Unraveling the Mystery of Phase Differences in Passive Circuits

[IssacBinary]

Im having a really hard time what the phase shifts mean in real life in a circuit. I understand them mathematically but not in physicality, mainly because I can't find any explanation.

Now I understand that as current flows through a capacitor voltage builts up. So at max current there is 0 voltage and then voltage builds and current decreases. I understand that, and the 90 degress lag the voltage has IN the capacitor.

I also understand reactance. In the case of the inductor. The self EMF it produces in the opposite direction of the voltage cancel out to leave a resultant voltage and thus a less current in the circuit making it seem like the inductor is acting as a resistor of a certain OHMs at a specific frequencies.

But I don't see how the self EMF from the inductor or the discharge from the capacitor, while it creates the effect of a resistor actually changes the phase of the current in the circuit with respect to the source. I see how it can be reduced...so the real part of impedance, but now how the phase part is create physically and why.

But like I said I understand the phase inside the capacitor and inductor, just not how it effects the over all picture.

 

[DragonPetter]

phase shift in the frequency domain is equivalent to a time delay in the time domain.

Phase is simply a delay from when a voltage/signal appears at an input to when it appears at the output. It can be confusing to interpret this sometimes, because often we talk of phase shift with continuous signals, and so you really lose a sense of the time aspect other than just a phase shift when the scale is -infinity to +infinity. So phase of a signal has to be considered relative to a reference signal, which is usually the input signal.

A good thing to think about phase is with group delay. When a signal enters a system, the phase response of the system will delay different frequency components at different times delays if the phase shift is not linear. Say a guitar and a bass drum sound hit a microphone at the same time, if the microphones phase is not linear over its frequency spectrum, the bass sound could pass through the microphone sooner than the guitar sound did, even if they both hit the microphone at the same time. You would hear this as a distortion in the sound.

Another way to think about phase is in control systems, if a controler's output is delayed too much (ie too much phase shift) vs. how quickly the input is changing, you will get instability problems because your system can't keep up with the changing input.

 

[DragonPetter]

Also I should mention, a capacitor's voltage lags behind its current by 90 degrees, and an inductor's voltage leads the current by 90 degrees.

This is the phase shift in a systems impedance that I'm talking about, but the idea is not strictly applied to just circuits.

I'm not exactly sure what I'm about to say is the correct way of thinking about this, so take it with a grain of salt: When the capacitor's current is leading its voltage, the resistor it is attached to on the other end will receive the current first and create a voltage across it, followed by the voltage 90 degrees later from the capacitor, and so then the voltage of the 90 degrees ahead current through the resistor created combined with the capacitor's 90 degree lagging voltage giving a net voltage across the resistor at all times, and so in a sense you have 2 voltages across the resistor, however their sum is always a real absolute voltage. This is why we represent impedance with complex numbers and phasors, because the real voltage is the magnitude of the complex number, but at any time some of the voltage is in the real part and some in the imaginary "frequency" part.

 

[IssacBinary]

Hmm it still doesn't really add up. I think there might be a lot of terms flying around that arnt properly explained / defined.

Let me give an example.

In the beginning this is what I thought.

A circuit that has an AC source and capacitor. (I know you need a resistor but bare with me)

Now, I thought that, you turn the AC on, the capacitor charges up and then you have a voltage produced by the capacitor coming out the other end that is lagging the source voltage.

So at the side before capacitor will have say 0 voltage and 0 current but on the other side of the capacitor in that part of the circuit will be max current and max voltage. And that this is what the phase shift means.

But then I found out I was wrong as current is the same through the whole circuit. So the current before and after the capacitor is the same.

So then this is when I realized about Reactance. As the voltage starts to decrease from the source (after the capacitor has been charged) the capacitor starts to discharge because the capacitor now has more voltage than what's being applyed to it at that time. This discharge as well as the existing voltage from the source combine together to make a resultant voltage.
This voltage intern creates a current and the current size is as if there was a resistor that has the value of the capacitors reactance.

One thing I don't like is when voltage and current is spoke about it leading and lagging beacuse they can't never lead or lag, there has to be a voltage to be a current, but in this case, the CAPACITORS voltage lags the SOURCEs current. and the current in the circuit is a combination of the 2 voltages.

So now that's all fine...but then the phase shift thing. current in the circuit has a phase shift with respect to SOURCE voltage. But what creates the OVERALL phase shift.

