Confusion about transformer and flux

[tasnim rahman]

Lets consider an ideal transformer in no-load state, i.e. with the primary side connected to a voltage source, and the secondary side in open-circuit. Now, this is what I believe; the voltage Vp in the primary produces a current, which lags the voltage by 90°, as would be in an inductor. This current produces an oscillating magnetic flux in the core, say øm, which is in phase with the current. This flux in turn induces a voltage in the primary coil, according to Faraday's law of electromagnetic induction, which is called the back-emf, and which lags the flux by a further 90°, and the applied voltage by 180°, and is equal to the applied voltage in magnitude. So, this back-emf voltage cancels out the effect of the applied voltage completely, with the effect that there should not be any current in the primary windings. Yet, magnetic flux øm, still seems to exist in such condition in the transformer core. But, how is this possible as flux can not be created without current, and if no current flows where does the flux come from? Confused. A little help needed. Thnx in advance.

 

[marcusl]

Your error is in saying "the voltage Vp in the primary produces a current, which lags the voltage by 90°". The current is in phase with the source or driving emf. The back-emf is out of phase as you say, so there is no cancellation. The net voltage across across the primary, when the secondary is open as you assume, is the vector sum of the resistive drop and the back-emf.

 

[tasnim rahman]

Thank you for that. But the primary coil behaves like an inductor. And in an inductor, the current lags the voltage by 90°. So, shouldn't the current in the primary lag by 90°??

 

[Born2bwire]

It does not behave as an inductor. There is no self inductance in an ideal transformer in the coil. An ideal transformer at the primary looks like the load impedance on the secondary scaled by the turn ratio. Any inductive behavior arises from leakage inductance.

 

[nasu]

In inductors, the 90 degree lag in the current is due to the induced emf.
Without induced emf it will be just a resistor so there will be no lag.

 

[marcusl]

It does not behave as an inductor. There is no self inductance in an ideal transformer in the coil. An ideal transformer at the primary looks like the load impedance on the secondary scaled by the turn ratio. Any inductive behavior arises from leakage inductance.

I think the primary looks inductive. Imagine unwinding the secondary--you are left with a coil, aka inductor.

The loop equations for primary and secondary are V=R_1 I_1+sLI_1-sMI_2 \\<br /> 0=R_2 I_2+sL_2I_2-sMI_1 according to common convention, where s is the Laplace transform variable and M is the mutual inductance. In this case, there is no secondary current so I_2=0 and V=R_1 I_1+sLI_1.
The presence of "s" in the last term gives the 90 degree phase shift.

 

[Born2bwire]

Oh we are doing an open circuit. Yeah.

 

[marcusl]

Actually self inductance is there even in the presence of a loaded secondary.

 

[sophiecentaur]

It does not behave as an inductor. There is no self inductance in an ideal transformer in the coil. An ideal transformer at the primary looks like the load impedance on the secondary scaled by the turn ratio. Any inductive behavior arises from leakage inductance.

One of the problems about discussing an Ideal Transformer (and any other 'ideal model' in Physics) is that it can give you unbelievable and counter-intuitive results. Frictionless bicycles will not work either.
The idea of a 'good' transformer involves infinite (very high) primary inductance. See what happens if you try to wind a mains transformer with the right ratio but only 50 turns in the primary.
If the secondary is unladed, no current can flow and it will have no effect on the primary current. Loading (resistive) the secondary will only have the effect of increasing primary current.

It is instructive to look at the 'equivalent circuit' of a real transformer (see this wiki link, half way down). The circuit behaves like an 'ideal' transformer, buried inside a whole host of other effective components that account for the shortcomings.

 

[tasnim rahman]

I think the primary looks inductive. Imagine unwinding the secondary--you are left with a coil, aka inductor.

The loop equations for primary and secondary are V=R_1 I_1+sLI_1-sMI_2 \\<br /> 0=R_2 I_2+sL_2I_2-sMI_1 according to common convention, where s is the Laplace transform variable and M is the mutual inductance. In this case, there is no secondary current so I_2=0 and V=R_1 I_1+sLI_1.
The presence of "s" in the last term gives the 90 degree phase shift.

