Unraveling the Mystery of Phase Differences in Passive Circuits

[I_am_learning]

So your telling me if you told someone

I = C d(Vc)/dt

they will be able to see what it means and just come up with the long paragraph explanation which you just told me? about the electrons moving, more there are the more push is needed etc..

I sure don't know anyone that does...

Yes. I also don't find people doing that. But my point is, it should be possible.
What I actually mean is -> the equation is actually a logical build-up of physical things we know.
We first developed columbs law, that's pretty physical isn't it? Two particles try to pull/push each other with force... . Then we use that law to develop the concept of Potential. Then Current. Then we establish that if we place them on the plates of a capacitor a voltage is developed, which is logically established to be Q / C (where C, capacitance is geometry thing). Are you following?

We also establish that due to repulsive nature of electrons, when we try to put more and more charges into a capacitor, it becomes more and more difficult, so when tried by a constant voltage, the current decreases exponentially. This physical phenomenum is written short-hand as
I = Imax(1-exp(-t/RC)).
Then, we establish that, if instead of applying constant voltage, if we apply a voltage that is also varrying (sinusodial) so its not trying to constantly push in more charge but also pulling out, then we can logically derive (which I can't do for the moment, but I think you would be able to do if you try), that although the frequency of current and the applied voltage match the phase wont.

In short, I mean to say that there is no short-cut explanation. If you try and find one, then it may not model all aspect of the real situation. You need to go back to basic theory and then build and accumulate the logical conclusions step by step. Thats why there are chapter1, chapter2 that talks about atoms,electrons,attraction,voltage etc before talking about the power-factor, phase-lag-lead in later chapters.

If you agree with rest of my view and are stuck at only the 'RED' part above, then I may try.

 

[IssacBinary]

Isn't there a serious snag with that argument? If they are just like resistors, then where does the phase shift come from? (Which was what his earlier questions have all been about)
I have a serious problem with any discussion of phase relationship and its 'causes' that doesn't either involve complex numbers or trigonometry - or both.

LIKE resitstors...LIKEEEEE. Like doesn't mean that are 100% the same.

And many posts back... several times I explained the LIKE.

Charges flowing off in one way (capacitor) or flowing in the opposite direction from induced voltages (inductor) and they are having to oppose the source current.

So 5V pushing one way, 8V from source opposite leaves a resultant in original direction of 3V.

THIS IS THE RESISTANCE part of impedance. Impedance is complex and that is the REAL PART.

I said that many times.

So we are working with the imaginary side.

Through the principle of superposition, we can combine these and we end up with a 45 degree overall phase shift.

GREAT! I remember someone saying what does superposition have to do with any of this when I mentioned superposition.

Now, imagine we continue to increase the frequency at which we change the battery output. The inductor's magnetic field keeps getting stronger and eventually it will push back much harder than the physical resistor, which doesn't change. In this case, almost all of the current will go through the resistor, and almost none through the inductor. While the very tiny bit of current going through the inductor still has 90 degrees of phase shift, the large amount of current going through the resistor dominates and we have an overall phase shift of about zero degrees by superposition.

While there is a almost zero phase the current will be almost zero as well right!?.
As a higher frequency increases the inductors reactance (resistance part) so the higher the reactance the less total current flowing in the circuit...so at very high f almost 0 current?..Just want to clear that up.

In this case, the current that goes through the inductor has 90 degree phase shift, but the current through the resistor has zero phase shift. Through the principle of superposition, we can combine these and we end up with a 45 degree overall phase shift

This was my main problem. I thought I understood what was happening 100%. Your post is very similar to my original attempt at explaining it on the FIRST page. But the main part that got me was the surer position part. Take a look at this graph

http://www.wolframalpha.com/input/?i=10sin+x,+8sin+(x+++(pi/2)),+10sin+x+++8sin+(x+++(pi/2))

The BLUE line with amplitude 10 represents source.
The RED line with amplitude 8 represents the opposing discharge from a cap or inductor. Its 90 phase shift. Just like my original thought and what you just said.

The BROWN line is the resultant of the 2 added together.

RED at 8 represents a quite high frequency, as its pushing back almost as much as source, so its high reactance.

I EXPECTED to see a bit of a phase shift..WHICH IS THERE. I am happy...BUTTT
The resultant is BIGGER than either red or blue...But it should be less? In my example amplitude of 2?

Changing the amplitudes you can see the resultant wave shifting left and right.
I THOUGHT THAT WAS IT!
But because the resultant amplitude is bigger that's why I thought there might be something else...

OR...have I just set up the graph wrongly?

But thanks Carl for your help. I can't believe its taken 7 pages to get here! ha

Again I am not looking for something to replace maths. I am looking something to compliment the formulas so we know PHYSICALLY what is going on to roughly cause what's happening.

We have non maths explanations for capacitors, inductors, reactance, current, resistors. So why not overall impedance?

We have already worked out the real part...how the cap or inductor acts like a resistor to effect the AMPLITUDES...due to the pushing back and resultants...but the overall phase?...why has a lot of people started jumping up and down and waving their arms saying "YOU CANT DO IT".

 

[sophiecentaur]

I thought Vectors Were Maths. Was I wrong? Isn't complex arithmetic Maths? If all that about imaginary and real isn't just a stealth way of getting some maths into the description then what is it?
I can see that you really don't want to drop this but can you really justify, not just one paragraph but page after page of 'explanation' which involves all sorts of Implied Maths - which may be beyond some people - and try to say that makes anything clearer or easier to understand? Whilst you say you can add volts across reactances to volts across resistance with simple arithmetic then there is no hope for you. Please don't try to confuse the uninitiated further than you have to. Do you not see that many of the readers of your posts may be taking such stuff as gospel and trying to use it to build up their world view of electronics?
I appreciate that you may want a 'pet' view of things. I have said that I have my own. But you do seem to be after a very 'special' and selective form of explanation which has to 'appear' not to be mathematical - when it is, in fact, loaded with implied Maths. I can't see what you are really after, here.

 

[IssacBinary]

Who said vectors where not maths?
You seem to be missing the point big time.

But, all I am going to say right now is this. If you see me saying wrong things and see my understanding as being so wrong and I've got no hope. Maybe try to help out and fix some of my understandings?...After all you are a teacher right?

 

[sophiecentaur]

OK
Don't subtract the voltages across resistors and capacitors by simple arithmetic. Do you understand that's not correct?

 

[IssacBinary]

Yes however,

The voltages I was referring to is the voltage that is coming out of the capacitor due to the charge stored in it, or voltage induced in the opposite direction in an inductor .

Does that make sense?

So your saying that is not right? You can't see it as the first box in my picture?

If that's the case what explanation can you give me to make me understand / fix it.

And also what could I do to adjust that graph if anything?

 

[sophiecentaur]

I see very little point in continuing this. You clearly have a very specific requirement for how you want to be 'taught' this and it is just not on my wavelength. It would seem that there aren't many others prepared to go along with you, either.
Teaching and learning involve two parties and, if someone wants to learn something then they may have to consider shedding some preconceptions if they want a result. As I'very said before, you seem to be demanding a very specific mix of maths and non-maths before you will be satisfied. I think you may need to compromise about that. I am not sure that you even know what it is that you want.