I see how it can create an impedance from the discharges = resultant voltages, but how and where does the TOTAL phase shift respect to source voltage in the circuit come into it / how is it created (qualitatively in the same style of my reactance explanation) ?

This is what I am trying to understand

 

[es1]

But what creates the OVERALL phase shift.

The physics of the capacitor does.

i=C*dv/dt

So using your example circuit let V on the source equal cos(w*t) where w is some frequency in radians.

i = C*d[ cos(w*t) ]/dt = -C*w*sin(w*t) = C*w*cos(w*t + pi/2) = -C*w*cos(w*t-pi/2)

Using this i it is then pretty easy to see why the Z of a cap is 1/jwC@-pi/2.

 

[IssacBinary]

es1 you seem to be explaining why the capacitors reactance is 1/wC. Which I understand fully, and also what it means in terms of what's happeining in the circuit.

Its not that I am talking about, its the whole impedance phase shift I am stuck with. What is the mechanics that's causing the current at any point in the circuit to be out of phase with the source current.

 

[es1]

the phase is the -pi/2 part it falls out directly from the math
do you understand why the mechanics of a cap is i=c*dv/dt?
that same mechanics will explain the phase

 

[DragonPetter]

es1 you seem to be explaining why the capacitors reactance is 1/wC. Which I understand fully, and also what it means in terms of what's happeining in the circuit.

Its not that I am talking about, its the whole impedance phase shift I am stuck with. What is the mechanics that's causing the current at any point in the circuit to be out of phase with the source current.

I think perhaps you could say that, in a capacitor, the electrons are polarizing the plates, or the energy is being transferred into the dielectric, before it can pass through. A capacitor is an energy storage device after all, which is the whole reason for it's reactance, and the actual mechanics of a capacitor is the transfer of charge between the plates of the capacitor. It takes time for the electrons to move between plates, and so the voltage is not instantaneous.

 

[IssacBinary]

I understand what's happening inside the capacitor and inductor. Also why the reactances are what they are and what it means.

So
As the voltage starts to decrease from the source (after the capacitor has been charged) the capacitor starts to discharge because the capacitor now has more voltage than what's being applyed to it at that time. This discharge as well as the existing voltage from the source combine together to make a resultant voltage.
This voltage intern creates a current and the current size is as if there was a resistor that has the value of the capacitors reactance. Without the capacitor there would be no opposing voltage and the current will as if that resistor isn't there.

Is that right? As in what is happening in the circuit and what reactance is?

What I am after is how by having the capacitor or inductor it effects the OVERAL phase shift. i.e total imepdance phase, in the circuit. Just like how the capacitor effects the overall current running in the circuit via the reactance explained by my analogy I am looking for a similar style of explanation in what causes the OVERAL phase shift respect to source voltage.

So the current at ANY point in the circuit - not inside the capacitor - what and how is the phase shift created / why.

Lets say we had a resistor at 10Ohms and Capacitor Reactance at 20Ohms.

Total impedance in the circuit would be 22.36Ohms at -63 degrees.

How does the -63 work. I know it means the current in the circuit will be leading source voltage by 63 degrees but HOW, what is happening to make it that / actually create a phase shift from source.

 

[es1]

The reactance explanation is kinda close, but not really right. The contradiction is there is no series R in the example but the discharge current is still creating a voltage.

I think you're hoping for an explanation that doesn't have any math but is purely physical and in the form A then B then C thus phase. I am not sure i can provide one because to me the two are not separable.

You also have to keep in mind phasors assume sinusoidal inputs. They break down if the input is something else.

To answer the original question, physically, for this circuit, phase is just the amount of time between max current and max voltage (or zero current and zero voltage, etc) when the input is sinusoidal.

For a resistor, max current and max voltage happen at the same time. Therefore is has zero phase.

For a cap to have a change in voltage there must be a change in charge, but a change in charge is a current. For a sinusoid, when max voltage happens the change in voltage (the derivative) is zero, so 0 current. Therefore max current and max voltage are not simultaneous for this device so there is some phase.

The physical dimensions of the plates, the dielectric, and the series resistor (which limits the current which changes the voltage of the plates), etc. all work together to determine what the phase difference is because what matters is how much charge needs to be moved to get the requisite dv and how long to I have to move it.

 

[IssacBinary]

I understand about the capacitor, and ITS graph. The graph of current and voltage in/across THE capacitor itself.

