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Voltage across inductor

  1. May 27, 2013 #1
    This is really not a homework, just need help to understand some concepts.

    In a circuit includes R, L, C in series, we apply Faraday's law:
    I*R + Q/C + 0 = - dψ/dt
    Here VL = ∫E*dl across the inductor is zero. I know this because the inductor has no resistance and according to Omh's law VL = 0. But if we consider the formula VL = ∫E*dl how can you know that E = 0?
    I mean that is there any way to explain the electric field between two ends of inductor is zero.
    If so, E = 0 how electrons move from one end to the other end of inductor?
     
  2. jcsd
  3. May 27, 2013 #2

    NascentOxygen

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    Inductance opposes change in current through it. Once the current reaches a steady value, and holds to that value, it is not changing so there is no voltage across the inductor (save for that necessary to overcome its wire's resistance). Current flows because there is nothing to oppose it (in an ideal inductor).

    Until the current reaches a steady level (this is determined in part by the rest of the circuit), there will need to be a voltage applied to the inductor to make its current change. That voltage = L. di/dit

    So if you want the current to increase linearly with time, di/dt = constant and this will happen if you hold a constant voltage across the inductor's terminals. Obviously, this can't go on forever, and eventually if you try to remove the voltage source (making it into open circuit) there will be a huge spark because L.di/dt is a very large negative value.
     
  4. May 28, 2013 #3
    Thanks for the reply.
    Let me express my question more detail.
    The formula for voltage between two points A and B is:
    VAB = integral from A to B of E*dl = ∫E*dl
    Now I want to apply the formula for ideal inductor:
    VL = ∫E*dl
    But according to Ohm's law, I can also write:
    VL = I*R = I* 0 = 0V
    From that I infer E = 0 along the inductor.
    But I wonder if there are others way to know that E = 0.
     
  5. May 28, 2013 #4

    NascentOxygen

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    ∫E*dl is not the equation to use, because the voltage across an inductor is not dependent on linear length

    I*R can be used to give that component of the total voltage due to resistance

    L. di/dit gives the voltage due to inductance, and that's what you are after
     
  6. May 28, 2013 #5

    rude man

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    Are you assuming dc voltage V? Looks like you are since you say V = R*I = I*0.

    In which case there is dc current through the inductor and even though the voltage across it is zero there is an electric field E in the inductor:
    j = σ E , j = current density vector, A/m^2, and σ = conductivity, S-m.
     
  7. May 28, 2013 #6
    Yes, I am considering DC.
    Voltage across the inductor is zero, VL = ∫E*dl = 0 but E ≠ 0 then to me it contradicts each other.
    How the equation ∫E*dl = 0 possible if E ≠ 0?
     
  8. May 28, 2013 #7
    I am applying Faraday's law for a circuit including R, L, C and a battery V.
    Integral in closed loop = -dψ/dt
    or:
    I*R + Q/C + ∫E*dl (across the inductor) = -dψ/dt
    In this case ∫E*dl = 0
    Now I don't know how this equation ∫E*dl = 0 possible while E ≠ 0.
     
  9. May 28, 2013 #8

    rude man

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    Your inductor has finite resistance. Therefore the voltage across it is not zero.

    Here are the relationships:

    j = current density, A/m^2
    i = current, A
    σ = wire conductivity, S-m
    E = electric field, V/m
    V = voltage across inductor, V = ∫E dL
    R = resistance of inductor = L/Aσ, ohms
    A = wire cross-sectional area, m^2
    L = length of wire, m

    We have
    j = σE = i/A
    σ = L/AR
    E = V/L
    Therefore i/A = (L/AR)(V/L)
    i = (L/R)E = V/R which you will recognize as Ohm's law.

    The only way for your inductor to have zero E and therefore zero V across it is if it was superconducting, where σ → ∞.
     
  10. May 28, 2013 #9
    Where did you get this equation, which you call Faraday's law, from?
    Faraday's law relates induced emf to rate of change of flux linkage.
     
  11. May 28, 2013 #10
  12. May 29, 2013 #11

    rude man

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    It's not zero, but the instructor considers the resistance of the inductor to be zero, so as I said, if you consider R = 0 then there is no E field. But in reality all wiring at room temperature has finite resistance so there is always some electric field.
     
  13. May 29, 2013 #12
    Can you explain why the electric field inside an ideal inductor is zero? And if it is zero, what cause electrons move through it?
     
  14. May 29, 2013 #13

    rude man

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    I did that in my previous post. I said if the resistance is zero then the E field is also zero. A superconducting coil would have zero resistance and zero E field. But at room temperature superconductivity has not (yet) been realized so in actuality every inductor has a finite resistance and therefore E field proportional to applied voltage. I am talking dc here.

    I am not qualified to describe the motion of charge in a superconductor. Maybe somebody else on this forum can ...
     
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