That means that the applied voltage and the back-emf actually cancel each other out. Right?

One of the problems about discussing an Ideal Transformer (and any other 'ideal model' in Physics) is that it can give you unbelievable and counter-intuitive results. Frictionless bicycles will not work either.
The idea of a 'good' transformer involves infinite (very high) primary inductance. See what happens if you try to wind a mains transformer with the right ratio but only 50 turns in the primary.
If the secondary is unladed, no current can flow and it will have no effect on the primary current. Loading (resistive) the secondary will only have the effect of increasing primary current.

It is instructive to look at the 'equivalent circuit' of a real transformer (see this wiki link, half way down). The circuit behaves like an 'ideal' transformer, buried inside a whole host of other effective components that account for the shortcomings.

I am sorry, but I seem unable to grasp the effect of infinite inductance. Does this mean that in an ideal transformer, magnetic flux can be produced from zero current? If so, that seems to explain it a bit. And, in no-load state the current in the primary is zero. Also, I did see the equivalent circuit for a real-transformer, and it was while trying to understand, I felt I needed to know the ideal one first.

 

[Dale]

It is instructive to look at the 'equivalent circuit' of a real transformer (see this wiki link, half way down). The circuit behaves like an 'ideal' transformer, buried inside a whole host of other effective components that account for the shortcomings.

This is a very good suggestion which is valid not only for transformers, but also any other real circuit element.

 

[sophiecentaur]

That means that the applied voltage and the back-emf actually cancel each other out. Right?
I am sorry, but I seem unable to grasp the effect of infinite inductance. Does this mean that in an ideal transformer, magnetic flux can be produced from zero current? If so, that seems to explain it a bit. And, in no-load state the current in the primary is zero. Also, I did see the equivalent circuit for a real-transformer, and it was while trying to understand, I felt I needed to know the ideal one first.

Could you be allowing confusion between cause and effect to get n the way of accepting this?
There is an analogy between Inductance and Mass. A floating oil tanker will produce (effectively) the same reaction force against a human, pushing it from the quay as a solid lump of concrete, fixed to the quay. A very high inductance will react against any change in current.
The back emf is L dI/dt and the mechanical reaction force is ma. In the limiting cases, there is neither current change nor acceleration.

 

[marcusl]

That means that the applied voltage and the back-emf actually cancel each other out. Right?

There is no "applied voltage." The voltage across an inductance is the sum of the drop across the inductor loss, which is in phase with the current, and the back emf, which is out of phase.

 

[sophiecentaur]

There is no "applied voltage." The voltage across an inductance is the sum of the drop across the inductor loss, which is in phase with the current, and the back emf, which is out of phase.

I can see you are trying to make a valid point here but, when you connect a voltage source to a transformer, there is 'an applied voltage', in the accepted sense of the expression. As with many of these kinds of threads, there is not enough use made of actual diagrams and Maths in the discussion. It's too easy for people to be talking at cross purposes without them.

 

[UltrafastPED]

I can see you are trying to make a valid point here but, when you connect a voltage source to a transformer, there is 'an applied voltage', in the accepted sense of the expression. As with many of these kinds of threads, there is not enough use made of actual diagrams and Maths in the discussion. It's too easy for people to be talking at cross purposes without them.

I agree - every circuit should have a clearly labeled diagram to "ground" the discussion! When we talk about circuits at work there is always a diagram.

 

[marcusl]

Sorry, I worded that terribly. What I meant is that one can't measure a separate "applied voltage." The voltage across the primary always includes the resistive and reactive drops.

 

[enorbet]

While I agree that Diagrams and Math lock down the issue compared to words which can be misinterpreted (the actual raison d'etre for diagrams and math ) it seems to me that the only misinterpretation here comes from using the word "transformer". A transformer primary presents three (3) somewhat interactive loads - DC Resistance, Inductive Reactance, and Reflected Impedance (from the secondary). As soon as we open the secondary and make it's effective impedance infinite, we no longer have a transformer. We only have an inductor. Looking at it as a simple idling inductive circuit, whether in the real world or an ideal one, I don't see any problem.

