# Mutual Induction: Exploring Questions & Answers

• Urmi Roy
Please please answer!In summary, the mutual inductance between two coils is equal regardless of the circuit. When current flows through the secondary, there is a flux that opposes the current induced by the primary, but this happens due to the counter-mmf generated by the constant voltage source.f

#### Urmi Roy

Hi,
I've been doing electromagnetism...and sure enough,I got to the part on mutual induction...and a thousand questions popped into my mind!

Firstly,suppose we have two coils wound side by side on a soft iron core and a ac power source is attached to one of them (primary).

Considering the ideal case,the flux linked to each is equal.

1.Why is the mutual inductance (M)of each coil equal?

2. Considering the mutual inductances of both coils in a circuit is equal...is this mutual inductance specific to a particular circuit?

3.When the current flows thorugh the secondary,there should be additional flux linked to the core due to this current...but it does not actually happen...Is this because as soon as the secondary tries to generate some additional flux,the primary generates an extra current,such that the secondary's flux is opposed?But,this would mean that in the primary,current flows not only due to the applied voltage,but also due to the induction business...I don't think that sounds right...

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The primary is connected to a constant voltage source. The flux is determined by the voltage value, frequency, area, & turns. This flux links the secondary almost completely, and the secondary voltage is determined by the flux, area, frequency, & turns. But the area, flux, & frequency for the secondary are almost identical to that of the primary. Only the turns differ, so that Vp/Np = Vs/Ns.

When a load is connected at the secondary, the current, Is, has an mmf and flux which tends to oppose that from the primary voltage source. This is a counter-mmf. But the constant voltage source feeding the primary outputs a current to oppose the counter-mmf, resulting in a counter-counter-mmf. In short, NpIp = NsIs, except for the primary exciting current.

Does this help?

Claude

If I understand your second question correctly:

The current induced in the second inductor ought to change the flux through the second inductor. Since flux is defined as the total magnetic field through a loop, it makes sense to consider the field caused by the loop itself in the flux term. Usually there is no difference when flux is defined as a function of time, or if the internal magnetic field is strictly defined as a function of time. Things get a little bit more complicated however when the current of one loop is given as a function of time, since you now need to worry about individual contributions to the magnetic field at any instant. Usually however you end up with a 1st order linear inhomogeneous differential equation in terms of the current through the coil being effected and that current's time derivative in terms of the time derivative of the changing current.

My experience with mutual inductance is that it depends only on geometry of the sources.
Here's a conceptual background on the derivation:

Flux can be expressed as the surface integral of magnetic field. Using stokes' theorem, this can also be expressed as the line integral of the vector potential around the surface.

The line integral of the vector potential is around the 2nd loop

the vector potential is evaluated from the 1st loop, and is dependent on the radial distance between each element of the two loops (since we are only evaluating the vector potential along the 2nd loop), on the current of loop 1, and on the length of loop 1.

Combining the two integrals, you find that the current through loop 1 is constant and so can be brought outside of both integrals. The remaining integral is purely geometric, and is the same regardless of which inductance you consider (Think of this like two clocks measuring time in different intervals "dt": one clock moves one interval "dt" after one full rotation of the other clock [sort of like a second-minute relationship]. You contain the same "group" of combination of position by switching the order of rotation; i.e. rotate the other clock one full rotation for every 'dt' in the one that was originally rotating)

Hi,

1.Why is the mutual inductance (M)of each coil equal?

2. Considering the mutual inductances of both coils in a circuit is equal...is this mutual inductance specific to a particular circuit?

3.When the current flows thorugh the secondary,there should be additional flux linked to the core due to this current...but it does not actually happen...Is this because as soon as the secondary tries to generate some additional flux,the primary generates an extra current,such that the secondary's flux is opposed?But,this would mean that in the primary,current flows not only due to the applied voltage,but also due to the induction business...I don't think that sounds right...

1. you mean "equal to each other". inductance is determined only by the geometry of the conductors. Ignore any Iron or other magnetic materials. It just complicates things beyond your ability to understand. Say we have two equal single turn coils. The way one calculates mutual inductance between them is to use the Neumann equation. (look it up)
This equation is a double integral around the two curves. The reason that M12 = M21 is that one can reverse the integrations between the first and second coils without changing the calculation. Thus the mutual inductance in either direction FOR A GIVEN GEOMETRY are the same.