From the start the most current is flowing "through" it as there is no charge stored to oppose it. But as it starts to charge it opposes the current thus current starts to decrease, but as this is because the charge in the capacitor is increasing the voltage across it is increasing. For DC anyway.

In AC its just every second there is a bit more current that's able to push a bit more charge into it and the difference in current going into it and the charge that's already in there is what allows the capacitor to continually build up, rather than flat lining in DC.

That I understand, and because of the way I understand it, I understand the maths involved, differentiation and the formulas, it sits into place and makes sense as it matches up.

Its all well and good remember the equations off by heart but for me if I don't actually understand or know what they mean in the circuit and what is happening its pretty pointless.

For the resistor as well, it just resists the current so its just going to have less voltage and current come out the other side but still in phase.

Ok so to fix my example I would just have a resistor before the capacitor, and that the voltage after the resistor is what's going to determin the max voltage across the capacitor, and once the voltage after the resistor is lower than what's current across the capacitor it will start to discharge and then the 2 voltages (waves) combine together to create a resultant voltage.

Now I would have though that its just the combination of the 2 waves that creates the phase. But I know that adding 2 sinusoids does NOT make another sinusoid so it can't be just the 2 waves overlapping creating a shifted resultant wave.

So I can see how it will lower the current in the circuit and that's what the reactance shows but again, how the resultant current in the circuit which has a value of i = v/z (z = sqrt of R*2 + Xc*2) (which shows how the capacitor effects the overall current in the circuit) actually has an overall phase shift with respect to the source voltage.

...On a side note, why is the resistor needed? why can't the current just be determined by the reactance alone because if there was no resistor z would just be the caps reactance...

 

[I_am_learning]

. But I know that adding 2 sinusoids does NOT make another sinusoid so it can't be just the 2 waves overlapping creating a shifted resultant wave.

I haven't much read other parts of your texts (I just skim read), But I was caught at this.

You actually DO get a sinusoid if you add two sinusoids (provided they have equal frequency).

Consider two sinusoid of equal frequency(say 1Hz) and amplitude. Align one of the peak of 1st sinusoid to time t = 0. Align the peak of the other sinusoid at t = 0.25.
So, we call the second sinusoid having a phase lead of 90 Degree (1/4th of Time Period Lead)

Now, add these two, points for points. You WILL get another sinusoid.
Thats why phasors are used. To simplify addition and subtraction of sinusoids with phase shifts.
Using phasor
A = 1<0
B = 1<90

C = A + B = sqrt(2) < 45. Simple.

Sorry if you already knew these.

 

[IssacBinary]

Ah damn what was I thinking. Your right, I overlooked the situation when they are in fact the same frequency.

I think that could possibly be it?

So its safe to say the discharge has the same frequency as the source.

Ok so maybe I've got it.

with have an AC source, resistor and capacitor

1) the capacitor can only charge up to the max voltage after the drop,
2) once the voltage after the resistor is less than what the capacitor is charged up to, the capacitor starts to discharge
3) the voltage produced from the capacitor is 90 degrees lagging the source voltage
4) the combination of the 90 degree lag smaller voltage going left + the original source voltage going right add together (superposition) to create a final resultant voltage in the circuit of a certain amplitude and phase.
5) but as it takes time for the capacitor to charge, the frequency of the ac source will effect how much charge the capacitor will have once it reaches its peak voltage. Thus meaning at higher frequencies less voltage will be discharged meaning the resultant waves amplitude will be effected less than at lower frequencies. This then shows why reactance is 1/wC. As the reactance is in Ohms, the capacitor will look like a smaller and smaller resistor...backing up the fact the resultant wave will be closer in amplitude to the source voltage than at lower frequencies.

 

[IssacBinary]

Ive made a few graphs using excel, and after adding wave 1 with amplitude of 10 and wave 2 of amplitude 6 with a 90 degree shift, I get a resultant wave that has an amplitude larger than 10.

However, changing the second waves amplitude does effect the resultants phase with respect to the first wave.

So what is wrong in my explanation? I am sure there's just one tiny piece I've forgot/missed out/haven't realized...hm

Any ideas?

 

[es1]

What was the result you expected?
A*sin(x)+B*sin(x+pi/2)=A*sin(x)+B*cos(x)
So no change of A or B should move the phase.

