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

[kmarinas86]

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?

 

[diazona]

As far as I know, it's simply because no physical device has ever been observed that produces a voltage drop proportional to $\mathrm{d}^3 Q/\mathrm{d}t^3$ (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.

 

[astrokreng]

I would like to clarify that there is no such thing as a "fourth" voltage drop. The concept of voltage drops is based on the three fundamental properties of a circuit - resistance, capacitance, and inductance. These properties determine how a circuit responds to changes in current and voltage.

The equation for voltage drop is V=IR, where V is voltage, I is current, and R is resistance. This equation holds true for all types of voltage drops, whether it is capacitive, resistive, or inductive.

In the case of capacitive voltage drop, the voltage drop is directly proportional to the charge (Q) stored in the capacitor. In resistive voltage drop, the voltage drop is directly proportional to the change in charge over time (dQ/dt). In inductive voltage drop, the voltage drop is directly proportional to the change in the rate of change of charge over time (d(dQ/dt)).

There is no need for a "fourth" voltage drop as the three existing types of voltage drops cover all possible scenarios in a circuit. The concept of "opposing electric fields" acting against acceleration is already accounted for in the equations for inductive voltage drops.

In conclusion, there is no need for a "fourth" voltage drop, and the theory behind this is based on the fundamental properties of a circuit and the equations that govern them. Any additional voltage drops would be redundant and not based on scientific principles.