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Iron core energy change and transformers vs. ohms law

  1. May 8, 2005 #1
    "Iron core energy change" and "transformers vs. ohms law"

    Ooh yes, I do have two question. So enjoy the meal all physics lovers! :smile:

    1. If the field is increased putting an iron core inside the circuit, then what about energy conservation? (I don't know much about H-fields.)

    2. Alright, now I do understand (really!) why a transformer will change both current and voltage and why power is conserved. But what about Ohms law then? According to Ohms law I will depend on U and vice versa. This applies to A.C. as well, right. So that does not make sense to me.

    Happy days to everyone!

  2. jcsd
  3. May 8, 2005 #2

    Andrew Mason

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    1. There is work required to insert an iron core inside a coil that is carrying current. That work results in an increase in the magnetic field self-energy. Energy is conserved.

    2. The inductive reactance limits current. If you increase current draw on the secondary, the inductive reactance in the primary decreases and the current increases. Ohm's law applies but you have to use the formula for impedance:

    [tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]

  4. May 9, 2005 #3
    But what about Lenz law and the law of inductance?

    Thanks for a good answer Andrew, although I don't quite get it.

    1. Ok, The field change will induce a current according to Lenz law if the core is there to start with. (So it is kind of like charging the magnet, although not quite.) Or it would require work to put it there if you try to plug it in while current is working. That's a good explanation for now.

    2. Yes, I also believe that the inductive reactance (or whatever :wink:) will reduce the current. Dut how could normal resistance do the same thing? The field change will induce a voltage, depending on the number of turns according to the law of inductance. And the field will be cancelled by the current, induced according to Lenz law and also depending on the number of turns. This must be so! For if the right amount of current was not induced, the magnetic field would not cancel. So the current won't care about the resistance, it seems.

    So that's my conceptual problem. Now over to the easy calculational problem: I thought you should calculate voltage and current using these laws:
    [tex]V_2 = V_1 \cdot \frac{N_2}{N_1}[/tex]​
    [tex]I_2 = I_1 \cdot \frac{N_1}{N_2}[/tex]​
    However, you seem to suggest that I should use the law of power conservation [tex](I_1 \cdot V_1 = I_2 \cdot V_2)[/tex] together with Ohms law [tex](U = Z \cdot I)[/tex]. Which one is correct? The problem is that I have a conceptual understanding of the former method, but not of the latter. So if the latter method is correct, my world will fall apart. :smile:
    Last edited: May 9, 2005
  5. May 9, 2005 #4

    Andrew Mason

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    If the secondary has no current, Lenz's law will limit current in the primary. But if current flows in the secondary, current flows in the primary. The induced current in the secondary actually reduces the current limiting effect in the primary.

    The problem is that induced current depends upon the load on the secondary. The load on the secondary determines the current in the primary. So, ultimately, current in the secondary and primary depends upon the load impedance in the secondary circuit: V/Z = I

  6. May 9, 2005 #5
    Aah that's the way it is. But still...

    This makes sense! And if there is a resistance in the secondary, the circuit won't be able to neutralize the field. In other words this is not an ideal transformer and the "un-neutralized" field will in some way lead to heat losses. Then an ideal transformer is defined as a transformer without resistance in the secondary circuit.

    But I still don't understand why, in the case of an ideal transformer, the field isn't neutralized both in the secondary and the primary. Why is all the neutralization located at the secondary? Why doesn't Lenz's law work at the primary? Magnetic flux should pass here as well.
  7. May 9, 2005 #6

    Andrew Mason

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    Don't forget: the current in the secondary induces an emf in the primary - except that this induced emf inhibits the back emf or reactance in the the primary (countering the Lenz' law effect).

  8. May 14, 2005 #7
    Problem solved

    Ok, I got it now. There will always be induction according to Lenz' law in the primary (why shouldn't that be?). My problem was that I thought this would lead to a heat loss. However this is not so, since a current reduction will simply reduce current in the primary, and therefore the effect will be lower to start with rather than being reduced. So this problem is solved now. The conclusion is that induction in the primary cannot lead to heat losses. Hope I got it right?

    I still have some questions, but I have to check the equations first.

    This is a really good forum! You get the bits and pieces you need to solve the problem. I have a much better understanding of transformers by now. Thanks for your help Andrew!

    (I had a little help running the Java transfromer at http://www.ngsir.netfirms.com/englishhtm/Transformer.htm by the way.)
    Last edited: May 14, 2005
  9. May 14, 2005 #8

    Andrew Mason

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    There will be some resistance in the primary so there will be some current and some losses even with an open secondary. But, ignoring that, with an open secondary the current and voltage in the primary would be 90 out of phase, so the energy consumption over a complete cycle is zero: ([itex]E_{nergy} = \int \vec V\cdot \vec I dt = V_{avg}I_{avg}cos(90) = 0[/itex]) where [itex]\vec V[/itex] and [itex]\vec I[/itex] are the vector (or phasor) representations of voltage and current.