2. It is specific to a given GEOMETRY as I said above. Change the shape or form (geometry) of the wires and the mutual inductance changes.

3. What you are talking about here is a transformer. And you have pretty much figured it out. But note that if you have a transformer and you suddenly load the formerly open secondary, that power being drawn from the secondary will need to be drawn from the primary source. So as you guess, the primary due to the induction is made to draw more current. So yes, it is right that the primary current is flowing not only due to the applied voltage but ALSO due to the "induction business". OK? That is correct. But don't forget that the "induction business" depends not upon currents and voltages but on the RATE of CHANGE of those things. Hence direct current does not flow through a transformer.

caraham:
The primary is connected to a constant voltage source.

Constant as in?...it is a sinusoidal wave of voltage that we apply,isn't it?

But the constant voltage source feeding the primary outputs a current to oppose the counter-mmf, resulting in a counter-counter-mmf. Claude

This is one point I would like to dwell upon...if I take for granted,that the primary resists the flux generated by the secondary by inducing an additional current in itself(on top of the current in it due to applied voltage)...
this current is a result of the induced emf...
but you say here,that "constant voltage source (i.e the applied voltage source)feeding the primary outputs a current "...which would mean that the additional current is due to the applied voltage itself...did I reason it out wrong here?

In short, NpIp = NsIs, except for the primary exciting current.

This derivation is done very simply ,even on wikipedia...it comes basically from the equating the voltage drop across the primary and secondary...

but I feel that this derivation doesn't tell the exact story because:

1. The fact that the faraday induction equation is used for the primary indicates that they are taking into account only the extra emf induced in the primary in order to oppose the flux due to the secondary...but then we need to take into account the applied voltage on the primary also...why haven't they done that?

2. In order to completely describe the situation,we would need to combine the effect of the primary,which at first induces some flux into the secondary,and the effect of the additional flux induced back by the secondary.

IHere's a conceptual background on the derivation:...

So basically,what you're doing,is to find out the magnetic field on the outer loop(second loop)...and it is found to depend upon the radial distance between them,length of loop1 and the current in loop one...then,since you're referring to a combination of
two integrals,I presume you are also finding out the magnetic field on loop 1 due to loop 2(which has it's own current)...then the magnetic field on loop 1 would also depend upon the radial distance between the coils,length of loop2 and the current in loop 2...but then how do you combine these two to get the result?
How does the clock analogy come in here?

Moreover,since both the coils carry current,wouldn't we follow some kind of superposition principle to combine their effects?

What you are talking about here is a transformer. And you have pretty much figured it out. ...does not flow through a transformer.

Thanks especially for confirming my idea...and ofcourse,for your further clarifications.

I tried looking up the Neumann equation...but I could ony find the Von Neumann equation...and it has something to do with Schrödinger equation...from the way it looks,I couldn't really figure out its sililarity with the mutual inductance equation...

caraham:

Constant as in?...it is a sinusoidal wave of voltage that we apply,isn't it?

This is one point I would like to dwell upon...if I take for granted,that the primary resists the flux generated by the secondary by inducing an additional current in itself(on top of the current in it due to applied voltage)...
this current is a result of the induced emf...
but you say here,that "constant voltage source (i.e the applied voltage source)feeding the primary outputs a current "...which would mean that the additional current is due to the applied voltage itself...did I reason it out wrong here?

This derivation is done very simply ,even on wikipedia...it comes basically from the equating the voltage drop across the primary and secondary...

but I feel that this derivation doesn't tell the exact story because:

1. The fact that the faraday induction equation is used for the primary indicates that they are taking into account only the extra emf induced in the primary in order to oppose the flux due to the secondary...but then we need to take into account the applied voltage on the primary also...why haven't they done that?

2. In order to completely describe the situation,we would need to combine the effect of the primary,which at first induces some flux into the secondary,and the effect of the additional flux induced back by the secondary.

Constant as in amplitude & frequency. Of course a sine curve (or cosine) is not truly "constant" at every point in time, but that is understood. "Constant" in the ac domain is understood as constant amplitude and frequency.

