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Basic Inductor Concept

  1. Aug 15, 2010 #1
    Consider a simple circuit of Real inductor exited by sinusoidal Source.

    What is getting me into trouble is this.
    1. What is the current through the inductor? Isn't it the same as that of resister,which is [tex]\phi[/tex] phase lagging with the applied voltage. (Current is same in a series branch)

    2. Considering that this inductor is the part of a transformers primary winding, the magnetic Flux Will also be [tex]\phi[/tex] phase lagging with applied field since Flux is directly proportional to the current (Assuming linear relation between B and H)

    But my teachers and my text-book says that in transformer, the magnetic Flux produced is perfectly 900 out of phase with the applied Voltage. My teacher said that only the 900 component of the current is used in producing the flux and the in-phase component will be simply lost in the resistive heating.

    So I want you to explain to me what this the phase of flux produced?
  2. jcsd
  3. Aug 15, 2010 #2
    If we neglect the small resistance r of the winding then flux is 90 deg lagging wrt voltage. Otherwise as you said it is lagging by phase [itex] \phi [/itex] only
  4. Aug 15, 2010 #3
    But How do you reply to someone saying this,
    "The current can be resolved into two component, one in-phase with the applied voltage and one out of phase. The in-phase current which flows through the internal-resistance of the coil is lost in the heating. The 900 phase current will only produce the flux."
  5. Aug 15, 2010 #4
    Moreover, consider this.
    We can also model the internal resistance of the inductor by choosing an equivalent parallel resistance like this

    In this case, we can clearly see that the current through the inductor and current through the resistor is different.
    Here, current through the inductor is exactly 900 out of phase with the applied voltage. Since the magnetic Flux is produced only by the inductor, the produced flux will also have exactly 900 phase lag with the applied voltage!

    Whats wrong here!!!? Somebody please, point out, this is driving me crazy!
  6. Aug 15, 2010 #5
    You generally can't model the resistance of the inductor as being in parallel. The reason is that we should expect the same exact current to go through both the inductor and the resistor.

    If we let Rs and Ls be our series model of the inductor and Rp and Lp be the parellel version, then we can only say that the models are equivalent if (Rs+Ls)=(RpLp)/(Rp+Lp)
  7. Aug 15, 2010 #6
    I din't mean to say that same value of L and R can be invariably switched between series and parallel model. I only wanted to point out that there is an 'equivalent' parallel model.

    My problem is that even if the model is equivalent in the respect that (Rs+Ls)=(RpLp)/(Rp+Lp)
    It doesn't seem to be equal in the respect of phase of flux being created. In series model, the Flux seems to be created in the same phase as the net current, in the parallel it seems to be created in the same phase as the current through the inductor.

    I hope this clarifies my problem.
  8. Aug 15, 2010 #7
    You could apply a DC voltage, give it time to settle, and directly measure Rs. It's a real thing. The resistor in the parallel (Rp) version is not a real thing. It's something that would be made up just to make the models mathematically equivalent. You would probably have to make up a new value for Lp as well so that isn't anything real either. In fact, the made-up parallel Lp might be capacitive instead of inductive just to make the math work out (maybe I'll work it out).

    In either case, the total current would be the same. In your parallel model, the sum total of the currents going through each branch would be the exact same as the current that goes into the series version.

    Think of the current that goes into the inductor as being a complex vector. There are many ways to add vectors of different phases to get another vector with another phase. There is no rule that they have to have the same phase.
  9. Aug 15, 2010 #8
    So, you mean to suggest the 'real' modelling is that with the series resistor and that the parallel resister modelling gives only the correct net current, its no 'real' thing.
    quite logical.

    So, your final answer would be that --> Since the series resistance model is the 'real' model, the current through the ideal inductor of the model would be [tex]\phi[/tex] phase lagging. So, the magnetix flux would also be [tex]\phi[/tex] phase lagging and not exactly 900 phase lagging.

    okay, thanks.
  10. Dec 25, 2010 #9
    I don't know if it is right thing to do (because this will bring this dead old thread to the top, for no special purpose) but I found the right answer to my own OP. (Posted this just for the sake of feeling goodness for myself.)

    All the things in the thread is good and true except for one thing, the teacher was correct!
    Actually, the teacher was talking about the resistance of the core, and not of the copper. With no load in the secondary, the primary current of the transformer is very small (only excitation current) and the copper resistance don't come into play.
    Due to non-linear nature of B-H curve, a phi phase lagging current will be required to produce the perfect 90 lagging flux in the core.
    So, to model these a core resistance Rm is placed parallel to the perfect inductor. In such model, the flux is produced only by the 90 degree component of the total current (i.e. current through the inductor) and the parallel component will be used up in the resistive heating of the core.
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