Why does current lag behind voltage in inductor?

[Bassalisk]

I've seen that this question has been asked but never answered because guy was silent on formulas.

I can say that i am familiar with formulas, and i know how to derive the equation for current, and that II/2 lag, but here is the thing... Why? i know that formulas say so... But can someone please try to explain this to me in more touchable and intuitive way ?


Thanks

 

[LotusEffect]

Inductors react against a change in current. Applying a voltage to its terminals will produce a current that will then create a magnetic field like any other wire. An inductor is called as such because its magnetic coupling is high. In other words, the magnetic field it creates will have an effect on the element itself.

If the magnectic field varies with time (which is the case when you just applied a voltage to the inductor terminals), an electric field will be induced within the wire the counter the change in magnetic field. This E-field will tend to "stop" the current that would normally be present without the inductance effect. Thus current "lags" behind voltage. Too quick of change in current creates an e-field that decreases dI/dt.

You can see an inductor like a wheel that is free to spin due to the flow of a river. If the flow changes too quickly, the wheel, due to its inertia, will tend to keep spinning for a moment. The same is true if the water of the river goes from steady to flowing in a short instant.

 

[Zryn]

http://en.wikipedia.org/wiki/Lenz%27s_law"

"An induced current is always in such a direction as to oppose the motion or change causing it"

So you apply your voltage to an inductor and the coil shape acts to oppose the change in current. Note that the voltage that changes the current comes first, and then the induced current (and flux) change. Also note that 90 degrees out of phase only works for ideal inductors.

Another way of looking at it is that if you apply your source voltage to an inductor, V(S) = sin(wt) then the current must be i = sin(wt)/R, so by the inductor formula:

V(L) = L di/dt = L/R d sin(wt)/dt = L/R cos(wt)

And we know that cos(wt) is 90 degrees ahead of sin(wt).

 

[Bassalisk]

So basically, we apply AC to a inductor, and then current "pushes" more and more through conductor, it induces more and more induced current? and that induced current is in some way pushing back the current that was originally going through inductor?
But the voltage on the inductor stays the same?

 

[haleycomet2]

http://en.wikipedia.org/wiki/Lenz%27s_law"

Another way of looking at it is that if you apply your source voltage to an inductor, V(S) = sin(wt) then the current must be i = sin(wt)/R, so by the inductor formula:

V(L) = L di/dt = L/R d sin(wt)/dt = L/R cos(wt)

And we know that cos(wt) is 90 degrees ahead of sin(wt).

So actually the back emf in the circuit is initially larger than supply voltage?

 

[atomthick]

The idea that current lags is only true only if you use alternative current, just like Zryn correctly expressed through formulas, because of the special relation between sin and cos. If you however apply a different variation of voltage let's say Aln(t) you will get a V(L) = A L/(Rt) which doesn't seem to be delayed but instead totaly changed.

You should think of the lag only as a special case (in alternative current) to help you remember the relationship between current and voltage.

 

[Bassalisk]

Ok i think i got it. Now i need someone with good knowledge of complex area applied to e engineering. I just learned from my classes that there is something called active and reactive resistance, power etc. I figured that active resistance is like a resistor. But i do not understand what reactive part is...
Thanks

 

[tiny-tim]

Hi Bassalisk!

In a resistor, voltage and current are dependent on each other, but in an inductor they are independent (so there is no resistance), and instead voltage is dependent on the rate of change of current: V = LdI/dt.

In a steady AC circuit, however, the rate of change of current is conveniently proportional to the current itself (and to the frequency), but 90° out of phase.

So the voltage is also 90° out of phase with the current.

I just learned from my classes that there is something called active and reactive resistance, power etc. I figured that active resistance is like a resistor. But i do not understand what reactive part is...

Reactive resistance (or reactance, X) is the "imaginary part" of https://www.physicsforums.com/library.php?do=view_item&itemid=303" …

Z = R + jX …

a reactive component (inductor or capacitor) has no resistance (ie voltage does not depend on current), but has a coupling between V and dI/dt (for an inductor) or between I and dV/dt (for a capacitor), and the power through it (V times I) in a steady AC circuit cycles with twice the frequency of the circuit itself, and averages zero … this zero-average power is known as reactive power, I think because you "get it back".