Nobody can answer every question involving the hows and whys of induction. But in response to your next question, I don't think of it as one current cancelling another in the primary. When first energized, the primary is connected across the CVS (constant voltage source). Because the primary has substantial inductance (air core and especially iron core), the current in the primary generates a magnetic flux which opposes the CVS flux per Lenz' law. Instead of thinking that the induced current cancels the source current, I view these as I just described. Any current will immediately create a counter mmf (magnetomotive force) which opposes the source mmf. Thus with secondary open, the primary carries only exciting current, which is relatively small, typically 5% or so of the full rated load current, or even less. The magnetizing component of exciting current is determined by the B-H loop of the core material (for air it is a straight line with slope of mu0) and the CVS amplitude & frequency.

When loaded at the secondary, a secondary current is established which generates an mmf which is oriented in opposite polarity to the original. This counter-mmf tends to reduce the core flux. A reduced flux results in a **reduced counter-emf** at the primary, which results in increased primary current. This primary current generates an mmf which opposes the counter-mmf of the secondary load current, hence it is **counter-counter-mmf**. This works this way because the source powering the primary is a CVS. As soon as the induced emf drops, the CVS outputs more current to the primary, restoring the mmf/emf which momentarily decreased. The flux is thus increased to very near the unloaded value, a bit less due to I*R drops in the winding resistance.

Does this make sense? BR.

Claude

cabraham:
Yes,this makes sense...only a small point in regard to the last line..the induced emf in the secondary..and hence the counter-counter emf is in phase with the applied voltage isn't it?

Also,since I think this part of the explanation is clear(as to what exactly happens)...please could you throw some light on the derivation of the inductor formula that I mentioned in the last part of post 6?

Yes, the counter-counter-emf is in phase with the input voltage The counter-counter mmf is approximately 90 degrees lagging wrt the input voltage source. It's hard to say which current/voltage is due to which flux/emf/mmf, etc.

The core flux with the secondary unloaded is due to both the magnetizing current and terminal voltage (from the CVS). Both are responsible for establishing core flux. The magnitude and harmonic content of the magnetizing current is influenced by the B-H characteristics of the core material. Neither Imag nor Vp are responsible for producing the other. Under ac conditions, the primary voltage & primary current (magnetizing) cannot exist independently. They mutually co-exist.

When a load is added, the additional current in the secondary results in a flux, an mmf, and an emf. All 3 are mutual, which one is more basic is a moot question. According to Lenz, the new flux opposes the original, and the flux starts to decrease. The unloaded condition resulted in the core flux and a counter-emf. The terminal voltage encountered a small leakage reactance, a small winging resistance, and a large magnetizing reactance. Or, the current is small, due to the counter-emf being almost equal to the terminal voltage. The difference appears across the leakage reactance & winding resistance.

The load current has a counter mmf which reduces the core flux. With a reduced core flux comes a reduced primary counter-emf. A larger voltage appears across the leakage reactance & the winding resistance. So the primary current increases. But this increased primary current has with it a emf, mmf, and flux. These all oppose the counter-flux/emf/mmf, making them *counter-counter-flux/emf/mmf*. So the flux increases which increases primary counter emf which eventually reaches equilibrium. The core flux is almost what it was unloaded, but the primary current has increased in proportion to the secondary load current. Conservation of power/energy holds fast.

What inductor formula?

Claude

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cabraham:
Thanks for your detailed answer, cabraham...even though it'll take a couple more reads to completely understand and realize it,I think I'm now beginning to see how all the factors come together!

By the formula,I simply meant NpIp = NsIs...a simple formula,but as I explained in post #6,I have some problems in reagrd to it's derivation(as done in wikipedia,and most textbooks,).

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Also, Couchyam ,could you please clarify the points that I mentioned about you derivation in post #7?

Well, really,it's been ages! Please could someone reply in place of Couchyam and cabraham?

I've been very swamped with work & homework, as I work full time & study for the Ph.D. part time. I have little free time.

The NpIp = NsIs is derived from Ampere's Law (AL). AL states that NI = closed loop integral of H*dl, where "*" is the dot product between vector H & incremental magnetic path vector dl. Thus the amp-turns NI needed to establish H, a magnetic field intensity, is given by AL. B, the magnetic flux density, is just mu*H.

