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Physical explanation of leading capacitive current (270 degrees!)Helpful Member!(5) 

bvc (Electrical) (OP)
11 Nov 12 17:23
Dear all,
Kindly correct my observations if wrong :
1) In any circuit, the current is produced by means of application of a voltage. Thus voltage is the cause and current the effect.
2) The phasors representing the various currents and voltage show the phase relationship between the components having the same frequency. Thus a phasor which is shown leading w.r.t another can be termed as leading the second phasor in time domain also. Thus, I can say that the first quantity say for example , a voltage is applied before the second quantity e.g current. In case of a resistance, I can say the current appears as soon as the voltage is applied. In case of an inductive circuit, the inductive current appears after a time delay of application of voltage as the inductor does not allow the current through it to change suddenly. In case of a capacitor, the current takes more time to reach steady state as it is an insulator so naturally, it must take more time for the current to flow through it after application of the voltage across it. So in this regard, is it fine to say that the current in case of a capacitor lags the voltage by 270 degree ( that is more time than inductor or resistor) ? I cannot fit a leading current with the time lag explanation as I don't understand how a current can flow through a capacitor without application of voltage ?

Waiting for the comments.

Thanks.
Helpful Member!  mikekilroy (Electrical)
11 Nov 12 17:59
bvc, I will start.... i would say you r looking at it backwards.... we had a saying in engineering school: ELI the ICEman.... E, voltage Leads I current in L inductor, but I leads E, voltage, in C capacitor....

You have a handle on L so you see & understand why I is 90 degrees BEHIND (lags) E voltage.

but u seem to be confusing urself on ICEman side..

Yes, you need some E as a forcing function to START the process of current flowing. But think of it this way instead: you have a 12v car battery hooked thru #16 clip leads to a 10,000 mfd capacitor. the #16 wire has resistance. when you first make connection, the capacitor looks like a short circuit, so the CLIP LEADS have the full 12 v dropped cross them - while the capacitor has 0 volts across it.... but big time current is flowing! hence, ICE - the I, current, Leads the E, voltage ACROSS the capacitor.... so does this not make sense to you that current flows BEFORE voltage builds up across the capacitor?

Lastly, dont confuse yourself with "270 degree" nonsense... the building voltage across the cap is the same 90 electrical degrees behind the current, not some 270 degrees before something....

hope this helps u visualize it.

73, AC8V
Skogsgurra (Electrical)
11 Nov 12 18:02
Since i=du/dt, there is a current as soon as voltage starts to change. Not very complicated. And forget about a capacitor is an insulator. That it is only for DC. Not for AC, not even for the initial edge when you apply DC.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

veritas (Electrical)
11 Nov 12 22:48
Also, current, whether DC or AC, never flows through a capacitor (or a healthy one at least!). With DC the cap gets charged and so current starts at an initial peak and drops down to just about zero when cap voltage = applied voltage.

With AC - on the positive half-cycle the current flow results in one-side of the capacitor being +ve charged and the other side -ve charged. This is reversed with the -ve half cycle. Note that the current charges the capacitor cyclically with current flowing from +ve plates through the source to the -ve plates and vice verse for the other half-cycle. No current ever through (ideal) capacitor. This is the mechanism by which charging current can flow in a transmission line or cable that is open at the load end.

With regards to lagging or leading, current immediately appears once voltage is applied and switch is closed. With both L and C there will be an initial transient current which settles to a steady state.
davidbeach (Electrical)
12 Nov 12 9:17
"No current flow through capacitors". Did you happen to run that one past Kirchhoff before posting? If you believe that, then how do you explain series caps in EHV transmission lines; thousands of amps (current) through the caps.
abrahamJP (Electrical)
12 Nov 12 12:00
Voltage and current are two dependant parameters.Without current,voltage donot exists and without volatge, current cannot exist.So I think,current and volatge lags/leads by a time delay doesnt mean that one exists witout other.Just zero crossing point of current and volatge waves are at different times
Helpful Member!  electricpete (Electrical)
12 Nov 12 12:38
All good answers. I'm not sure any further is required but here' my response to op:

Quote:

1) In any circuit, the current is produced by means of application of a voltage. Thus voltage is the cause and current the effect.
There is no reason to identify current or voltage and cause or effect. The terminal relationships describing a linear passive element (V = I*R, V = L * di/dt, I = C * dv/dt) apply regardless of which variable is considered an independent variable.

