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AC Coils

  1. Oct 30, 2005 #1

    cepheid

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    *Sigh*. I'm embarrassed to ask this question because it seems like a silly question. But at the same time, if I don't ask, I dont know how else I'll find out the answer.

    I have some documentation for work that says, "The relays are 110 V AC coils..." Believe it or not, that snippet is all you need. The actual context isn't important. I've read about relays, seen photos, and seen one in real life. From what I understand, they are remote switches -- electromechanical devices consisting of a solenoid that actuates an armature magnetically. If the relay is normally open, then the armature closes a contact, making an electrical connection.

    Here is my question: for this application, isn't a constant magnetic field required? Especially if the contact is supposed to remain closed as long as there is current in the coil? If so, isn't DC required? Sure, I understand the applications of AC coils in situations in which one wishes to induce an EMF (e.g. in a transformer). But, I don't understand how a relay with an AC coil would work. Just to make sure I wasn't totally off, I reviewed solenoids:
    The magnetic field of of our tightly-wound, ideal, infinite solenoid with n turns per unit length and a steady current I flowing through it is given by:

    [tex] \mathbf{B} = \mu_0 n I \mathbf{\hat{z}} [/tex].......[1]

    inside the solenoid. z is the longitudinal coordinate. Now, when I tried to think of what would happen in a solenoid supplied with an AC voltage, I got muddled up. If the voltage supplied was of the form:

    [tex] v_s(t) = V_0 \cos(\omega t + \phi) [/tex]

    then at first, the solenoid will try to draw a current:

    [tex] i(t) = I_0 \cos(\omega t + \phi) [/tex]

    with the relationship between I0 and V0 depending on the resistive (real) part of the coil's impedance, RL I.e.:

    [tex]I_0 = \frac{V_0}{R_L} [/tex]

    But whoah! The presence of a time-varying current in the coil and the corresponding changing magnetic flux through the coil induces an opposing EMF given by:

    [tex] v_L(t) = L\frac{di(t)}{dt} [/tex].......[2]

    So, since the EMF is always opposing, I'm guessing that the current at any instant is reduced, in the following manner:

    [tex] i_L = \frac{v_s - v_L}{R_L} [/tex].......[3]

    But wait!!! From [2], this alteration in the current alters vL. Which alters iL. Ad infinitum?!?!? ARRRRRRRRRRRGHH! I can't figure it out. Maybe one of you can explain it to me. But you know what? For the moment, I don't care. I'm just going to assume that in the steady state, the current through a solenoid supplied with an AC voltage can be expressed in the form:

    [tex] i(t) = I \cos(\omega t + \phi) [/tex]

    In which case (getting back to the issue at hand), from [1] the magnetic field in the coil would be given by:

    [tex] \mathbf{B} = \mu_0 n i(t) \mathbf{\hat{z}} [/tex]

    [tex] = \mu_0 n I \cos(\omega t + \phi) \mathbf{\hat{z}} [/tex]

    The magnetic field also varies sinusoidally, and so it keeps switching direction, back and forth. That doesn't strike me as being very useful for actuating a switch! So what gives? :frown:
     
    Last edited: Oct 30, 2005
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  3. Oct 30, 2005 #2

    Integral

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    That's a lot of work and I have not checked it all. The key to a solenoid is that the direction of the current does not matter. The lowest energy state is when the metal core is completely inside the coils. So when you apply a current, any current, the metal core is drawn in. Perhaps if you did an energy analysis you could make some since of what is happening.
     
  4. Nov 3, 2005 #3

    cepheid

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    Thank you Integral. I will look into it. How can find more information about the second part of my question (when I got sidetracked), which was, how does a solenoid respond to some driving signal?
     
  5. Nov 3, 2005 #4

    krab

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    Your equations are correct for individual imductor or resistor, but you have both at the same time, so the equation is
    [tex]Ldi/dt+Ri=v[/tex]
    Now use your assumed form for v, and solve for i. You will find that i has the same frequency, but not the same phase [itex]\phi[/itex] as the phase of v.
     