 

[IssacBinary]

What are you asking me to compromise?

I said I can do the maths, I can do the equations, I can do the questions. I can get to the end result.

It sounds like your compromise is, ... if you can do it, just get on with it, and stop trying to work out what it all means and why. If you can do it that's all you need.

So what preconceptions are you saying I need to get rid of?

Im not demanding anything, I am just asking for a physical "long paragraph" explanation on how an overall phase shift is created in a circuit. Not an explanation of the phase in a cap or inductor, those I know and people have said a few times, but when we put them in the cicuit and work out the overall phase shift. That is what I would like explained.

Thats all I am asking.

I can do the maths, I can do the equations, but them alone arnt going to make you understand what's happening...and if you say they can then please tell me.
Please explain to me a long paragraph explanation on what the formula means / how overall phase shift is created but general theory such as "this gets bigger so this gets smaller, this goes that way then it goes this way".

Thats all I am asking.

I don't want to have to just blindly accept that overal phase is -tan(Xc/R). Yes it works, but what is happening

 

[I_am_learning]

As I'very said before, you seem to be demanding a very specific mix of maths and non-maths before you will be satisfied.

Thats how I was feeling too.
He accepts talking about phasor sums but denies talking about differentials.

 

[I_am_learning]

Hi, IssacBinary,
I am finding it interesting to talk to you.
For me the complete explanation of any physical phenomenum is when everything is eplained down to the basic laws i.e. everything is explained in terms of Newtown's law of motion, Max-well's equations and Lorentz force law. Sometimes, its easier to explain things interms of other equivalet law such as the Columb's law and gauss law and I accept that.
Even more, I even accept things when they are explained in terms of 'pseudo law' (like, in Capacitor Current leads voltage by 90 degree) bacause 'pseudo law' has already been logically derived from the basic law.
Whats your criteria for complete explanation?
You must accept some basic laws, there is no other way? Which laws are you ready to accept?

 

[sophiecentaur]

What are you asking me to compromise?

I said I can do the maths, I can do the equations, I can do the questions. I can get to the end result.

It sounds like your compromise is, ... if you can do it, just get on with it, and stop trying to work out what it all means and why. If you can do it that's all you need.

So what preconceptions are you saying I need to get rid of?

Im not demanding anything, I am just asking for a physical "long paragraph" explanation on how an overall phase shift is created in a circuit. Not an explanation of the phase in a cap or inductor, those I know and people have said a few times, but when we put them in the cicuit and work out the overall phase shift. That is what I would like explained.

Thats all I am asking.

I can do the maths, I can do the equations, but them alone arnt going to make you understand what's happening...and if you say they can then please tell me.
Please explain to me a long paragraph explanation on what the formula means / how overall phase shift is created but general theory such as "this gets bigger so this gets smaller, this goes that way then it goes this way".

Thats all I am asking.

I don't want to have to just blindly accept that overal phase is -tan(Xc/R). Yes it works, but what is happening

You are, actually being Very Demanding because you require an answer that Exactly fits some picture, in your head, of how the answer should read. The only answer of that kind would be one that you invent for yourself.

I suggest that you make two lists, one with all the maths (implied or explicit) that you are prepared to admit into an 'acceptable' answer and the other with the maths that you are not prepared to be in there. You will see that you are already allowing loads of maths in there and that the few bits you don't want involve the consequences of complex arithmetic and the Argand diagram and a few basic definitions.

I have scanned through some of your past posts and find a number of inexact terms used.These are not helping you in finding a better understanding.
Here are a few:
"electrons moving":- you must be aware that the electrons may only be 'moving' (that is the mean drift) less than an atomic radius during a cycle of AC. This model has little use in these discussions and need not be involved at all.

"I have said many times I UNDERSTAND the phase inside the capacitor":- this is a meaningless statement and doesn't make it clear whether you mean the phase relationship between the current and volts or something else.

"THIS IS THE RESISTANCE part of impedance. Impedance is complex and that is the REAL PART.
I said that many times.
So we are working with the imaginary side .:- what 'we' are dealing with is a complex value - not a real or imaginary part.

the voltage that is coming out of the capacitor :- volts don't 'come out of' a component. They appear across it - they are a Potential Difference between two points.

What "blind acceptance' is involved with calculating the phase angle in terms of X and R? You seem happy that differentials (rates of change of one variable relating to the value of another variable) are involved. Going from that to the result of differentiating as sine function and a bit of vector addition is no worse than working out the speed of an object if you know the distance and the time. Would THAT be too mathematical and would you need motion to be described in verbal terms of 'how fast' you go and how far?

Could you allow yourself the luxury of a "paragraph of non-mathematical explanation" if you were to have any hope of getting to grips with QM? (I am not talking of Pub-Conversation level understanding here)

"just get on with it, and stop trying to work out what it all means' is not my message at all. My message is keep at it but don't expect anyone else to make that particular conceptual leap for you; it is far too specific a problem. All the tools are there for you to do this yourself; it's a personal thing. There are no 'really's in Science.

 

[NascentOxygen]

The voltages I was referring to is the voltage that is coming out of the capacitor due to the charge stored in it, or voltage induced in the opposite direction in an inductor .

Does that make sense?

Could I suggest that you get into good habits right from the word "go" in your physics studies? Adopting correct terminology is vitally important. Others have pointed out your use of the word "resistance" where referring to reactance. Because the word "resistance" is precisely defined in electronics, feel free to come up with something synonymous but not already taken, perhaps "resisting effect" would suit your purposes?

You also speak of "the voltage coming out of the capacitor". In electronics, we refer to current flow into or out of a capacitor. The voltage is described as being "on the capacitor" or "across the capacitor" (more precisely, across the capacitor plates). If you are meticulous in how you apply these key phrases, observing how one applies to current and the other to voltage, then others will be able to better appreciate your explanations.

As a student, that means you will be best placed to earn the marks you deserve for endeavoring to understand the circuit's operation.

 

[wbeaty]

@I am learning
Well put. The suggestion in a previous post that you can look upon a circuit as a flywheel would never have been made if Maths had been used. Maths involves actual numbers and would have revealed the sort of value for the angular momentum involved with 1/4000 of the mass of the wire and speeds of a few mm/second being involved. A Flywheel??

Yes of course a flywheel. I've seen this analogy in several undergrad physics and EE texts. Capacitors are analogous to springs, inductors are analogous to moving mass (to flywheels,) and if you hook a spring up to a flywheel you obtain an oscillator with a known frequency.

A superconductor energy-storage ring connected in a closed circuit? It's much like flywheel energy storage. Breaking the circuit of an SC ring will give you a violent "inductive discharge," and it's analogous to sticking a piece of wood into the spokes of a spinning bicycle wheel. BANG!

But of course with circuits, the "flywheel" is extremely massive, and is spinning very slowly. (The analogous "mass" is part of the b-field, while the electrons' mass is tiny and irrelevant.)

