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Sign of inductive voltage?

  1. Mar 7, 2007 #1
    Hi,
    Let L be an inductance, then you find the following formula in textbooks (high school level):

    U_ind = - L dI/dt

    When you actually do calculations for circuits, you see that the minus sign is wrong.

    So why do they put it there?
     
  2. jcsd
  3. Mar 7, 2007 #2

    marcusl

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    The minus sign is certainly not wrong, but it can be tricky to sometimes to draw the circuit voltages and currents properly. You can always figure out the sign from Lenz's Law as follows: imagine that the induced emf drives a current through the coil and surrounding circuit. The virtual field produced in the coil by that current must oppose the field that actually exists, telling you if you got the emf in the right direction.
     
  4. Mar 7, 2007 #3
    I think it is definitely wrong. Yes marcusl, the field inside the coil opposes that in the source. But you could as well say that of a resistor. But I have never seen a minus sign in R = U/I.
     
  5. Mar 7, 2007 #4

    marcusl

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    Edit: removed this post, not relevant to the solution. (Sorry!)
     
    Last edited: Mar 7, 2007
  6. Mar 7, 2007 #5

    berkeman

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    The minus sign just depends on how the circuit is drawn. The way I figure out inductive kickback waveforms is just from remembering how a flyback circuit works, or how a boost DC-DC works, or how a low-side relay drive transistor behaves upon opening up.

    In all of these, the top of the inductor is connected to the + power supply rail, and the bottom of the inductor is connected to ground through a transistor operating as a saturated switch. When the transistor is turned on, a current builds up in the inductor (or transformer primary winding). The current may end up being limited by the coil resistance, as in a relay, or it may keep on building until the switch is turned off.

    When you turn off that low-side switch, you get a positive spike in voltage at the bottom of the inductor. That's how I remember the polarity of the L di/dt kickback.
     
  7. Mar 7, 2007 #6
    marcusl,
    sorry, no offence meant! Just trying to understand.
    Well imagine you have L and R in series. You switch on the source. Then the current is
    I(t) = U/R (1 - exp (-R/L t)).
    OK?
    I can only derive this using U_ind = + L dI/dt, with a plus sign instead of minus. But the minus sign is there in the textbook, so my question is why.
     
  8. Mar 7, 2007 #7

    berkeman

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    It's got to be the way the circuit is drawn, or else a typo in the book. If you put a positive voltage across an inductor (+ at top, - at bottom), then you will definitely get a positive downward current building in the inductor per the usual equation

    [tex]V = L \frac{di}{dt}[/tex]
     
  9. Mar 7, 2007 #8

    marcusl

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    The minus sign was chosen by convention. A series circuit is assumed to have voltage sources or emfs, on the one hand, and passive devices on the other. A voltage V is dissipated in a resistance R with the "counter-emf" in R defined in the way you mention.

    For an inductive circuit let's consider an example of a battery, switch, R and L in series. The "counter-emf" in the inductor seems to be in the same direction as in the R, as you point out, BUT it's considered by convention to be a source instead, just like the battery. Since it reduces the battery voltage (as it must to satisfy Lenz's Law), it must carry a negative sign. This negative sign carries through to the solution that you wrote down for the current
    1 - exp (-tR/L)

    There's a very clear discussion in Reitz and Milford, Foundations of EM Theory, p. 246 (1960), if you can find a copy in your U library.
     
  10. Mar 8, 2007 #9
    Thank you marcusl and berkeman.
    There's the negative sign in many standard textbooks and even websites such as wikipedia, so I guess it's not a typo. Rather they treat the coil as a source.
    Still seems strange to me ...
     
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