The peak of the result should be higher than both A and B.

http://www.wolframalpha.com/input/?i=Plot[10*Sin[t]+++6*Sin[t+++\[Pi]/2]+,{t,+0,+4+\[Pi]}]

 

[IssacBinary]

http://www.wolframalpha.com/input/?i=10sin+x,+6sin+(x+++pi(7/9)),+10sin+x+++6sin+(x+++pi(7/9))

Ive plotted 3 graphs.

AC source with amplitude of 10
Capacitor that gets charged up to 6 and its discharge is out of phase by 140 degrees.

You can see the resultant wave has smaller amplitude than the source and also by changing the phase shift of the discharge it will change the phase shift of the resultant relative to the AC source.

Another example.
http://www.wolframalpha.com/input/?i=10sin+x,+10sin+(x+++pi(7/9)),+10sin+x+++10sin+(x+++pi(7/9))

!Explanation Fix!

Ive tweaked my theory / explanation a bit.

1) the capacitor can only charge up to what the capacitors time constant let's it to at a given frequency. Higher frequency will mean it has less charge thus less voltage across it because it has less time to charge up. So capacitor is charged to something less than AC source.
2) once the voltage after the resistor is less than what the capacitor is charged up to, the capacitor starts to discharge
3) the voltage produced from the capacitor is 90 degrees lagging the source voltage
5) Due to higher frequencys meaning the capacitor has less charge and less voltage, for it to be able to discharge it needs to wait longer for the AC source to meet the capacitors voltage and go less. This would mean there would be MORE than a 90 degree lag with the discharge. As say the AC source peaked up 10V and the capacitor only charged to 2V then it has to wait to almost the end of the cycle for it to be able to discharge (when AC source reaches back down to 2V and less). This makes the phase shift closer to 180 and not 90 (in this case).
4) the combination of the capacitors discharge smaller voltage going left with over 90 degree lag + the original source voltage going right add together (superposition) to create a final resultant voltage in the circuit of a certain amplitude and phase.

How does that sound?

Please bare in mind all this I am coming up with is on my own (with the help of you guys) as anywhere else I look it is just equations and I am trying to rewrite it in physical happenings.

So any help on fixing my "theory" or pointing out where I am going wrong if at all would be helpfull.

 

[sophiecentaur]

The whole thing just rests on the fact that the Voltage on a Capacitor is proportional to the Charge in it and the Current that can flow into it depends upon the Voltage across the series resistor.
What is wrong with just allowing that to be applied to find what happens when there is a sinusoidal voltage connected to the resistor? The Maths describe what happens so much better than arm waving (and in about three lines, too). It's just as valid a description as the subjective,"physical" one and cannot be misinterpreted. Taking particular instances of Voltages and Frequency does not prove the general case and, although useful for plotting graphs, spreadsheets and simulations in general are only aids to displaying what happens; they don't explain anything.

 

[wbeaty]

Would it be possible to check out my thread

I understand the concept of phase but not how it comes about. What CREATES it, the mechanics behind it. How. Everywhere I look just gives equations.

This comes close to Feynman's "What I cannot create, I do not understand." If you can SEE all the machinery inside the components, get a feel for the interconnected phenomena, and even poke at it until you know the behaviors at limits ...then you understand it. You uploaded the entire thing into your head. That's the physicists' secret trick: rules-of-thumb and estimation skills. Then, once you know how it all works, you can whip out the math tools and obtain some exact values. (And, if you do it this way, then you'll be half way to expert status: you'll have a good idea about how things in the circuit will change if you adjust things like that or like that. But still you'll need the math to get exact values of the changes.)

But I don't see how the self EMF from the inductor or the discharge from the capacitor, while it creates the effect of a resistor actually changes the phase of the current in the circuit with respect to the source.

Ah, that's because it doesn't *change* an existing current. Instead, without a capacitor connected, there was no current there to begin with. It doesn't shift a current, instead it acts like a load which is drawing a lagging current. Was that your problem? Another problem: we can't suddenly connect a capacitor across an ideal voltage source or across a square wave without creating infinite current.

But back up a little. It might help to ditch the sine waves at first. Instead, visualize what happens when we connect a capacitor in series with a resistor, then apply a square wave voltage source across the two. (Assume that the capacitor voltage started out zero.)