  6. Nov 3, 2005 #5

    cepheid

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    Right, right. Of course. If the impedance were purely reactive, the current would be 90 degrees out of phase. So, my question is, is my attempt to look at things "stepwise" in error? In other words when supplied with an AC voltage, will the coil immediately respond by drawing a current i(t) in such a way that the equation:

    [tex] v_s(t) = L\frac{di(t)}{dt} + Ri(t) [/tex]

    is satisfied?
     
    Last edited: Nov 3, 2005
  7. Nov 4, 2005 #6

    pervect

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    Given v_s(t) which will be something like the function below (idealized)

    v_s(t) = 0, t<0
    v_s(t) = sin(wt), t>0

    you have a differential equation for i(t) which will always be satisfied. This solution can be written as the sum of a steady-state part plus a transient part. The total solution is determined by the boundary condtions, which are given by v_s(t), and the fact that i(t) = 0 t<0.

    So far you have not actually computed the transient solution, but it doesn't matter except for educational purposes.

    http://hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html

    might help you out some, you might try reading some about differential equations too.
     
  8. Nov 7, 2005 #7

    cepheid

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    :redface: Actually, I've done plenty of differential equations, the latest course being in PDEs, with the heat equation, wave equation, Laplace's equation, Poisson's eqn etc. many of which let to transient and steady state solutions that depended of course on the boundary conditions. We even did some Sturm-Liouville boundary value problems at the end. It's just that my brain regressed so much over the summer holidays as usual, and I forgot how to solve this simple 1st order linear ODE. In fact, I'm still having trouble with it. Here's my work:

    [tex] V_0 \cos(\omega t) = L \frac{di}{dt} + Ri [/tex]

    [tex] \frac{V_0}{L} \cos(\omega t) = \frac{di}{dt} + \frac{R}{L}i [/tex]

    The choice of integrating factor is obvious:

    [tex] \frac{V_0}{L} e^{\frac{R}{L}t} \cos(\omega t) = e^{\frac{R}{L}t}\frac{di}{dt} + \frac{R}{L}e^{\frac{R}{L}t}i [/tex]

    [tex] \frac{V_0}{L} \int{e^{\frac{R}{L}t} \cos(\omega t)\,dt} = e^{\frac{R}{L}t}i [/tex].............[1]

    This integration by parts was an ordeal! But I did it. Let me cut to the chase:

    [tex] \int{e^{\frac{R}{L}t} \cos(\omega t) \,dt} = \left(\frac{R^2}{R^2+\omega^2 L^2}\right) \frac{L}{R}e^{\frac{R}{L}t} \left[\cos(\omega t) + \frac{\omega L}{R} \sin(\omega t)\right] + C [/tex]

    subst. into [1] and solve for i(t):

    [tex] i(t) = \frac{V_0 R}{R^2 + \omega^2 L^2} \left[\cos(\omega t) + \frac{\omega L}{R} \sin(\omega t)\right] + \frac{V_0}{L}Ce^{-\frac{R}{L}t} [/tex]

    Solve for C using the I.C. that i(0) = 0

    C = -LR / (R2 + ω2L2)

    Which yields:

    [tex] i(t) = \frac{V_0 R}{R^2 + \omega^2 L^2} \left[\cos(\omega t) + \frac{\omega L}{R} \sin(\omega t)\right] - \frac{V_0 R}{R^2 + \omega^2 L^2}e^{-\frac{R}{L}t} [/tex]

    I can see what you mean about the transient and steady state solutions, but I'm not sure if it is totally correct. I don't expect anyone to slog through all that as well, but if you already know the answer for an RL circuit, maybe you could check it against that? I'm also not happy with the solution in that form. I wanted to try to convert the sinusoidal part so that it would be expressed entirely in terms of cosine, to see how it related to the forcing function (the phase difference). I tried to use complex exponentials to do that conversion, but I got lost somewhere. How would you change it into that form?
     