    What happens physically is that the energy is put into increasing the magnetic field on half the cycle and the energy is regained when the field decreases. However, since the core is iron (a conductor) little induced eddy currents occur in the core and this causes heat losses (look up hysteresis). This is minimized by laminating the core so there is a stack of insulated iron sheets instead of a block of iron. There are still losses though. So there is still a heat loss even with no resistance in the primary.

    That is a very good site. Animations are a very good way to visualize these things.

    BTW, your persistence is commendable.

  10. May 26, 2005 #9
    Annoying infinitesimalities in differential equations

    Hi again!

    Interesting things said about hysteresis. Why didn’t anyone tell me before?

    I told Andrew I would look at some equations and then come back. The equation I’m concerned about is the differential equation for the RL-circuit:
    [itex]u = R \cdot i + L \cdot \frac{di}{dt}[/itex]​
    This equation says that the inductor will hinder the effective voltage [itex]R \cdot i[/itex] to build up. But to me this equation doesn’t make sense, although I would agree it is correct. Let me explain this in the case of a positive DC voltage.

    First of all, from the physical (not mathematical!) viewpoint, why wouldn’t the term [itex]L \cdot \frac{di}{dt}[/itex] exceed the voltage? Because then we would have a negative voltage over the inductor causing the current to flow backwards (or at least decrease). But this is an absurdity since the current has to always increase to keep the term positive. I have no complaints this far.

    The second question is: why would the current increase at all. If you solve the equation you will (of course) find an increasing current. The funny thing is that, according to me, the equation doesn’t give a physical reason why the current would increase. Looking at a certain time the effective voltage is [itex]R \cdot i[/itex], which is enough to drive a current [itex]i[/itex]. It is not enough to drive a current [itex]i+di[/itex]!

    Now my idea is that in real life the term [itex]L \cdot \frac{di}{dt}[/itex] would be somewhat smaller than [itex]u - R \cdot i[/itex]. Then that would allow for current to build up, since you then have a “net voltage”. This can be done if the differential equation is modified:
    [itex]u = R \cdot i + (1+k) L \cdot \frac{di}{dt}[/itex]​
    where k<<1. The constant k would probably be extremely small. But at least this modified equation doesn’t contradict Newton’s second law: a particle will need a force (voltage/meter*charge) in order to accelerate. So the second equation would make sense to a physicist, but an engineer wouldn’t care.

    Now what is so interesting about this? I don’t know if I’m right. But I do feel that some equations in physics are very effective in hiding the conceptual aspects. They are almost impossible to understand as they stand, hiding important physics in an infinitesimal (infinitely small) corner. Is this right? Has anyone else thought about this?

    Last edited: May 26, 2005
  11. May 27, 2005 #10

    Andrew Mason

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    It can exceed the applied voltage. That is exactly what happens when a DC current through the coil is interrupted. What you cannot have is more power from the induced voltage than from the power source.

    When the DC circuit is closed, the applied voltage causes current to flow and the induced back-emf causes current to flow in the opposite direction. The rate of change of current cannot produce an induced voltage that causes more current to flow through the resistance than the applied voltage causes to flow. If you could do that you would be producing more energy than you are delivering to the circuit.

    The physical reason is that it starts at 0 when the circuit is open and ends up as I = V/R. But it is not instantaneous. As current is increasing, the magnetic field of the inductor increases and induces a back emf in the coil which limits the rate at which current builds up. So it takes a finite amount of time to build up.

    The effective voltage across the resistance depends on the voltage across the inductor, which depends on the inductive reactance (which is a function of the rate of change of current). When the current is 0 (ie at the very beginning), the voltage across the resistance will be 0 and the voltage across the inductor will be maximum. Current through the inductor will lag the applied voltage by something less than 90 degrees. As the current builds up, the voltage across resistor increases and the voltage across the inductor decreases. Eventually, the current reaches a stable level (I=V/R) and there is no further inductive reactance (di/dt=0). The coil, at that point, is simply a wire carrying a steady current and producing a steady magnetic field.

    Last edited: May 27, 2005
  12. May 28, 2005 #11
    More complaints about incomplete differential equations

    That's a good point. I never thought about that one. However, I think I do have a good understanding of the dynamics of the inductor now, having thought about them for a long long time :smile:. I have a picture about how the fields are hindering current to flow. What I'm trying to understand right now is how to understand if the differential equation is complete or if it describes an ideal situation.

    My suggestion was that it wasn't complete and that you would need a damping term [itex]k \cdot \frac{di}{dt}[/itex] to make it more realistic. For suppose the electrons were extremely heavy. Then they wouldn't accelerate very easily. And the acceleration of electrons is the same thing as increasing current. Then current would need more time to build up.