When unloaded, a xfmr draws exciting current, which consists of magnetizing current, and loss current to cover hysteresis & eddy current losses. NpIp is determined by AL, where l is the flux path length.

A load is added. A secondary current is drawn. AL still holds fast. For a given core flux, we need the same value of net amp-turns. But the secondary current in the load, has an amp-turn value of NsIs, which tends to counter the original amp-turns. This results in a counter-emf/mmf so that the flux is reduced, and the CVS at the primary encounters less counter-emf. The result is an increase in primary current, or NpIp goes up until the difference between NpIp & NsIs is equal to closed loop integral H*dl.

Does this help?

Claude

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Thanks cabraham...I didn't realize that you were so tied up...infact I was a little frightened to see that you had replied after such a long time...I thought you might scold me for being so persistant!

If I understand your second question correctly:

The vector potential is evaluated from the 1st loop, and is dependent on the radial distance between each element of the two loops (since we are only evaluating the vector potential along the 2nd loop), on the current of loop 1, and on the length of loop 1.

Combining the two integrals, you find that the current through loop 1 is constant and so can be brought outside of both integrals. The remaining integral is purely geometric, and is the same regardless of which inductance you consider

I've been thinking this over...I found the expression for the vector field potential due to the magnetic field inside a conducting loop...but now it reamins for me to inderstand which 2 integrals Couchyam was taliking about...and when we equate these two integrals,if only the current due to loop 1 can be taken out of the integral, what about the current due to loop 2?

I read through cabraham's post,and I understood it!

The only thing perhaps a little fuzzy,still is as to why the NpIp doesn't exactly equal NsIs...but I think I've more or less figured that out too...it's because of the resistive loss in the primary coil...

Now,if anyone could take a look Couchyam's derivation and explain the points I mentioned in the post just above...

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It may be helpful to see the question addressed from a circuit theory point of view.

Here are a couple of pages from a circuit theory text by Skilling, showing a proof of the equality of the mutual inductances of a pair of coupled coils. At the top of Page2 is a description of the case of a somewhat idealized transformer. Some equations from a previous page (Page1) are referenced.

The footnote at the bottom of Page2 is for a general case.

#### Attachments

Thankyou very much, The Electrician, for responding.

So, what I gather from the pages is,for an ideal transformer,the self inductances are in the square of the ratio of the number of turns,whereas the mutual inductances L12 and L21 are equal.

However,the only proof to the latter seems to be given by considering that the total flux produced by either coil has to be equal to that produced by the other...based on this,from the expression of mutual inductance,we reason out,quite easily,that L12 =L21 .
The only thing is,during operation of a transformer,current is supplied to only one coil at a time...and that one operating coil produces some amount of flux,according to the number of turns and current supplied...but is it really necessary that when the power is now supplied to the other,it should produce the same flux?

Also,it states that the mutual inductances of the coils should be equal,even if the transformer isn't ideal...how can that be? After all,in case of a non ideal transformer,each of the coils has its own coupling factor k1 and k2,which are of different values (depending on the geomtery and shape of the coils)...so now,the flux produced by one and linked to the other are not the same..

Thankyou very much, The Electrician, for responding.

So, what I gather from the pages is,for an ideal transformer,the self inductances are in the square of the ratio of the number of turns,whereas the mutual inductances L12 and L21 are equal.

However,the only proof to the latter seems to be given by considering that the total flux produced by either coil has to be equal to that produced by the other...based on this,from the expression of mutual inductance,we reason out,quite easily,that L12 =L21 .
The only thing is,during operation of a transformer,current is supplied to only one coil at a time...and that one operating coil produces some amount of flux,according to the number of turns and current supplied...but is it really necessary that when the power is now supplied to the other,it should produce the same flux?

Also,it states that the mutual inductances of the coils should be equal,even if the transformer isn't ideal...how can that be? After all,in case of a non ideal transformer,each of the coils has its own coupling factor k1 and k2,which are of different values (depending on the geomtery and shape of the coils)...so now,the flux produced by one and linked to the other are not the same..

Yes, of course k1 and k2 are not usually equal, but mutual inductance is equal despite the inequality. A good illustration is a coaxial cable. All of the flux due to current in the shield surrounds the shield, and surrounds the center conductor as well. So the coupling from shield to center, ksc, is unity.

