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Inductor Voltage Drop

  1. Oct 12, 2011 #1
    This is not a homework question.

    Why is the voltage drop on an inductor opposite the electron flow?

    "When the current through an inductor is increased, it drops a voltage opposing the direction of electron flow, acting as a power load. In this condition the inductor is said to be charging, because there is an increasing amount of energy being stored in its magnetic field. Note the polarity of the voltage with regard to the direction of current:"

    http://www.allaboutcircuits.com/vol_1/chpt_15/1.html" [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 12, 2011 #2

    berkeman

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    It's best just to think in terms of the differential equation that relates voltage and current for an inductor:

    [tex]v(t) = L \frac{di(t)}{dt}[/tex]

    As long as you keep that equation in mind, the polarities make sense.
     
    Last edited by a moderator: May 5, 2017
  4. Oct 12, 2011 #3
    I think you have forgotten a minus sign in front of the derivative.

    Qualitatively speaking this happens due to conservation of energy. Since the current is increasing the energy stored in the enclosing magnetic field is increasing too. This energy cant come for free the electron flow has to pay its price against an opposing electirc field which comes from the opposing voltage (thats what the minus sign says actually, that the voltage drop is opposing a change in current. If the current is increasing then the voltage will be such as to try to oppose this increase, hence oppose the electron flow, if the current is dereasing the voltage will be such as to oppose this decreasing hence help the electron flow.
     
    Last edited: Oct 12, 2011
  5. Oct 12, 2011 #4

    berkeman

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    Nope. Why do you say that?
     
  6. Oct 12, 2011 #5

    berkeman

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    EDIT -- BTW, the i(t) is the standard "positive" current, not the flow of electrons. Is that what confused you?
     
  7. Oct 12, 2011 #6
    But if the current say is increasing, hence the derivative of current is positive, the voltage has to oppose this increase hence to be negative. If current is decreasing the derivative will be negative and the voltage has to oppose this decrease hence be positive.
     
  8. Oct 12, 2011 #7

    berkeman

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    This is why I said to use the differential equation as your guide, so folks don't get confused trying to talk their way through it.

    http://en.wikipedia.org/wiki/Inductor

    The easiest way to think about it is that the voltage across the inductor affects the current through the inductor, according to the DE above.
     
  9. Oct 12, 2011 #8
    Ok so the minus sign should be there if we were taking i(t) to be the flow of electrons? (and not the conventional current which we consider opposite to the flow of electrons). Seems somehow i got confused by Lentz Law. Is Lentz Law true only for the flow of electrons and not for the conventional positive current?
     
  10. Oct 12, 2011 #9

    berkeman

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    Lentz' Law doesn't involve current explicitly. The two equations are consistent.

    [tex]EMF = -N \frac{d\Phi}{dt}[/tex]

    [tex]v(t) = L \frac{di(t)}{dt}[/tex]
     
  11. Oct 12, 2011 #10
    To derive the formula [tex]v(t) = L \frac{di(t)}{dt}[/tex] we start from [tex]EMF = N \frac{d\Phi}{dt}(1)[/tex] and also we use Ampere's law to express [tex]\frac{d\Phi}{dt}[/tex] as a function of [tex]\frac{di(t)}{dt}[/tex]. Why do we neglect the minus sign in (1) in this derivation?( Or the minus sign goes away by saying that the electron flow is minus (opposite) conventional current??)
     
  12. Oct 12, 2011 #11
    Suppose a circuit has a total resistance R and a self inductance L.

    At some instant t=0 a switch is closed connecting the circuit across a battery of constant emf, E, but of negligable internal resistance.

    The (conventional ) current in the circuit before the switch is closed is zero.

    Some time afterwards it must be E/R by ohms law.

    To study the growth of the current use Kirchoffs law

    "The sum of the products of the current and resistance taken around a closed circuit is equal to the sum of the electromotive forces round that same circuit, taken with their proper sign."

    (Note this version of Kirchoffs law will not fail you when considering inductance. The guys at MIT love showing how the normal equation set to zero fails)

    If the flux linkage is [itex]\phi [/itex] in the positive direction

    then the induced emf is [itex] - \frac{{d\phi }}{{dt}}[/itex]

    So by Kirchoffs law

    [tex]E - \frac{{d\phi }}{{dt}} = Ri[/tex]

    In this example the flux linkage is cause only by the current in the circuit itself so

    [tex]\phi = Li[/tex]

    Thus

    [tex]L\frac{{di}}{{dt}} + Ri = E[/tex]

    does this help?
     
    Last edited: Oct 13, 2011
  13. Oct 13, 2011 #12
    Ok thanks Studiot that was enlightning. However the version of Kirchoff's 2nd law you apply to derive this is equivalent to the standard version ("the algebraic sum of voltage drops around a closed loop equals zero")??? This version doesnt seem to work if there are capacitors in the circuit.
     
  14. Oct 13, 2011 #13
    You can indeed do the same thing with a capacitive circuit.

    For a circuit with a single resistance and capacitance (and switch as before) applying Kirchoff yields

    [tex]E = Ri + \frac{1}{C}\int {idt} [/tex]

    Notice I have not included inductance and capacitance. To see why look at the capacitance equation, which contains an integral and the inductance one which contains a derivative and ask yourself how you would solve the combination.

    [tex]E = Ri + \frac{1}{C}\int {idt} + L\frac{{di}}{{dt}}[/tex]

    This is why the complexor equations were introduced

    E =RI + jωLI for inductance

    and

    E = RI - jI/ωC for capacitance

    Now you can readily combine these.

    go well
     
  15. Oct 13, 2011 #14

    sophiecentaur

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    It isn't. It is opposite to the CHANGE in current.
    Forget about the 'electron' bit - it only confuses people - with the possibility of double negatives and a pointless complication using particles. You will notice that all the replies have used current and not electron flow - for a very good reason.
     
  16. Oct 13, 2011 #15

    sophiecentaur

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    The confusion with the sign of the induced voltage and the fact that it is a 'back emf' would be resolved if someone had actually drawn a diagram, showing the Inductor in a simple circuit with some defined 'directions' on it. This induced voltage acts to 'prevent a change'. That 'change' must be the result of some external applied PD.
    I think it may be asking too much for someone- a beginner, who may already be confused, to keep to the rigorous definition of induced emf in isolation. Then those damned electrons come into it and make matters even worse. I am SOOOO glad they never messed with electrons when I was taught the basics of electrical theory.
     
  17. Oct 13, 2011 #16

    berkeman

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    :biggrin:
     
  18. Oct 13, 2011 #17

    sophiecentaur

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    You've been there too?
     
  19. Oct 13, 2011 #18

    berkeman

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    [hijack]

    Even worse for me was the concept of electrons and holes in semiconductor physics. I was okay until they started saying that the "holes" were real things with positive charge. Ack. :tongue2:

    [/hijack]
     
  20. Oct 13, 2011 #19

    sophiecentaur

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    Those little sods actually behave as if they were really there! We've got a lot to learn, I think.
     
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