How do eddy currents in transformers affect power loss?

[Jimmy87]

Hi pf,

Could someone please explain to me why the back emf on the primary caused by eddy currents does not result in power loss. From what I have been reading about transformers eddy currents have two effects:

1) Producing heat (P=I^2R)
2) Producing a flux which opposes the primary flux

I am happy with the first point which DOES result in an energy loss but the second point is not mentioned as a source of power loss. This doesn't add up to me because if eddy current are creating a flux that opposes the primary flux then this will cause less emf on the secondary so surely the product of voltage x current is less so less power? So to be more specific, how does the back emf from the eddy current not cause a power loss to the system?

Thanks

 

[Charles Link]

Hi pf,

Could someone please explain to me why the back emf on the primary caused by eddy currents does not result in power loss. From what I have been reading about transformers eddy currents have two effects:

1) Producing heat (P=I^2R)
2) Producing a flux which opposes the primary flux

I am happy with the first point which DOES result in an energy loss but the second point is not mentioned as a source of power loss. This doesn't add up to me because if eddy current are creating a flux that opposes the primary flux then this will cause less emf on the secondary so surely the product of voltage x current is less so less power? So to be more specific, how does the back emf from the eddy current not cause a power loss to the system?

Thanks

I think your analysis is correct. I'm no expert on transformers, but from everything I've been able to piece together, they often put laminations in transformer iron (look at the side of a transformer -you will see the stacks of thin layers)=layers with plastic insulation to block the eddy currents so that little or no reverse current flows. The EMF's in the iron are offset by electrostatic buildup from only very minute eddy current flow, where it immediately encounters the plastic lamination, and thereby eddy currents are virtually eliminated.

 

[Jimmy87]

I think your analysis is correct. I'm no expert on transformers, but from everything I've been able to piece together, they often put laminations in transformer iron (look at the side of a transformer -you will see the stacks of thin layers)=layers with plastic insulation to block the eddy currents so that little or no reverse current flows. The EMF's in the iron are offset by electrostatic buildup from only very minute eddy current flow, where it immediately encounters the plastic lamination, and thereby eddy currents are virtually eliminated.

Thanks but what about the back EMF from the secondary coil would this not also reduce the flux from the primary coil and surely you cannot ignore that?

 

[Charles Link]

Thanks but what about the back EMF from the secondary coil would this not also reduce the flux from the primary coil and surely you cannot ignore that?

Again, you are correct. If current flows in the secondary, magnetic fields are generated that would reduce the EMF=-d(BA)/dt from the magnetic fields created from the current in the primary.

 

[Jimmy87]

Again, you are correct. If current flows in the secondary, magnetic fields are generated that would reduce the EMF=-d(BA)/dt from the magnetic fields created from the current in the primary.

Thanks so would this back Emf cause a power loss to the system? It never seems to be listed as a form of power loss but I can't see why because if there is less voltage on the secondary then surely there is a power loss?

 

[Charles Link]

Thanks so would this back Emf cause a power loss tot he system? It never seems to be listed as a form of power loss but I can't see why because if there is less voltage on the secondary then surely there is a power loss?

If the primary windings and currents far outweigh the secondary windings and currents, any observed voltage change would be minimal. In the power to your home, if you turn on enough appliances, I think you might observe a slight reduction in the voltage at the outlets.

 

[Merlin3189]

Now Charles has kicked off, I can join in.
If we have a transformer which DOES have eddy currents (Charles seems to be discussing transformers with little or no eddy currents), then I think both your conclusions are correct.

... if eddy currents are creating a flux that opposes the primary flux then this will cause less emf on the secondary ...

This comment I don't agree with.
The primary current depends on the primary emf and the flux. With no secondary load and (if we can imagine this for a moment) no eddy currents (which I am taking as equivalent to another secondary load), the primary emf establishes a primary current which creates flux sufficient to create a back emf equal to the primary emf. This flux is the magnetising flux and in a good transformer is produced by a fairly small current.
Now, when a secondary load current flows, its flux opposes the magnetising flux and would reduce the net flux. But, if the net flux falls, the primary emf will simply drive more current into the primary, until the full magnetising flux was restored - since only then would the back emf equal the primary emf. So the effect of having a secondary current flow is simply to increase the primary current flow (in the proportion required by the turns ratio), but to keep the flux the same.
IMO eddy currents can be considered simply as another secondary load. They tend to reduce the net flux, but as with any secondary load that simply allows the primary to pass more current and restore the full magnetising flux.

