# CURRENT TRANSFORMER MAGNETICS

Discussion in 'Physics' started by b.shahvir, Jan 11, 2009.

1. ### b.shahvir Thread Starter Active Member

Jan 6, 2009
444
0
Hi Guys,

I am grappling with a peculier (but somewhat common) problem with Current Transformer (CT) magnetic circuit. I will be very grateful if someone could help me out with my problem/s.
I will ennumerate my query in points for clarity as given below;

1) In an ideal Voltage Transformer (VT), the magnetic flux developed in the core is constant at all loads. But the magnetic flux in the core of a CT reduces as soon as a 'Burden' (load) is connected to the secondary winding of a CT! Hence, why is the magnetic flux in a CT core not constant (with connected load, also) as is the case with a VT ? Why does this reduction in magnetic flux occur in case of a CT, since a CT basically resembles an ideal VT as far as the core magnetic circuit is concerned ?

2) I am looking for a phasor diagram for an UNLOADED CT (open circuited secondary) but i am unable to find one yet. I will be extremely grateful if someone could provide me with one or guide me to a suitable link.

3) The NO-LOAD primary current (excitation current) 'Io' in an ideal transformer (VT or CT) is the vector sum of Magnetizing current 'Im' and no-load Losses current ( iron + Cu) 'Iw'. But the no-load or on-load magnetic flux 'ψ' developed (visible) in an ideal transformer core is in phase with the Magnetizing current 'Im' only (which itself is in quadrature with the applied primary voltage 'Vp').
Hence my query is; What happens to the magnetic flux component of 'Iw' which is in phase with 'Vp' ? Why is it not 'visible' in the transformer magnetic core ? Where does it disappear ? Please explain.

Thanks & Regards,
Shahvir

Last edited: Mar 9, 2009

Jan 6, 2009
444
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3. ### leftyretro Active Member

Nov 25, 2008
394
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Standard commerical current transformers are never operated with a open secondary (danger, danger!) as with the very high turns ratio used the secondary winding insulation will break down. They must always have a load attached.

Lefty

4. ### loosewire AAC Fanatic!

Apr 25, 2008
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The 110 or 220 volts source is metered on demand,the transformer is sealed. What does he mean by open secondary.

5. ### BillO Distinguished Member

Nov 24, 2008
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There is really no difference between what you call a ‘current transformer’ and a ‘voltage transformer’. They are both still transformers. It is more the application that defines the difference. The effects on core flux and field is more due to the way the transformer is utilized by the application.

An ideal transformer would have zero resistance in both the primary and secondary windings and there should be 100% coupling between them.

In a current transformer application the load on the secondary would have zero impedance. There would be no flux in the core as the field created by the primary would be 100% damped by the field created by the secondary.

The same transformer used in a voltage transformer application is just the opposite. There would be zero load on the secondary and there would be no damping effect on the core flux from the secondary.

In real-life situations, the flux in the core will vary with loading, the efficiency of the transformer and the resistance and other impedances in the primary and secondary circuits.

Most real applications for transformers that approach the ideal current or voltage applications are in instrumentation. Other transformer applications, such as signal, impedance matching and power, fall somewhere in between these ideals and secondary load characteristics certainly have an effect on core flux.

6. ### b.shahvir Thread Starter Active Member

Jan 6, 2009
444
0
Dear All, Thanks very much for your replies.

1) My question is theoretical (academic interest) in nature and not practical. The magnetic flux in the core of 'voltage transformer' (VT) remains constant irrespective of the value of load connected to the secondary (whether at no-load or on-load). However, the same is not true for a 'current transformer' (CT), where the flux is neurtralized by the load current. I want to know what are the factors which make a CT behave differently than a VT, magnetically! I have considered winding resistances too.

2) It is known that the magnetic flux in the core is due to the magnetizing current (Im) which is in quadratue with applied primary voltage 'Vp'. At no-load, If the magnetic flux 'visible' in the core is only due to the quadrature component (Im) of primary current (Io), then what happens to the magnetic flux of the in-phase component (Iw) of the primary current? Why is it not 'visible' in the core? This problem can also be considered analogous to magnetic flux conditions present in a practical inductor (containing series winding resistance).

Kind Regards,
Shahvir

Last edited: Jan 19, 2009
7. ### BillO Distinguished Member

Nov 24, 2008
985
136

First, you keep mentioning ideal and then things like resistance and other non-ideal concepts, so its difficult to know what you are arfter.

