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Self-Inductance calculation help

  1. Nov 28, 2007 #1
    1. The problem statement, all variables and given/known data

    When the current in a circuit containing a inductor in parallel with a galvanometer and battery rises (after closing a switch), why do I see a maximum voltage across the inductor that quickly diminishes to a steady voltage?

    I thought during a current rise the emf is less due to self-inductance?

    Or does the voltage jump then fall to steady when the current stops rising and there is no self-inductance?

    Please help me understand what's going on.
     
  2. jcsd
  3. Nov 28, 2007 #2

    George Jones

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    There a couple of parts to this.

    First, what's an expression for the voltage across a pure inductance?
     
  4. Nov 28, 2007 #3
    I'm not sure I understand what you're asking me
     
  5. Nov 28, 2007 #4
    Are you referring to Faraday's law?
     
  6. Nov 28, 2007 #5

    George Jones

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    Yes, which gives the magnitude of the voltage across and inductor as

    [tex]L \frac{dI}{dt}.[/tex]
     
  7. Nov 28, 2007 #6
    I'm confused as to exactly which voltage this represents...

    Would this be the measured voltage across the inductor?
     
  8. Nov 28, 2007 #7

    George Jones

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    Yes, assuming the inductor had zero resistance.
     
  9. Nov 28, 2007 #8
    Alright thanks I understand that bit now.

    Is there a circumstance where the voltage would drop to a value > 0 when the current became steady?
     
  10. Nov 28, 2007 #9

    George Jones

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    Yes. A real inductor is made of wire that doesn't have zero resistance. A real inductor is modeled by a resistor in series with an ideal inductor. What happens when the current becomes steady?
     
  11. Nov 28, 2007 #10
    V = IR for a real inductor?

    I'm trying to interpret some dad for a lab:

    <conditions>: <maximum>/<steady> (voltage)

    laminated bar on E-core, close switch: 18/15
    laminated bar on E-core, open switch: -6/0

    so there is a toroidal coil on the middle post of the "E" core and a laminated core goes across the top to complete the magnetic circuit

    Does this make any sense?
     
  12. Nov 28, 2007 #11

    George Jones

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    Yes, this the voltage across a coil of resistance R when a steady DC current I goes through it.

    I didn't actually seeing what went on in the lab, but I think this makes sense. When the switch is opened, dI/dt is negative, and the steady-state current is zero since the switch is open.
     
  13. Nov 28, 2007 #12
    Ahh that make sense.

    Thanks a lot for your help.
     
  14. Nov 28, 2007 #13
    Could you verify one more thing for me?

    If a iron bar is stuck to a core of a coil and a current is put the coil how does a piece of paper affect the amount of hysteresis of magnetism in the iron bar?

    I think the paper would be a material of lower permeability so it should lower the hysteresis and the strength of the magnetization. Am I correct?
     
  15. Nov 29, 2007 #14

    George Jones

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    I haven't run into the term hysteresis since I was an undergrad, which was more years ago than I care to admit.

    Fair warning.

    I, too, think so.

    I'm not at all sure, but I think this increases hysteresis ("lagging" between B and H), which increases flux loss.

    Someone who knows this stuff better than do I should comment.
     
    Last edited: Nov 29, 2007
  16. Nov 29, 2007 #15
    Thanks for your help.

    I just noticed you're in Saint John. I lived in Fredericton but I'm in Ontario going to Waterloo right now. Small world. :)
     
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