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Inductor initial condition

  1. Apr 26, 2015 #1

    cnh1995

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    If the switch in an RL dc circuit is closed at t=0, the current i in the inductor rises exponentially. But at t=0, i=0 because the inductor generates a back emf equal to the applied voltage at t=0. So in absence of any current,how does the inductor come to know that the switch has been closed? How does it come to know the exact magnitude of applied voltage?? The initial current is 0 due to initial back emf but the initial back emf is due to what??
     
    Last edited: Apr 26, 2015
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  3. Apr 26, 2015 #2

    berkeman

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    Never anthropomorphize inductors -- they hate it when you do that. :smile:

    Are you familiar with the differential equation relating the voltage across an inductor to the current through it? It pretty much says it all:

    [tex]v(t) = L \frac{di(t)}{dt}[/tex]
     
  4. Apr 27, 2015 #3

    cnh1995

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    Yes I know all the math related to further response of inductor. But initially, just after switching, what produces back emf? There should be a rate of change of current..(which there is..V/L). That means some change(although very small) should take place at t=0. Through the changing current, inductor will "know" when to generate back emf..But at t=0, when i=0 how does it come to know? Yeah I know it's kinda stupid question:-p But my real problem is I can't "visualize" the derivative function, although I'm good at solving them on paper..Books say current "tries to increase". I can visualize increase in current but how to visualize "trying to increase"...?
     
  5. Apr 27, 2015 #4

    davenn

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    when you close the switch (t = 0), current will flow through the circuit and slowly rise to its maximum value at a rate determined by the inductance of the inductor. This rate of current flowing through the inductor multiplied by the inductors inductance in Henry’s, results in some fixed value self-induced emf being produced across the coil as determined by Faraday’s equation ...

    v(t) = Ldi(t)/dt

    that Berkeman stated before


    Dave
     
  6. Apr 27, 2015 #5

    anorlunda

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    I think the OP is confusing himself by the contradiction of an ideal step function. At t=0, the voltage is 0, but at t=0, the voltage is 1.

    This riddle resolves itself in real life. There is no such thing as a perfect step function in real life. There is no such thing as a circuit element that does not have nonzero R and C and L (except superconductors have zero R). As the switch closes, a spark will jump across the contacts and start current flowing some billionths of a second early, so thst the event at t=0 is actually smeared across a short period before and after 0.

    We use idealized components in analysis because they make things easier. But if we imagine them existing in real life, they lead to contradictions. To avoid confusion always say to yourself, I have a circuit with an approximately pure R and C and L, and a nearly ideal voltage source and a nearly ideal switch.
     
  7. Apr 27, 2015 #6

    jim hardy

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    Is this as Berkeman suggested math problem as opposed to a physics problem?

    F = MA = MdV/dt =Md2X/dt2
    anybody who's ever paddled a canoe or handled a small boat around a dock 'feels' this intuitively.

    That's what a derivative is, rate of change ,

    It is useful to connect an analog ohm-meter across the terminals of a huge inductor and feel it oppose current. I have one that takes a full second before the needle starts to move. It'll really "bite" when you remove the ohm-meter lead.

    EMF = LdI/dt = Ld2Q/dt2 becomes intuitive when you've felt it this way.

    it knows something that we do not - what is electromagnetism.

    old jim
     
  8. Apr 27, 2015 #7
  9. Apr 27, 2015 #8

    cnh1995

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    The behavior of this circuit is intuitive but the physics behind it is really not...At least for me..
     
  10. Apr 28, 2015 #9

    jim hardy

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    Perhaps
    at the instant of switch closure when there appears a changing electric field along the length of the coil,
    there appears a changing magnetic field too, per Maxwell, that gets things rolling.

    curl E = -dB/dt

    we're taught the electric and magnetic fields are inseparable.

    i'm not a reliable source for applications of Maxwell's equations. Just trying to "prime the pump" here.

    see http://www.wwheaton.com/waw/mad/mad7.html

    and http://en.wikipedia.org/wiki/Maxwell's_equations

     
  11. Apr 28, 2015 #10

    cnh1995

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    Yeah...That's what I too had thought..That explains how it 'knows' the exact value of the applied voltage...But its not discussed in any book so I was not sure..Now since you too think so, I can stick to it..Thanks:smile::smile:
     
  12. Apr 28, 2015 #11

    cnh1995

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    The reason why I think so is that the circuit elements communicate through the electric field signals only...So I thought the above mentioned maxwell's equation must be operating here..Similar equation exists for the displacement current in charging capacitor (E is in differential term and B is integral term)..So I thought that might have some connection with this..
     
  13. Apr 28, 2015 #12

    jim hardy

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    bear in mind that i am not expert in vector calculus. So cross check your intuitive feel against formulas, and with other denizens of PF.


    My understanding of wave propagation came from a good high school electronics teacher who introduced us boys to transmission lines and antennas from the perspective of amateur radio.
    When i hit the fields course junior year at college i was able to struggle through the math only because each term in those fearsome equations i could relate to something we'd physically measured in high school. We'd equipped our classroom open wire transmission line with a trolley and measured E & B fields along it with various terminations.
    Of course in high school we'd not been exposed to math beyond operator j and rectangular to polar conversions (by slide rule, back then).
    But we understood SWR's and Smith charts.

    So - I've never done that on a wound inductor. Hence my prefacing the post with "perhaps" .
    There are people here who are fluent in Maxwell's equations.

    One of PF's core values is
    and higher math is one of mine.
    I feel academically unqualified to participate at PF, but do offer practical help where i can. They're good enough to tolerate me.

    Let us know what you figure out !

    thanks, old jim himself
     
  14. Apr 28, 2015 #13

    cnh1995

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    I'll do crosscheck..
    (P.S.-Academically unqualified??? Take a look at my CT magnetics thread:smile::smile:)
     
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