Don't think so. Equal and opposite with respect to an arbitrary internal reference point is not the same as equal and opposite in the outside world. Also, while "opposite phase" is defined as 180 degrees in balanced audio, that is not automatically true everywhere else in this universe. (Standard disclaimer: these conditions do not necessarily apply on Mars.)
.....
In post #1 the TS confuses voltages with respect to an internal reference point, with external voltages measured with respect to something (anything) else. After 20 posts, the two still are not the same.

ak

Dear DGElder, DrAI, NSASpook, and AnalogKid

All the quite valuable discussion you put in here for the benefit of less knowledged people like me is quite admirable and awe inspring really. Especially DGElder and DrAI is amazing. And we have gotten to possibility of 180d phase difference between capacitor plates relative to absolutely equal balanced material point if I am not reading it all wrong.

As AnalogKid, Jony130, CMartinez mentioned that I must take measurements relative to some other reference, I made a simple new simulation below. (Proteus does not provide differential graphs, and my free DoCircuits account is limited to 10 parts only. Excuse my cheapness pls).


Tried making a +10V reference here relative to earth ground, with minimal current flow. As people mentioned that meters need a closed circuit to measure anything, otherwise we would get just 0V. Which I really did on bench. AC source ampl is 10Vpp. Both caps are 1uF, top resistor 100ohm, bottom resistor 20MOhm. Leftmost DC voltmeter is VMeter00, next is Vmeter01. Hopefully in correct order, some simulation curves for 1Hz, 10Hz, 50hz, 100Hz, 1kHz, 10kHz, 50kHz, 75kHz, 100kHz are below.

Starting with 1Hz:




10Hz:



50Hz:


100Hz:


1Khz:


10KHz:


50KHz:


75KHz:



100KHz:


Now relative to both 10V line and ground we see no phase shift in any of these frequencies. Just for some fresh breath from high math could someone care to take a stab at what's going on here. I did not yet test this on bench, and yes simulators do have limitations. This one warns about entering virtual resistors on (+) leads of each voltmeter.

For bench test I will not use a signal generator. But generate sine wave from an op amp circuit, so everything including PC meters will be totally independent from power lines, if that helps. Will share results here.

With my sincere respects to masters here.

 

Dear DGElder, DrAI, NSASpook, and AnalogKid

All the quite valuable discussion you put in here for the benefit of less knowledged people like me is quite admirable and awe inspring really. Especially DGElder and DrAI is amazing. And we have gotten to possibility of 180d phase difference between capacitor plates relative to absolutely equal balanced material point if I am not reading it all wrong.

As AnalogKid, Jony130, CMartinez mentioned that I must take measurements relative to some other reference, I made a simple new simulation below. (Proteus does not provide differential graphs, and my free DoCircuits account is limited to 10 parts only. Excuse my cheapness pls).

View attachment 110202
Tried making a +10V reference here relative to earth ground, with minimal current flow. As people mentioned that meters need a closed circuit to measure anything, otherwise we would get just 0V. Which I really did on bench. AC source ampl is 10Vpp. Both caps are 1uF, top resistor 100ohm, bottom resistor 20MOhm. Leftmost DC voltmeter is VMeter00, next is Vmeter01. Hopefully in correct order, some simulation curves for 1Hz, 10Hz, 50hz, 100Hz, 1kHz, 10kHz, 50kHz, 75kHz, 100kHz are below.

Starting with 1Hz:

View attachment 110206


10Hz:
View attachment 110208


50Hz:
View attachment 110210

100Hz:
View attachment 110213

1Khz:
View attachment 110207

10KHz:
View attachment 110209

50KHz:
View attachment 110211

75KHz:
View attachment 110212


100KHz:
View attachment 110214

Now relative to both 10V line and ground we see no phase shift in any of these frequencies. Just for some fresh breath from high math could someone care to take a stab at what's going on here. I did not yet test this on bench, and yes simulators do have limitations. This one warns about entering virtual resistors on (+) leads of each voltmeter.

For bench test I will not use a signal generator. But generate sine wave from an op amp circuit, so everything including PC meters will be totally independent from power lines, if that helps. Will share results here.

With my sincere respects to masters here.

You're taking this well beyond the original thread topic and questions; this will only muddy the conclusions. If you wish to pursue it please remove your post and start a new thread. We have 123 posts on the phase of voltages on a capacitor in a simple RC circuit, I think the thread has run its course.