 

[Bassalisk]

Inductors react against a change in current. Applying a voltage to its terminals will produce a current that will then create a magnetic field like any other wire. An inductor is called as such because its magnetic coupling is high. In other words, the magnetic field it creates will have an effect on the element itself.

If the magnectic field varies with time (which is the case when you just applied a voltage to the inductor terminals), an electric field will be induced within the wire the counter the change in magnetic field. This E-field will tend to "stop" the current that would normally be present without the inductance effect. Thus current "lags" behind voltage. Too quick of change in current creates an e-field that decreases dI/dt.

You can see an inductor like a wheel that is free to spin due to the flow of a river. If the flow changes too quickly, the wheel, due to its inertia, will tend to keep spinning for a moment. The same is true if the water of the river goes from steady to flowing in a short instant.

@Lotus if he sees this.

Can you explain this for the capacitors in AC circuits too? I think goes ahead by some value? again same question why? I know the equations.

Thanks

 

[Bassalisk]

Hi Bassalisk!

In a resistor, voltage and current are dependent on each other, but in an inductor they are independent (so there is no resistance), and instead voltage is dependent on the rate of change of current: V = LdI/dt.

In a steady AC circuit, however, the rate of change of current is conveniently proportional to the current itself (and to the frequency), but 90° out of phase.

So the voltage is also 90° out of phase with the current. Reactive resistance (or reactance, X) is the "imaginary part" of https://www.physicsforums.com/library.php?do=view_item&itemid=303" …

Z = R + jX …

a reactive component (inductor or capacitor) has no resistance (ie voltage does not depend on current), but has a coupling between V and dI/dt (for an inductor) or between I and dV/dt (for a capacitor), and the power through it (V times I) in a steady AC circuit cycles with twice the frequency of the circuit itself, and averages zero … this zero-average power is known as reactive power, I think because you "get it back".

I think i got it. Thanks. So reason why this reactive resistance is attached to imaginary part, is because in principle it doesn't exist, in form of Ohm resistance, but this new kind of resistance due to inductance?

 

[tiny-tim]

Can you explain this for the capacitors in AC circuits too?

In an inductor, V is proportional to dI/dt, so (in a steady AC circuit) V leads I by 90°.

In a https://www.physicsforums.com/library.php?do=view_item&itemid=112", I is proportional to dV/dt, so (in a steady AC circuit) I leads V by 90°.

EDIT: (I didn't notice your second post )

So by coupling u DON'T mean depending?

No, they mean the same thing, I was just using a different word.

V depends on dI/dt. ​

 

[Bassalisk]

In an inductor, V is proportional to dI/dt, so (in a steady AC circuit) V leads I by 90°.

In a https://www.physicsforums.com/library.php?do=view_item&itemid=112", I is proportional to dV/dt, so (in a steady AC circuit) I leads V by 90°.

EDIT: (I didn't notice your second post )


No, they mean the same thing, I was just using a different word.

V depends on dI/dt. ​

Yea i understood and edited. Thanks ! :D

 

[tiny-tim]

So reason why this reactive resistance is attached to imaginary part, is because in principle it doesn't exist, in form of Ohm resistance, but this new kind of resistance due to inductance?

(It's all measured in ohms, of course, but yes …)

yes, reactive resistance isn't the V/I resistance we're used to, so it has to be dealt with separately.

As to why (and how) it's imaginary, that's a mathematical trick (which only works in steady AC circuits) … see the PF Library on https://www.physicsforums.com/library.php?do=view_item&itemid=303"

 

[Bassalisk]

(It's all measured in ohms, of course, but yes …)

yes, reactive resistance isn't the V/I resistance we're used to, so it has to be dealt with separately.

As to why (and how) it's imaginary, that's a mathematical trick (which only works in steady AC circuits) … see the PF Library on https://www.physicsforums.com/library.php?do=view_item&itemid=303"

Will do thanks tim !

 

[sophiecentaur]

So actually the back emf in the circuit is initially larger than supply voltage?

Not higher but just and opposite - like the upward reaction force of the chair you're sitting on - never big enough to throw you upwards.

 

[Jiggy-Ninja]

I think i got it. Thanks. So reason why this reactive resistance is attached to imaginary part, is because in principle it doesn't exist, in form of Ohm resistance, but this new kind of resistance due to inductance?