Quote:

2) The phasors representing the various currents and voltage show the phase relationship between the components having the same frequency. Thus a phasor which is shown leading w.r.t another can be termed as leading the second phasor in time domain also.
OK so far.

Quote:

Thus, I can say that the first quantity say for example , a voltage is applied before the second quantity e.g current. In case of a resistance, I can say the current appears as soon as the voltage is applied.
I guess so. In the case of resistor, you can say i(t) = v(t) / R

Quote:

In case of an inductive circuit, the inductive current appears after a time delay of application of voltage as the inductor does not allow the current through it to change suddenly.
i(t) = 1/L * int(v(t)dt
current starts at zero and increases as integral of voltage.
in steady state, current lags voltage by 90 degrees.
I = 1/L * V/(j*w) where I and V are phasors

Quote:

In case of a capacitor,
Why don't you just swap the role of current and voltage from your inducator example. Apply a current.
v(t) = 1/C * int(i(t)dt
voltage starts at zero and increases as integral of current
in steady state, voltage lags current by 90 degrees.
V = 1/C * I/(j*w) where I and V are phasors

=====================================
(2B)+(2B)' ?

Helpful Member!  veritas (Electrical)
12 Nov 12 16:56
davidbeach - a capacitor has a dielectric between the two parallel plates. Ideally no current should ever be able to flow through the dielectric as it is meant to be an insulator to effect charge separation. This is the basic physics of a capacitor. Current flow through the dielectric (more than a tiny leakage current) implies that the capacitor has failed. Thus at the physical level current does not flow through the capacitor but it gets cyclically charged and dischaged at the frequency of excitation if you consider the instantaneous current.

One can express the r.m.s. value of this current and in general it is expressed or thought of as a current flowing through the capacitor - whilst at the physical level actual current flow is as described above.

Seen from another angle - how would you otherwise explain transmission line charging current if the line is open at the remote end?
bvc (Electrical) (OP)
12 Nov 12 17:42
Thanks to all experts who gave their precious time for answering my basic query. After going through the various comments, what I understand is as follows:

1) Since i=C*du/dt, there is a current as soon as voltage starts to change. The fact that capacitor acts as an initial short for applied voltage is in fact a pointer that the current actually leads the voltage. Additionally proved mathematically as
" v(t) = 1/C * int(i(t)dt voltage starts at zero and increases as integral of current
in steady state, voltage lags current by 90 degrees ".
2) Capacitor acts as an insulator only for steady state DC. There is capacitor charging/discharging current flow through external circuit even at the rising / falling edge of dc voltage. As for AC voltage it is continuously getting charged and discharged depending on the positive and negative cycle.

I think, when I explain in terms of phasors, I should say which phasor peaks before the other. That is in case of capacitance, I can say the current phasor peaks before the voltage phasor ( or simply current leads the voltage !!). As correctly pointed out by Mr. Abraham JP, the current and voltage cannot exist without one another. Only their relative magnitudes and phase angles at any instant are different in case of capacitor or inductor.

Finally for Mr. Veritas and Mr. Davidbeach discussion, I have the following comment:
I read in a physics forum that in a capacitor, the current will only flow through it if the dielectric breaks down( as pointed by Mr. Veritas).
Current is termed as the total charge ( coulombs) passing through a given part of a circuit in a second. When we apply a voltage across the capacitor, there is polarisation of the molecules within the capacitor resulting in a net positive charge on one plate and negative on the other. That is within the capacitor there is no smooth electron flow (or conductive current due to applied voltage) just like the conductor in external circuit ( i remember something called electron gas here) !! Thus when we look at the overall effect, should we not say there is a charge flow in the external circuit and hence overall a current flow in the circuit (correct me if I am wrong)? As regards, Mr. David beach remark about 1000 A flowing through a series compensated TX line is a fact and I think can be explained based on the same analogy of charge moving to the positive plate and negative charge to another thus resulting in a current in the external circuit. I think a series compensated line can be considered as a series LC circuit L being line inductance.