    Last edited: Nov 7, 2005
  9. Nov 7, 2005 #8

    cepheid

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    Solution Method # 2 (Phasors set to kill!) :rolleyes:

    Why can't I just do this? Let:

    [tex] \overline{V} = V_0 e^{j \omega t} [/tex]

    [tex] \therefore \, v_s(t) = Re[\overline{V}] [/tex]

    [tex] \overline{V} = \overline{Z}\overline{I} [/tex]

    [tex] \overline{I} = \frac{\overline{V}}{\overline{Z}} [/tex]

    [tex] = \frac{V_0 e^{j \omega t}}{R + j \omega L} [/tex]

    [tex] = \frac{V_0 e^{j \omega t}({R - j \omega L})}{R^2 + \omega^2 L^2} [/tex]

    Expanding out the first factor of the numerator using Euler's equation, and then multiplying it all out, and then not bothering to calculate the imaginary part of it since we don't need it, I get:

    [tex] i(t) = Re[\overline{I}] = \frac{V_0 R \cos(\omega t) + V_0 \omega L \sin(\omega t)}{R^2 + \omega^2 L^2} [/tex]

    Looks about the same, except that this is only the steady state solution. Why??? :confused: Also, can you please help me get the solution in the final form (cosine only) in both methods?
     
    Last edited: Nov 7, 2005
  10. Nov 7, 2005 #9

    krab

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    2 questions, each answers the other. You end up with the steady-state solution because you are starting with a steady-state voltage.
     
  11. Nov 7, 2005 #10

    cepheid

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    Wasn't I starting from a steady state voltage in the first solution attempt too? I don't really follow. I started with V0cos(wt).

    EDIT: Nevermind. I get it. The differential equation gives you a solution for i(t) right from t = 0, and therefore it takes into account what happens when that sinusoidal (in the steady state) voltage is first 'turned on', jumping immediately up to a value of V0, before beginning to oscillate. Calculating the current from the voltage and the impedance, on the other hand, assumes a steady state, and just says, if your voltage is this and your impedance is this, then your current is this (without worrying about how your voltage got to be "this", or the fact that the expression for the impedance is really only true in the AC steady state, once the voltage is like "that"). A terribly-worded explanation, I know, but it makes sense to me now.

    Also, is it possible to get even just a quick explanation of how to convert those sinusoidal terms as I has asked before?
     
    Last edited: Nov 8, 2005
  12. Nov 8, 2005 #11

    krab

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    You have
    [tex]A\cos\theta+B\sin\theta=C\cos(\theta+\phi)[/tex]
    and want to find amplitude C and phase [itex]\phi[/itex], right? Just expand the right side and you get
    [tex]A=C\cos\phi,\ B=-C\sin\phi[/tex]
    Now can you solve it?
     
  13. Nov 8, 2005 #12

    pervect

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    Usually, I would just use the Fourier or Laplace transform to get the steady-state solution. This would be, using the EE notation for the Fourier transform

    v = (jwL+R)*i, or i = v/(jwL+R)

    (Laplace transforms would replace the jw by an s but would otherwise be identical).

    This is equivalent to your solution (I think) which has been re-written to be of the form

    v(jwL-R)/(R^2 + w^2L^2) to give you the real and imaginary parts. (An alternate would be to express the solution in terms of magnitude and phase).

    The basic approach/result using Laplace transforms is very similar to your "phasor" approach.

    There is a way of getting the transient solution from the Laplace transform as well, but I've forgotten it :-(. I would generally just find the transient solution from the linear equation,

    to the linear differential equation

    L di/dt + R = 0

    which I know to be of the form

    K*e^(-at)

    which gives

    -L a e^(-at) + R e^(-at) = 0 or -La = R, a=-R/L

    The value of K is adjusted to meet the initial boundary conditions, i.e. i(t)=0

    Now we just add the linear solution to the previously found steady-state solution and adjust the constant K in the linear solution to meet the boundary condtions.

    In your solution, the boundary condition is just that i(0)=0. You can see that your solution meets this requirement, and that this requirement i(0)=0 is what determines the value of K.