... so surely the product of voltage x current is less so less power? ...

IMO the error here is to assume the current is fixed. The eddy current tries to reduce the flux, so the primary current increases to restore the flux.

... how does the back emf from the eddy current not cause a power loss to the system?...

Energy is lost as heat via eddy currents. (I presume your I²R here refers to the current and resistance of the iron, not the copper.) So where does this energy come from? It comes from the primary passing more current at the same emf. This is just the same as a secondary load. Where does its power come from? Again from increased primary current x same primary emf.
Since, I have argued, the net flux remains constant, the secondary emf remains constant. This is the case whether a separate secondary winding draws some current, or an eddy current provides the equivalent of another secondary load. Then, since the secondary emf is unchanged, the secondary current and power remain unchanged. All loads (secondaries and parasitic eddies) draw their power from the primary current x primary emf.

Just a small caveat: this is a little oversimplified. As the primary current increases to give power to secondary loads and to eddy current losses, then the I²R losses in the primary increase. The effective primary emf driving the magnetising current is reduced by IR, so in fact the magnetising flux DOES fall a little due to any load and therefore so does the secondary emf.
Since the eddy current losses are there before you take any secondary load, they are maximum when there is no load. As you start drawing secondary load, the small fall in magnetising flux actually reduces the eddy current losses.

 

[Jimmy87]

Now Charles has kicked off, I can join in.
If we have a transformer which DOES have eddy currents (Charles seems to be discussing transformers with little or no eddy currents), then I think both your conclusions are correct.

This comment I don't agree with.
The primary current depends on the primary emf and the flux. With no secondary load and (if we can imagine this for a moment) no eddy currents (which I am taking as equivalent to another secondary load), the primary emf establishes a primary current which creates flux sufficient to create a back emf equal to the primary emf. This flux is the magnetising flux and in a good transformer is produced by a fairly small current.
Now, when a secondary load current flows, its flux opposes the magnetising flux and would reduce the net flux. But, if the net flux falls, the primary emf will simply drive more current into the primary, until the full magnetising flux was restored - since only then would the back emf equal the primary emf. So the effect of having a secondary current flow is simply to increase the primary current flow (in the proportion required by the turns ratio), but to keep the flux the same.
IMO eddy currents can be considered simply as another secondary load. They tend to reduce the net flux, but as with any secondary load that simply allows the primary to pass more current and restore the full magnetising flux.

IMO the error here is to assume the current is fixed. The eddy current tries to reduce the flux, so the primary current increases to restore the flux.

Energy is lost as heat via eddy currents. (I presume your I²R here refers to the current and resistance of the iron, not the copper.) So where does this energy come from? It comes from the primary passing more current at the same emf. This is just the same as a secondary load. Where does its power come from? Again from increased primary current x same primary emf.
Since, I have argued, the net flux remains constant, the secondary emf remains constant. This is the case whether a separate secondary winding draws some current, or an eddy current provides the equivalent of another secondary load. Then, since the secondary emf is unchanged, the secondary current and power remain unchanged. All loads (secondaries and parasitic eddies) draw their power from the primary current x primary emf.

Just a small caveat: this is a little oversimplified. As the primary current increases to give power to secondary loads and to eddy current losses, then the I²R losses in the primary increase. The effective primary emf driving the magnetising current is reduced by IR, so in fact the magnetising flux DOES fall a little due to any load and therefore so does the secondary emf.
Since the eddy current losses are there before you take any secondary load, they are maximum when there is no load. As you start drawing secondary load, the small fall in magnetising flux actually reduces the eddy current losses.

Wow, that is one of best answers I have ever had - thank you! Based on what you have said could you tell me if the following is correct/incorrect because this is from my textbook:

If an iron core was unlaminated then the eddy currents in the iron core would become larger. The flux produced by the larger eddy currents would oppose the primary flux. This would decrease the rate of change of flux through the secondary coil so it would be observed that the emf across the secondary coil would fall if the iron core was unlaminated.

Is this right? Going on what you said I would have said this was incorrect because although there is an opposing flux from the eddy currents, this will increase the current in the primary coil which will bring the primary flux back to its normal value and therefore the emf across the secondary should be unchanged?