For question 1)  in and ideal transformer used in a voltage application, as the load on the secondary increases, the current draw on the primary increases which compensates for the damping field coming from the secondary. Hence the core flux remains constant. This requires that not only the transformer be ideal, but the voltage source on the primary must also be ideal (no resistance). In an ideal current application, there is no more current available to the transformer from the source and the secondary has an infinite load (0 resistance) so the core flux is fully damped.

For question 2)  Im really not sure what you are after. An ideal transformer with no load will behave exactly the same as an ideal solenoid. Also, what kind of core are you using? Phasing will be dependant on the core material. In and ideal, air core transformer with no load, the flux is in phase with the current going through the coil.

Last edited: Jan 20, 2009
8. ### b.shahvir Thread Starter Active Member

Jan 6, 2009
444
0
Dear BillO,

Thanks for your reply. I am very much grateful. I am trying to draw an analogy between a VT and a CT, magnetically. You are right! i created a confusion by considering resistances pertaining to an ideal transformer. Let me clarify;

CASE 1 :- What i have understood from your explanation is that, if an ideal transformer has a resistance (or impedance) in series with it, then the magnetic flux in the core is dampened (neutralized) due to the voltage drop across the said impedance, when a load is connected to the secondary winding of the above transformer.

Hence, a CT can be considered analogous to a VT with an impedance in series with it (system load), resulting in the dampening of the core magnetic flux when a 'burden' is connected to its secondary winding. Please correct me if i am wrong!

Now my query is; if the magnetic flux in the core of a loaded transformer reduces due to the dampening effect of series impedance, then the iron losses should also reduce! This means that iron losses would vary with the amount of load connected to the secondary winding of a VT. But according to the E.E. textbooks, the magnetic flux in the core of a VT remains constant, at all loads. Please elaborate.

CASE 2 :- Consider a R-L series inductance coil with a normal iron core. The total coil current 'Ic' will be the vector sum of 'Ir' (resistive component) and 'Im' (magnetizing current). This is analogous to a transformer at no-load wherein the no-load excitation current 'Ie' is the vector sum of 'Iw' (Cu+iron loss current) and 'Im' (magnetizing current).The magnetic flux present (visible) in the iron core is only due to magnetizing current 'Im' (in either case).
Hence, my query is; What happens to the magnetic flux due to the resistive component 'Ir' or 'Iw' (in either case) ? Why is this flux not 'visible' (available) in the iron core ? Where does it disappear ? Please explain.
Thanks once again.

Kind Regards,
Shahvir

Last edited: Jan 23, 2009
9. ### BillO Distinguished Member

Nov 24, 2008
985
136
My last response on this.

Again, ideal means no losses. No core losses, no electrical resistance (copper losses) and no field losses. So if you are talking about ideal, you cannot also talk about resistance and other losses.

In a ideal CT application, the load on the secondary is a short cricuit - infinite conductance - zero resistance.
1) – My last answer still stands. Magnetically, the feild produced by the primary is exactly opposed by the feild produced by the seconday. If you don’t understand it, show it to someone that does understand it and maybe they can explain it to you further, but I can’t think of any other way to state it.
2) – Okay, I have a better understanding of what you are asking. This is obviously NOT an ideal situation. The reason that Ir and Iw do not contribute to the core flux is that they represent losses. They represent the energy that goes into producing heat in the windings and the core. Since they are ‘used’ in producing heat, they cannot also contribute to the magnetic energy in the core.

Last edited: Jan 24, 2009
10. ### b.shahvir Thread Starter Active Member

Jan 6, 2009
444
0
Dear BillO,

Thanks very much.

CASE 1 :- I have understood your explanation. Please do not misunderstand me! What i have interpreted from your explanation is that the deviation of a practical transformer (CT or VT) magnetically from an ideal transformer is due to the presence of resistance (impedance) in series with the windings.

Hence, due to this resistance (impedance), the magnetic flux in the core reduces resulting in the reduction in iron losses in the core. Hence, my only intention was to re-verify with you whether my interpretation of your explanation is correct or not! Hence, i request you to comment on the same.

CASE 2 :- In this case, too, i have understood your explanation. However, a purely resistive conductor curent 'Ir' will also develop a magnetic field/flux around it (even though it is used in producing heat/losses).

Hence, according to me, this magnetic field/flux due to 'Ir' should also be 'visible'/available in the magnetic core surrounding the resistive conductor. Please comment/elaborate on the same.
Sorry for trouble!

Kind Regards,
Shahvir

Last edited: Jan 26, 2009
11. ### studiot AAC Fanatic!

Nov 9, 2007
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The minute you start to move from an ideal model of a transformer to the real world you must not only take into account winding resitance, which is usually small anyway, but also model the true magnetic circuit in the core.