"Imaginary" is just a name, don't read too much into it. i ¹ is just as "real" as any other number.

I don't think that i has any special physical significance like other constants such as \pi, it's just a bit of a mathematical trick that makes it easy to do math with waves. In addition to its amplitude, the waves' phase is important as well. Since reactance is 90 degrees out of phase with resistance, it is attached to the imaginary pat of the equations.

¹What do you do when you need to start a sentence with a lowercase constant? Uppercase would be technically wrong, but lower case just looks weird.

 

[LotusEffect]

The idea that current lags is only true only if you use alternative current, just like Zryn correctly expressed through formulas, because of the special relation between sin and cos. If you however apply a different variation of voltage let's say Aln(t) you will get a V(L) = A L/(Rt) which doesn't seem to be delayed but instead totaly changed.

I don't know what you just did there but if you apply V(t) = A ln(t) to an inductor, the current will be I(t) = 1/L * (t ln(t) - t). Current in an inductor is the integral of voltage.

@Lotus if he sees this.

Can you explain this for the capacitors in AC circuits too? I think goes ahead by some value? again same question why? I know the equations.

Thanks

You mean use some sort of analogy like the one of the spinning wheel for an inductor? I can't find one, but what I can tell is that in a capacitor, voltage "lags" vehind current because voltage across a capacitor terminals depends on the total amount of charge that accumulated on the plates. If you quickly change the current flowing through the capacitor, it takes a short instant for the charges to "leave" the cap. So a capacitor stores energy in the form of accumulated charges and an inductor stores energy in the form of accumulated magnetic field.


Impedances are a very neat trick to calculate currents and voltages in an ac circuit. When dealing with signals that all have the same frequency, it's clever to express them and impedances using phasors. Phasors are just complex numbers, and when expressed in polar form, they make calculations of rms values and phase angle of signals very easy.

 

[Jiggy-Ninja]

You mean use some sort of analogy like the one of the spinning wheel for an inductor? I can't find one,

A https://ccrma.stanford.edu/~jos/fp3/Mechanical_Equivalent_Capacitor_Spring.html" .

 

[tiny-tim]

Hi Jiggy-Ninja!

I don't think that i has any special physical significance like other constants such as \pi, it's just a bit of a mathematical trick that makes it easy to do math with waves. In addition to its amplitude, the waves' phase is important as well. Since reactance is 90 degrees out of phase with resistance, it is attached to the imaginary pat of the equations.

Yes, j (or i) describes the physical attribute of being +90° out of phase.

¹What do you do when you need to start a sentence with a lowercase constant? Uppercase would be technically wrong, but lower case just looks weird.

Patient to doctor: it hurts when I do that!
Doctor to patient: then don't do that!​

Then don't start a new sentence! …

"Imaginary" is just a name, don't read too much into it … i is just as "real" as any other number. ​

(alternatively, use j ! )

… a rubber diaphram.

From the PF Library on https://www.physicsforums.com/library.php?do=view_item&itemid=112" …

Fluid mechanics analogy:

In fluid mechanics, a capacitor is analogous to a diaphragm blocking a pipe.

Water cannot pass through the diaphragm.

But pressure suddenly applied on one side will make the diaphragm bulge, so that water continues to flow "into" the diaphragm on one side, and "away from" the diaphragm on the other side, until the diaphragm reaches its maximum bulge for that pressure.

The volume of the bulge, divided by the pressure, is the capacitance.

The energy stored is half the volume of the bulge times the pressure.

The displacement current is rate at which the volume of the bulge is increasing (and is zero once the maximum bulge is reached).​

 

[Bassalisk]

U guys should get paid seriously XD Thanks, i got everything clear now!

 

[Jiggy-Ninja]

Hi Jiggy-Ninja!

Hi fictional-character-from-a=Charles-Dickens-story-who-has-a-terminal-illness!

Yes, j (or i) describes the physical attribute of being +90° out of phase.

What I'm getting at is that conceptually, \pi is pretty easy to understand. It's a ratio of lengths, and can be very easily visualized. (I'm a very visual thinker)

What \sqrt{-1} has to do with being 90° out of phase is a leap of understanding that I have yet to make. All I know is that it apparently works.

