Struggling to understand the voltage drop across an inductor

Discussion in 'General Electronics Chat' started by epsilonjon, Apr 1, 2011.

  1. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    Hi.

    I understand that the self-induced emf across an inductor is \epsilon=-L\frac{di}{dt}. Now with reference to the circuit below, assume that the current is increasing. Then the emf across the inductor will be in the direction show, opposing the increase in current. Why then is the voltage across the inductor V_{ab}=L\frac{di}{dt} , i.e. in the other direction?

    I don't understand why V_{ab}\neq\epsilon ?

    If someone could explain it to me I would be most grateful, as i've been trying to understand it all afternoon!! :confused:

    Many thanks,
    Jon.

    [​IMG]
     
    Last edited: Apr 1, 2011
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    The point is that the coil emf acts in the direction around the loop opposing the current i.

    If the generator in your diagram has its positive terminal uppermost, then it is acting to aid the assumed clockwise current direction. The coil emf also has its positive pole uppermost for increasing i, but this is acting in the opposite (anticlockwise) direction around the loop.

    Quite apart from any consideration of electromagnetism, this has to be so to avoid violating conservation of energy.
     
  3. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    But how do you know the coil has its positive pole uppermost for increasing i? I would have thought that if the induced current in the coil was to be in a direction opposite the original current, you would have the positive pole at the bottom of the inductor and the negative pole at the top; then current would flow from bottom to top?

    Sorry if this is obvious and i'm being dumb, I just want to understand it properly.

    Cheers, Jon.
     
  4. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Who said that Vab is not equal to e?

    If you observe carefully, the direction of current is opposite to the direction of voltage. Thus, the minus sign in the equation is correct.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I suspect a lot of these uncertainties arise from matters of convention and terminology.

    If the lower terminal of the inductor were more positive than the top with respect to an induced emf, then this would tend to drive the current out of the lower terminal - thereby assisting the flow of current in the original direction under the influence of the external source. This would not be the case.

    Lenz's law [related to your ε=-Ldi/dt] often seems to confuse rather than aid our understanding with respect to induced voltages. Text books seem to explain the negative sign on the premise that we will understand the significance - namely that the induced emf always opposes the change that is causing the magnetic flux to change in the inductor.

    In relating the commonly used e=Ldi/dt to the alternative ε=-Ldi/dt how can we sensibly reconcile the 'apparent' discrepancy. It's important to firstly note that for the same physical conditions that 'e' and 'ε' have equal absolute values or magnitudes.

    We generally write the induced emf is e=Ldi/dt
    What do we imply when we write this?

    I usually take the meaning as ...

    "If the magnitude of the current is increasing in an inductance then the polarity of the induced emf is such that the terminal into which the [conventional] current flows is at a higher potential than the other inductor terminal (where the current exits).
    Alternatively, if the magnitude of the current is decreasing in an inductance then the polarity of the induced emf is such that the terminal into which the [conventional] current flows is at a lower potential than the other inductor terminal."

    The alternative statement ε=-Ldi/dt would be equally valid if the above statement were re-worded as ....

    "If the magnitude of the current is increasing in an inductance then the polarity of the induced emf is such that the terminal out of which the [conventional] current flows is at a lower potential than the other inductor terminal (where the current enters).
    Alternatively, if the magnitude of the current is decreasing in an inductance then the polarity of the induced emf is such that the terminal out of which the [conventional] current flows is at a higher potential than the other inductor terminal."

    So it's a matter of convention as to what we assume the written equation actually means when we relate that equation to a physical 'reality'.
     
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