A ferrous core is made of conductive material, with a significantly higher resistivity than the copper windings.

This conductive core material is subject to a varying magnetic field, therfore it currents develop in it.

These are known as 'eddy currents'.

Eddy currents passing through a relatively high resistance material produce a relatively high level of resistive heating. This fact is the bane of core design and responsible for the air gap in the E an I construction. It is also the reason for laminated construction of cores.

Most of the losses are due to eddy currents. Powerline transmission transformers often contain water cooling coils in their cores to help combat this.

The grain structure (metallurgical) of the core material also affects these currents.

12. ### b.shahvir Thread Starter Active Member

Jan 6, 2009
444
0
Dear studiot,

What i have interpreted from you reply is that the 'eddy currents' developed in a non-ideal magnetic core are responsible for neutralizing/dampening the magnetic flux produced in the magnetic core by the resistive component current 'Ir' of the total winding current. Hence, as a result of this neutralizing/dampening effect of the eddy currents, the magnetic field/flux due to resistive current 'Ir' is not 'visible' or present in the core. Could you please elaborate further on this concept for clarity ?

Kind Regards,
Shahvir

Last edited: Jan 26, 2009
13. ### b.shahvir Thread Starter Active Member

Jan 6, 2009
444
0
Dear All,

Let me consider 'non-ideal' conditions;
Consider a purely resistive conductor (or coil) carrying a purely resistive current 'Ir'. Obviously, 'Ir' would be in-phase with applied voltage 'V'. Consider it to be surrounded by an air-core (no iron losses). The purely resistive current 'Ir' will develop a magnetic field/flux 'ψr' around it which will magnetize the air-core surrounding it.

Hence, can i consider the purely resistive current 'Ir' to be 'magnetizing current' for all practical purposes, since it magnetizes the air-core surrounding it with magnetic flux 'ψr' ?

In this case, the magnetic flux 'ψr' would be in phase with 'Ir' which is real and not in quadrature with 'V' (reactive) in nature, as is the case normally with any magnetizing current! I humbly request you all to comment on the same.

Thanks & Regards,
Shahvir

Last edited: Jan 27, 2009
14. ### studiot AAC Fanatic!

Nov 9, 2007
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515
I have been reviewing the posts in this thread.

It is true to say that many textbooks are either very advanced (and complicated) or pass over details of transformers very quickly.

This is because transformers are actually very advanced and complicated devices. It is in fact impossible to draw a phasor diagram for the currents in an iron cored transformer because of hysteresis. This phenomenon makes the current waveform non sinusoidal. You can only draw phasor diagrams for quantities that have a linear relationship eg harmonically related sinusoids.

I assume you are studying and it would be very helpful to know what subject and at what level (eg 1st year university electrical engineering or whatever ). this will help pitch answers at the correct level.

I also assume that given a few pointers you can look up a few things for yourself.

So let us clarify some basics.

An ideal transformer has zero resistance in its windings.

The purpose of a 'core' is to concentrate the magnetic flux where it is wanted i.e. cutting the secondary coils. An air 'core' does not do this and cannot strictly be called a core.

Transformers made with only air as the linking mediem are very inefficient especially at low frequency. Consequently they are used at higher frequencies where many other considerations come into effect. One such is the capacitance between the individual coils on a winding.

Transformers with suitable material (usually ferrous) in a core, concentrate the magnetic flux into this core. However if you have studied the hysteresis curve you will know that the induced current is a non linear function of applied voltage, and further is a different function for increasing and decreasing flux.

'Air cored' and ferrous cored transformers are therefore considered separately in advanced texts, but you seem to mix them indiscriminately.

So please confirm that you have a hold on these basic ideas so that we can proceed appropriately.

Last edited: Jan 28, 2009
15. ### b.shahvir Thread Starter Active Member

Jan 6, 2009
444
0
Dear Studiot,

Thanks for your reply. Let us forget transformers for a while. My confusion is related to the earlier mentioned post. I'll repeat again.

Normally in AC circuits, the magnetic flux surrounding a conductor is in-phase with the quadrature (wattless/reactive) component 'Im' of the total coil current, say 'Ic'. 'Im' is also known as magnetizing current; meaning its function is just to magnetize the core...agreed!

Now coming to the earlier post by BillO, wherein he has stated that resistive current 'Ir' cannot contribute to producing magnetic flux in the core since it is 'used' in producing heat (losses) and hence, the magnetic flux due to 'Ir' is not present/visible in the core. So, technically, 'Ir' cannot be termed as a 'magnetizing current'...accepted theoretically!