 

Dear DGElder,

Its the very same circuit you have been analyzing. Only a 10V line added between that and ground. I am still only interested in the original question only. How and why do we see same polarity voltage on both sides of a capacitor under AC. Not asking anything different. Just applying what people have suggested. Feel free to disregard/ignore post #122 if you think its not relevant at all.

Again, thanks for all the wonderful analyses you made. I will print them all and study.

 

Dear DGElder,

I am still only interested in the original question only. How and why do we see same polarity voltage on both sides of a capacitor under AC. Not asking anything different.

That question has been answered, qualitatively in many posts and with mathematical rigor. If you don't understand it by now you never will. Making more complicated circuits only complicates the question.

 

OK, master.

 

Here is an example of calculated and measured results.


View attachment 110205

--

I'm puzzled by your measured results.

You haven't indicated what node on the circuit of post #121 is your reference node. That would be where you have connected the ground clips of your scope probes. Is it the bottom of the resistor R?

Assuming it is the bottom of resistor R, how can the amplitude of the channel 1 and channel 2 signals be so nearly equal?

Could it be that the scope capture is of the circuit shown in post #119 rather than post #121?

I wired up the circuit you show in post #121; 100 nF in series with 1000 ohms. Using the bottom of R as my reference I applied 2V P-P @1kHz to the top of the capacitor.

Channel 1 (yellow) shows Va and channel 2 (green) shows Vb; I didn't bother to show Vb-Va. My result is quite different from what you show in post #121. Would you have any idea why the difference?

 

I'm puzzled by your measured results.

You haven't indicated what node on the circuit of post #121 is your reference node. That would be where you have connected the ground clips of your scope probes. Is it the bottom of the resistor R?

Assuming it is the bottom of resistor R, how can the amplitude of the channel 1 and channel 2 signals be so nearly equal?

Could it be that the scope capture is of the circuit shown in post #119 rather than post #121?

I wired up the circuit you show in post #121; 100 nF in series with 1000 ohms. Using the bottom of R as my reference I applied 2V P-P @1kHz to the top of the capacitor.

Channel 1 (yellow) shows Va and channel 2 (green) shows Vb; I didn't bother to show Vb-Va. My result is quite different from what you show in post #121. Would you have any idea why the difference?




View attachment 110221

Thanks for your interest.

The source voltage and probe measurements are with respect to ground. The AC signal is supplied via 2 channels (180 deg out of phase) of a function generator. So the only ground point in the circuit schematic is inside the Vs source - which shares earth ground with the o-scope. (It is not necessary to connect the scope ground clips to anything, the scope measures with respect to ground regardless -unless it is a portable)

My hook up acts as a balanced supply, yours as a single ended supply, thus the difference. Your yellow probe is measuring the source and green is measuring the voltage across the resistor. Since the voltage on the resistor is in phase with the current the phase difference you see on your scope is equal to theta.

It is interesting that my phi = 2*theta result is actually easy to obtain without all the complicated math if one can just intuit (is that a verb?) that |Va| and |Vb| must be equal. In the phasor diagram you can see that |Va| and |Vb| with |Vr| would form an isosceles triangle making the angle phi and its relation to theta easy to see. I confess I didn't realize this till after I did all the math, but in retrospect I probably should have known. Still it was a nice exercise for the gray matter.

The Vb-Va trace isn't relevant to the verification of my math. I put it there because a poster got stuck on the idea that the phase of the voltage across the capacitor (presumably with reference to source) is the same thing as the phase difference between Vb and Va. The trace shows it isn't.

 

Thanks for your interest.

The source voltage and probe measurements are with respect to ground. The AC signal is supplied via 2 channels (180 deg out of phase) of a function generator. So the only ground point in the circuit schematic is inside the Vs source - which shares earth ground with the o-scope. (It is not necessary to connect the scope ground clips to anything, the scope measures with respect to ground regardless -unless it is a portable)

My hook up acts as a balanced supply, yours as a single ended supply, thus the difference. Your yellow probe is measuring the source and green is measuring the voltage across the resistor. Since the voltage on the resistor is in phase with the current the phase difference you see on your scope is equal to theta.

It is interesting that my phi = 2*theta result is actually easy to obtain without all the complicated math if one can just intuit (is that a verb?) that |Va| and |Vb| must be equal. In the phasor diagram you can see that |Va| and |Vb| with |Vr| would form an isosceles triangle making the angle phi and its relation to theta easy to see. I confess I didn't realize this till after I did all the math, but in retrospect I probably should have known. Still it was a nice exercise for the gray matter.

