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Quick conceptual question about inductance

  1. Oct 9, 2009 #1
    Ok so i understand that after a switch in a circuit is closed for awhile the inductor acts like a battery, however, what happens after the switch is opened after being closed for awhile.
    A review question asks whether the energy:

    - is converted to sound and spark
    - remains stored in the inductor
    - goes into driving the current across the resistor

    I want to say it is stored in the inductor, but i know that after the switch is left open for awhile, the emf is zero. So i don't know whether the energy is considered to be stored in the inductor.
     
  2. jcsd
  3. Oct 9, 2009 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. Why do you say that the inductor acts like a battery? I hadn't heard that analogy before, and it sounds wrong unless I'm missing some subtlety.

    What is the equation that relates the current through an inductor and the voltage across it? And what about a real inductor limits the build-up of current when a constant voltage is put across it?

    And what resistor are you alluding to in the problem statement above?
     
  4. Oct 9, 2009 #3
    Thanks..

    Maybe I am wrong. I had the impression that while the switch is closed an inductor takes on a charge and after the switch is closed, the inductor can keep the current flowing until the charge dissipates.

    the circuit consists of a battery in series with a resistor in series with an inductor.. and then another resistor in parallel with the inductor.
     
  5. Oct 9, 2009 #4

    berkeman

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    Staff: Mentor

    Where is the switch in the circuit?

    Current through an inductor stores energy in the magnetic field of the inductor. The current cannot change instantaneously (just as the voltage across a capacitor cannot change instantaneously). The equation that I was hoping you would write out for the inductor is:

    [tex]V(t) = L \frac{dI(t)}{dt}[/tex]

    That means that the voltage across the inductor is equal to the inductance L multiplied by the time rate of change of the current through the inductor. The other way to write this equation is in the integral form:

    [tex]I(t) = \frac{1}{L} \int {V(t) dt} [/tex]

    So if you have a constant voltage across an ideal inductor, the current climbs without limit. In the circuit you describe, the series resistor will limit the current...
     
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