But the flux due to center current completely surrounds the center conductor, but incompletely surrounds the shield. Thus the coupling coefficient from center to shield, kcs, is less than unity. So ksc = 1, but kcs < 1.

The mutual inductance, Lm, however is defined as the flux in 1 circuit due to current in the other. The flux enclosing the shield due to center current is that portion which surrounds the shield. The flux enclosing the center due to shield current is still that which surrounds the shield.

Although the k's are unequal, you can verify that the mutual inductance, Lm, is equal to the self inductance of the shield, Ls. If Lc is the center conductor self inductance, Ls is the shield self inductance:

Lm = Ls.

Remember that Lc > Ls. The center conductor has a greater self inductance since its "height", magnetically speaking, is shorter. Thus Lc = phi_c/Ic > Ls = phi_s/Is. The center conductor produces more flux than the shield for the same current value. Ic = Is, since each conductor carries current for the same network.

The larger flux of the center partially links the shield. The smaller flux of the shield completely links the center to the same amount. Hence Lm = Ls, even though kcs does not equal ksc. Does this make sense, or did I make matters worse?

Claude

Yes, of course k1 and k2 are not usually equal, but mutual inductance is equal despite the inequality...
Claude

Well,I've pretty much understood the reasoning...but here,you say Lm = Ls ...firstly,I suppose this has to be true for only the case above,as this is not found to be true in general...secondly
the fact that the mutual inductances are equal isn't exactly clear from the above post(the ethos seems to be in proving Lm =Ls)...but I think it might be because
Lm of core=Nc(k*phi_s)/Is Lm' of core =Ns(k'*phi_c)/Ic...and you said k' was less than unity and Ic=Is...since Nc>Ns (these are the magnetic heights,which is analogous to turns in coils,I suppose)...then (k*phi_s)<(k'*phi_c)...but k'<k...so phi_s>phi_c...but that's the opposite!What's wrong with my reasoning?

Oh...and something more...,I was doing a sum just yesterday from my book,in which it said that the k's for two coils placed in proximity are equal...is this an assumption made for an ideal case?

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Coming to think of it...at the beginning of this thread,we proved that the net flux in the transformer core is that which is caused by the input voltage,in the primary coil...this was the case when there was only one source of energy,the input voltage attached to the primary...but in your case of the coaxial cable,the core and shield are independantly given current by the source...this is analogous to the case where both the transformer coils are attached to a voltage source...so how would we calculate the net flux in the transformer core(or equivalently,how would we calculate the net flux shared between the core and the shield)?

I did some googling,and I found a webpage that describes how indeed we would calculate the net flux in a case as I described above...by using the superposition principle...but cabraham doesn't seem to have brought in the superposition principle in the case of his coaxial cable...
Could someone please get me out of this infinite tangle and get this over and done with?

I did some googling,and I found a webpage that describes how indeed we would calculate the net flux in a case as I described above...by using the superposition principle...but cabraham doesn't seem to have brought in the superposition principle in the case of his coaxial cable...
Could someone please get me out of this infinite tangle and get this over and done with?

The way I understand coaxial cable would be the shield acts as a short-ckted secondary (cylindrical conductor). Hence, giving a separate voltage source (to shield) does not arise! The net flux in the core would then be only due to the primary conductor. I would not be able to help u with equations, but it would be helpful if u make it a bit clear as to what are u exactly looking for.

Regards,
Shahvir

Thankyou,B.Shahvir!

So what I understand is that my understanding of the coaxial cable as the core having a separate voltage source from the shield is indeed wrong,the source of power for both of them is the same i.e the alternating applied voltage source...that's sorted.

However,I should be more clear as to what exactly I'm looking for.
The thought that is presently nagging me is as what would happen if we indeed connected a separate voltage source to the secondary!

In that case,we would have to calculate the net flux shared between the two coils...but then...either of the two coils has a certain amount of its own generated flux(generated in response to the applied voltage)which it shares with the other coil...then how would we calculate the net flux in the transformer core?
(we can't just add them up.)

This is where the 'superposition principle' seems to comes in...but I don't really understand how we should go about using it.

Here's a sum that might illustrate what exactly I want ...

