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There are voltage drops based on Q, dQ/dt, and d(dQ/dt); what about d(d(dQ/dt))?

  1. Jun 28, 2009 #1
    Capacitive voltage drops are based on Q.
    Resistive voltage drops are based on dQ/dt.
    Inductive voltage drops are based on d(dQ/dt).

    However, wouldn't there be a logical progression from here? Isn't there a voltage drop for d(d(dQ/dt)), that is to say, some kind of property of the circuit that resists alterations in the acceleration of charge?

    Think of it this way:

    Capacitance is to position, in the same way that
    Resistance is to motion, in the same way that
    Inductance is to acceleration, in the same way that
    ??? is to changes in acceleration

    This can be likened to how the various results one may get from different vehicles depending on how hard and swifly one presses the power pedal. Alternatively, it can looked upon this way:

    1) Some cars will build their acceleration almost instantaneously (e.g. electric cars).
    2) Others will take more time to get that same acceleration (e.g. equivalent gasoline-powered cars).

    I'd imagine that if a charge is prevented from accelerating, there must be something to act against that acceleration. But this can go both ways to where if a charge is prevented from additional deceleration, then something may act against that deceleration. It would be as though a opposing field were generated simultaneously (but not equally matching) upon the application of a field of opposite polarity. So would the various conditions that determine the rate change and the acceleration of rate change of the presence of these "opposing electric fields" be parameters underlying "other" voltage drops unique from V_R, V_L, and V_C? The fourth one, in particular, would analogous to "mechanical impedance", where a smooth application of torque on a motor may result not in long term angular acceleration but simply an accumulated drop in angular velocity (while the introduction of that torque would be what is necessary to cause that angular acceleration). However, would such voltage drops be separate from resistive, inductive, and capacitive voltage drops, or are those three categories of voltage drops already inclusive of these considerations?

    If a "fourth" voltage drop such as this doesn't exist, what is the theory behind that conclusion?
    Last edited: Jun 28, 2009
  2. jcsd
  3. Jun 28, 2009 #2


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    As far as I know, it's simply because no physical device has ever been observed that produces a voltage drop proportional to [itex]\mathrm{d}^3 Q/\mathrm{d}t^3[/itex] (or any higher derivative). I can't think of any physical reason why one couldn't exist. But I'm not an expert on circuit components.
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