If the square wave suddenly goes positive 1.0V, then a charging current appears in the circuit, and the 0Vdc capacitor voltage begins rising towards +1.0V. OK so far? If the cap value is small and the square wave period is far longer than RC, then the capacitor voltage will climb very close to +1.0V. But what happens if instead the cap value is large, and the square wave quickly switched negative? In that case the capacitor current reverses, and the voltage across the capacitor starts falling towards -1.0V. If the period of our square wave is lots faster than the RC time constant, then the capacitor voltage never had any chance to approach 1.0V before the square wave reverses it's polarity again.

In that case, the waveform of the capacitor voltage will be a triangle wave.

OK so far? Two voltage signals: square wave in, triangle wave out. Fast square wave, and a slow-charging long RC constant.

Now ask yourself this: is that triangle wave in phase with the square wave? (Go and change both waves a bit, so they look closer to sine waves.) Nope, they're not in phase. The peaks of the triangle wave don't line up with the middle parts of the square wave. Instead they line up with the rising and falling edges of the square wave.

So, if both waves are morphed into sinusoids, then we'd see that the capacitor voltage lags behind the drive voltage by ~90deg.

Why? Obviously it's because the drive waveform turns on first, then the capacitor voltage rises later. And the capacitor voltage hits its max value exactly when the square wave reverses it's polarity.

NOW the AC math tools come in. Given R, C, and jw values, you can show that the output phase lag is very close to 90deg for large RC and high frequency drive voltage.

Then go to http://www.falstad.com/circuit/ circuit simulator. Set up the same squarewave capacitor circuit and watch the charges flowing inside the schematic connections. I used the given 40Hz square wave, 100ohm resistor, but changed to 300uF cap. Add analog outputs to the drive and to the capacitor, and turn them on as scope displays. You can even add a SPDT and a 40Hz sinewave source, so you can easily switch back and forth from square driver to sine.

When you understand all that, then set the cap value to 1u and see how the phase lag now gets very close to 0deg.

 

[IssacBinary]

This comes close to Feynman's "What I cannot create, I do not understand." If you can SEE all the machinery inside the components, get a feel for the interconnected phenomena, and even poke at it until you know the behaviors at limits ...then you understand it. You uploaded the entire thing into your head. That's the physicists' secret trick: rules-of-thumb and estimation skills. Then, once you know how it all works, you can whip out the math tools and obtain some exact values. (And, if you do it this way, then you'll be half way to expert status: you'll have a good idea about how things in the circuit will change if you adjust things like that or like that. But still you'll need the math to get exact values of the changes.)

Perfect perfect perfect! That is exactly my thoughts on science and engerneering and its what I've been trying to say to people. Learning equations is going to get your only so far but understanding mechanically what's happening or what things actually mean is what its about, and if I am doing a degree (which I am) I don't want to be spending 4 years and all this money just so I know where to apply an equation I want to know WHAT is happening and how. Otherwise I could just go and buy a formula book.

And that once you understand the concepts, the maths behind it just sits on top and slots in, and when you do understand it you don't have to blindly remember 100s of formulas as you understand what's happening which helps retain it all.

Sometimes when I start to ask questions, I understand it may not be 100% clear but in their explanations you wonder if they know them self what's happening...

Anyway,

Ah, that's because it doesn't *change* an existing current.

Well I didn't mean change like that, I meant it would make the current behave differently than if the capacitor wasn't there.

I understand your square wave example.

I understand what is happening INSIDE a capacitor or inductor.

You turn the circuit on, current flows through the capacitor at the beggining, as it is uncharged there is no voltage across it so there is nothing opposing the current. As current passes through it, charge builds up inside create a voltage across it which starts to push back and opposes the current.

One the capacitor is fully charged it will be the same voltage as the source because the source is only strong enough to push its own amount. If it was to push 15v onto the capacitor when the source was only 10v, where is the extra 5V coming from.

Obviously charging isn't instantaneous. So the faster the frequency the less time the capacitor has to actually charge up, thus less charge is stored in the capacitor and there will be a smaller voltage across it. This would then mean there is less of an opposition to the current.

So high frequency means lower opposition, also larger the capacitor the longer it takes to charge up.

So this then "proves" 1/wC is its reactance.

This is the kind of explanations I am after.

This also shows that inside the capacitor there is a 90 lag. Max current at begging means no voltage across it. As time goes on and voltage increases current goes down, when there's max voltage there is no current.

90 lag.