Patient to doctor: it hurts when I do that!
Doctor to patient: then don't do that!​

Then don't start a new sentence! …

Completely useless, smart-*** answer to a half-serious question? You and I are going to get along just perfectly.

(alternatively, use j ! )

I ran into i first, so that's what I use by habit. No point in having two symbols for the same weird number.

Or, just go with Japanese. Some of the hiragana are quite pretty. I especially like み for some reason. I'll use that for my complex numbers instead.

U guys should get paid seriously XD Thanks, i got everything clear now!

Alas, don't I know it. I'm basically an unpaid teacher's assistant in my classes. Helping other people who are confused is a near-irresistible impulse for me.

 

[Bassalisk]

Well, in mathematics Complex numbers C are extension of Real numbers.

Complex numbers have special way of multiplying and adding together such that u have a consequnce: when u take a square root of -1 is i. Now that is NOT a definition of i. Definition of i is that i is for complex numbers what 1 is to real numbers.

Real definition of i is long and we would have to go in calculus analysis.

And there is a reason for j. In electricity u have I for current, sometimes written in small letters i. U don't want to get mixed up, that why some ppl use j instead of i.

Complex numbers come in pairs. By the time engineers thought of using complex numbers, they were fully developed.

I don't think that there is some direct connection between what u said, Jiggy, I don't even think it could be. Think about it, complex numbers were mainly made to cover that area where u can't do square roots of negative numbers, those were first reasons for making complex numbers.

Engineers came and said, heeey we can use that etc etc. I only find analogy in reactive resistance being bound to complex area. Why? because it is not REAL resistance, like due to conductivity of matter.

It is some new kind of resistance that it needed to be distinct from ohmic resistance.I am too kinda guy that wants to visualize everything. That's why I have so many questions that are mainly related to theory, not formulas.

 

[Bassalisk]

Another question:

When we have a current going through inductor, at first we have a change in magnetic field. That change self-induces a current that opposes original current. Now, when current is fading, another change in flux, will that change now try to push(in the same direction as current) the original current?

 

[Jiggy-Ninja]

Another question:

When we have a current going through inductor, at first we have a change in magnetic field. That change self-induces a current that opposes original current. Now, when current is fading, another change in flux, will that change now try to push(in the same direction as current) the original current?

That change self induces a voltage that opposes the change.

And yes, as the current fades, the flux will collapse and induce a voltage that tries to keep pushing the current forward.

 

[LotusEffect]

Another question:

When we have a current going through inductor, at first we have a change in magnetic field. That change self-induces a current that opposes original current. Now, when current is fading, another change in flux, will that change now try to push(in the same direction as current) the original current?

Exactly, as the spinning wheel analogy suggests!

 

[Bassalisk]

Physics is phun!

 

[cabraham]

I would suggest that the reason current lags voltage in an inductor has to do most w/ energy. An ideal inductor would possesses no capacitance or resistance. Changing the voltage on an inductor requires no work. But changing the current requires work, since inductor energy is W= (L*I^2)/2. It takes time for energy to change, the rate of said change is power. To change current instantly would require changing energy instantly, which is infinite power. The discussion would then be theological, not physical.

Since time is needed for an energy change, time is needed for a current change in an inductor, but not for a voltage change since no work is done. A capacitor is the complement. The current can change quickly since the energy is not changing. But a capacitor cannot have its voltage changed instantly because that would require an instant energy change & infinite power.

Of course, real world inductors have capacitance & vice-versa. But at low frequencies a good inductor stores much more energy in its inductance vs. its capacitance. I would advise any interested parties to read peer-reviewed texts & articles for reference info. All this talk about current "being pushed" is utter nonsense. Current isn't "pushed".

Claude

 

[LotusEffect]

This is a good post and great addition to the discussion^^.

But btw, we all know that current isn't "pushed" :P. But sometimes using familiar expressions help understand the problem :). We also often say that toasters and ovens "pull" a lot of current!

 

[epenguin]

¹What do you do when you need to start a sentence with a lowercase constant? Uppercase would be technically wrong, but lower case just looks weird.

i wonder if this looks better?

 

[Bassalisk]

All this talk about current "being pushed" is utter nonsense. Current isn't "pushed".

Claude

When you first started to study electricity they threw u gradients and divergences and u understood it all instantaneously?