 

[sophiecentaur]

It depends on how near you want to get to an accurate model.
Merely disconnecting the secondary load doesn't alter the presence of the secondary and itself and mutual capacitance. The transformer doesn't 'become' and inductor. You would need to rebuild it for that to be true.

 

[Juan Largo]

I worked as an engineer for an electric utility for over 30 years, specializing in transformer operation and maintenance, so I think I can answer this question.

Looking at the transformer at no load:

The magnetizing current in a sense does lag the primary voltage by 90°, but the current isn't purely sinusoidal because there are a lot of odd harmonics (magnetic flux isn't a linear function of current). But you're right, there is a back emf that is essentially 180° out of phase with the primary voltage, so the back emf almost cancels the primary voltage. The back emf will have odd harmonics also, so it isn't purely sinusoidal.

The residual emf -- the primary voltage minus the back emf -- is equal to the the emf that is required to force magnetizing current to flow through the primary circuit (the rest of the system connected to the primary winding). Since the magnetizing current is typically very small, the residual emf is small also. In an "ideal" transformer, you can assume the magnetizing current is zero.

 

[enorbet]

It depends on how near you want to get to an accurate model.
Merely disconnecting the secondary load doesn't alter the presence of the secondary and itself and mutual capacitance. The transformer doesn't 'become' and inductor. You would need to rebuild it for that to be true.

After reading your comment (as well as your sig line ) I must concur. I had made the error of assuming we were talking about low frequencies where self and mutual capacitance are negligible. In practice, as frequencies climb it becomes harder and harder to distinguish an inductance from a capacitance and this does indeed become a real concern... for example at microwave frequencies. Back on track for "ideal" now, thanks.

 

[sophiecentaur]

After reading your comment (as well as your sig line ) I must concur. I had made the error of assuming we were talking about low frequencies where self and mutual capacitance are negligible. In practice, as frequencies climb it becomes harder and harder to distinguish an inductance from a capacitance and this does indeed become a real concern... for example at microwave frequencies. Back on track for "ideal" now, thanks.

I think that the two schools of Transformers are showing their differences in approach in this thread. Apart from when I have bought the occasional off the shelf mains transformer kit (roll yer own) or a ready made special, I have not been involved with 50Hz stuff. My experience has been much more in the RF field where the parasitics have been much more relevant - many transformers being operated at resonance even. Different things are relevant to me, compared with the Power specialists - that's all.

 

[tasnim rahman]

Could you be allowing confusion between cause and effect to get n the way of accepting this?
There is an analogy between Inductance and Mass. A floating oil tanker will produce (effectively) the same reaction force against a human, pushing it from the quay as a solid lump of concrete, fixed to the quay. A very high inductance will react against any change in current.
The back emf is L dI/dt and the mechanical reaction force is ma. In the limiting cases, there is neither current change nor acceleration.

Sorry everyone, I have been away from this thread for a very long time, what with quizzes and exams being held almost everyday, at the university, and also due to some other circumstances. Well anyways, I am finally back here, and am surprised that this thread has not been closed yet. And so, I would like to thank the admins for this.

@sophiecentaur. I think I understood the analogy for back emf: that, for the same back emf, the one with higher inductance will show a less change in current, compared to the one with low inductance showing a greater change in current. And that, for the limiting case of infinite inductance, the change in current will be zero. Right?? But, what I wanted to know was about the equation ø = L I, and L = ø / I. Here for L to be infinite, I has to be zero, irrespective of the value of ø. Which seems to mean to show that for infinite inductance, L , any magnetic flux, ø seems to be produced from zero current, I. Right? And, I think I understood that the back emf was the voltage drop across the inductor itself, i.e. L dI/dt, which is obviously equal to the supply voltage, V_AC, as seen from the picture. Am I understanding it right?? Would anyone kindly verify.

