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Inductor in dc

  1. Dec 20, 2007 #1
    will there be any current in a circuit with an ideal dc battery and an ideal inductor?
     
  2. jcsd
  3. Dec 21, 2007 #2

    chroot

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    Yes. The inductor is fundamentally just a piece of wire, after all. It is essentially a short circuit at dc.

    - Warren
     
  4. Dec 25, 2007 #3
    L di/dt=V( by Kirchoffs law).this would mean current is forever increasing infinitely. is this practically possible?
     
  5. Dec 25, 2007 #4
    What is the long-term di/dt in a dc circuit?
     
  6. Dec 25, 2007 #5

    chroot

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    There is no reason to assume that the current will increase without bound. In reality, unavoidable resistances elsewhere in the circuit (or in the wire itself) will limit the current.

    - Warren
     
  7. Dec 26, 2007 #6

    rbj

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    not practically. the reason is that, practically, one would need to include the internal resistance of both the inductor, L, and the voltage source, V. practically, it is not possible to hook up an ideal inductor to an ideal voltage source, only those inductors and voltage sources that have some small, but non-zero internal resistance. so you more accurately have an RL series circuit connected to the voltage V.
     
  8. Oct 23, 2009 #7
    "L di/dt=V( by Kirchoffs law).this would mean current is forever increasing infinitely. is this practically possible?"

    This means that the volatage accross the inductor is proportional to the change in inductor current over change in time multiplied by the inductance.

    In a DC circuit the (steady state) the impedance Z = jwL where w is the angular frequency or 2Pif or 6.28 times the frequency (6.28f).

    When f=0 (DC case), Z=0. So there is no impedance in the inductor and it acts as a short.

    If the inductor acts like a short, there is no change in current and therefore di/dt =0. so essentially it says that for a DC circuit the voltage accross the inductor =0 as there is no change in current.

    However saying this I do understand your frustration. If we look at V=IZ and then solve for I we get I = V/Z. If Z=0 then I = infinity. However, we understand this to mean that the inductor acts as an ideal resistor = 0 ohms (offering no resistance) and that the voltage drop accross the inductor is ideally zero. So if you just had a voltage source and an inductor, it means you would short out the voltage source and high amounts of current would flow. If you had an inductor in series with a resistor it means that it is just like you only had the voltage source and the resitor in the circuit and the current would be equal to V=IR or I=V/R (so no infinite current).

    So getting back to the I=V/Z delima. You have to look at it like this: in a DC circuit with an ideal inductor (no resistance), the current will be at it's maximum value i.e. the inductor is not trying to limit the current through the circuit (it just acts like a wire).

    Hope this helps and I hope it is not too long of an explaination.
     
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