And because of the usual resistivity of copper, this electron-flywheel is sitting on a pile of sand, or perhaps molasses! If spun, it grinds to a halt almost instantly. Stick a magnet pole into the center of a copper ring. The induced current starts spinning in a circuit, but it halts within microseconds. Add an iron core and you can get this time constant up to a second or so (and there's an Exploratorium exhibit which demonstrates this by using a laminated split-core and a massive aluminum ring made of 1" rod.)

 

[wbeaty]

I can see how they act like a resistor. If its got 5Volts going one way and 8Volts still pushing against it, its going to look like an overall of 3Volts.

Nope Issac, this is still wrong. If you hook an ideal capacitor up to a voltage source, there is only one voltage. No subtraction. When you change that power supply voltage (by twisting the knob on your ideal regulated supply,) there will be a current in the capacitor. No resistors involved.

Remember the capacitor analogy with the iron sphere? Rubber membrane across the center, with both sides full of water? In that analogy, if you slightly increase the pressure difference, it makes the rubber get stretched more. More pressure, more deflection of the rubber. For every value of voltage, the rubber will get deflected by a different amount. OK?

What happens if you slowly and constantly increase the pressure difference across the water-capacitor?

 

[IssacBinary]

I understand I may be using the wrong terms at the wrong times and places, so ill make sure I do my best to keep everything in correct terms.

Ok I am going to try and go through all my understandings and hopefully we can pick it apart and rebuild what I need to finally get my original question sorted out.

In my subtracting voltages example. This is my thinking and reasoning..

If you connect 2 batteries in series. Let's say 8 Volts and 3 Volts and their polarity's are in the same direction you now effectively have 1 battery / voltage source of 11Volts.
However if you reverse the 3Volts battery so now its opposing the 8Volts you effectively have 1 battery / voltage source of only 5 Volts.

The 11 Volts and 5 Volts is what I am referring to as a "resultant" voltage.

So...

If you hook a capacitor to a battery, the capacitor will charge up to the voltage of the battery.

If you then remove the battery and connect the two ends of the capacitor it will discharge through the side it was charged.
So the voltage across the capacitor is in the opposite direction of the voltage across the battery.

As we know caps block DC. But if you have and LED after the CAP and you connect a battery, for a split second the LED will light, then will very quickly get dim and go off.
Because when you first flick the switch there is no voltage across the cap, so there is no opposition.

But the cap starts building its charge. So as its voltage increases (in the opposite direction) they start to cancel each other out.

Then when the cap is fully charged it has the same voltage across it as the battery.

This is when the circuit has no current flowing.

As there is 10V from left to right (across source) and 10V now across the cap going right to left.
Its like a balancing act, nothing can move as they are perfectly pushing against each other.

So a resultant of 0V and thus that's why there is 0amps flowing in the circuit. = Caps block DC.

This resultant voltage in the circuit can be worked out by using the capacitors reactance which is its opposition measure. In DC which is 0 frequency you effectively have 1/0 = infinite. So basically a very high "opposition" measure in ohms. So you could replace a capacitor with a resistor of the same ohms and the effect in the circuit will be the same.

So,

Lets start with just this for now, then ill move on to the next bits after this is verified and or fixed.

Thanks

 

[wbeaty]

You also speak of "the voltage coming out of the capacitor". In electronics, we refer to current flow into or out of a capacitor.

No, if we're being precise, we refer to charges flowing through a capacitor. Or we refer to the current in the capacitor.

"Since a current is a flow of charge, the common expression
'flow of current' should be avoided, since literally it means
'flow of flow of charge.' "
-Modern College Physics. Sears,Zemanski,Richards,Wher
​

Hah! :)

Actually "flow of current," versus "flow of charge" is a major pet peeve of mine because as a physics student I personally experienced disruption of my entire understanding of simple circuits. Once I finally attained an intuitive grasp of the basics, I could finally see my prior misconceptions. The main offender among these was my belief that "current" could flow, as if there was a substance called "current" which could move around inside of wires. (No, the 'substance' is called charge. Electric current is a flow of charge. Charges can flow along, but currents just appear and vanish, same as in rivers.)

 

[I_am_learning]

I understand I may be using the wrong terms at the wrong times and places, so ill make sure I do my best to keep everything in correct terms.

Ok I am going to try and go through all my understandings and hopefully we can pick it apart and rebuild what I need to finally get my original question sorted out.

In my subtracting voltages example. This is my thinking and reasoning..

If you connect 2 batteries in series. Let's say 8 Volts and 3 Volts and their polarity's are in the same direction you now effectively have 1 battery / voltage source of 11Volts.
However if you reverse the 3Volts battery so now its opposing the 8Volts you effectively have 1 battery / voltage source of only 5 Volts.

The 11 Volts and 5 Volts is what I am referring to as a "resultant" voltage.

So...

If you hook a capacitor to a battery, the capacitor will charge up to the voltage of the battery.

If you then remove the battery and connect the two ends of the capacitor it will discharge through the side it was charged.
So the voltage across the capacitor is in the opposite direction of the voltage across the battery.

As we know caps block DC. But if you have and LED after the CAP and you connect a battery, for a split second the LED will light, then will very quickly get dim and go off.
Because when you first flick the switch there is no voltage across the cap, so there is no opposition.

But the cap starts building its charge. So as its voltage increases (in the opposite direction) they start to cancel each other out.

Then when the cap is fully charged it has the same voltage across it as the battery.

This is when the circuit has no current flowing.

As there is 10V from left to right (across source) and 10V now across the cap going right to left.
Its like a balancing act, nothing can move as they are perfectly pushing against each other.


So a resultant of 0V and thus that's why there is 0amps flowing in the circuit. = Caps block DC.

This resultant voltage in the circuit can be worked out by using the capacitors reactance which is its opposition measure. In DC which is 0 frequency you effectively have 1/0 = infinite. So basically a very high "opposition" measure in ohms. So you could replace a capacitor with a resistor of the same ohms and the effect in the circuit will be the same.

So,

Lets start with just this for now, then ill move on to the next bits after this is verified and or fixed.

Thanks

Ok that's quite fine. But a few points. To be precise its not precise to say - "Caps Block DC". As you have yourself clarified, DC current flows for sometime (or longtime?). So you should be wary of such generalizations.
Also, you can't get complete modelling by replacing the Capacitor by its equivalent reactance. This too you yourself have verified because the reactance comes out to be infinity so no current should flow. But current flows for sometimes initially. So, be wary of the generalization this time also. It works only for 'steady state condition' (Know this term?)

 

[sophiecentaur]

These ideas tend to chase their tails. Someone 'knows' the theory of a complicated phenomenon well. They use a simplification or generalisation in an informal way, to get over the idea quickly to someone who can't get the full theory. Next thing you know, the (attactive) generalisation gets used by someone, who doesn't know basic theory, to explain the original sophisticated idea to someone who knows even less. And thus the Science Myth is born and arguments rage on forums like this one.

e.g. You can replace a capacitor with a resistor if you want to explain phase. Excuse me.