 

[Merlin3189]

As you say, IMO that is not correct, because of the implication that the eddy current directly reduces the flux in the secondary. Your final sentence states my general position on this. As I mentioned in my final paragraph, the flux does fall a bit, due to the primary coil not being ideal (zero resistance), so that the flux cannot be fully maintained. So there is some truth in their conclusion, though IMO their logic is flawed.

Now I am not an expert here. The only transformer theory I studied officially was at A-level, probably using textbooks and teachers like yours. However I have been interested in electronics as a hobby for a long time and am a radio amateur, so I have needed or wanted to look at these things a bit more closely over the years. Much more recently I have been provoked to look even harder at transformer theory by discussions on PF and I must admit, my understanding, particularly on the subject of magnetising current, has changed a lot in the light of that. So take what I say as an opinion and evaluate it on the strength of the arguments.

I have found these articles by Rob Elliot very interesting.Power Supplies - Transformers and they are the main source of my current understanding (if such it be!) of magnetising flux.

A real expert on PF who seems to know an awful lot about transformers is Jim Hardy. It might be worth looking through his posts (there are a lot!) to see if he has discussed this.
Edit: I have taken my own advice and found a thread with some comments on this topic.
I also noticed there The Electrician is also keen on transformers.

 

[Jimmy87]

As you say, IMO that is not correct, because of the implication that the eddy current directly reduces the flux in the secondary. Your final sentence states my general position on this. As I mentioned in my final paragraph, the flux does fall a bit, due to the primary coil not being ideal (zero resistance), so that the flux cannot be fully maintained. So there is some truth in their conclusion, though IMO their logic is flawed.

Now I am not an expert here. The only transformer theory I studied officially was at A-level, probably using textbooks and teachers like yours. However I have been interested in electronics as a hobby for a long time and am a radio amateur, so I have needed or wanted to look at these things a bit more closely over the years. Much more recently I have been provoked to look even harder at transformer theory by discussions on PF and I must admit, my understanding, particularly on the subject of magnetising current, has changed a lot in the light of that. So take what I say as an opinion and evaluate it on the strength of the arguments.

I have found these articles by Rob Elliot very interesting.Power Supplies - Transformers and they are the main source of my current understanding (if such it be!) of magnetising flux.

A real expert on PF who seems to know an awful lot about transformers is Jim Hardy. It might be worth looking through his posts (there are a lot!) to see if he has discussed this.
Edit: I have taken my own advice and found a thread with some comments on this topic.
I also noticed there The Electrician is also keen on transformers.

Great thanks Merlin, I will look through these posts. I haven't studied transformers in much detail yet and a few of the things you said that confused me the more I read over your posts. If there is no load on the secondary you said the back Emf of the primary cancels the applied Emf on the primary (at least for a good transformer). How does this work exactly because are you saying that if there is no load on the secondary and you apply say 8V to the primary, a voltmeter across the primary will in fact read zero volts? How does the transformer equation (i.e. ratio of turns to voltage) hold then? Or is it more of a case that when there is a secondary load the voltage in the transformer equation refers to the net voltage you get? And also the equation therefore only applies when there is a secondary load?

 

[Jimmy87]

I just found this document searching google (I typed 'transformers back emf') which also seems to support what you said Merlin and anyone reading this post who is in my position may find it useful.

 

[Merlin3189]

... If there is no load on the secondary you said the back Emf of the primary cancels the applied Emf on the primary (at least for a good transformer). How does this work exactly because are you saying that if there is no load on the secondary and you apply say 8V to the primary, a voltmeter across the primary will in fact read zero volts? How does the transformer equation (i.e. ratio of turns to voltage) hold then? Or is it more of a case that when there is a secondary load the voltage in the transformer equation refers to the net voltage you get? And also the equation therefore only applies when there is a secondary load?

The power supply (mains usually) applies an (AC) emf to the primary winding, causing a current (AC) to flow, causing a magnetic flux (AC) in the core. The alternating flux produces a back emf (AC), opposing the applied emf. When the flux reaches a level where the back emf equals the applied emf, then there is no net emf to increase the current further, so we have reached a steady state current and flux - the magnetising current and flux.
If I said that the back emf "cancels" the applied emf (did I?), then perhaps I should have said "opposes equally" or somesuch. Both emfs are still present, equal and opposite, and will be measured by a meter.