Being pushed and pull, of course we do not mean that literally. To push and pull u need a force, which u do not have in inductor. Key word is analogy my friend.

 

[sophiecentaur]

I've seen that this question has been asked but never answered because guy was silent on formulas.

I can say that i am familiar with formulas, and i know how to derive the equation for current, and that II/2 lag, but here is the thing... Why? i know that formulas say so... But can someone please try to explain this to me in more touchable and intuitive way ?


Thanks

Science has always tried to arrive at descriptions of processes which do not rely on the "intuitive" because intuition potentially leads you into unfounded conclusion. Analogies are great but the greatest analogy is surely Maths. If, as you say, you are familiar with the formulae and what is implied by expressions like "time derivative" then you won't improve much on that. A lot of forum discussions involve stretching analogies further and further to fit something which Maths describes pretty well. It's as if people want the understanding without all the real sweat that is necessary.
Don't put the maths aside. Hang on to it- it's your friend.

 

[Bassalisk]

Math rox, I have a pretty good math. But in my life, physics > math always. Why?

Because math is too abstract, its all raw numbers. I don't think scientists came up with an answer purely by shooting formulas.

All i say is that some equations must have a thought background.

 

[sophiecentaur]

No no. Maths is not just "raw numbers". A sub set of Maths which we call Arithmetic is Numbers but the Greater Maths is about describing relationships between quantities and describes these relationships like no ordinary sentence or arm waving can. If you really want to get into Science then there's no alternative, I'm afraid. Everything else can only be second best. Naturally, you have to pick the right symbols and operations to describe a scientific idea. But that is just part of the deal. I can't think of a single advancement made in Physics in the last 100 years that didn't involve some Maths. Why do you think that is?

 

[Bassalisk]

No no. Maths is not just "raw numbers". A sub set of Maths which we call Arithmetic is Numbers but the Greater Maths is about describing relationships between quantities and describes these relationships like no ordinary sentence or arm waving can. If you really want to get into Science then there's no alternative, I'm afraid. Everything else can only be second best. Naturally, you have to pick the right symbols and operations to describe a scientific idea. But that is just part of the deal. I can't think of a single advancement made in Physics in the last 100 years that didn't involve some Maths. Why do you think that is?

True

 

[sophiecentaur]

And, to continue my rant (indulge an old man), the very fact that we can use maths to describe these relationships, from setting up descriptions of processes in mathematical models, means that we can see parallels in the many areas of Science. This is where the no-Maths enthusiast may pick up and run with an analogy but getting the final conclusion wrong because there wasn't exact equivalence between the two models (which the Maths would have shown up). It all becomes just an 'interesting' set of instances which are fun to talk about in the pub - but that's all.

 

[Pagedown]

I have a question related :

Why capacitors blocks DC signal while passes AC signal? I know the formula Xc=1/jwC. If frequency is 0 then the impedance is inf. But why in theory?

Does inductors block DC current too?

 

[sophiecentaur]

The reason that a capacity "blocks" DC is that charge will flow through it (i.e. it will charge up) until it is fully charged, after which, no more current can flow. How soon it reaches this point depends upon the series resistance of the circuit. This even happens when a switch os opened; a tiny current will flow until the two contacts of the switch have miniscule charges on each (but only for a nanosecond or less) and the total circuit voltage is dropped across the contacts.
Bearing in mind that an inductor consists of an unbroken length of wire, it can hardly "block" DC. What will happen is that, eventually, the voltage drop across it will go to zero (as the Magnetic field reaches its final value and stops changing).

 

[cabraham]

When you first started to study electricity they threw u gradients and divergences and u understood it all instantaneously?

Being pushed and pull, of course we do not mean that literally. To push and pull u need a force, which u do not have in inductor. Key word is analogy my friend.

I have no beef w/ "analogies" but sometimes bad habits can be developed using analogies. When I had gradients & divergences thrown in my face, sure I struggled, & it took time to understand them, but I'm much better off for it now. Had I been presented w/ "push, pull, etc.", I may have developed misconceptions that would have taken years to erase, if ever.

I've noticed that the "pushing/pulling current" heresy is very hard to get people to let go of, so I avoid it 24/7. Although the laws of physics & the associated math are tough at 1st glance, a beginner is better off struggling for a while. Once the concept is understood, they can go on to more advanced viewpoints & not struggle.