"2 coils A and B lie in parallel planes.Coil A has 150 turns and coil B has 120 turns.
55% of the flux produced by coil A links coil B. A current of 6A in coil A produces 0.05 mwb,while the same current in coil B produces 0.08mwb. Calculate the mutual inductance and coupling coefficient."

My doubts:
1. Firstly,as cabraham said in his last post,the coupling factors 'k1' and 'k2' for the two coils should not be the same..so why do they say " calculate the coupling coefficient"?

2. It says "55% of the flux produced by coil A links coil B" ...so would the coupling coefficient for A be 0.55?

3. As it says in the sum,both the coils are energised...so we need to calculate the net flux shared i.e the mutual flux (which would require a superposition principle) but in my book,the say:
"mutual flux=55% of 0.05mwb(flux produced by A)"...but this doesn't take into account the flux produced by B!

(One important point to note: they want us to calculate the mutual inductance ...which means that they know that the net flux in the transformer core is not due to only A or B...but the question remains...how do we calculate the net flux in the transformer core when both the coils are energised,and taking into account their individual coupling factors)

Thankyou,B.Shahvir!

So what I understand is that my understanding of the coaxial cable as the core having a separate voltage source from the shield is indeed wrong,the source of power for both of them is the same i.e the alternating applied voltage source...that's sorted.

However,I should be more clear as to what exactly I'm looking for.
The thought that is presently nagging me is as what would happen if we indeed connected a separate voltage source to the secondary!

In that case,we would have to calculate the net flux shared between the two coils...but then...either of the two coils has a certain amount of its own generated flux(generated in response to the applied voltage)which it shares with the other coil...then how would we calculate the net flux in the transformer core?
(we can't just add them up.)

This is where the 'superposition principle' seems to comes in...but I don't really understand how we should go about using it.

The question is...whether the 2 voltage sources are out of phase. If the 2 voltage sources are in phase, then the 2 fluxes will cancel out in opposite direction (in case of core type Xmer) and the net flux within the core will add up to zero. However, in case of shell type Xmer, if both the windings are on the same limb, the both fluxes will add up in same direction (2 equivalent coils in parallel with addition of flux). Also, observe equivalent ckt. of Xmer, especially magnetizing reactance branch in parallel with the voltage sources responsible for flux creation.

An interesting case is that of induction motor during regenerative braking when rotor acts as a separate voltage source secondary. The primary (stator) voltage is in quadrature with secondary (rotor generated) voltage. Hence the fluxes are in quadrature with each other and do not directly add up by superposition principle. The rotor flux is immediately neutralized by reverse stator flux (primary current fed back to supply via stator i.e. reverse Xmer action). However, net flux in air gap is in same direction as before and is due to stator magnetizing current only.

The question is...whether the 2 voltage sources are out of phase. If the 2 voltage sources are in phase, then the 2 fluxes will cancel out in opposite direction (in case of core type Xmer) and the net flux within the core will add up to zero.

So what I understand is,suppose we have the two coils of a transformer,we apply separate voltages to the coils,then at any instant,the net flux in the core would be what we get by adding the flux by each of the coils as per superposition principle.

In that case,if the two applied voltages are not exactly in phase,or exactly out of phase,we would have to perform this kind of addition for each instant separately,otherwise we couldn't know.

The thing is,NpIp=NsIs...here,Ip and Is are obviously not the same,as we have two different voltage sources attached...so in each coil,we would have to consider the current due to
1. the voltage source attached to it.
2.the mutually induced current.
3. the self induced current,

only then we could make NpIp=NsIs...hence,the transformer core would contain flux due to all of these factors...and so my simple reasoning of superposition principle fails!

So what I understand is,suppose we have the two coils of a transformer,we apply separate voltages to the coils,then at any instant,the net flux in the core would be what we get by adding the flux by each of the coils as per superposition principle.

In that case,if the two applied voltages are not exactly in phase,or exactly out of phase,we would have to perform this kind of addition for each instant separately,otherwise we couldn't know.

The thing is,NpIp=NsIs...here,Ip and Is are obviously not the same,as we have two different voltage sources attached...so in each coil,we would have to consider the current due to
1. the voltage source attached to it.
2.the mutually induced current.
3. the self induced current,

only then we could make NpIp=NsIs...hence,the transformer core would contain flux due to all of these factors...and so my simple reasoning of superposition principle fails!