 

[sophiecentaur]

Yes of course a flywheel. I've seen this analogy in several undergrad physics and EE texts. Capacitors are analogous to springs, inductors are analogous to moving mass (to flywheels,) and if you hook a spring up to a flywheel you obtain an oscillator with a known frequency.

A superconductor energy-storage ring connected in a closed circuit? It's much like flywheel energy storage. Breaking the circuit of an SC ring will give you a violent "inductive discharge," and it's analogous to sticking a piece of wood into the spokes of a spinning bicycle wheel. BANG!

But of course with circuits, the "flywheel" is extremely massive, and is spinning very slowly. (The analogous "mass" is part of the b-field, while the electrons' mass is tiny and irrelevant.)

And because of the usual resistivity of copper, this electron-flywheel is sitting on a pile of sand, or perhaps molasses! If spun, it grinds to a halt almost instantly. Stick a magnet pole into the center of a copper ring. The induced current starts spinning in a circuit, but it halts within microseconds. Add an iron core and you can get this time constant up to a second or so (and there's an Exploratorium exhibit which demonstrates this by using a laminated split-core and a massive aluminum ring made of 1" rod.)

This flywheel thing is interesting. It is an absolutely lousy 'physical' analogy, bearing in mind the values of the quantities involved BUT the Maths of both systems are pretty much identical. SO, by Isaac's argument, I think, it shouldn't be used as an explanation because, physically, it's not close enough but the Maths are spot on. The 'flywheel' thing is, in fact, a magnetic field - not electrons buzzing around.

 

[wbeaty]

I understand I may be using the wrong terms at the wrong times and places, so ill make sure I do my best to keep everything in correct terms.

Nope, that's not the problem.

When you connect a capacitor to an ideal 6-volt battery, you get infinite current. It lasts for zero time. The capacitor voltage jumps up to 6v instantly. OK? Then to get rid of the infinities we add a resistor.

BEEEP wrong. Big mistake.

Ok I am going to try and go through all my understandings and hopefully we can pick it apart and rebuild what I need to finally get my original question sorted out.

The problem appears when you add the resistor. Or the LED.

That phase-shift you're having trouble with? I understand it pretty well, no math needed at first. From my perspective I can clearly see that it's impossible to understand this stuff if we add a resistor to the circuit. The resistor totally derails our thinking; sends us up a conceptual dead end. We have to avoid the resistor. Bad, evil, naughty resistor, corrupts our minds and derails our search for teh Holy Grail. A SHRUBBERY! Ignore that shrubbery.

A totally different approach is required.

Instead, do it with an ideal capacitor and an ideal variable-DC voltage supply. Or do it with the iron-sphere water capacitor connected to an ideal water pump: a weird special pump which creates constant pressure. No resistances are needed (everything has infinite conductivity, zero friction.)

 

[IssacBinary]

It seems like a few people are jumping the gun. In my explanation I didnt mention anything about phase. I said I wanted to break it down and we go through bit by bit.

So in that post I was just clearing up the opposition part first.

But can you see how the explanation didnt have any maths, but I am not saying I can use that explanation to design a circuit. Also its not a "wrong" explanation like what people have as their "pet idea" to help them get through the concepts.

But yes, it is the phase part that is what I am having trouble with. So wbeaty if you could explain your understanding without any maths like you say then please go ahead.

"Caps Block DC". As you have yourself clarified, DC current flows for sometime (or longtime?). So you should be wary of such generalizations.

It seems this generalization is taught everywhere.

Caps block DC and pass AC
Inductors block AC and pass DC.

But as I have clarified myself it does allow a bit of DC through at first. But it is was just a generalization, but those do seem to be in all the textbooks and also at school and university...

Again when I said about switching out a resistor I was referring to its opposition being the same (after the initial charging). Again I wasn't referring to any of the phase shift stuff just yet.

Im guessing steady state just means after the transitional stages of charging and discharging, when everything has levelled out.

When you connect a capacitor to an ideal 6-volt battery, you get infinite current.

It also has to be an ideal capacitor as well though right?

But yes I can see that, but also slightly not. As a wire and capacitor has negligible resistance there is no resistance in the circuit to slow the current so it would be infinite. However wouldn't it still need time to charge the capacitor to 6Volts due to...
Oh yes I see, the infinite just overcomes any oppositions and time constants.

So yes I can see,

I would like to move on to my other bits of understanding without getting to off course again if that's ok. So could we keep to fixing these bits of knowedge first then we can move forward.

 

[Studiot]

I hope I haven't wasted an hour reading this thread, but 120 posts for something so simple?

What is phase?

Well 'phase' by itself is almost meaningless.

Phase difference is a way of measuring the time difference between two specific events.

The events are the maximum values of a pair of changing quantities. These quantities may be the same eg both voltage or they may be different eg one voltage and one current.

Design engineers often want to use a math free (except perhaps some simple arithmetic ) understanding of how a particular circuit works. That is what does what and when.

We call this a 'walking through analysis' and I will illustrate with a very simple circuit.
If you do much circuit analysis you will often come across the phrase or explanation

"It is impossible for the voltage to change instantaneously across a capacitor so any change imposed on one plate will instantaneously be reflected by the other."

for some action of the circuit. It is very useful to achieve understanding.

In stage 1
I have shown a voltage source and a capacitor connected through a switch.
Let us assume the capacitor is totally discharged and the switch is open.
There is zero voltage across the capacitor and no currrent flows in the circuit.

In stage 2
The switch is closed instantaneously.
The voltage at the top plate takes on the voltage of the source.
The voltage across the capacitor remains zero since it cannot change instantaneously.
But a large current starts to flow. The current is at its maximum.

In stage 3
The current gradually diminishes
The votlage across the capacitor increases as the paltes become charged.

In stage 4
As steady state is reached
The capacitor is now fully charged and the voltage across it equals the source voltage, so the voltage is at its peak.
The current has fallen to zero.

Now the maximum current occurs before the maximum voltage, so we say the current leads the voltage.

go well

 

[IssacBinary]

I hope I haven't wasted an hour reading this thread, but 120 posts for something so simple?

What is phase?

Well 'phase' by itself is almost meaningless.

Phase difference is a way of measuring the time difference between two specific events.

The events are the maximum values of a pair of changing quantities. These quantities may be the same eg both voltage or they may be different eg one voltage and one current.

Design engineers often want to use a math free (except perhaps some simple arithmetic ) understanding of how a particular circuit works. That is what does what and when.

We call this a 'walking through analysis' and I will illustrate with a very simple circuit.
If you do much circuit analysis you will often come across the phrase or explanation

"It is impossible for the voltage to change instantaneously across a capacitor so any change imposed on one plate will instantaneously be reflected by the other."

for some action of the circuit. It is very useful to achieve understanding.

In stage 1
I have shown a voltage source and a capacitor connected through a switch.
Let us assume the capacitor is totally discharged and the switch is open.
There is zero voltage across the capacitor and no currrent flows in the circuit.