FWIW, I use the term "current drawn". I'll say that this motor draws 30 amps at start up & 5 amps steady state. As long as we understand that this is colloquial & not literal, all is well.

A person who sees current as being "pushed" seldom gives up such heresy. They will not progress to the next level. When semiconductor physics is discussed, or energy bands, conduction in semi & metals, Faraday's induction law, etc., that darn "pushing/pulling" heresy keeps rearing its ugly head, & efforts to dispel it are usually not successful.

Trying to explain electrical concepts w/ fluids, and/or "pushing/pulling something" etc., results in bad habits & misconceptions being developed. These myths hinder the student from advancing to deeper understanding of things.

I would recommend, based on my 1/3 century of EE practice, to learn things from peer reviewed texts. If the math is too advanced, there are books that cover the math as well. To really understand circuits at the micro level requires the understanding of e/m fields. But to learn e/m fields, one must learn the math. The concept of vectors, phasors, integrals including line, surface, & volume integrals, ordinary, total, & partial derivatives, as well as curl, divergence, gradient, & Laplacian operators, is needed to fully appreciate what is involved.

Having said that, if time does not permit such effort, I concede that an analogy could be helpful. But I urge all to not push the analogy too far, & do not view the analogy as an equivalent to the e/m problem under scrutiny. View the analogy as a means of covering something complicated by invoking something we understand from everyday life. The analogy is not the law. Please do not invoke simplified analogies as if they were law, since they are not law.

Otherwise, analogies can indeed be helpful. I hope I've made my point.

Claude

 

[Bassalisk]

I have no beef w/ "analogies" but sometimes bad habits can be developed using analogies. When I had gradients & divergences thrown in my face, sure I struggled, & it took time to understand them, but I'm much better off for it now. Had I been presented w/ "push, pull, etc.", I may have developed misconceptions that would have taken years to erase, if ever.

I've noticed that the "pushing/pulling current" heresy is very hard to get people to let go of, so I avoid it 24/7. Although the laws of physics & the associated math are tough at 1st glance, a beginner is better off struggling for a while. Once the concept is understood, they can go on to more advanced viewpoints & not struggle.

FWIW, I use the term "current drawn". I'll say that this motor draws 30 amps at start up & 5 amps steady state. As long as we understand that this is colloquial & not literal, all is well.

A person who sees current as being "pushed" seldom gives up such heresy. They will not progress to the next level. When semiconductor physics is discussed, or energy bands, conduction in semi & metals, Faraday's induction law, etc., that darn "pushing/pulling" heresy keeps rearing its ugly head, & efforts to dispel it are usually not successful.

Trying to explain electrical concepts w/ fluids, and/or "pushing/pulling something" etc., results in bad habits & misconceptions being developed. These myths hinder the student from advancing to deeper understanding of things.

I would recommend, based on my 1/3 century of EE practice, to learn things from peer reviewed texts. If the math is too advanced, there are books that cover the math as well. To really understand circuits at the micro level requires the understanding of e/m fields. But to learn e/m fields, one must learn the math. The concept of vectors, phasors, integrals including line, surface, & volume integrals, ordinary, total, & partial derivatives, as well as curl, divergence, gradient, & Laplacian operators, is needed to fully appreciate what is involved.

Having said that, if time does not permit such effort, I concede that an analogy could be helpful. But I urge all to not push the analogy too far, & do not view the analogy as an equivalent to the e/m problem under scrutiny. View the analogy as a means of covering something complicated by invoking something we understand from everyday life. The analogy is not the law. Please do not invoke simplified analogies as if they were law, since they are not law.

Otherwise, analogies can indeed be helpful. I hope I've made my point.

Claude

I agree, i must emphasize also that i am first year at EE and i simply do not have the tools to understand all immediately. U also must know that I also have an urge for deeper understanding, but u must take steps. U start with analogies and end up with most deepest thoughts.

I remember my first view of orbits in atom, quantum numbers, how our teachers described it. Now, totally different view.


Sometimes math equations can be very hard to understand immediately.

But all in all, I get what you are saying.