Now coming to my query; A resistive filament (eg. the one used in incandescent lamps) can be considered as a purely resistive conductor, for all practical purposes. The current 'Ir' produced by it will be in-phase with the applied voltage 'V'. Also, this purely resistive current 'Ir' develops a magnetic flux around it (according to the basic laws of electricity).
So, my simple question is; Why can't 'Ir' be considered as a magnetizing current since it has the abillity to magnetize its surrounding core/medium ? Why is the magnetic flux due to 'Ir' not considered to be present/visible in the core/medium surrounding the purely resistive filament ? Where does this magnetic flux due to 'Ir' disappear ? It has to be neutralized by something if it is not considered to be present/visible in the surrounding core/medium! Please comment/elaborate on the same.

Kind Regards,
Shahvir

Last edited: Jan 28, 2009
16. ### studiot AAC Fanatic!

Nov 9, 2007
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This is not so.

In particular do you know what hysteresis is?

17. ### BillO Distinguished Member

Nov 24, 2008
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136
Shahvir,

In a real transformer primary, the energy transferred to the field and available to be coupled to the secondary is somewhat less than the energy put into the transformer. Im represents that part of the energy that is used to build the magnetic field in the core. If Ic were the total current through the primary then Ic=Im+Ir+Iw. Ir and Iw represent the energy lost. Ir is lost to wire resistance, Iw to core losses, which are, as studiot pointed out, predominantly eddy currents. Both go into heating up the transformer, not into the core field and not available to be coupled to the secondary. The field produced in the core is proportional to Im, not to Ic.

In your light bulb analogy, the filament of the bulb is certainly not an ideal resistive element. It does produce a magnetic field and that field would be due to energy not turned into photon radiation (heat and light) and other minor energy sinks.

It looks like this needs to be explained to you from first principles, and studiot has offered to do that. You need to read his questions, think about them and respond accordingly.

18. ### b.shahvir Thread Starter Active Member

Jan 6, 2009
444
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Dear BillO,

Your explanation indicates that the magnetic field due to 'Ir' is not available in the surrounding medium since it is neutralized by the eddy currents existing in a ferrous core around it. But, what if the filament is surrounded by an air-core. In that case, eddy currents and hysteresis losses would be absent!
So, what phenomenon dampens/neutralizes the magnetic flux due to 'Ir' in an air-core medium ? (Let us forget about transformers when relating to this particular query).

Also, could you elaborate when you say "Ir is lost to wire resistance or used up in producing heat". How can energy be lost or destroyed in producing heat when 'Ir' component of 'Ic' still flows into the filament and develops a magnetic field around it?

Dear Studiot,

My understanding of hysteresis is the delay in the rate of change of the magnetic polarities established in a ferrous core, when subjected to AC magnetic fields. I have poor knowledge about hysteresis losses in general, and would be grateful if you could throw some light on this concept.

P.S. I know my queries sound a bit ridiculous technically (as a result of which you guys would obviously end up doubting my veracity), but it is to gain a refreshing insight and in-depth knowledge into the particular E.E. topics of interest which has compelled me to put forth my doubts/queries in front of experienced experts like you. I have full faith that you guys will not disappoint me!

Thanks & Regards,
Shahvir

Last edited: Jan 29, 2009
19. ### studiot AAC Fanatic!

Nov 9, 2007
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A pity you didn't indicate the level you are working at, as I will have to start with the most basic.

At the rate of one post/reply per day this will take a long time.

So let us start with a straight length of resistance wire, such as used in lamp filaments, and place it in a vacuum.

Pass a current, I, through it and there will be heating due to the resistance of the wire.
Because there is resistance you will need an energy source to kepp this current flowing. Energy will be lost by the energy source and gained, as heat, by the rest of the universe.

There will also be a magnetic field surrounding the wire in accordance with Faraday's laws.

Do you understand the direction and shape of this field?

This field will interact with the wire, again in accordance with Faraday's laws.

Can you predict the interaction?

If I now bend a section of the wire at right angles, can you now predict the interaction?

If I now loop some of the wire into a coil can you predict the interaction?

If I now make several loops what is the interaction?

Is the resistance of the coiled wire any different from the straight?

So will there be any change to the heat generated?

Finally, for now, if I point out that the permeability of

vacuum is 1.266 E-6 H/m
air is 1.256 E-6 H/m
Soft iron 5 E-2 H/m

What does that say about field strengths in air, vacuum and iron?

Am I justified in saying that air and vacuum are essentially the same?

20. ### BillO Distinguished Member

Nov 24, 2008
985
136
Arrrgh!

I have to bow out of this, sorry.

Studiot, you are a better man than I, sir!