In stage 2
The switch is closed instantaneously.
The voltage at the top plate takes on the voltage of the source.
The voltage across the capacitor remains zero since it cannot change instantaneously.
But a large current starts to flow. The current is at its maximum.

In stage 3
The current gradually diminishes
The votlage across the capacitor increases as the paltes become charged.

In stage 4
As steady state is reached
The capacitor is now fully charged and the voltage across it equals the source voltage, so the voltage is at its peak.
The current has fallen to zero.

Now the maximum current occurs before the maximum voltage, so we say the current leads the voltage.

go well

Sorry studiot but it seems like you completely missed the point of this thread. And it seems like you haven't read my posts.

Almost every page I have stated I understand the phase shift that is happening across a capacitor. The voltage across the capacitor lags current through it by 90 degrees.

This is NOT what this thread is about.

Im asking for an explanation that explains physically (like your explanation) what is happening to create an OVERALL phase shift in the circuit as a whole.

Actually, I don't think you did read the whole thread! Did you not see my picture? Or even my circuit example with impedance worked out.

Its the phase part of impedance I want to understand, what is happening to create this.

I can't believe I've just had to explain myself again...

 

[Studiot]

In your rather rude response you clearly show that you haven't understood.

Where did I say the phase difference in my example is 90 degrees?

In your initial post you talk about a phase change.

That is not the same as a phase difference.

A simple capacitor cannot produce a phase change.

A phase change implies that there is an initial phase difference and that the change alters this phase difference.

There is a phase change between the voltage at the base and collector of a transistor.
There is a phase change for a traveling wave reflected at a boundary.

 

[IssacBinary]

I didnt mean to come across as rude. Sorry.

In your drawing and explanation you said basically.

Zero voltage and large current
Max voltage and zero current

which is pretty much implying the 90 degrees shift.

In my example
https://www.physicsforums.com/showpost.php?p=3377063&postcount=19

Its the -56.3 LAG I want explained. But what I want explained is what is happening to cause THIS overall phase shift in the circuit. The mechanics behind it

 

[Studiot]

Zero voltage and large current
Max voltage and zero current

I did indeed say that.

But look at what else I said.
How long does is take for the circuit to get from stage 2 to stage 4?

What does that make the phase difference?

 

[IssacBinary]

Studiot, I hope you accept my appologies.

I think that one line

How long does is take for the circuit to get from stage 2 to stage 4?

might be the answer!

So back to my example.

If the cap was at 8V and the source was 10V there would only be 2V at that point in time.

This is the opposition effect.

Also when there is max voltage across the cap that's when there is no current in the circuit.
When there is 0 Volts across the cap that's when there is max current in the circuit.

But as it takes TIME for the cap to charge, the graph of the cap might not match the graph of the voltage produced from the source. Its this difference which is causing the overall phase shift in the circuit?

So if we have a 10V AC at 0.1Hz. If the cap was to get fully charged at 5 seconds (half a cycle of the voltage source) there would be no phase shift in the circuit as a whole as the voltage and current have already reached 0 at that point.

But if it was to become fully charged at around 3 seconds. Even though the source is producing a voltage (if the cap wasnt there) because it is fully charged no current can flow so its going to make the graph shifted to the left a bit.

Is that correct?

Is that an explanation of the mechanics of the overall phase shift in a circuit?! :O :) ! ?!

 

[Studiot]

Here are two further illustration to ponder.

I have done the first one, which is a transistor.

The working of a transistor are such that when we supply a voltage signal to the base at point A an increase in base voltage corresponds to an increase in base current, so the voltage and current are in phase. The phase difference may not be exactly zero but it is low.

At the output, point B, an increase in base current and voltage result in an increase in collector current. But this means a decrease in collector voltage.
So at the output the current and voltage are now out of phase. ie the phase difference has changed or shifted.

This is what is meant by a walking through analysis.

Now you claim to be fully conversant with phase shift for capacitors so can you do a similar analysis for the second example, which is a simple capacitor?

What is the change in the phase difference between input and output?

Edit I have to knock off for a couple of hours now. I will look in again later

 

[IssacBinary]

Studiot, Did you see my last post? I don't know if you perhaps missed it as you was typing out your post still when I had submited mine...

Anyway.
For your example...

Would the output have 0 phase different? As current is the same throughout the circuit. You couldn't have 2 different currents either side of the cap.

 

[I_am_learning]

Hi, IssacBinary.
Don't feel bad but I want to ask you a question,
Currently are you making any effort in trying to get the answer by yourself OR all your effort are currently only for trying to come-up with a Post that clearly tells what you want to know?
The reason I ask this is because, you seem very good at examining things (Yeah, really, if not who would continiue this thread so long. People like you reply --"Yeah thanks, I got it" after few long replies and move-away when actually they haven't) What I want to suggest is you start out fresh from Chapter 1. Understand Voltage, Current, Charges, Resister and Capacitor. Pherhaps you yourself can put the pieces together. Every tools is in front of you.

 

[IssacBinary]

I don't really understand that line fully,

Currently are you making any effort in trying to get the answer by yourself OR all your effort are currently only for trying to come-up with a Post that clearly tells what you want to know?

Sounds like there's something missing. But I am trying to fix my problem and come up with an explanation myself so yes I am putting effort in. I can't expect to just sit here and hope someone just comes up with it. Its already shown that is going to be very unlikely.

With regards to starting fresh that's what I would like to do and have started.

https://www.physicsforums.com/showpost.php?p=3395984&postcount=115

And you said that post was all fine.

However, I am waiting on someone to just confirm or dis-confirm this post.

https://www.physicsforums.com/showpost.php?p=3396255&postcount=127

Then ill move onto to explain my next part of understanding.

 

[wbeaty]

But yes, it is the phase part that is what I am having trouble with. So wbeaty if you could explain your understanding without any maths like you say then please go ahead.

Here I think is the key you've been missing:

When the iron-sphere water-capacitor is connected across a source of water pressure-difference, the rubber membrane will have a certain unchanging deflection.

If you now start changing the water pressure source, the rubber membrane starts changing its deflection. A water current appears. Very important: you'll find that the value of current is constant.

If OK, please say! :)

The upshot: pressure doesn't cause current. Instead, a changing pressure causes a constant water flow.

Note that there are no resistors involved. We could be using perfectly frictionless water, and everything would still work the same.

Now the capacitor...
When a capacitor is connected across a voltage-based power supply, it "draws a current." Or said more conventionally: the capacitor's current is proportional to the slope of changing supply voltage. If the supply voltage isn't changing, then the capacitor current remains zero. Voltage doesn't cause current. Capacitors aren't like resistors. Instead, changing voltage causes the capacitor current.

So, two important principles:

1. Give a capacitor an unchanging current, and its voltage will constantly change until it explodes.
   
2. Give a capacitor a smoothly-changing voltage, and the capacitor current will be constant.

OK? To understand the phase stuff, first you must completely understand #2 above. Number 2 is the central critical concept. There's no point to going on to the phase explanation until you get No. 2 under your belt. It has to become intuitively obvious.