Once again, thanks to u all for your time and expertise !
Bvc
Skogsgurra (Electrical)
12 Nov 12 17:57
Oh, why do you not just accept what is being said in the text-books? Do you think that you are the first to think about these things? And, do you think that your thinking is more correct than others? I am not impressed.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

veritas (Electrical)
12 Nov 12 18:50
Skogsgurra - I found your last posting mean, unfounded and out of place. bvc - as any other engineer - has every right to question as it goes a long way to internalise the theory and in so doing reach a better understanding of the principles involved.
Skogsgurra (Electrical)
12 Nov 12 21:50
You may find it mean. It probably is. But it this is Engineering Tips. Not a chat where well-founded scientific and engeneering truths are questioned.
Yes, most of us are boring engineers. But that is what makes products work and work safe. Ideas like no current in a capacitor belong somewhere else.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

electricpete (Electrical)
12 Nov 12 22:11
I think it's a good discussion if you're in a playful mood. And at the same time I understand Gunnar's impatience in rehashing things that are already well known.

fwiw, I understood bvc to say there is current in the circuit attached to the capacitor, but within the capacitor, it is not a traditional current associated with movement of charged particles. That sounds roughly right to me. The current that flows in a capacitor is a different nature and is given a special name: Maxwell's displacement current. From circuits standpoint, the distinction is not important, we only care about what happens in the parts of the circuit attached to the capacitor.

=====================================
(2B)+(2B)' ?

Skogsgurra (Electrical)
12 Nov 12 23:11
Right.
My main objection is that there are references to unidentified physics fora ("I read in a physics forum that in a capacitor, the current will only flow through it if the dielectric breaks down") and such references confuse and start "philosophical" discussions that are not needed on an engineering site.

Another example is the eternal power factor discussions that are going on in many places where beer and froth and other fruitless parallels obscure the simple fact that there is a difference in time, which influences the momentary product of voltage and current. Every such "workshop floor" analogy obscures thinking and makes young (maybe also old) people unsure, while a simple and correct explanation would have added to their knowledge.

I agree that bvc has every right to think and say whatever he thinks is right. But the virtue of questioning well established facts may be good in politics - not in engineering. There are site rules to that effect.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

sibeen (Electrical)
12 Nov 12 23:23
we had a saying in engineering school: ELI the ICEman.... E, voltage Leads I current in L inductor, but I leads E, voltage, in C capacitor....

Never heard of that one, I was taught 'CIVIL'.

Capacitor I lead V lead I for L(inductance).
Skogsgurra (Electrical)
13 Nov 12 3:26
I risk being called mean again. But that kind of mechanical memory aids is what I mean with "workshop floor knowledge".

If such letter combinations are needed to remember things, I do not think that there is a good understanding of the physics at all and then it will be difficult to apply the knowledge in real situations.

It is a basic insight that changing voltage across a capacitor leads to a current proportional to capacitance and voltage rate of change. It is also a basic insight that voltage across an inductor is proportional to inductivity and the current's rate of change.

Not very difficult. Gives immediate insight into the working of circuits and there is no need for letter combinations or home-brew "philosophy".

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

veritas (Electrical)
13 Nov 12 6:15
Skogsgurra - I beg to differ with your sentiments. The textbooks are written by subject experts as one would expect. However, with all due respect to all authors of engineering texts, I sometimes find that the way a given topic is presented in a textbook does not always make understanding and grasping the subect matter a straightforward process. The reader is then required to seek further assistance to clarify and aid understanding.

Personally I would prefer an engineer who questions and seeks to understand rather than accept everything presented at face value. I also differ with your point regarding certain untouchable engineering basics. Many years ago I had a mentor who often question me on many of the basics which made me think long and hard and question assumed givens. These were enormously fruitful as I always came out the wiser on the other side as I still do today.

If the beer and froth causes confusion then does that not imply that there was improper understanding from the start? I believe that there is much to be gained by time and again visiting familiar territory as there nearly always something new one comes to appreciate. And I have no problem if an engineer like bvc, reformulates the basics, or not so basics, in his own words for then it really becomes apparent whether he understand or not.

I have been in this game for more than 2 decades and can still be floored by a seemingly innocuous question by a beginner. It does not shame me neither do I cringe from it as I treasure every opportunity to learn or to teach.

Anyway, I can ramble on for hours still, but you get my drift.
Skogsgurra (Electrical)
13 Nov 12 6:26
Yes, I get your drift. Absolutely. And I agree that thinking about things are good.