I am 200% fine with the workings of the capacitor as its own unit, its lag, its reactance. etc

200% fine. (Unless there's a mistake in my explanation)

What it is that I am stuck on is when you put it into your circuit.

The TOTAL IMPEDANCE PHASE part

So let's say a 2k Ohm resistor in series with 3K Ohm reactance capacitor.

The total impedance "resistance" part will be.

= sqrt ( R^2 + Xc^2) = sqrt (13k) = 3605 Ohms.

3.6K Ohms is the total impedance and is as if you replace the capacitor and resistor with one 3.6K resistor.

The current will just be I = V/Z using Z as 3.6k.

BUT

The next part

-tan^1(Xc/R) = -tan(3k/2k) = -56.3 LAG

So total Impedance will be

3.6k Ohms and -56.3 phase shift.

ITS THE -56.3 PART I want to understand

Now I understand the 90 degree shift INSIDE the capacitor.

And I am trying to find an explanation similar to my capacitors explanation at the top that will explain how the capacitor and resistor work together and what's the "mechanics" inside this circuit that is causing the OVERALL -56.3 Phase shift in the circuit.

If the capacitor WAS NOT there and we just used 1 single 3.6k resistor there would be NO overall phase shift in the circuit. So what is happening to cause this?

This is quite clearly NOT 90. The 90 Degrees is what's happening INSIDE but the 56 is the OVERALL picture of the current at any point in the circuit and its phase shift with respect to the SOURCE voltage.

Whats the pushing and pulling, the mechanics, the physical properties WITHOUT any maths to explain the OVERALL / TOTAL / RESULTANT phase shift in the current with respect to source voltage.

Unless I am completely miss understanding what the phase part of a circuits TOTAL impedance means?I really appreciate everyones time, I know I may be a pain but I really feel strongly about how electronics and science and anything technical in general should be learnt, and its these explanations that are just lacking in every place I seem to look. Hopefully this won't just help me but anyone else that comes to work with capacitors and inductors and wants to understand the phase shifts in the circuit OVERALL.

 

[sophiecentaur]

You say you understand the phase lag "inside" the capacitor. There is NO phase lag when a capacitor is connected to a voltage source. It is only when there is some resistance involved that there is a phase lag. With no series resistance, the current flows intantaneously and the capacitor charges to its appropriate value as soon as the voltage is changed.

The two previous posts both go to show what a waste of time it is to discuss simple circuits by waving your arms around. Just do the Maths and you get a perfectly understandable answer which means the same thing to everyone. Electrical theory is not like Social Science or Art Appreciation, both of which are invaluable in their own context. Anyone who wants an explanation which doesn't involve the Maths is wasting their time.

 

[wbeaty]

You say you understand the phase lag "inside" the capacitor. There is NO phase lag when a capacitor is connected to a voltage source.

I think he means the -90deg capacitor current phase wrt the capacitor voltage.

Capacitor equation I(t) = C* dV(t)/dT, the 1st derivative of sin(t) has what value of phase wrt sin(t)?

 

[es1]

Capacitor equation I(t) = C* dV(t)/dT, the 1st derivative of sin(t) has what value of phase wrt sin(t)?

This was the explanation given in post #5 and was deemed insufficient.
And I think that fact that we're arriving back at it makes Sophiecentaur's point.

 

[es1]

IssacBinary,

I think you would agree that many circuit can exhibit "phase" and not all of those circuits include capacitors. So just to make sure we're all talking about the same thing, if you had to give a general definition of phase that could be applied to any circuit, what would that definition be?

 

[IssacBinary]

Hi Es1,

Im really not sure how else to explain what I am after.

I have to say though, I completely disagree with the point made by sophiecentaur.

Anyone who wants an explanation which doesn't involve the Maths is wasting their time.

Firstly I am not asking for an explanation which doesn't involve maths. I said I UNDERSTAND the maths behind it. Like I just showed in my post #19. If I was given a question I could work it out.

What I meant by doesn't involve maths was, an explanation that uses physical happenings.

Once you understand what is happening the maths makes a lot of sense and just sits on top of the explanation.

Anyone can learn the maths and do it, but to understand why its happening or what the maths mean in terms of what's happening and why is another level.

I think if you read my post #19 you should see what I mean.

At the beginning I given an explanation of the 90 lag inside the capacitor and also an explanation of a capacitors reactance using NO maths.

