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Instant change of current in an inductor

  1. Nov 13, 2008 #1
    Ok, here is a goofy question. Say you have an inductor that is connected in a simple DC circuit and has been there for a "long time". Now if you had a switch that you could move at any speed that you wanted to and could be any distance from the terminal that you wanted it to be what would happen when you disconnected it? Since an inductor resists any change in current would it be able to create any voltage you wanted it to by opening that switch as far as you wanted to. I believe theoretically the answer is yes. You could use the resistance in the air and the distance for the resistive part of the time constant and that would tell you the amount of time that the voltage would exist. Does this mean that when an inductor is removed from a circuit it could never store energy like a capacitor could?
     
  2. jcsd
  3. Nov 13, 2008 #2
    you can never get that infinite voltage spike by opening the switch because the switch would just arc.

    i think the energy storage analogy would be to short the ends of the inductor (you open the ends of the cap to store energy). the problem is, the flux induces a current in the coil, and that coil also has resistance, so your energy dissipates.
     
  4. Nov 13, 2008 #3
    This is pretty much exactly what's done in experiments with superconductors, though. They induce a current into a superconducting loop (which has some inductance), and then come back years later and see if it's still flowing.

    Inductive energy storage as a practical alternative to capacitive energy storage still hasn't made many inroads, though.
     
  5. Nov 13, 2008 #4
    Even if the windings of the inductor are superconducting, there is always a capacitance. If the inductor is carrying current for a long time, then the switch is opened, in an attempt to quickly interrupt the current, the energy will discharge into the winding capacitance.

    In an analogous manner, a perfect cap with superconducting plates and leads is charged to a steady dc voltage. If shorted with a zero ohm termination, superconducting, what happens? The self-inductance of the cap limits the current. Energy is transferred from the electric field between the plates, to the magnetic field in the closed loop.

    Does this answer the question?

    Claude
     
    Last edited: Nov 14, 2008
  6. Nov 13, 2008 #5

    berkeman

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    Staff: Mentor

    In fact, this technique is used in "flyback" topology DC-DC power supplies, like the ones that generate the high voltage for the picture tube in (ancient) TVs. The capacitance is explicit in these supplies, and is sized to give the peak output voltage that is rectified to generate the output high voltage.

    But as cabraham says, if you have just a real inductor alone, the peak output voltage will be determined by the parasitic capacitance of the inductor -- the energy stored in the inductor current gets transferred into the peak voltage across the parasitic capacitance, and then that energy is transferred back into a reverse current through the inductor, and so on. There is also a DCR (parasitic resistance) associated with the inductor, and that is the R element in the lossy RLC resonant circuit of the real inductor all by itself.

    The arcing effect mentioned will come into play if the peak voltage across the parasitic capacitance exceeds the arc-over voltage across the (partially) open switch... the race is how quickly the gap opens on the switch versus the RC resonance half-period on the first ring-up.
     
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