 

[Studiot]

Now the capacitor...
When a capacitor is connected across a voltage-based power supply, it "draws a current." Or said more conventionally: the capacitor's current is proportional to the slope of changing supply voltage. If the supply voltage isn't changing, then the capacitor current remains zero. Voltage doesn't cause current. Capacitors aren't like resistors. Instead, changing voltage causes the capacitor current.

So, two important principles:

Give a capacitor an unchanging current, and its voltage will constantly change until it explodes.

Give a capacitor a smoothly-changing voltage, and the capacitor current will be constant.
OK? To understand the phase stuff, first you must completely understand #2 above. Number 2 is the central critical concept. There's no point to going on to the phase explanation until you get No. 2 under your belt. It has to become intuitively obvious.

I would like to strengthen what Bill says in post#132, while he is still online so this will be in two parts.

'Voltage' refers to two different physical quantities.
Mixing these up is a frequent cause of confusion.

Voltage is a measure of potential and of potential difference. these are not the same and not interchangeable.

This is not the only area of physics where this happens.
For example Newton-metres are both a measure of work and of moment.
Here there is even more difference.

So when we are discussing circuits we need to distinguish carefully whether we are talking about the potential at some single point in the circuit (as Professor Lewin does for instance) or the potential difference between two points.
Many apparent paradoxes can be created if you do not do this.

 

[NascentOxygen]

No, if we're being precise, we refer to charges flowing through a capacitor.

Actually, we NEVER refer to charges flowing through a capacitor. For the simple reason that they don't! Haven't you studied the theory of dielectrics yet? The dielectric in a capacitor is an insulator. No one says charges flow through a capacitor. To think that they do highlights a poor understanding of the physics of capacitors.

I did originally write "current flowing" because that's the customary phraseology, but I decided it would be more helpful to the OP's concept of circuit operation if I emphasised it as current "flow". I stand by the redundancy, and believe it assists in imparting a clearer concept of electronics fundamentals, your own conniptions notwithstanding.

Actually "flow of current," versus "flow of charge" is a major pet peeve of mine because as a physics student I personally experienced disruption of my entire understanding of simple circuits.

I'm sorry to hear of the turmoil in your early student life, and am pleased to learn that you have managed to unlearn the misconceptions that hindered your progress. Such scars can be long-lasting.

Perhaps your misconception that charges flow through a capacitor is yet proving a major block to deeper understanding?

 

[Studiot]

So here is part 2, building on what I have already said but using the diagram where I have labelled various points around the circuit.

This may appear elementary but try to read it carefully because I think it is the key to your difficulty.

At the outset all points A, B, C, D, E are at the same potential.
this could be zero or have some value say 1 or 100 or 100,000 volts.

The potential difference at all points is zero.

Now introduce some source of voltage say +v between C and D.

Instantaneously
There is a potential difference of v between C and D.
There is zero potential difference between D and E.
There is a potential difference of v between E and A.
There is zero potential difference between B and C.
and crucially
There is zero potential difference between A and B.


So there is a potential difference between D and A.

Consequently there is a current flowing from D to A.

But I stress again at this instant there is zero potential difference between A and B.

So there is no opposing voltage by the capacitor as you proposed in earlier posts.

And A is at the same potential as B.

After some time the current establishes a potential difference between A and B, which are now at different potentials.

Are you with the story so far?

 

[IssacBinary]

Ok, I think we need to have a bit of structure here haha.

It seems like there's bit of information coming in from all over the place but what we need to do is work through it step by step, and I am positive doing it this way we will be able to come up with an explanation together.

Like I_am_learning said. I think we should start from basics of what I know, we build it up, and after I explain each section and build up you guys could fix my holes which could be causing the big problems (if there is any) / or just so we can go through and come up with a good explanation together.

It seems like everyone agrees and is fine with this section.

https://www.physicsforums.com/showpost.php?p=3395984&postcount=115
This is only referring to the opposition effect at the moment. If that's all ok ill move on to the next part.

BUT

I still would like someone to either confirm or deny my thought here.
https://www.physicsforums.com/showpost.php?p=3396255&postcount=127

After we have cleared those 2 posts up then ill move on and hopefully we can reach the end goal together.

But before we do ill reply to the latest posts.

@wbeaty

Give a capacitor a smoothly-changing voltage, and the capacitor current will be constant.

Yes I am 100% fine with that. Heres my "explanation" to that.

Lets say source starts at 10V and increases in 10V increments. Cap is uncharged.
Charge flows into the cap with a "force" of 10V. This repeals electrons on the other plate and we have X charge come out the other end of the cap.
The cap is now charged to 10V.
The source increases to 20V. Charge flows into the cap with a "force" of 20V but as there is already 10V across the cap only 10V of "force" is going to the cap due to the 20 and 10 cancelling each other down.
Due to this, it creates the same amount of current coming out the cap as there was for the initial 10V.
The source increases to 30V but as there is 20V already across the cap, again a net "force" of 10V is going into the cap and thus the same amount of current is coming out.

Give a capacitor an unchanging current, and its voltage will constantly change until it explodes.

To be able to give the capacitor an unchanging current the voltage step needs to increase on each step.
So from 10V to 20V then to 40V then to 80V for example. As this will overcome the cancellation due to the voltage built across the cap and will be able to keep the current constant.

This fits in with the cap exploding and my previous explanation because the net force going into the cap is going to be increasing causing the voltage across it to continually increase.

Until BAM.@NascentOxygen

I may be aware I may have said it but I am 100% fully aware that current is a FLOW of charge and thus a flow of current is a non correct term. A flow of current would basically be describing an acceleration of charge. (i.e Miles per hour per hour)

@Studiot

Yep, I am 100% A OK with that.Honestly I can't see anything wrong with the 2 explanations I've just given, but I am happy to accept there may be errors.

So can we clear them up, and also the posts in the 2 links at the top then I can move onto the next section is need be.

Thanks :)

 

[Studiot]

So there is no opposing voltage by the capacitor as you proposed in earlier posts.

@Studiot

Yep, I am 100% A OK with that.


Honestly I can't see anything wrong with the 2 explanations I've just given, but I am happy to accept there may be errors.

How are these two statements compatible?

Since you clearly don't want to read what others have written (this is the second time)

I withdraw from the thread.

 

[IssacBinary]

I don't understand what your saying is wrong.

Your saying Instantaneously there is zero potential difference between A and B.

So that means there is no potential difference across the capacitor to oppose the source voltage.

Which is what you said.

Can you explain what isn't compatible?

 

[NascentOxygen]

Also when there is max voltage across the cap that's when there is no current in the circuit.
When there is 0 Volts across the cap that's when there is max current in the circuit.

True, sort of. If you are considering a DC source, then you can speak of "charging". But usually with an AC source, we don't. There is a short-lived transient when AC is first applied to a passive circuit, but that quickly dies away and all we are usually interested in is the steady-state AC condition. In the steady state, the capacitor is continually changing its state of charge, and over one cycle it averages out to zero volts. Getting back to your question, maximum instantaneous current in the series R-C circuit will be when the voltage difference between the instantaneous source voltage (a sinusoid) and the instantaneous capacitor voltage (also a sinusoid) is a maximum. This maximum need not necessarily coincide with the peak of the source sinusoid, nor with the negative maximum of the capacitor voltage.