    It should be obvious, I hope that any solution to

    (eq=0) added to a solution to (eq=v(t)) gives a solution to (eq=v(t))

    ?
     
    Last edited: Nov 8, 2005
  14. Nov 8, 2005 #13

    cepheid

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    Sure. I should have seen it before:

    [tex] C = \sqrt{A^2 + B^2} = \sqrt{R^2 + \omega^2 L^2} [/tex]

    [tex] \phi = \tan^{-1} \left(\frac{\omega L}{R}\right) [/tex]

    [tex] \therefore \, i(t) = \frac{V_0}{\sqrt{R^2 + \omega^2 L^2}} \left[\cos \left(\omega t - \tan^{-1} \left(\frac{\omega L}{R}\right) \right) \right] - \frac{V_0 R}{R^2 + \omega^2 L^2}e^{-\frac{R}{L}t} [/tex]

    Is that the correct, final solution?
     
  15. Nov 9, 2005 #14

    cepheid

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    ^Well hopefully somebody will confirm that that is correct. But in the mean time, I wanted to try to address my other question:

    The best I could come up with was this:

    Trying to pull an iron core out of a solenoid with current I, we have the following work done for a small displacement of the core (z is the longitudinal axis):

    Fapp is the applied force, F is the magnetic force on the core:

    [tex] dW = F_{app}dz = -Fdz [/tex]

    [tex] \rightarrow F = -\frac{dW}{dz} [/tex]

    The total energy stored in the system is:

    [tex] W = \frac{1}{2}LI^2 [/tex]

    which represents the work done to overcome the induced emf and produce the final magnetic field as the current increased from 0 to I in a coil of inductance L. Assuming that the current in the coil remains constant as the core is moved,

    [tex] F = -\frac{dW}{dz} = -\frac{1}{2}\frac{dL}{dz}I^2 [/tex]

    After I did this I realized something. Magnetic forces do NO work. Damn! So my whole argument is flawed??? Well, even if it were true, it doesn't tell you much about how the force changes with z, because we don't know the inductance L(z) as a function of the position of the core. In order to find it out, we need to know the flux through the air part and the ferromagnetic part too, which depends on the position of the core, and the field in the core, which depends on the magnetization of the core and the permeability of the core, which is further complicated by the fact that ferromagnetic materials aren't linear, they don't have a constant magnetic susceptibility [itex] \chi_m [/tex]. So if you ask me, this problem is waaaay too difficult. And I'm still only looking at the DC case! I couldn't find a comprehensible discussion of this anywhere. All I really want is a theoretical way to show that

    1. The core is always drawn into the solenoid when there is a current in the coil. (EVEN for time varying currents!!!)

    2. The force on the core increases with distance (as it is drawn in).

    Any help would be much appreciated.
     
  16. Nov 9, 2005 #15

    krab

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    This is one of the most pervasive myths in physics. It's not true. Why would anyone believe it? How do electric motors do work? They use magnetic fields. What you have with a solenoid is basically the same thing as an electric motor. The difference is that it operates linearly and so cannot repeat.
     
  17. Nov 9, 2005 #16

    cepheid

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    Because it's in the textbook, and the subtleties of situations in which magnetic forces appear to do work are thoroughly explained. That having been said, I don't know how to do the same with the current situation, and proceed with the analysis I've been trying to do here. Again, I'm just trying to mathematically answer the questions:

    "why is the ferromagnetic core always drawn into the coil, even if a time varying current runs through that coil?"

    "why does the force increase with position as the core enters the coil?"

    If anyone could show me the way, and help me with post #14, it would be much appreciated.
     
    Last edited: Nov 9, 2005
  18. Nov 9, 2005 #17

    Integral

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    Here is a scan of a few pages from Elementary Theory of Electric and Magnetic Fields by Cheston.

    I aplogize for the orientation, but beleive that this should help you understand what is happening.
     
  19. Nov 10, 2005 #18

    cepheid

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    Much appreciated. Thanks for going through the trouble! :smile:
     
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