Question about ideal transformers and inductors

Question

Recently I was studying about alternating currents and transformers and I got confused about some of the concepts. I tried a lot of sources but none of them seemed to address the problem.

I understood that in a purely inductive DC circuit, the EMF of the source is exactly equal to the back EMF from the inductor at all times (due to KVL) and the current increases linearly with time.

In case of a purely inductive AC circuit, the net EMF is still \$0\$ at all instants and the current follows a sine curve and lags behind voltage by \$\pi/2\$ rad.

Please tell me if I've got something wrong in my understanding or misinterpreted something. I'll be very grateful.

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Now my problem:

In an ideal transformer, if the secondary coil is open, the primary coil just behaves like an ideal iron-core inductor since the secondary coil has no effect. Hence the current in the primary coil must be there lagging behind the voltage by 90 degrees.

What I think is there will be a changing current flowing in the primary coil just enough to maintain the necessary back EMF. Here are a few articles I read.

1. http://www.electricalunits.com/transformer-on-no-load/
2. https://www.electronics-tutorials.ws/transformer/transformer-loading.html
3. this Quora post (please see Steven J Greenfield's comment below)

If that's the case the back EMF has to be equal to \$-V_0\sin\omega t\$ which is precisely opposite to the applied voltage and that's the case of ideal inductors as I mentioned in the beginning. I think that this is what should really happen neglecting losses and other non-idealities. Again I may be wrong so please correct me.

But I also found some posts talking about the current in the primary coil being zero.

Here are a few posts which talk about this:

1. this EE SE post
2. this EE SE post
3. this Quora post

Moreover if the current were always zero in the primary coil under no load, that would mean \$\frac{\text dI}{\text dt}\$ is 0 so no back EMF would be induced and it'd be a short since there would be no resistance.

Actually I'm confused about whether or not the primary current at no load will be zero.

Please tell me where I've gone wrong and if something I mentioned is wrong.

Answer 1

Lets's discuss only about what you said "I have understood"

Not sure how you consider those those historical non-measurable quantities named EMF, but the result is ok for DC: There's linearly increasing current in an ideal coil which is connected to a constant DC voltage.

There's no need for the EMF. If some mechanism, say an ideal voltage source forces an ideal inductor to have voltage U between it's terminals, the current through the inductor has changing rate U/L where L=the inductance. This is how inductors work, otherwise it's something else than an inductor. No EMF needs to be imagined to exist.

With AC voltage source the same law can still be used. If the voltage has sinusoidal form, the current which changes with rate = U/L is also sinusoidal, but lags 90 degrees behind the voltage.

Some of us surely want to keep concept EMF alive because one can feel he's more systematic in that way. With EMF one can have a logical reason for the existence of current and measurable voltages between circuit nodes. In addition many clever looking laws can be formulated with EMF such as "in any closed circuit loop the sum of voltage drops must be equal with the sum of EMFs".

But an alternative way is to discard EMF and to formulate how some ideal components work with measurable voltages. Practical components can well be modelled as circuits which are made of ideal components. I applied that thinking when I wrote how inductors work.

For transformers we can write a pair of equations for primary and secondary currents and voltages. The equation pair states how it works and gives to us the foundation to find how currents and voltages in the whole circuit must vary in the time. There's no Federal nor other laws that force us to call the induced voltage components EMFs and tease us to include term EMF to our talks and writings before we can write the total law for a component.

I heve learned to discard the EMF because everything which happens in circuits is finally a consequence of how electric and magnetic fields happen to settle as well inside the parts and around them. In Maxwell's equations there's no separate "motive electric field" and "drop electric field", there's electric field as a place and time dependent vector field.

Answer 2

There are a lot of confusing posts out there about this. Part of this comes from people not being clear which component of current they’re talking about:

• The reactive current, 90 degrees out of phase with the primary voltage. It’s real, it’s measurable, it has important impact when designing the transformer, but it transfers no power.

• The current in phase with the primary voltage, which transfers power to the secondary via the usual np:ns ratio.

Both are there, both are real, both make up the total current. But because they’re independent, it’s a useful approximation to talk about them separately or even to ignore one.

Answer 3

Consider an ideal inductor with a sinusoidal voltage source, \$\small v_S=V\:cos\: \omega t\$. The induced EMF must exactly match this, therefore we have:

$$\small L\frac{di}{dt}=Vcos\;\omega t $$

solving the differential equation for \$\small i\$ gives:

$$\small i=\frac{V}{\omega L}sin\:\omega t$$

This is the magnetising current, which is also present in a transformer.

Not that the voltage and current are \$\small 90^o\$ out of phase so the there is no power transfer.

Answer 4

I understood that in a purely inductive DC circuit, the EMF of the source is exactly equal to the back EMF from the inductor at all times (due to KVL) and the current increases linearly with time.

In DC steady-state, all currents are constant, so there're no more back EMF induced in inductors (their voltages are zero). I don't understand why you said that current increases linearly with time (in DC transient state, for RL networks, the current increases/decreases exponentially, not linearly).

In an ideal transformer, if the secondary coil is open, the primary coil just behaves like an ideal iron-core inductor since the secondary coil has no effect. Hence the current in the primary coil must be there lagging behind the voltage by 90 degrees.

No, if the secondary is open, the primary won't behave just like an ideal inductor. Why? Since the transformer is ideal, all the flux from one coil passes to the other coil. If the secondary is open, no current flows through it. This requires that there's no primary current either. You can see this analytically with the equation \$N_1 i_1(t) = N_2 i_2(t)\$; if \$i_2(t) = 0\$ then \$i_1(t) = 0\$. An ideal transformer is not just a pair of magnetically coupled inductors, but a pair with unity coupling coefficient.

If you're going to read Quora to learn, be very careful; I suggest rather to read textbooks which are safer. For the most detailed explanation, perhaps physics books about electromagnetism are best (specifically the chapters about magnetic fields and electromagnetic induction). To complement, read electric circuit textbooks.