But as it takes TIME for the cap to charge, the graph of the cap might not match the graph of the voltage produced from the source. Its this difference which is causing the overall phase shift in the circuit?

They are certainly related. If the capacitor is able to "nearly fully charge" right up to the AC source voltage, then that means the series resistor must be comparatively small. And a small series resistor does result in nearer to zero phase shift (of capacitor voltage relative to the source voltage).

So if we have a 10V AC at 0.1Hz. If the cap was to get fully charged at 5 seconds (half a cycle of the voltage source) there would be no phase shift in the circuit as a whole as the voltage and current have already reached 0 at that point.

But if it was to become fully charged at around 3 seconds. Even though the source is producing a voltage (if the cap wasnt there) because it is fully charged no current can flow so its going to make the graph shifted to the left a bit.

Is that correct?

Is that an explanation of the mechanics of the overall phase shift in a circuit?! :O :) ! ?!

If it takes the capacitor a long time to fully charge, this means the series resistor must be limiting the current in a major way so indicates that the resistance must be relatively large. And a large resistor causes the phase shift in the capacitor voltage relative to the source voltage to be greater. It also means that the amplitude of the AC voltage across the capacitor will be much smaller than the AC source voltage, because the current is small.

Summarising: in a series R-C circuit, supplied an AC signal, if R is comparatively low, then the circuit is primarily capacitive and the voltage across the capacitor reaches close to the voltage of the source. The current through the resistor (and the capacitor) leads the source voltage by close to 90^o. On the other hand, if R is comparatively high, then it limits the current in a major way. The current will be small, and causing the sinusoidal voltage across the capacitor to be correspondingly small. The circuit will be primarily resistive, and the current from the source (which flows through R and C) will show closer to 0^o phase shift relative to the source voltage.

Regardless of what AC circuit it is connected in, the voltage across a capacitor will always lag by exactly 90^o the current through the capacitor.

 

[sophiecentaur]

@Isaacb
Are you conversant with Kirchoff's Laws?
Do you understand that the 'way' to describe what is happening is NOT to use the word "Force". Volts are not a force - they are a Potential Difference and the Potential produced by the source will be equal to the energy drop on the way round the circuit. How can you hope to 'understand' this if you insist on using your own naive descriptions.
Just what are you after on this thread? Do you want to invent your own, personal, Physics to describe what goes on?
Just get a book, start at the beginning, with some rigorous definitions of the quantities involved and go through things step by step. There is no point in saying, yet again, that you "Understand the Maths" and you "Understand what is happening 'inside' a Capacitor". I think that you do not, in fact, understand it. If you really understood it, there would be no problem and you wouldn't be wanting a home-spun description of the process. You keep refusing to take on board the correct ideas, from a number of people who (believe me) know about these things and you insist on your own models. I really don't see how it can work for you.
You are chasing your tale on this one.
Do you know the saying "You can lead a horse to water but you can't make it drink."?

 

[I_am_learning]

Ok, if you find talking about Force and motion easier then look at this.

The bottom image is the equivalent mechanical 'circuit'. Ok now analyze this circuit. (It must be lot easier for you to understand what's going on here).
The Applied Sinusoidal Voltage is EQV to Applied Sinusoidal Force (using that crankshaft)
The Capacitor is equivalent to the Spring
The resister is equivalent to the friction with ground
The current is equivalent to dx/dt i.e. the velocitiy of the tip of the spring.
Net Charge flow into the capacitor is eqv to the distance x moved by tip of the spring.
Ask if something isn't clear.

 

[wbeaty]

Actually, we NEVER refer to charges flowing through a capacitor.

Oh yes we do. But to be nit-pickingly correct we say "path for current is through the capacitor." (It's always a good idea to be an extreme nitpicker when writing a grammar textbook. Also electronics texts.)

The current path through capacitors is the entire basis of the engineering equations for capacitors. The current in the dielectric is Maxwell's displacement current, and it's identical to the value and direction of current in the capacitor terminals. (And ironically for this thread, it's the one place where the term "current" doesn't equal the term "charge flow.) In the iron-sphere water capacitor, would we say that water flowed through the capacitor? That's not too wrong, but only rubber was "flowing" in the center. Better to speak of paths for current.

dV/dT = I*C

V = 1/C * intgl( I dt )

...with I being the current through the capacitor.

• The path for current is through a conductive material
  
• The path for current is through an inductor
  
• The path for current is through a capacitor.

It's a conceit of the lower grades to pretend that capacitors are somehow different than all other components. But once you go through your undergrad networks course, you'll have that wrong idea beaten out of you. (Or sometimes perhaps not, since I've met engineers who still believe that capacitors differ from all other components in *not* presenting a continuous current path to any circuit in which they're connected.)

I'm sorry to hear of the turmoil in your early student life,

I wasn't clear enough. After curing my misconceptions, I encountered large numbers of people who had the exact same problems. I also learned that the physics education community is aware of this "flow of current" problem. It's mostly discussed by word-of-mouth, but it does appear in Sears/Zemanski.

 

[wbeaty]

With regards to starting fresh that's what I would like to do and have started.

https://www.physicsforums.com/showpost.php?p=3395984&postcount=115

And you said that post was all fine.

Actually that post is very NOT fine. What's written is OK, but it only covers half the story. And your problem is in the other half. (The other half is above, in my post about the iron-sphere analogy.)

 

[NascentOxygen]

Actually, we NEVER refer to charges flowing through a capacitor.

to be nit-pickingly correct we say "path for current is through the capacitor."

Ah, the clatter of rapid back-peddling. And I see you're now acknowledging that we never refer to "charges flowing through a capacitor." (Well, not once we've been shown up as wrong, anyway.

it's the one place where the term "current" doesn't equal the term "charge flow.)

Well, well.

I wasn't clear enough.

Not at all. Your misconception was abundantly clear. But I seem to have set you right, at last. We never refer to charges (sic) flowing through a capacitor.

 

[IssacBinary]

Actually that post is very NOT fine. What's written is OK, but it only covers half the story. And your problem is in the other half. (The other half is above, in my post about the iron-sphere analogy.)

Like I said that post was only to explain the opposition side of things. The modulus of the total impedance complex number. (Is this what you mean when its only half of it?)

Do you understand that the 'way' to describe what is happening is NOT to use the word "Force". Volts are not a force

I do understand that. However the only reason I used the word "force" was just to make it a bit clearer (which didnt work).

To describe what is coming out the other side. Kind of like a Newtons cradle. So 10V source when the cap is at 0V would have 10Volts worth of current coming out the other end.

If there's 20V going in and the cap is now at 10Volts you would still only have 10Volts of current coming out the other end. Due to the net result of the 20Volts having to overcome the 10Volts across the capacitor which is in the opposite direction and will be causing an opposition.

