# Confusion about transformer

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: 133
 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.
PF Gold
P: 11,919
 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: 64
 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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