by tasnim rahman
Tags: ideal transformer, magnetic flux, no load
 P: 11 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.
P: 190
 Quote by 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 its self 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.
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PF Gold
P: 12,256
 Quote by enorbet 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.
P: 70
 Quote by sophiecentaur 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, VAC, as seen from the picture. Am I understanding it right?? Would anyone kindly verify.
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 P: 70 Please. May I have some help here, in clearing this thing out. @sophiecentaur
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P: 10,767
 Quote by 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.
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
P: 70
 Quote by ehild 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.