simple elementary circuit theory with standard circuit elements, split power AC sources with center grounds that no one here disagrees with.

 

ozsavran,

You seem to think you have made a revelation to the board; as nsaspook said: this is elementary. Even if you knew very little about electronics, given the perfect symmetry of the circuit any other result would make no sense

This all started by your confusion, caused by the idea that voltage is an absolute, not a relative measurement, which is why you could not understand the results you got in post #7. This, post #42, simulation combined with your simulation results in post #6 and #7, just prove what I said two days ago in post #16.

"You can say that the plates at any instant in time have equal and opposite charges, but their voltages with respect to ground are not, in this circuit [post #6 and 7], equal and opposite at any instant in time and are not 180 degrees apart.

What you can say is that with respect to the point half way between the two plates of the capacitor the voltages on the capacitor plates are equal and opposite. And in an AC circuit their voltages with respect to that midpoint are 180 degrees apart. This can be inferred from the physics of capacitors and the o-scope waveforms, but you can not directly measure it in your circuit."






.

 

ozsavran, in simulation in post #42 your two oscilloscope do not measure the capacitor voltage. You can even replace the cap with resistor and you still will see this "180d out of phase". Because this 180d has nothing to do with the capacitors.
This arrows show what you are really measure (Osc0 ch1) Vmeasured = Vac1 - VR1.... AC voltage minus voltage drop across R1 resistor.



We have a similarly situation for R0 and AC2

Try read this
http://www.ittc.ku.edu/~jstiles/312/handouts/312_Introduction_package.pdf (start at page 3)

And sorry for my not the best English

 

Dear Elder,

Again, that's only how you see it. Only thing I revealed in reality is simple simulation that proves DrAI analayses, descriptions. not yours.

Its true I was confused and suspicious, hence my little elementary question here. You helped me understand and solved my confusion. DrAI simply confirmed and explained it. Then some of you guys gut things confused etc. If one goes over your comments, he sees you argued on both sides of the issue.

Simple fact is 90d phase difference is between current and voltage over a cap. Voltages on each side has to be 180d apart by definition. If correct point of reference and direction is observed for measurement.

I would be delighted to see an experiment from you where you show to me voltages across an AC capacitor is 90d apart. That is voltages on each plate of a capacitor.

You are right that this is a verysimple, elementary, and now fully solved and understood issue. My sincere thanks for everybody's time and effort to explain it to me.

 

Dear Jony130,

Those resistors r1 and r2 both same value. Their existence simply reduces voltages supplied tocapacitors. nothing else. You can see where they reduce the voltage from scope outputs.

 

Those resistors r1 and r2 both same value. Their existence simply reduces voltages supplied tocapacitors. nothing else. You can see where they reduce the voltage from scope outputs.

Sure yes, you right . But my point was that this "180d phase shifts" has nothing to do with the capacitors. The capacitor do not "create" this 180d phase shift.

 

I was only talking about this mystical mid-point zero voltage for a 180 phase shift from a capacitor. 1/2 the total non-zero potential across the capacitor is not zero.

Hi,

I am still not sure what you mean. If you have 1/2 the potential across 1/2 the element and the other 1/2 potential across the other half of the element then there is NO reason on earth why you cant call that central point zero.
In circuit analysis we can call any node zero. In a given circuit element we can call any physical spot (that is small relative to the application) zero volts. For example, a copper bar 10 feet long and 2 inches wide and 1 inch thick, if we apply a 10v DC voltage across the bar and call a spot along the cross sectional area internal surface zero, then each end of the bar will be a different polarity. Even a spot on the 2 inch wide surface somewhere could be called zero. Heck, a spot 1 foot from one end could be called zero, and the two ends are still opposite in polarity.

 

Neat but this demo is simple elementary circuit theory with standard circuit elements, split power AC sources with center grounds that no one here disagrees with.

Hi again,

Ok then tell me where the standard circuit elements are in figure 3 of this next diagram. (figure 3 is the one on the far right and the other two are just there for reference). Simulation environment: MicroCap.

 

ozsavran, in simulation in post #42 your two oscilloscope do not measure the capacitor voltage. You can even replace the cap with resistor and you still will see this "180d out of phase". Because this 180d has nothing to do with the capacitors.
This arrows show what you are really measure (Osc0 ch1) Vmeasured = Vac1 - VR1.... AC voltage minus voltage drop across R1 resistor.

View attachment 109908

We have a similarly situation for R0 and AC2

Try read this
http://www.ittc.ku.edu/~jstiles/312/handouts/312_Introduction_package.pdf (start at page 3)

Hi,

Isnt the whole point though to see what happens with a capacitor?
We can use a sliding contact resistor to show that we can call anywhere on the surface of the element zero volts.

My point is that whenever we have a gradient we have a progression from 0v to some higher voltage, and between 0v and that higher voltage we have a range of voltages and we can call anywhere in there zero volts.

 

Sure yes, you right . But my point was that this "180d phase shifts" has nothing to do with the capacitors. The capacitor do not "create" this 180d phase shift.

Hi again,

Dont you have to show some proof of that?

In other words, you have to show that two nodes with 180 degree phase shift do not act like the capacitor with a charge, or changing charge. That is, that the two nodes do not influence an external circuit in the same way. I would say using DC would be easier and acceptable i think.

 

ozsavran,

I find some of your comments disingenuous. First off, I did not take both sides of the argument. Second, I never said the voltages across a capacitor is 90d apart.

"I would be delighted to see an experiment from you where you show to me voltages across an AC capacitor is 90d apart.
That is voltages on each plate of a capacitor."

Is it voltage across the capacitor or on each plate of a capacitor? And if on each plate, with respect to what? Voltage is relative! I don't know how else to say it.