But I also have experienced (first employed by ASEA/ABB in 1962 and then by Siemens - seems to be five decades of experience in the forefront of electric development) how bad understanding is conserved by all these simplified mnemonic "rules" that flourish on the workshop floors. Once I sit down with one of these guys with paper and pen and describe in correct and fundamental terms how things work, I get immediate and good results.

I can't do that here. So I sometimes need to be harsh in order to get things across.

I can also go on for hours. And weeks.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

sibeen (Electrical)
13 Nov 12 6:36
Skogsgurra, in my defense I was taught that simple mnemonic back in '79. I was also taught "bad boys..." as well. It was a tool so students wouldn't be too confused. I was more responding to mikekilroy, as I'd never heard his one before. Maybe a North vs Southern hemisphere thing, and in that I'm assuming mikekilroy is from the northern latitudes.
mikekilroy (Electrical)
13 Nov 12 7:26
So I guess to expand the "no current flows THROUGH a capacitor" into "none of the current flows INSIDE the wires" would be a bad thing now, huh? OK, I won't go there from this SOUTHERN Ohio Town up north....
electricpete (Electrical)
13 Nov 12 8:26

Quote:

none of the current flows INSIDE the wires
I haven't heard that one. Can you explain what you mean?

=====================================
(2B)+(2B)' ?

bvc (Electrical) (OP)
13 Nov 12 11:40

Well, it is nice to know the various view points by industry experts. Myself, I am only 9 years into Electrical engineering work. Some two weeks back , I was going through the "Performance and design of AC machines" by MG Say, for certain information on transformers and I was forced to go back to the no load phasor diagram. Was slightly shocked to know that my understanding of the transformer phasors was wrong. I never read this book ( or for that matter any text book of such standards) during my engineering school days. So some of the links were missing in my understanding of things . Never my intentions to challenge well established facts but only to understand them and that too in the best possible way , not by blindly memorising any formula or phasor diagram! My mentor used to say "never hesitate to beg for knowledge ".

Anyway thanks to everyone out here for their patience and time. I have got my clarifications.

Cheers !!
Helpful Member!  Skogsgurra (Electrical)
13 Nov 12 11:48
I think that I should apologize for being too outspoken. You have the right attitude!

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

bvc (Electrical) (OP)
13 Nov 12 13:22
Thanks ! Even the 270 degree stuff was not one of my own making but rather it was conveyed during a training session by an old Engineer ( well past his 60's) some 3 years back.
I used to be a lazy bone during my college days -:(. With the passing of time, I am sure I will become a better Engineer.
mikekilroy (Electrical)
13 Nov 12 17:56
electricpete,

I do not want to get into trouble with discussing a topic of how current flows OUTSIDE wires and get banned here.....but just a teaser....

In defense of our friend bvc, schools, including engineering universities, teach the 'bump' theory - that electrons travel at the speed of light down the inside of the wire to the load by bumping into each other..... they do this as it is easiest to convey the CONCEPT of current flow - just as they teach current flows THRU a capacitor (insulator). but we also know that electron flow inside the wire is at a SLOW rate of only a few inches per second! So how does this instantaneous energy get from the source to the load?

Energy does not travel in the wire at all but travels outside the wire in the surrounding TWO fields 90 degrees to each other: magnetic field and electric field and this happens near instantaneously. The only energy inside the wire is due to losses from these fields cutting across it..., there are some really good discussions of this elsewhere if one wants pursue it.


electricpete (Electrical)
13 Nov 12 19:39
mike - I get your gist. If we look for where the power is transmitted electromagnetically (associated with the Poynting vector S=ExH), it is not within a perfect conductor because the fields are zero there. Or for that matter even in a real conductor the radial fields are approximately zero. So the power is transmitted electromagnetically in the associated fields which may lie between conductors or between a conductor and a shield. It is not such a surprise that the E and H fields extend outside the conductor. What you said originally was a little different... that the current flows outside the conductor. But maybe I'm splitting hairs. Good discussion.

=====================================
(2B)+(2B)' ?

electricpete (Electrical)
13 Nov 12 19:46
I complicated the whole thing by mentioning perfect conductor, where current flows on the skin. Let me remove that distractor. I should've stuck with real conductor. Radial fields are zero inside the real conductor, so the power transmitted (longitudinally) by a set of real conductors lives in the radial fields outside the conductors.