Thanks

 

[Bassalisk]

I would suggest that the reason current lags voltage in an inductor has to do most w/ energy. An ideal inductor would possesses no capacitance or resistance. Changing the voltage on an inductor requires no work. But changing the current requires work, since inductor energy is W= (L*I^2)/2. It takes time for energy to change, the rate of said change is power. To change current instantly would require changing energy instantly, which is infinite power. The discussion would then be theological, not physical.

Since time is needed for an energy change, time is needed for a current change in an inductor, but not for a voltage change since no work is done. A capacitor is the complement. The current can change quickly since the energy is not changing. But a capacitor cannot have its voltage changed instantly because that would require an instant energy change & infinite power.

Claude

Can u explain this approach more? Why u don't energy to change voltage but energy to change current.

I am confused about this in AC. If you change the potentials, wouldn't you increase current as well?

 

[haleycomet2]

Not higher but just and opposite - like the upward reaction force of the chair you're sitting on - never big enough to throw you upwards.

but if the supply voltage is sin t,and the current is (sin t)/R,and if

a)the back emf induced is just as the supply voltage but in opposite direction,then the back emf should be -sin t(or slightly lower due to resistance,but still sine curve), isn't??

b) if we consider by the differential equation,v=L dI/dt,then the back emf will be L(cos t)/R.Therefore the current must be sometimes flow against the supply voltage when cos t>sin t !?

OR,the hypothesis that current is proportional to supply voltage is wrong?

 

[sophiecentaur]

The current flowing around the circuit at any time will be governed by the supply volts minus the back emf. This will never be in antiphase with the original supply volts. Are you neglecting the existence of some resistance in the circuit?

 

[haleycomet2]

If the current is in phase with the supply voltage(sine curve),then the back emf will be differentiation of current(L di/dt),which is cosine curve --then the emf will be not in phase with supply voltage ?!
ps:I think that resistance will only affect the amplitude of curve but remains same trend.

 

[sophiecentaur]

The resistance will limit the back emf to less than the supply volts, that's all. Without resistance in the coil or the supply the problem is not soluble. It becomes another one of those questions involving irresistable forces and immovable objects.

 

[cabraham]

The resistance will limit the back emf to less than the supply volts, that's all. Without resistance in the coil or the supply the problem is not soluble. It becomes another one of those questions involving irresistable forces and immovable objects.

But the supply has already been stipulated as an ac source, sinusoidal waveform. Even w/o any resistance, i.e. superconducting windings, the inductor presents a reactance w/ no resistance. If the ac voltage source value is V, & the inductive reactance is X (X = omega*L), then I = V/jX. No problem arriving at a solution here.

Were you thinking a dc voltage source across the inductor? In that case the current would ramp upward towards infinity if no resistance were present. BR.

Claude

 

[sophiecentaur]

Yes, you're right about that. I haven't really been paying attention.

This is a simple question if one is prepared just to do the sums (assuming you can differentiate). Too many words have been expended on this and the Maths has been ignored. 'Verbal' explanations so often manage to fall down holes. (Just what I did, in fact.)

 

[haleycomet2]

What i can't understand is if I=V/jX =1/jX(sin t),then back emf will be Ldi/dt=L/jX(cos t),which is 90 degree out of phase of supply voltage.Doesn't back emf is always nearly equal (or at least in phase) to the supply voltage??
ps: jX =reactance right??
Thanks a lot.

 

[sophiecentaur]

X is reactance (a real value).
Impedance (Z) = R + jX (a complex value)

 

[sophiecentaur]

There will be a problem in calculating what you call "back emf" unless you introduce some resistance, somewhere. If you don't, the voltage you will measure across the inductor will be the same as the source voltage (because the constant voltage supply will 'insist' that it is that value. Back emf won't appear anywhere.
The 'back emf' will only show itself by its effect on the current that flows through the source resistance and, hence, some voltage drop at the terminals of the inductor.
If you work out the current through the circuit, it will be V/(R+jX) and work from there, you will get one 90degree phase change and then another - giving a back emf which is in antiphase.

 

[haleycomet2]

Referring to the graphs below(with or without resistance),the sign of current sometimes is in opposite with the voltage.Does this happened when back emf is greater than supply voltage or how to explain this?

ps:these pictures are originally from http://www.allaboutcircuits.com/vol_2/chpt_3/2.html