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I won't to avoid mentioning Fourier or Laplace. I would like a simple way of deriving this. Perhaps, i can say something about the energy at higher frequencies being attenuated and relate to Coulombs law and electric feild. Any suggestions?

2007-11-28 05:07:19 · 7 answers · asked by Paine 6 in Science & Mathematics Engineering

Thank you all. My question stems from precisely the point made by one of the contributors. All to often books gloss over the concept and just state it by definition. I was hoping to avoid this and make it as easy as possible for the students. The students should be fine with differentiation. I suppose a mixture of complex numbers and differentiation are the best way forwards - it is easy to demonstrate the phase change this way... The increased impedance at higher frequency is not necessarily so intuitive to students though. A Maxwellian approach would be above the level of the students and Ansatz function is a new one to me, sounds intresting (a function based on no underlying principles i read at Mathworld)!

2007-11-28 07:18:44 · update #1

7 answers

Generally, the 'beginners' books give the impedance of a capacitor as simply: 1/(2*pi*f*C), with a phase angle of -90 degrees, and the reader is asked to accept that without question.

To get into, "why is it that way", or the derivision of the impedance, requires the same knowledge Fourier or Laplace had, that is of calculus and complex variables.

There are plenty of decent books that deal with complex impedances without the use of Laplace transforms. Schaums Outline Series on Electric Circuits (Joseph Edminster) does a great job. It is old, but some version of it should still be in print. In this book, Laplace and Fourier transforms are not even mentioned until the last 2 chapters.

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2007-11-28 05:34:42 · answer #1 · answered by tlbs101 7 · 0 0

What level are you teaching to?

The analogy I've seen which I think makes the most sense, is that if electiricity was water in a pipe, a capacitor would be a light and somewhat flexible but unbreakable membrane crossing the pipe. A continuous flow of water (i.e. DC current) gets stopped by the membrane, while fast ripples in the water (i.e. high-frequency AC current) barely even notice the membrane since it just tries to push it back and forth quickly instead of going through.

For a real capacitor, DC is stopped because fundamentally there is no conductive path through the capacitor. But high-frequency AC can get through since it just causes an oscillating charge on one side, which causes an oscillating electric field in the capacitor, which cases the electrons on the other side to move.

2007-11-28 05:21:25 · answer #2 · answered by Michael T 4 · 1 0

Fourier transformations are not needed to explain a capacitor. You have to start with Maxwell's equations (the part that you need for a simple plate capacitor) and work out the differential equation for the current and voltage on the capacitor. Then you solve this equation with a complex Ansatz (complex exponential function) and you are done.

Of course, this assumes your students know how to deal with complex numbers. If they don't, this is a bad place to start explaining them.

2007-11-28 05:45:05 · answer #3 · answered by Anonymous · 1 0

In principle, you can't, by the way Coulomb's Law
is stated. Since it's an isolated charge system.
In practice, the simplest way is show that
the complexness is a property of the dialectric,
rather than the capacitor.

2007-11-28 06:29:25 · answer #4 · answered by Anonymous · 1 0

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2016-12-30 05:58:09 · answer #5 · answered by barksdale 4 · 0 0

To show impedance and how it changes over frequency you can use the equation for Xc and show examples of how as Xc approaches 0 and as Xc approaches infinity. Of course there is also the who phase angle subject to approach, which is probably best described with Trig.

2007-11-28 05:28:35 · answer #6 · answered by EE dude 5 · 1 0

graphically, compare the voltage and current through a Resistor an Inductor and a capacitor for sinusoidal excitation.

Then use the following

V= I R

I = C dV/dt

V = L dI/dt

I'm assuming that your pupils are familiar with differentiating sines and cosines.

2007-11-28 05:29:57 · answer #7 · answered by frothuk 4 · 1 0

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