    Now suppose the electrons are as ultra-light as they are in reality. Then they would still hinder current to build up. And the damping term would still be valid, although the constant k would be small. (Having appreciated it in my head it seems to be small.) This applies to the equation [itex]u = R \cdot i[/itex] as well. Otherwise current would build up instantaneously, revealing an infinite force.

    You could also need terms to take heat loss into consideration.

    To solve this new differential equation could be messy, or it could be just as easy. But it shouldn't be done really. The purpose of the equation is to show that there is a connection between mechanics and electricity. In the original differential equation there is no such connection. Isn't this a valid point? To understand any differential equation you would have to know what it doesn't take into account. This is often left unanswered by physicists. That is my point.

  13. May 28, 2005 #12

    Andrew Mason

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    The resistance term provides the damping.

    Free electrons have somewhat random motion in a conductor. While electrons moving through the conductor having a group velocity or drift velocity, electron motion it is a somewhat complicated statistical concept. So I don't think you can say that the acceleration of electrons can be equated to increasing current.
    The drift velocity of electrons in a conductor is amazingly small, due to the enormous number of free electrons in a conductor. Zapperz can give you a much better explanation on the actual mechanics of electron motion in a conductor than I can.

  14. May 28, 2005 #13


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    But you have, nevertheless, some misconceptions. First, understand what the different terms mean.

    u is the applied voltage (it is a number fixed by the power supply/battery) not some "effective voltage"; iR is the voltage drop across the resistor; and Ldi/dt is the voltage drop across the ends of the inductor.

    The fact that the electric field generated by the battery is a conservative field (ie : the work done by the field on a charge is path independent, or the work done over a closed loop is zero) requires that the voltage gain provided by the battery equal the total voltage drop across various elements in the circuit. This concept is contained within Kirchoff's Voltage Law.

    At t=0, there is no current and hence no drop across the resistor. So, the entire potential drop is across the inductor, or Ldi/dt = u. Since u > 0, di/dt > 0. This tells you that i must increase from its initial value of 0, causing a potential drop to appear across the resistor. For the equation to work (which it must, for the reason stated in the previous paragraph), as iR increases, Ldi/dt must decrease, so as the keep the sum constant. Since Ldi/dt started out equal to u and has henceforth only decreased, it can not exceed u (in theory).

    The equation never intends to give a physical reason why the current increases. The equation (as many others) merely tells you how to find the current at a certain time. The reason why the current increases is that the net force on the electrons is non-decreasing over timescales that are large compared to electron-phonon scattering times. In the absence of the inductor, when u=0, i=0; and as u instantaneously changes to some non-zero number (by throwing a switch), the current too reaches a non-zero value "instantaneously" and stays steady at that value. So, the current increases simply because of the appearnce of a driving force - the electric field generated by the battery. Thus, it is seen that the damping force from the resistor has an extremely short time (virtually zero) constant. This is indeed true and this time constant would be of the order of the relaxation time between scattering off the atoms in the resistor. This number is of the order of nanoseconds, and hence may be neglected as far as the big picture is concerned. This, however, is what you are talking about when you speak of the effective mass of the electron having a role in the damping (not the Ldi/dt term). Now you know why this contribution is not important to the behavior over longer time-scales.

    When an inductor is included in the circuit, there is an addition retarding force coming from the induced magnetic field whose value is proportional to the rate of change of the field/current (and hence disappears in the steady state). This slows the current growth over time scales that are much bigger than the relaxation (or scattering) time.

    From how you are using this term, it would be unphysical to have a "net voltage". This would require that the work done by the electric field over a closed loop be non-zero, resulting in a violation of energy conservation.
    Last edited: May 28, 2005
  15. May 29, 2005 #14
    My misconceptions

    I sure do. When I woke up this morning I realised I had a false picture of Lenz' law. I guess the reason I hung onto it was that the picture was kind of fancy. So I had to modify it slightly. But the picture is still fancy :smile:.

    I have to agree this was a misconception. It is much better to talk about voltage drops. Then there is no greater difference between resistors and inductors.

    That is why I "complain".

    I don't know how to define scattering time or how electron-phonon scattering works. (Never studied solid state physics.) But if you say the relaxation time is of the order of nanoseconds, that's a very good thing to know!

    This is not a misconception. I proposed this "net voltage"~"net force" to overcome resistance during the relaxation time.
    (Really, I wrote about acceleration. But this was a naïve thought, as Andrew pointed out. It's a good thing you don't have to use your real name in this forum :blushing::smile:.)

  16. Nov 6, 2010 #15
    Re: "Iron core energy change" and "transformers vs. ohms law"

    You can say that I am a complete beginner when it comes to physics. This query by Order made me curious and I did some research on the subject. I was able to find out that the best way to conserve energy is by putting an iron core inside a coil that is carrying current. By doing so we will be able to increase the strength of the magnetic field and this in turn results in energy conservation. A lot of research is going on regarding this concept.
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