For Voltages on each plate with respect to ground as you did in your original circuit, you merely need to change your resistor to near zero ohms to get a near 90d phase difference. And the phase difference will be near 0 degrees apart if you change it to a very high value - which your first attempt in the original circuit approximated. And every phase angle in between those extremes is realizable with varying resistor sizes. You have the simulation setup, do the experiment yourself.

 

Dear Guys,

We do have to give it to DrAI here. He is seems the best grounded in knowledge/experience, and most balanced in establishing logic amongst us all in this fundamental but critical little discussion we are fleshing out here. Below, I simulated the experiment he suggested on page 2 of this thread.


View attachment 109904 View attachment 109905

Please do note that the 2 AC sources are not in opposite polarization, or 180d out of phase. Its the oscilloscope setup that makes it seem that way at first glance. Since the ground is in the middle of 2 in-synch AC sources, their measurements are in reverse directions. If we were to connect AC sources serially, but put earth ground to only to lower source, scopes would show same phase overlapping AC outputs. If there was a differential probe facility in this simulator, I could show it. Perhaps someone could do this experiment with real equipment, and share pictures? Even a PC connected multimeter can show the voltage wave between outer pins of AC sources in this setup.

Now my conclusions are:

1- Confusions stem from where to take the reference from and the direction of measurement. That simple.
2- Midpoint of combined capacitors show 0 Volts relative to earth ground, as I humbly suspected and DrAI kindly confirmed.
3- Yes, even planet Earth should have a non-zero voltage potential relative to some other point in universe. But we just don't have to give a damn in this little case here.
4- If what you refer as "ground" in your circuit is not earth, and have a voltage difference relative to earth, then your capacitor midpoint will not be at 0V relative to earth. But it will be relative to your circuit's reference ground. Lets not mix things up.

This discussion would be a lot more fruitful for us all if we present our inputs as actual physical experiments instead of personal views from this point on. Especially ones with good lab equipment. Actual physical reality is the only thing we can ever gravitate to agree upon. As Crutschow's cool quote above nicely points. I will use that one, thanks.

Looking forward to see a few little experiments, if you want to continue on this thread please.

Hi,

That's very kind of you to say. You started this wonderful thread though so you had the bright idea first

I am very convinced that you have the interpretation correct. True, there are other interpretations, but the one we are concerned with here must be correct also.

As further proof, i show in this simple simulation demonstration that the battery can also be viewed as a differential source, ie not a single ended output but a balanced differential output which the capacitor, at the heart of it, is too.
Note the polarity of the first two, then the last one with NO resistors.
Simulator: MicroCap environment.



If that does not put an end to the disbelief then probably nothing will

 

Hi again,

Ok then tell me where the standard circuit elements are in figure 3 of this next diagram. (figure 3 is the one on the far right and the other two are just there for reference). Simulation environment: MicroCap.
View attachment 109911

If the voltage source in figure 3 was a 9 volt transistor battery floating in space and with your nice Fluke 87 meter you connect one end to ground (no need for it to be at infinity) and the other meter lead to the - terminal are you telling me you would really read on the meter -4.5 volts and then +4.5 when moved to the + terminal?

 

Mr. Al,

Your arguments are circular. You presume axiomatically that the midpoint of the capacitor is at zero volts and then you prove it by showing in various examples that the voltages on the plates are equal and opposite to each other with respect to that point and therefore the midpoint must be zero.

You can call any arbitrary point zero. The point is if you want to discuss the physical behavior of a system as evidenced by voltages then you need to define your voltage reference point first or you end up with silly arguments like this one.

 

Dear Jony130,

Those resistors r1 and r2 both same value. Their existence simply reduces voltages supplied tocapacitors. nothing else. You can see where they reduce the voltage from scope outputs.

Sure yes, you right . But my point was that this "180d phase shifts" has nothing to do with the capacitors. The capacitor do not "create" this 180d phase shift.

Then where does it come from in your opinion? Resistor? Nope. AC sources? Nope they are on same phase exactly. Why not do your own construction or simulation instead. This is objective physics fact finding underneath here. Not any subjective human constructions like religion or politics.

Please understand, I sincerely am not targeting you or anybody else or his personal opinion or pride in particular. Have no energy for that. But so much confusion here among "experts" is so very very meaningful, and very telling to me. As always and as everywhere, there is 90% smoke where 10% fire in electronics. So many people just memorize and repeat things and brag about it out of proportion. Its so off-putting, and an impediment to dissemination of true engineering mind set. Which is honestly getting to the bottom of what you are talking about. My sincere apologies if this offends anybody. Bragging professors who really did not really comprehend crap about what they are talking, did severely ruin my early engineering enthusiasm and my young career years. Not surprised to see the same old facade/charede still alive with even more meaningless jargon added.

Next week or so, I may repeat the simulation with real instruments, and post pictures here, if that is still needed. If I am not kicked out of here that is.

Lets all lighten up, experiment more, criticize and ridicule a lot less guys. Please.

 

Then where does it come from in your opinion? Resistor? Nope. AC sources? Nope they are on same phase exactly. Why not do your own construction or simulation instead.

Try it yourself first. Replace C0 and C1 with two 100R resistor and show me the results.

 

"further proof, i show in this simple simulation demonstration that the battery"

That is a circuit simulator not a physics simulator. When you disconnect the battery it is not a circuit anymore and the results are not meaningful.

Try it in the lab as nsapook suggested.

 

Next week or so, I may repeat the simulation with real instruments, and post pictures here, if that is still needed. If I am not kicked out of here that is.

Lets all lighten up, experiment more, criticize and ridicule a lot less guys. Please.

No one is going to kick you out but the tally on the crackpot index is increasing with diatribes about 'professors' and 'experts'.