 

[tasnim rahman]

Please. May I have some help here, in clearing this thing out. @sophiecentaur

 

[ehild]

Lets consider an ideal transformer in no-load state, i.e. with the primary side connected to a voltage source, and the secondary side in open-circuit. Now, this is what I believe; the voltage Vp in the primary produces a current, which lags the voltage by 90°, as would be in an inductor. This current produces an oscillating magnetic flux in the core, say øm, which is in phase with the current. This flux in turn induces a voltage in the primary coil, according to Faraday's law of electromagnetic induction, which is called the back-emf, and which lags the flux by a further 90°, and the applied voltage by 180°, and is equal to the applied voltage in magnitude. So, this back-emf voltage cancels out the effect of the applied voltage completely, with the effect that there should not be any current in the primary windings. Yet, magnetic flux øm, still seems to exist in such condition in the transformer core. But, how is this possible as flux can not be created without current, and if no current flows where does the flux come from? Confused. A little help needed. Thnx in advance.

Start from the basics: The magnetic field of a coil is B=μNI/l and Amper's Law: The change of the flux Φ produces back emf in a coil: emf= -dΦ/dt. Φ= B A, emf= -μNA/ldI/dt. L is defined as the coefficient of dI/dt, so you have the equation of the back emf : emf= -L dI/dt.

If you have a loop with an ideal ac generator with emf E=E₀cos(ωt) and an ideal inductor with inductance L, E-LdI/dt=0 according to Kirchhoffs Law. So dI/dt = E₀/L cos(ωt) and integrating, you get I=E₀/(ωL)sin(ωt) +C. Initially the current is zero, so C=0 and I=E₀/(ωL ) sin(ωt)
The current is a sin function, lags the voltage by 90° and has the amplitude of E₀/(ωL ). It is not zero.

ehild

 

[tasnim rahman]

Start from the basics: The magnetic field of a coil is B=μNI/l and Amper's Law: The change of the flux Φ produces back emf in a coil: emf= -dΦ/dt. Φ= B A, emf= -μNA/ldI/dt. L is defined as the coefficient of dI/dt, so you have the equation of the back emf : emf= -L dI/dt.

If you have a loop with an ideal ac generator with emf E=E₀cos(ωt) and an ideal inductor with inductance L, E-LdI/dt=0 according to Kirchhoffs Law. So dI/dt = E₀/L cos(ωt) and integrating, you get I=E₀/(ωL)sin(ωt) +C. Initially the current is zero, so C=0 and I=E₀/(ωL ) sin(ωt)
The current is a sin function, lags the voltage by 90° and has the amplitude of E₀/(ωL ). It is not zero.

ehild

Thank you, and I think I have already come to that conclusion, in a recent post. That, the back e.m.f. is actually a manifestation of the drop of the supply voltage across the inductor, and as such is equal to the supply voltage, with the current lagging behind the voltage by 90°. But, another question that just came to mind is that, while the back emf is equal to supply voltage in magnitude, for an inductor, is it in-phase or out of phase by 180°, to the supply voltage, for an inductor, and consequently for the primary side of the ideal transformer in no-load condition. For I have read that in that case, the back emf although equal, is out of phase by 180° to the applied voltage.

 

[sophiecentaur]

Thank you, and I think I have already come to that conclusion, in a recent post. That, the back e.m.f. is actually a manifestation of the drop of the supply voltage across the inductor, and as such is equal to the supply voltage, with the current lagging behind the voltage by 90°. But, another question that just came to mind is that, while the back emf is equal to supply voltage in magnitude, for an inductor, is it in-phase or out of phase by 180°, to the supply voltage, for an inductor, and consequently for the primary side of the ideal transformer in no-load condition. For I have read that in that case, the back emf although equal, is out of phase by 180° to the applied voltage.

Yes. That's why it is called 'back' emf; it subtracts from the supply voltage and, in a good transformer, it produces a very small net voltage.

 

[tasnim rahman]

Thank you very much. I think I get it now. But if there is any other problem I will just post it here. Thanks again.