The Vb-Va trace isn't relevant to the verification of my math. I put it there because a poster got stuck on the idea that the phase of the voltage across the capacitor (presumably with reference to source) is the same thing as the phase difference between Vb and Va. The trace shows it isn't.

Hello again,

You've added a second phase to the circuit here so that makes your circuit a two phase circuit. That makes it a little different, but perhaps still workable. Note that if you change the phase of source 2 you change the result of the phase, which can not happen with a single phase circuit.

What you did and i am sure you will agree with this part, is you put a probe on A and a probe on B, and A was (for reference) 0 degrees and B was 180 degrees with respect to A.

But then you conclude that since the sine at A measures 0 degrees and the sine at B measures 180 that the phase across the cap must be 180 degrees. So lets do that and see what happens...
Va=sin(wt+0)
Vb=sin(wt+pi)
Now let's see what the cap voltage is:
Va-Vb=sin(wt)-sin(wt+pi)

and it turns out that doing that subtraction we get:
sin(wt)-sin(wt+pi)=2*sin(wt)

which is at zero degrees but twice the amplitude.
So the phase of 2*sin(wt) is in phase with sin(wt) so there is no phase shift.

Also, if we connect 10 resistors in series with the single source and look at the voltage at the top, then 1 resistor down, then 2 resistors down, etc., when we get to the last resistor on the bottom we see a much reduced amplitude but no reversal of phase. That's even if we use 100 resistors, 1000 resistors, etc. When we add another phase however, we see a change in phase as we go down past ground. We can make the phase anything we want.

Calculation:
CapPhase=-atan(2*w*R*C)
take the limit as w approaches zero we get:
CapPhase=-atan(0)=0 rads=0 degrees.

More straightforward method using two equal value resistors, one on top and one on bottom of the cap, and a single source...
zC=1/(s*C)
D(s)=R+R+zC
Va(s)=(R+zC)/D
Vb(s)=R/D
Vc(s)=Va(s)-Vb(s)=(R+zC-R)/D=zC/D
Vc(s)=zC/D=1/(2*s*C*R+1)
Vc(jw)=1/(2*j*w*C*R+1)
r=realpart(Vc(jw))=1/(4*w^2*C^2*R^2+1)
i=imagpart(Vc(jw))=-(2*w*C*R)/(4*w^2*C^2*R^2+1)
Ph=atan2(i,r)=-atan(2*w*C*R)

keeping in mind this is the two resistor circuit. I used two resistors to isolate the cap from either supply terminal, but the results will be the same with one resistor just drop the '2' inside there.

If you like next we can look at this from other standpoints:
1. Measurements with the source right across the cap.
2. Distant lossless transmission line measurement.
3. Light beam communications measurements, possibly.
4. Phase locked loop measurements.

 

OK, master.

Hi,



I just wanted to let you know that i think you may have posted TOO much information in that one previous post and that makes it very difficult to go over. If you can summarize your intent in that post with 100 scope pictures i think we might be able to make some sense out of it. Also, maybe say what you were trying to prove there.

 

TheElectrician,

There is another difference in our circuits.

Because my source is a balanced source the maximum voltage it puts across the circuit is 2V, your single ended source puts a max of 1V. I have two poles each with a peak voltage of |Vs|. The +side of my source is |Vs|sin(wt), the negative side is -|Vs|sin(wt). Double your voltage and you will have the same current. You can see that is the case if you look at my voltage definitions, but my schematic could have made that clearer.
(I have since changed Vs symbol)

After doubling your voltage: your circuit and my circuit are identical as far as the components in the circuit are concerned. That is to say the current through the components are identical, the voltage drop across each component is identical and the voltage phase relationships across the components are identical. The only difference is the choice of the reference for measuring single point voltages. You are measuring from the point in my circuit which equals -|Vs|sin(wt) with respect to my ground. So if you added |Vs|sin(wt) to all your measurements you would get the same values I got for Va and Vb. Or if I subtracted |Vs|sin(wt) from all my measurements I would get your voltages for Va and Vb. However, the difference in our ground choice makes no difference in the measured voltage across a component, so if you did the Math function on your scope for Vb-Va, your trace should match my trace for Vb-Va.





.

 

TheElectrician,

There is another difference in our circuits.