Using this explanations its then makes sense and you understand what's happening, so THEN when you do the maths its makes it a lot easier as you know what your working with and why the numbers are what they are etc.

...The bottom half of my post is showing what I am trying to understand.

What is causing the overall phase in the circuit. At the end of the equations it gets to an overall lag of -56.3.

What is happening to cause the lag to be that at any point in the circuit. What is the physical happening that is causing it.

I think wbeaty is on the same level as me and possibly can help me out, but just need him to see what I am after. ha

 

[f95toli]

What I meant by doesn't involve maths was, an explanation that uses physical happenings.

Once you understand what is happening the maths makes a lot of sense and just sits on top of the explanation.

Anyone can learn the maths and do it, but to understand why its happening or what the maths mean in terms of what's happening and why is another level.

That is not how physics or even engineering works. The bottom line is that if you understand the math you understand the problem; trying to look for a "physical reason" in this case is rather pointless. This is true for almost everything in physics.

Note that this does not mean that there is no "deeper" explanation for what phase is; but that explanation would inevitably involve more math. "Phase" is actually quite a "deep" concept, and in order to go beyond circuit theory and understand what is happening in terms of transport theory (which I suspect is what you are after) you need some rather advanced solid state physics (beyond anything you'd learn as an undergraduate at university) which in turn requires a fair bit of quantum mechanics (which no one understands without math). Example: if you work with superconducting electronics the phase of the circuit is directly connected (it is essentially the same thing) to the phase of the pairing wavefunction of the copper pairs of the condensate.

 

[IssacBinary]

So your telling me you don't agree with

This comes close to Feynman's "What I cannot create, I do not understand." If you can SEE all the machinery inside the components, get a feel for the interconnected phenomena, and even poke at it until you know the behaviors at limits ...then you understand it. You uploaded the entire thing into your head. That's the physicists' secret trick: rules-of-thumb and estimation skills. Then, once you know how it all works, you can whip out the math tools and obtain some exact values. (And, if you do it this way, then you'll be half way to expert status: you'll have a good idea about how things in the circuit will change if you adjust things like that or like that. But still you'll need the math to get exact values of the changes.)

Also,

Lets say someone who has no idea of what electricity is and says, "What is voltage?"

what your saying is an equivalent of giving an answer of
V=IR

that doesn't help to explain what voltage is.


I said I can do the maths, I can do the questions. Like I already did in my post #19.

Impedance was 3.6k Ohms and -56.3 phase shift. Fine.

But WHAT is causing a total phase shift in the circuit of 56.3.

But what your saying is, that that should be fine? If you can work it out you know what is happening?

Im looking for an explanation to what is causing the 56.3. It means at any point in the circuit there is a lag of 56 between the source voltage and current.

But I am just looking for a domino effect type of explanation on how the current in the circuit as a whole can have a phase shift.

 

[f95toli]

I think you misunderstood what Feynman meant. I am not saying that you should not understand the problem, what I am saying that the way to understand it is to understand the math and the mathematical models (including their limitations). I often work with problems in quantum mechanics and I frequently use "mental pictures" of what is going on, but the "picture" in my head has of course nothing do to with what is going on in the real world. A typical example would be to visualize the time evolution of a two-level system by using the Bloch sphere. If you've understand the sphere, you've understood a LOT about the physics of e.g. electrons. Another example would be to use Feynman diagrams whcih is another "visual tool".
This is what Feynman meant (note also that Feynman was a proponent of "shut up and calculate" school of thought)

I seriously doubt that there are that many engineers or even physicists that really understand what voltage "is". It is another "deep" concept which at even the basic level require to have some understanding of things like Fermi surfaces etc. That said, one can of course almost always get away with thinking about it as the potential energy; but phase is much trickier.

 

[IssacBinary]

The voltage thing was an example.

You could say the same thing with force. "What is force?" F=ma.

But it doesn't help to understand / visualise what force is.

But saying something like, "The amount of energy in a movement" then it makes sense.

But I was just using these as examples.

Im not asking for explanation of what PHASE is. Its just the time difference between the same points on 2 different waves for example.

Im asking what creates the overal phase shift in the circuit.

If you read my explanation of the capacitors reactance and capacitors phase shift (inside). then you will see the type of explanation I am after.

So in my example -56.3 means that anywhere in the circuit the current is out of phase with the source voltage by 56.3 degrees?

