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Confused about Inductors

  1. Oct 30, 2007 #1

    cepheid

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    I thought I understood inductance. It's sort of an inertial property in circuits. Physically, it accounts for the fact that "it takes work to do stuff", meaning that if you put a voltage across a coil of wire, the current through that coil can't jump up to whatever steady state value it will attain instantaneously. That would mean an infinite rate of change of current, which, for non-zero inductance (always the case) could only be provided by an infinite voltage. Inductors oppose changes in the current across them. They produce back emf that tries to oppose the change in magnetic flux across the coil that would result from said change in current. This is all a result of Faraday's law and can be expressed as follows:

    [tex] \mathcal{E} = -L\frac{di}{dt} [/tex]

    Where the curly E is the back emf being referred to, and the the negative sign indicates that the emf opposes whatever change in flux is occurring. So far so good. Now, if you open up an electrical engineering textbook on basic circuit theory, it will tell you that if the instantaneous current through an inductor is given by i(t), then the instantaneous voltage across it v(t) is given by:

    [tex] v(t) = L\frac{di}{dt} [/tex]

    I'm confused about the signs. What is the "voltage across the inductor?" How is this different from the back emf produced by the coil when you try to change the current flowing through it?
     
    Last edited: Oct 30, 2007
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  3. Oct 30, 2007 #2

    cepheid

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    Please, any insights would be appreciated.
     
  4. Oct 30, 2007 #3

    rbj

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    if you define positive current going into the "+" terminal of the inductor, the derivative of the current is proportional to the voltage of the inductor (where L>0 is the constant of proportionality). the back emf has the same polarity for increasing current (so it creates an opposing voltage to increasing current).
     
  5. Nov 1, 2007 #4

    cepheid

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    So you're saying that for our fictitious positive charges approaching what from their point of view is the "near" terminal of the inductor, if they are doing so at an increasing rate, that near terminal will become positive relative to the far terminal, thereby slightly repelling the positive charges that are trying to approach it. And if the near terminal is positive relative to the far terminal, then there is a conventional "voltage drop" of Ldi/dt "across" that inductor. Yet, if you were to express this as an emf, it would be negative Ldi/dt, because it would be an emf opposite in polarity to the positive terminal of the battery that those charges came from (or to whatever other source is responsible for starting to drive the charges around the circuit in that direction in the first place). As a result, when considering this voltage as an emf, we note that it is tryng to impede the flow of charge being driven by the source.

    Fine, my question is, as opposed to the charges "approaching" the near terminal of the inductor, what about the charges already "inside the inductor" so to speak, when the back emf arises? Wouldn't the polarity as you have described it be such that it would tend to want to make them GO in the direction of the current that is just ramping up? That is the issue that had me so confused as to what polarity this "back" emf should have in the first place.
     
    Last edited: Nov 1, 2007
  6. Nov 1, 2007 #5

    rbj

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    i dunno what you mean by "near" terminal and, since it isn't the semantics we use to describe the device, i don't think i'm going to oblige using it.

    the "+" terminal is the terminal that is not the "-" terminal. the voltage (potential difference) between the "+" and "-" terminals is defined as positive if a positive charge hanging out at the "+" terminal has more potential energy than it would have at the "-" terminal. in other words, if the voltage between "+" and "-" is greater than zero, positive charges hanging out at the "+" terminal are "uphill" compared to what they would be if the same positive charges were hanging out the "-" terminal. that is the only meaning i will ascribe to the notion of assignment of polarity to the terminals of a device.

    now, if the current flowing through an ideal inductor is constant, the charges at the "+" end have the same potential energy as the charges at the "-" end, so the voltage difference is zero. but if you were to increase the current (this current is defined as positive flowing into the "+" terminal and that is the same current flowing through the inductor from the "+" terminal down to the "-" terminal), that would require that the charges at the "+" terminal have more potential energy than those at the "-" terminal. that is because of this "back e.m.f." you mentioned. that back e.m.f. is lined up in the same direction, so if the current going into the "+" terminal is increasing with time, the charges making up that current have to be uphill at the "+" terminal.

    a hydrodynamic analog would be a cylindrical chamber with an axel going down the axis of the cylinder and (virtually massless) fan blades attached to that axel. water flows in at one end of the cylinder and out the other. the rotational speed of those fan blades will reflect the amount of current (liters per second) of water flow through the chamber. if the current is one direction the axel will turn one way and if the direction of the current reverses, the axel will spin the other way. if the current increases in speed the spin of the axel will increase. also, if i attached the axel to a motor, i could make a pump and force water through this chamber (rather than just measuring the movement of water). now imagine instead of attaching the axel to a motor, we attach the axel to a flywheel. that's a hydrodynamic inductor. the flywheel "likes" to spin at a constant rotational velocity. if you want to increase the speed of spinning, that will require torque which will require force on the fan blades which will require a pressure differential in the chamber from one side of the fan to the other. the back e.m.f. is the rotational inertia of the flywheel (as translated to the surfaces of the fan blades) that resists increasing (or decreasing) its speed of spin.
     
    Last edited: Nov 2, 2007
  7. Nov 1, 2007 #6

    marcusl

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    This discussion from an old thread on the same topic might be helpful:

    The minus sign was chosen by convention. A series circuit is assumed to have voltage sources or emfs, on the one hand, and passive devices on the other. A voltage V is dissipated in a resistance R with the "counter-emf" in R defined in the way you mention.

    For an inductive circuit let's consider an example of a battery, switch, R and L in series. The "counter-emf" in the inductor seems to be in the same direction as in the R, as you point out, BUT it's considered by convention to be a source instead, just like the battery. Since it reduces the battery voltage (as it must to satisfy Lenz's Law), it must carry a negative sign. This negative sign carries through to the solution that you wrote down for the current
    1 - exp (-tR/L)

    There's a very clear discussion in Reitz and Milford, Foundations of EM Theory, p. 246 (1960), if you can find a copy in your U library.
     
  8. Nov 3, 2007 #7

    cepheid

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    Marcusl

    Your explanation that the inductor is conventionally thought of as a source, and since it is opposite in polarity to the driving source, it must be a "negative" emf -- that answers my question, thank you. Hence, the "emf provided" by the inductor and the "voltage drop across" it are equal and opposite by definition, even though they both describe the exact same difference in potential between one terminal and another.

    rbj. I know what potential difference is, so I'm sorry you had to write that paragraph. I realize that I was using made up non-standard terminology, but I thought it was fairly clear what I meant. Evidently not. Again, I was speaking from the "point of view" of the charges. Therefore, given that they are approaching the inductor through the wire from a certain direction, the near terminal would simply be the one closest to them, i.e. the one they would encounter first, on the side of the inductor from which they were approaching. Obviously, then, the far terminal would be the one on the far side, or opposite side of the inductor from the approaching charges. I realize that this is bad way to describe things, because the line of charges that constitute the current doesn't "start" or "end" anywhere. All the free electrons everywhere in the wire begin moving in the net direction when the emf is applied.
     
    Last edited: Nov 3, 2007
  9. Nov 3, 2007 #8

    marcusl

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    Glad I could help!
     
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