Because my source is a balanced source the maximum voltage it puts across the circuit is 2V, your single ended source puts a max of 1V. I have two poles each with a peak voltage of |Vs|. The +side of my source is |Vs|sin(wt), the negative side is -|Vs|sin(wt). Double your voltage and you will have the same current. You can see that is the case if you look at my voltage definitions, but my schematic could have made that clearer.

After doubling your voltage: your circuit and my circuit are identical as far as the components in the circuit are concerned. That is to say the current through the components are identical, the voltage drop across each component is identical and the voltage phase relationships across the components are identical. The only difference is the choice of the reference for measuring single point voltages. You are measuring from the point in my circuit which equals -|Vs|sin(wt) with respect to my ground. So if you added |Vs|sin(wt) to all your measurements you would get the same values I got for Va and Vb. Or if I subtracted |Vs|sin(wt) from all my measurements I would get your voltages for Va and Vb. However, the difference in our ground choice makes no difference in the measured voltage across a component, so if you did the Math function on your scope for Vb-Va, your trace should match my trace for Vb-Va.





.

Hello again,

Why on earth would you want to do Vb-Va ?
Phase is not measured like that either, it has to be Va-Vb.
See post #129.

 

TheElectrician,

There is another difference in our circuits.

Because my source is a balanced source the maximum voltage it puts across the circuit is 2V, your single ended source puts a max of 1V. I have two poles each with a peak voltage of |Vs|. The +side of my source is |Vs|sin(wt), the negative side is -|Vs|sin(wt). Double your voltage and you will have the same current. You can see that is the case if you look at my voltage definitions, but my schematic could have made that clearer.

After doubling your voltage: your circuit and my circuit are identical as far as the components in the circuit are concerned. That is to say the current through the components are identical, the voltage drop across each component is identical and the voltage phase relationships across the components are identical. The only difference is the choice of the reference for measuring single point voltages. You are measuring from the point in my circuit which equals -|Vs|sin(wt) with respect to my ground. So if you added |Vs|sin(wt) to all your measurements you would get the same values I got for Va and Vb. Or if I subtracted |Vs|sin(wt) from all my measurements I would get your voltages for Va and Vb. However, the difference in our ground choice makes no difference in the measured voltage across a component, so if you did the Math function on your scope for Vb-Va, your trace should match my trace for Vb-Va.

Mr. Electrician;

To verify what I said above I setup the same circuit but with a single ended supply just as you did, grounded at the bottom of my schematic. Except I doubled your source voltage to match mine. You can see in the traces below that the scaled values of Vb and Va match yours with the same phase relationship, phi or theta. And you can see the voltage of Vb-Va still matches my original Math trace [Vb-Va] with the balanced supply.


Single Ended Source:





Original Setup with a Balanced Source:




When measured with respect to ground and
with a symmetric or balanced circuit PHI = 2(THETA),
with a single ended source PHI = THETA

 

My Dear Elder,

Well, sure I see your particular point there. The capacitor midpoint itself can be at +100V relative to actual Earth ground of your building. Which is generally a meter long galvanized steel rod you literally hammer down into ground somewhere in your basement here. For all practical intents and purposes we can assume earth as the absolute ground, final reference point of 0V for all everything else.

In my second circuit below, I made a 10V pp AC source by adding two 5V AC sources in series and putting earth ground between them. So there is no ground relative to anything, but just physical real earth ground. Simulation including both scopes and voltmeters show 2 sides of the capacitor at 180d voltage phase difference. It has to be.

Perhaps one or both of you guys take the time to simulate this same circuit yourself or setup a little actual experiment to show us where I am wrong in my little simulation?

@ozsavran:
The problem with this entire thread is that you are creating examples that are outside the purvue of your original question. Your lack of understanding of phase and of capacitors is evident. And unfortunately nobody else stopped you right there. You need to understand exactly what a capacitor does in a DC environment first-- because you don't. Your thread and question says as much.

 

My Dear BobaMosfet,

Why not be a hero, and enlighten us with your obviously "deep down" understandings, in a step by step fashion for us mere mortals please. So we get something actually useful out of your highnesses mind. Like this:

1 : A cap does this under DC:
2 : This is what phase is:
.... ....
.... ....
n-1 : this is why we can see same polarity voltage on both sides of a cap under AC.
n : this is what electrons in a cap are doing exactly under AC.

C'mon please do not be shy. We will try to be worthy of your deep insights...

Rest assured, I will satisfactorily close down every loose ended, dubious comment in this thread for good, for honest readers who are really trying to understand something, not just memorize ready lines. Don't you worry. You guys have given me enough motivation already.