inductor: v=L(di/dt)

Discussion in 'Homework Help' started by bug13, Mar 22, 2012.

  1. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    [​IMG]

    according to v=L(di/dt), when J1 changes from on to off, I can understand that a huge back emf is generated by the inductor.

    my question is, when J1 change from off to on, inductor will briefly oppose the change in current from zero to some magnitude, why? Isn't it supposed to give a huge opposite emf as well?

    (it doesn't explain in the DC\Inductor section of this site)
     
  2. mlog

    Member

    Feb 11, 2012
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    Forget about the back EMF for a moment. The important thing to remember is that because of the magnetic field, the inductor opposes an instantaneous change in current through it. So the current remains the same at the instant of switching (no matter whether switch opens or closes). When that current (if there is a current) flows through a resistance, it will cause a voltage across the resistance.
     
    Last edited: Mar 23, 2012
  3. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    The current is ZERO right before the switch closes. At the instant after it closes (t=0+) the current must also be zero since an inductor will NEVER allow an instantaneous change in current.
     
  4. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    After reading your reply, I think I can understand it now, just like you said, at the instant after it closes(t=0+dt), the current remain zero as the inductor will oppose the current change at that instant.

    but when J1 switch off, physically disconnect the circuit (no loop), the current have to drop to zero at that instant, hence a huge di/dt, an huge back emf is produced, is that correct?
     
  5. panic mode

    Senior Member

    Oct 10, 2011
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    if at time t you have current I through inductor, then just after you turn off the switch, current will still be the same and continue in the same direction. if the circuit is open, (large impedance), you will get voltage spike.

    this is because inductor MUST release it's energy (cannot store it long term like capacitors). result is that voltage will climb until the current can continue at same level - even if it means reaching levels sufficient to ionize air.
    ionized mater is conductor and that is when we observe spark.

    the more current was flowing through and the larger inductance, the more energy was stored. spark is just manifestation of that energy being released.

    you can think of inductor being like a flywheel, it takes some effort to get it going - and also to stop it. when you try to spin the flywheel, you spending energy. when the wheel is spinning and you try to stop it, wheel is releasing it's energy which can be used to do some work.
     
  6. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Thanks for your explanation, I can understand it now, and I like your flywheel metaphor :)
     
  7. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    Correct. And electrons will jump the air as the switch contacts open. That's why the points in a car burn over time.
     
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