 

[es1]

Im not asking for explanation of what PHASE is. Its just the time difference between the same points on 2 different waves for example.

But this is the key to the problem. Why follows directly from what.

You say you understand why Q=CV. So then you understand i=Cdv/dt.

i.e., For a cap to have a change in voltage there must be a change in charge, but a change in charge is a current. This is all the physical understanding you need, everything follows from it.

Let's pick the max as the point in your definition.

Then the "thing" that creates phase, using your definition, is that max i and max dv/dt do not occur at the same time for this input waveform.

For a sinusoid, when max voltage happens the change in voltage is zero, so there is 0 current in the cap. Therefore max current and max voltage are not simultaneous for this device, so there is some time difference, so there is some phase.

If you want to know how big the time difference is, you have to do the math.

 

[IssacBinary]

Right, but now we are back at a capacitor.

Im asking for an explanation that uses the cap to create an overall phase shift in the circuit.

The shift in the cap is 90 degrees but the total impedance - overall phase shift - in the circuit is 56.3.

What and how is causing the overall of 56.3

 

[es1]

The what is the fact the i is being restricted by the R. As the cap changes V the R changes V too.

 

[wbeaty]

That is not how physics or even engineering works.

Nope. That's exactly how physics works. See the book "The Character of Physical Law," the section about Babylonian approach to physics. A bit of it is online in various places:

http://www.google.com/search?q=babylonian++feynman

But in double-E, we only need the math models. To do design, we have zero need to understand the internal physics of components. An equation is enough, and "understanding" is not the goal. So perhaps the real problem is that this is the EE section of bbs.

It seems that asking physics questions here just makes everyone annoyed and defensive, since it's not their area of expertise, so they don't know how to answer.

 

[wbeaty]

What and how is causing the overall of 56.3

How about this.

You're OK with the physics behind a capacitor having 90deg leading current, right? (not lagging.) And a resistor has 0deg in-phase current? You understand the physics behind both of these?

Well, if you hook a resistor in series with a capacitor, and connect this whole thing to a sinewave voltage generator, the circuit will draw a current that's somewhere BETWEEN 0deg and -90deg phase. And by changing the resistor and capacitor values, you can adjust the current phase to anything you want, between 0 and 90.

Where does the non-zero phase come from? From the capacitor. (After all, it's the only thing in there which can contribute a phase-shift!)

 

[IssacBinary]

Thanks for backing me up wbeaty.

But I think your right, its probably the wrong place to be asking. But to me its not like I am asking for a super in depth on quantum, just what's happening in the circuit.

EE guys seem to be fine in explaning what's happening in a capacitor, with the electrons, repealing one of the other plate, they build up bla bla bla...but that's more than just saying "90 degrees" so I don't really see the problem with my question.

So, this is what I am come up with so far, an AC source, resistor and capacitor all in series...

1) Higher the frequency the shorter the time the capacitor has to charge up, meaning less voltage across the capacitor at its maximum.
2) once the voltage after the resistor is less than what the capacitor is charged up to, the capacitor starts to discharge
3) Due to higher frequencys meaning the capacitor has less charge and less voltage, for it to be able to discharge it needs to wait longer for the AC source to meet the capacitors voltage and go less. This would mean there would be MORE than a 90 degree lag with the discharge. As say the AC source peaked up 10V and the capacitor only charged to 2V then it has to wait to almost the end of the cycle for it to be able to discharge (when AC source reaches back down to 2V and less).
4) the combination of the capacitors discharge smaller voltage going left + the original source voltage going right add together (superposition) to create a final resultant voltage in the circuit of a certain amplitude and phase.

Thus a 56.3 shift as per my example. Which is dependant on frequency and capacitance to calculate reactance which is point 1, 2 and 3, and reactance I had explained without maths in a few posts before...

I realize this explanation is not 100% spot on, but I think its a long the right path but needs to be tweaked...

Hopefully this makes people realize what kind of explanation I am after.

 

[sophiecentaur]

Right, but now we are back at a capacitor.

Im asking for an explanation that uses the cap to create an overall phase shift in the circuit.

The shift in the cap is 90 degrees but the total impedance - overall phase shift - in the circuit is 56.3.

What and how is causing the overall of 56.3

The 'reason' for the arbitrary looking phase shift is that the voltage across the supply is shared between C and R and the R governs the amount of current flowing into the capacitor and the voltage across it.