A sort of analogy (with apologies to any real physicists & engineers for the sloppy language):
If you had two 12V batteries connected +ve to +ve and -ve to -ve, their emfs would oppose each other, but we would still measure the 12 V emf across their terminals. In that case, no current at all would flow, so not quite the analogy I want. So we now replace one of the batteries with a resistor (or conductor) of 12 Ω resistance. There is no "back emf" from the resistor with no current flowing, so the 12V battery is able to push current through the resistor. Once the current reaches 1A, the "back emf" from the resistor, given by Ohms law, is 12V in the opposite sense to the applied emf. There is then no net emf to push more current through the resistor and it stays at 1A. A meter across the resistor (which is equally across the battery) will read 12V.

So your unloaded transformer with 8V applied to the primary will have some current flowing, causing flux, generating 8V back emf and your meter will measure 8V across the primary terminals. The current will be (nearly) 90^o out of phase with the voltage, so no (little) power is going into the transformer. (Ideally, 90^o and zero power, but in reality there are losses and that is power, so not quite 90^o)
The secondary emf is in phase with the primary back emf, because they are both caused by the same flux. Therefore their sizes are in proportion to their turns.

When you now connect a load to the secondary, some current flows. Let us stick to a resistive load for simplicity, so that current is in phase with the emf and opposes, or tries to reduce, the flux in the core. But reducing the flux would reduce the primary back emf, so that it was no longer equal and opposite to the primary applied emf. So the supply can now push more current into the primary until the flux is restored to that value which generates the full back emf. This extra current must be exactly that required to balance the secondary current: the sum total of (extra primary current x primary turns) + (secondary current x secondary terms) must be zero, because there is no change in flux. Primary extra current is in phase with secondary current, otherwise their contributions to the flux would not exactly cancel, and so it is also in phase with the primary emf. Therefore this extra current does represent real power flowing into the transformer.

This is why the transformer equation holds. The role of the (ideally constant) magnetising flux is to generate the primary and secondary emfs proportional to their turns. The magnetising current and flux are then ignored, because in an ideal transformer they are 90^o out of phase with the emf and do not represent real power. . If and when a secondary current flows, it will be balanced by a matching primary current - in inverse proportion to their turns. Since the flux does not change, neither do the primary nor secondary voltages (unless as a consequence of supplying more current, the supply to the primary drops its voltage.)

 

[Jimmy87]

The power supply (mains usually) applies an (AC) emf to the primary winding, causing a current (AC) to flow, causing a magnetic flux (AC) in the core. The alternating flux produces a back emf (AC), opposing the applied emf. When the flux reaches a level where the back emf equals the applied emf, then there is no net emf to increase the current further, so we have reached a steady state current and flux - the magnetising current and flux.
If I said that the back emf "cancels" the applied emf (did I?), then perhaps I should have said "opposes equally" or somesuch. Both emfs are still present, equal and opposite, and will be measured by a meter.

A sort of analogy (with apologies to any real physicists & engineers for the sloppy language):
If you had two 12V batteries connected +ve to +ve and -ve to -ve, their emfs would oppose each other, but we would still measure the 12 V emf across their terminals. In that case, no current at all would flow, so not quite the analogy I want. So we now replace one of the batteries with a resistor (or conductor) of 12 Ω resistance. There is no "back emf" from the resistor with no current flowing, so the 12V battery is able to push current through the resistor. Once the current reaches 1A, the "back emf" from the resistor, given by Ohms law, is 12V in the opposite sense to the applied emf. There is then no net emf to push more current through the resistor and it stays at 1A. A meter across the resistor (which is equally across the battery) will read 12V.

So your unloaded transformer with 8V applied to the primary will have some current flowing, causing flux, generating 8V back emf and your meter will measure 8V across the primary terminals. The current will be (nearly) 90^o out of phase with the voltage, so no (little) power is going into the transformer. (Ideally, 90^o and zero power, but in reality there are losses and that is power, so not quite 90^o)
The secondary emf is in phase with the primary back emf, because they are both caused by the same flux. Therefore their sizes are in proportion to their turns.

When you now connect a load to the secondary, some current flows. Let us stick to a resistive load for simplicity, so that current is in phase with the emf and opposes, or tries to reduce, the flux in the core. But reducing the flux would reduce the primary back emf, so that it was no longer equal and opposite to the primary applied emf. So the supply can now push more current into the primary until the flux is restored to that value which generates the full back emf. This extra current must be exactly that required to balance the secondary current: the sum total of (extra primary current x primary turns) + (secondary current x secondary terms) must be zero, because there is no change in flux. Primary extra current is in phase with secondary current, otherwise their contributions to the flux would not exactly cancel, and so it is also in phase with the primary emf. Therefore this extra current does represent real power flowing into the transformer.