@I_am_learning
That spring and crank seem to be fairly ok.
So the voltage across the capacitor is the energy in the spring. Whatever is left after the crank has overcome the friction is what goes into the spring.

Same as bigger the resistor the less the cap gets charged up to.

So the phase shift (which is my original question) is created by the spring pushing back.

So without the spring when the wheel is at 90 degrees, the shaft is horizontal and thus there is no current as there is no horizontal movement.
But with the spring, even though the source is at 0 Volts (and 0amps) the spring is actually pushing back, and depending on the friction determines how long it lasts for.

So this pushing back is causing the tip to move which means there is a current, even though the source is at 0.

So this is what's causing the overall phase shift between source voltage and current in the circuit?

@wbeaty
Could you take a look at the explanations I gave to your 2 points please. Bearing in mind what I said above about me using the term "force".

@NascentOxygen
Your post before your last one, when commenting on my phase explanation, it slightly sounds like what you said agrees with what I said. I just want to check fully.

Also if anyone else could comment on what I had said in that post. As it sounds similar to the crank shaft?

Lastly, as studiot has withdrawn from this thread. Can anyone point me to what he was referring to?
He just said the 2 things I said are no compatible and I am not reading but didnt actually give me anything to check on or references.

I am reading every post and I am trying to take it all in and piece it together and fix what I know and hopefully answer my question. I do think we are close.

One thing could I ask for is that when someone replys with a bit of information would it be possible to quote what I have said that your fixing so I can see what goes with what. As sometimes I am reading and I am not sure what part goes with what if that makes sense.

It may not seem like it but I am picking bits up.

One thing that is cleared up for me is that the voltage drop across the resistor and across the capacitor add up to the source voltage. I knew the phrase but its been put into perfective now.

 

[Studiot]

One thing that is cleared up for me is that the voltage drop across the resistor and across the capacitor add up to the source voltage. I knew the phrase but its been put into perfective now.

Take care here that may not be the case if you also have an inductor in circuit.

 

[I_am_learning]

That spring and crank seem to be fairly ok.
So the voltage across the capacitor is the energy in the spring.

You need to follow the analogy strictly.

No, sorry. Voltage across the capacitor (q/c) is the Force Developed in the spring (kx). Energy Stored in Capacitor (1/2*q^2/C) is equivalent energy stored in spring (1/2*k*x^2).
(C EQV to 1/k)

Same as bigger the resistor the less the cap gets charged up to.

I hope you understand that its the case only for AC. In DC, how large be the resister the Cap gets charged upto the source voltage. In ac the cap don't get charged to anything, its voltage is varying continuously. But It has a peak, which will decrease if we increase the resister.

In the original circuit, The phase shift occurs between the phase of applied voltage and the phase of current flowing.
So, in this mechanical analogy, current is equivalent to the velocity of the tip of the spring. (dx/dt). So, by phase shift we mean here that the velocity isn't maximum at the same time when the force applied is maximum. The applied Force-Maximum and Velocity-Maximum occurs at different times. (they both are varying sinusoidally)

So the phase shift (which is my original question) is created by the spring pushing back.

So without the spring when the wheel is at 90 degrees, the shaft is horizontal and thus there is no current as there is no horizontal movement.
But with the spring, even though the source is at 0 Volts (and 0amps) the spring is actually pushing back, and depending on the friction determines how long it lasts for.

So this pushing back is causing the tip to move which means there is a current, even though the source is at 0.

So this is what's causing the overall phase shift between source voltage and current in the circuit?

Of course its because of the spring and its pushing that's creating phase shift.
If you remove the spring, the Force Maximum and Velocity Maximum occurs at the same time. Thats all.

Caution: There is one pitfall in that analogy. You shouldn't imagine the crankshaft as rotating at constant speed and hence applying sinusoidal force to the spring. That would only provide sinusoidal Velocity (which would be equivalent to sinusoidal current source not a voltage source). Forget about this mechanics, just remember sinusoidal force is being applied and you will be fine.

 

[sophiecentaur]

I am learning.
That diagram of a mechanical circuit is not correct if your crank wheel is to be a voltage source. The 'friction patd' you have drawn appears in an electrical circuit as a resistor in parallel with the generator an has no effect. The 'correct' place to put the frictional component would, I think, have to be in series with the push rod. Something like a piston with holes in a hydraulic cylinder (with no mass, of course) perhaps.

Analogies are full of pitfalls, alas.

 

[IssacBinary]

Im not looking for an analogy to answer my question. However they might help get there.

@sophiecentaur
I think you think I am looking for an analogy, and this is where your saying I can't find one because everyone might have their own different ones...and that even if it gets them through it, it might be wrong.

However I am not saying don't share.

But ultimately I am just looking for a basic explanation. Not an analogy but also something that doesn't rely on maths phrases.

Like the geometry example a few pages back.
https://www.physicsforums.com/showpost.php?p=3381555&postcount=56

My explanation wasn't an analogy and it wasn't relying on maths to explain what was happening (and as far as I can see its not wrong)

Hopefully that might help people see what I am after a bit more.

I hope you understand that its the case only for AC. In DC, how large be the resister the Cap gets charged upto the source voltage. In ac the cap don't get charged to anything, its voltage is varying continuously. But It has a peak, which will decrease if we increase the resister.

Yes sorry, when I say it gets charged less when talking about AC I mean that it has a less max voltage.

 

[DailyDose]

Perhaps I can take a stab at this?

I have gone through a few pages of this..."blog"...and I feel the same way you do about people perhaps not understanding what it is exactly you are asking. However, I did like a few answers. My favorite quote was...

If someone were to ask about voltage, you wouldn't just say V=IR.

Something of that nature.

And you're right, btw. V=IR only describes Voltage's direct relationships. You would want to say something about Voltage being a magnitude of potential energy. To a non-EE that might make sense.

Anyways, back to your "what is phase." Physics definition of phase is:
"a particular stage or point of advancement in a cycle; the fractional part of the period through which the time has advanced, measured from some arbitrary origin often expressed as an angle (phase angle), the entire period being taken as 360°."

A more simple definition:
"to schedule or order so as to be available when or as needed."

Try to think of it as an assembly. I know, what the crap? Just bare with me. Things are happening in certain orders. And much of that has been well explained. Granted, everything is happening very very fast. But if we were to slow it all down, we'd see things happening in phase. Current passes through cap first. Cap starts charging to meet voltage. Polarization changes once Cap reaches maximum voltage then discharges. You know this. I know you know this. But the phase shift is a representation of what's happening in order. That's all it is. We use it like a clock to tell us what the circuit is doing at such and such a time. So...what is status of a car being manufactured at 4:00pm, we would say, what is the status of this cap at 50 degrees. I'll probably get slapped around for that analogy, but it's the best I could come up with. The formulas point to current leading, but we already know that current will be "first." I wouldn't say phase is happening inside the Cap. Phase is what's happening in the circuit altogether. The order that current and voltage are changing with respect to the frequency.

Now, if everyone wouldn't mind, please tell me what I said wrong.