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Capacitors Are Never 2nd Order

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Posted by Richard Ruquist on November 12, 2002 13:22:06 UTC

Here is what Ray said:

***If we charge several capacitors in parallel through serial resistors, such that each following capacitor charges at a rate determined by the charge level on the previous one, we'll find that the first capacitor charges on a 1st Order curve, and each succeeding one on a 2nd Order curve, with the succeeding curves becoming progressively more "S" shaped. There is a LAG in the transfer from capacitor to capacitor. ***

This is an easy problem to solve, and although the curve for any given capacitor experiences reduction of decay, the decay rate never increases or as required for 2nd order, reverses. It is equivalent to a heat conduction problem. Temperature never reverses in such problems, as is required for 2nd order. "S" shape is not sufficient for 2nd order.

Not even the charge on each capacitor, going from capacitor to capacitor, has an "S" shape. The exponential curve is maintained from capacitor to capacitor. The reason is that there is no LAG in going from capacitor to capacitor. You need induction to get a lag. If you are talking about an experiment where there is induction, then you get second order. But you never get it from pure first order with one unknown.

The train example can be 2nd order because of elasticity in the car couplings. But this effect cannot be described by first order curves without introducing extra unknowns.

For example, Maxwell's equations, which yield both decay and wave like curves, are written as first order equations, but of two unknown variables, E and B. But your example of a first order equation or curve for the capacitor has only one unknown. Therefore it is a pure first order curve as a function of time. You can never turn such a curve into second order.

yanniru

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