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Pure inductor (only) in AC circuit

  1. Mar 27, 2008 #1

    suppose we have a sinusoidal AC source connected to only a pure inductor of self inductance L ( no resistance at all ) ..then emf induced in the inductor is L * dI/dt and is equal and opposite to the applied emf. but if the induced emf opposes the applied emf , then how can there be a current at all in the circuit and in the inductor ?

  2. jcsd
  3. Mar 27, 2008 #2
    When the AC current reverses itself an emf will develop across the inductor opposing the flow. Onto the next cycle, another emf will develop impeding the flow again. So in a sense an inductor represents a sort of resistance to AC signal. We call it reactance (units in ohms), and it depends on the frequency of AC.

    At low frequencies, inductors exhibit little resistance to AC and pass all current.

    At high frequencies, inductors exhibit very high resistance and look like an open circuit.
  4. Mar 28, 2008 #3
    hey, waht-

    I think the term "back emf" is poor terminology, but it's standard usage. It's only used with inductors and motors, that I know of. As you say, it means an inductor develops a voltage opposite the source. But so do a resistor and capacitor.

    An inductor also stores energy and is a source of emf--but so also, a capacitor.

    On top of that, remove the voltage source and the inductor produces a forward emf in the sense of the source.

    A motor with inertia, behaves more like a capacitor than an inductor with the removal of an emf source, so the terminology seems to have developed to put a mental picture behind this spooky magnetic stuff.
  5. Mar 28, 2008 #4
    i think i sort of got it. . thanks.but any more illuminating explanations are welcome.
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