This is why the transformer equation holds. The role of the (ideally constant) magnetising flux is to generate the primary and secondary emfs proportional to their turns. The magnetising current and flux are then ignored, because in an ideal transformer they are 90^o out of phase with the emf and do not represent real power. . If and when a secondary current flows, it will be balanced by a matching primary current - in inverse proportion to their turns. Since the flux does not change, neither do the primary nor secondary voltages (unless as a consequence of supplying more current, the supply to the primary drops its voltage.)

Excellent - thank you Merlin that helps a lot. Just one final point I have been thinking about which is why there is always a back emf on the primary (i.e. it never goes away)? I have been reading a lot around inductors (as I have not learned about these yet) after reading your first responses to this thread. If you have a DC supply with a resistor and an inductor (so an LR circuit I think they call it) then the inductor opposes the current and it takes time to reach its steady state value and the bigger the inductor the longer it takes. They are very similar to RC circuits which we have done a lot of. When an inductor reaches its steady state value then there is NO back emf and the inductor literally has no role in the circuit and if you take an ideal inductor (zero resistance) then it is like the inductor has disappeared and you are left with a steady current an a resistor. This is not the case on the primary circuit of a transformer because you always have a back emf as you said earlier. No-one seems to explain this but is this simply because you have AC therefore the magnetic flux through the inductor is always changing hence there is always a back emf? Also, by steady-state (which you mentioned) do you mean steady-state AC because if you consider half a cycle of AC then you would get a slowly increasing current to a maximum at which point the back emf goes to zero but then the AC switches and a guess it repeats itself (as oppose to the current staying at a maximum and back emf staying at zero as in the case of DC)?

I also found another source which discusses opposing flux from eddy current and it is in agreement with what you discussed about the primary coil increasing its current to compensate. It seems to say that thinking that eddy current flux opposes the primary flux is a big violation of energy conservation. It argues that eddy current flux could never decrease the primary flux and argues this in terms of violation of energy conservation (and Kirchoff's law). They said that if the primary flux was actually reduced then the rate of change of flux through the primary would be reduced which means the back emf would fall. Since the primary is supplied with a fixed voltage then this mismatch would violate conservation of energy. They went on to say that the sum of the emf's around the primary would then be non zero. Would you agree with this?

Once again thanks for all your help.

 

[lychette]

the back emf is not caused by eddy currents ! eddy currents flow in the transformer (iron) core as a result of emfs induced in the iron core. The iron core experiences changing magnet flux linkage and this is what produces induced emfs.

 

[Charles Link]

I think I can add what may be a couple of helpful inputs. Merlin is correct in earlier comments with his assessment that the system is voltage driven, so that flux stays constant in this system even when secondary currents and eddy currents generate fields that would reduce the flux. The currents in the primary increase when this occurs, thereby maintaining the magnetic flux. It would appear that these secondary currents along with eddy currents decrease the impedance seen by the voltage source of the primary coil. As Merlin pointed out, a good transformer (with few eddy currents) and no load draws very little current. ==>> (It has a rather high impedance and and thereby small I=V/Z.) If a good iron core is replaced by an iron core with more eddy currents, the impedance of the transformer (seen by the primary voltage source) will be lower and it will draw more current. Also, when a load is placed in the secondary circuit, e.g. an electrical appliance, more current flows in the secondary, and the impedance seen by the voltage source of the primary will decrease, so that there is increased current in the primary. (All currents and voltages here are of course AC (i.e. f=60 Hz), along with the fields. Any reverse EMF's are a result of the changing magnetic flux. When the amplitude of the flux level is constant with a constant voltage amplitude, the time derivative of the flux also maintains the same level. )

 

[Merlin3189]

..why there is always a back emf on the primary (i.e. it never goes away)?
... When an inductor reaches its steady state value then there is NO back emf
...on the primary circuit of a transformer because you always have a back emf
... is this simply because you have AC therefore the magnetic flux through the inductor is always changing hence there is always a back emf?
Also, by steady-state... do you mean steady-state AC because if you consider half a cycle of AC then you would get a slowly increasing current to a maximum at which point the back emf goes to zero but then the AC switches ...