=====================================
(2B)+(2B)' ?

electricpete (Electrical)
13 Nov 12 19:56
All of which goes to prove, you can make anything more complicated than it needs to be.

=====================================
(2B)+(2B)' ?

DRWeig (Electrical)
14 Nov 12 11:54
Occam's razor, Pete. 2thumbsup

Best to you,

Goober Dave

Haven't see the forum policies? Do so now: Forum Policies

veritas (Electrical)
14 Nov 12 15:49
Skogsgurra - thank you and thank you also for all your posts. I've read many of them and they've been very helpful. Keep it up.
Helpful Member!  Bronzeado (Electrical)
15 Nov 12 10:17
Hi folks,

Really, I enjoyed this discussion. Back to fundamentals!

I think the confusion from the ideias as because some of you thinks physically and others mathematically.
Regarding to bvc starting question, if the angle between the voltage and current applied on a capacitor
is 90 in advance or 270 behind, it is happens only on steady state condition, with the phasor theory been applied (maths).

However, as Skogsgurra said, as soon the voltage is applied on a capacitor, a current in it is generated (physics).

About current flowing through a capacitor, physically it does not, but mathematically it may depending on the model applied to solve an engineering problem. The same thinking may be applied to current outside the lines.

At the end of the day, all of you are right!

Best Regards,

Herivelto S. Bronzeado
Brasília, Brazil
http://www.linkedin.com/profile/view?id=46319837&a...

bvc (Electrical) (OP)
18 Nov 12 0:23
The same electrical phenomenon can be explained based on Circuit Theory or Field Theory ( may be some more theory I am not aware of) giving rise to different explanations.
ScottyUK (Electrical)
18 Nov 12 12:00

Quote (ePete)

All of which goes to prove, you can make anything more complicated than it needs to be.

I seem to recall that in school, up until the age of maybe 14 or 15 years, teachers insisted that it was impossible to take a square root of a negative number. Then one day in high school 'i' appeared and we realised that we had been lied to for years, only for 'i' to become 'j' in electrical principles class a few years later... and then having gotten to grips with the 'j' operator we are confronted by the 'h' operator in a bid to confuse us even further, on the off chance that Messrs Laplace, Dirac, et. al had not already already succeeded in that goal...

PHovnanian (Electrical)
18 Nov 12 13:30

Quote:

we are confronted by the 'h' operator in a bid to confuse us even further,

I understand the root of the 'i' vs 'j' schism.

But 'h'? Why? Is someone still trying to use Fortran?
ScottyUK (Electrical)
18 Nov 12 16:12
You're more than familiar with the 'j' operator that is used to denote a 90° phase shift relative to a reference vector. The 'h' operator does the same except it denotes a 120° shift. Mathematically it is the complex cube root of 1. I'm sure you can see its immediate usefulness in 3-phase analysis by avoiding the awkward -1/2 ± j1/√3 term which occurs when the 'j' operator is used to describe a 3-phase system.

Fortran? Now that is something I haven't had nightmares about for a long time. wink

electricpete (Electrical)
18 Nov 12 17:08
I think it's interesting someone chose the words "real" and "imaginary".
Imaginary almost sounds like "fictional"
A complex subject.

=====================================
(2B)+(2B)' ?

xnuke (Electrical)
18 Nov 12 17:47
Scotty

Most of our textbooks on this side of the pond seem to use "a" instead of "h".

electricpete, nice pun.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
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sibeen (Electrical)
18 Nov 12 21:07
Most of our textbooks on this side of the pond seem to use "a" instead of "h".

That applies in the antipodes as well. When Scotty mention the 'h' operator I had no idea what he was on about. When he described it I went, hold on , that's 'a' or alpha.
ScottyUK (Electrical)
19 Nov 12 1:59
I wonder if it is a British thing? I have seen the 'a' operator in texts and never really questioned it; just assumed that it was similar to 'i' versus 'j' between the mathematics world and the engineering world.
PHovnanian (Electrical)
19 Nov 12 13:22

Quote:

I wonder if it is a British thing?
I think its British, or European. Like 'u' for voltage instead of 'v'.

I've heard 'a' used on this side of the pond. Although that might have something to do with my proximity to Canada, eh?

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