Exactly!... for in-phase quantities superposition (arithmatic addition) is applicable, whilst for out of phase quantities vector addition is applicable. Hence AC quantities have more of vector mathematics than simple arithmatics as most of the AC quantities are out of phase with each other due to presence of reactive components in the AC system. This is applicable for both 3 phase & 1 phase AC system. Also, the type of magnetic circuit available affects magnetic flux distribution & measurements.

Hope this helps.

Regards,
Shahvir

Regarding the latest questions, my response is verified in many textbooks, which anyone can obtain. The "coupling coefficient" can be called "k". If k12 represents the coupling *from 1 to 2*, and k21 represents coupling from 2 to 1, then the overall coupling is given by:

k = sqrt (k12*k21).

As far as the coax cable goes, I don't think that the shield is a shorted turn. If we sketch an iron core xfmr, add a shorted secondary turn, and examine the direction of the core flux, we will see that the shorted turn is oriented normal to the flux. But the coax shield is along the flux of the center conductor, not normal.

The mutual inductance Lm, does equal the shield self inductance Ls, and Henry Ott of Bell Labs derives this relation in his highly acclaimed book "Noise Reduction Techniques In Electronic Systems". This is true for a general coaxial cable. When we say "in general" I presume that coax cable is under discussion. If "in general" refers to other configurations besides coax, then different relations are encountered.

In general, if 2 coils mutually interact, then k = sqrt (k12*k21), and Lm = k*sqrt(L1*L2). This can be derived but it is involved. An advanced fields text might have the derivation with illustrations. With grad school I have no time to derive it. Maybe in June when things slow down I might have time. Best regards.

Claude

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Regarding the latest questions, my response is verified in many textbooks, which anyone can obtain. The "coupling coefficient" can be called "k". If k12 represents the coupling *from 1 to 2*, and k21 represents coupling from 2 to 1, then the overall coupling is given by:

k = sqrt (k12*k21).

Actually it's me who seems to be getting mixed up.I had been thinking all this time that k depends exclusively on the individual coils themselves (no. of turns,length,etc.)...so we would have different k values for each...
also,as you said in an earlier post, the k values don't have to be the same for the coils,I thought that confirmed my idea,
However,you say,(like in my book )that k = sqrt (k12*k21).,which means there is only one value of k for a particular transformer...I can't really explain to myself why there is a unique value of k!

(As 'The Electrician' knows,my book often contains mistakes,so I was wondering if the fact that k has a unique value as stated in my book is one of those mistakes,however,you've just confirmed that it's not.)

The mutual inductance Lm, does equal the shield self inductance Ls

I got that...then I tried to prove how the mutual inductances could be equal for both the core and the shield(and without success ofcourse)...
"Lm of core=Nc(k*phi_s)/Is Lm' of core =Ns(k'*phi_c)/Ic...and you said k' was less than unity and Ic=Is...since Nc>Ns (these are the magnetic heights,which is analogous to turns in coils,I suppose)...then (k*phi_s)<(k'*phi_c)...but k'<k...so phi_s>phi_c...but that's the opposite!"

As far as the coax cable goes, I don't think that the shield is a shorted turn. If we sketch an iron core xfmr, add a shorted secondary turn, and examine the direction of the core flux, we will see that the shorted turn is oriented normal to the flux. But the coax shield is along the flux of the center conductor, not normal.

The mutual inductance Lm, does equal the shield self inductance Ls, and Henry Ott of Bell Labs derives this relation in his highly acclaimed book "Noise Reduction Techniques In Electronic Systems". This is true for a general coaxial cable. When we say "in general" I presume that coax cable is under discussion. If "in general" refers to other configurations besides coax, then different relations are encountered.

In general, if 2 coils mutually interact, then k = sqrt (k12*k21), and Lm = k*sqrt(L1*L2). This can be derived but it is involved. An advanced fields text might have the derivation with illustrations.
Claude

I agree. Also, mutual induction are compex quantities & depend on plane of co-incidence resulting in lesser net magnitude of flux linkage.

Could someone clear the confusion about the k values as I stated in post 32#, please.

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