(I've tried to edit to just the points t I'll try to answer.) Yes, you're right, both about the sloppy language and the proper explanation. I've been very casual in my terminology -partly becuse I'm really not too hot on the proper mathematical formulation, partly to try to be simple and understandable.
Just as we're talking about an AC applied emf = V sin(ωt) , AC magnetising current = I sin(ωt - 90^o) and AC back emf = E sin(ωt), I am thinking of an alternating flux with a constant peak amplitude and frequency ω. (I won't say something like flux = F sin(ωt - 90^o) because I'm pretty sure it generally isn't a sinusoid in iron cored transformers.) By constant flux or constant emf, I was simply referring to the amplitude of the AC quantity.
When the supply is first connected to the transformer there are significant transient effects lasting many cycles, so these nice simple AC currents and voltages only apply once it has settled to a steady state. Similarly when a load is connected there will again be a transient stage until order is restored and I am steering clear of these stages, when there are indeed significant flux changes (both in AC amplitudes and DC flux.)

Regarding the paragraph on eddy current the energy argument is a new one to me. I'll give it some thought, but some of these explanations seem to get a bit convoluted and don't make it any clearer to me. Certainly the voltages around the primary loop must sum to zero, though one has to be careful with phase in AC circuits. I'm not sure why this is a conservation of energy issue, but I'd be interested to read their full argument.
I've also had a read of the link in your post #1, thanks. There are several bits that puzzle me, but I think it is similar to my view. Again I need to digest it a bit. It's a shame they don't give their undergraduate explanation, as I think a nice algebraic explanation is what I need to add to my portfolio.

Thank you for starting the thread. Trying to explain things to other people makes me think again about them and often helps clarify ideas that I'd just accepted without too much question. This time I was prompted to revisit Rod Elliot's site to get the link for my earlier post and I discovered several new additions since my previous visit. I also re-read the transformer article and looked more closely at some bits I'd only skimmed before. So I'm sure I've got as much out of this thread as you.

 

[cnh1995]

I am thinking of an alternating flux with a constant peak amplitude and frequency ω. (I won't say something like flux = F sin(ωt - 90o) because I'm pretty sure it generally isn't a sinusoid in iron cored transformers.)

Since the voltage waveform is sinusoidal, shouldn't the flux be also sinusoidal? According to Faraday,
E=dΦ/dt or Φ(t)=∫E(t)dt. So, I believe the flux has to be sinusoidal in steady state and it is the magnetizing current that should be non-sinusoidal.

 

[lychette]

If the emf applied to the primary coil is E = E Sin(ωt) and the primary is considered to be R + L in series then the voltages across R and the coil (L) are related by
E² = (V_r)² + (V_L)² and I = ICos(ωt)
V_R = I x R
and V_L = I x X_L

 

[Merlin3189]

Yep! Quite right. Trying to be a bit pedantic, I put my foot right in it!

In fact, your picking me up on this is very helpful. I think I have been confusing these two in my own mind. Now you have brought it to prominence, I think it will stay clearer in my memory. Thank you very much.

It does of course screw up some of my phasor diagrams: I'm not sure how you do a phasor for that spikey current. Lots more food for thought now!

 

[cnh1995]

It does of course screw up some of my phasor diagrams: I'm not sure how you do a phasor for that spikey current.

I believe the spikey current waveform is a result of the non-linear B-H curve of the core material. If it is operated in the linear region of the B-H curve, magnetizing current will be sinusoidal. This is possible only in ideal transformers. So, I guess magnetizing current's phasor is drawn only in case of ideal transformer or by approximating the practical waveform to sinusoudal waveform. I'm not sure how it's done since we have not come to that yet in our curriculum..

 

[Merlin3189]

I see Lychette posted while I was replying to cnh1995's post.
As cnh1995 says, the B-H curve is not linear. As the flux grows, the iron starts to saturate, so that increasing increments of current are needed to make equal increments in flux. The current has to increase to whatever value (and at whatever rate) is needed to make dΦ/dt = E -IR , which makes the current waveform very non-sinusoidal.
This diagram from here shows the effect (though I've seen much spikier current waveforms)

As far as the phasor goes, I guess we work with the fundamental phasor and assume the harmonics just follow along! (Though I suppose the mathematicians will be able to prove it works.)