# Confusion about transformer

1. Jun 18, 2014

### 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.

2. Jun 18, 2014

### 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.

3. Jun 18, 2014

### 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°?? :tongue:

4. Jun 18, 2014

### 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.

Last edited: Jun 18, 2014
5. Jun 18, 2014

### 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.

6. Jun 18, 2014

### marcusl

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 $$V=R_1 I_1+sLI_1-sMI_2 \\ 0=R_2 I_2+sL_2I_2-sMI_1$$ 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.

7. Jun 18, 2014

### Born2bwire

Oh we are doing an open circuit. Yeah.

8. Jun 18, 2014

### marcusl

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

9. Jun 19, 2014

### sophiecentaur

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.

10. Jun 20, 2014

### tasnim rahman

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.

11. Jun 20, 2014

### Staff: Mentor

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

12. Jun 21, 2014

### 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.

13. Jun 21, 2014

### marcusl

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.

14. Jun 22, 2014

### sophiecentaur

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.

15. Jun 22, 2014

### UltrafastPED

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.

16. Jun 22, 2014

### 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.

17. Jun 23, 2014

### 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.

18. Jun 24, 2014

### 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.

19. Jun 24, 2014

### 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.

20. Jun 24, 2014

### 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.