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You'll have to forgive me if this question is either extremely elementary or does not make sense — I am a civil engineer with little E/M physics background beyond high school education.

I am working on a project that involves the harvesting of energy from what is effectively a huge Faraday flashlight or linear alternator. (A Faraday flashlight is an application of the linear alternator, correct?) When the apparatus is shaken (for a finite period of time), a magnet moves through a coil, inducing an e.m.f. in the coil. The magnitude of the induced e.m.f. in the coil, as a function of coil geometry, magnet strength, and magnet position, can be found in this paper (equation 20).

I'd like to quantize the maximum amount of electrical energy produced in this process that can be stored in a capacitor or battery bank. As far as capacitors are concerned, using E=0.5CV^2 would be nice, but two issues present themselves:

  1. There is no maximum value of C, implying that an infinite amount of energy could be stored in the capacitors from this finite process, something which is a clear violation of the Law of Conservation of Energy.
  2. As the shaking is arbitrary, so too is the voltage created in the coil through Faraday's Law of Induction. Since V changes with time, I have no idea what value to use for it.

Considering this, how would one go about measuring the amount of energy stored in a capacitor fed by an arbitrary voltage source?

I have no education with regard to batteries, but I understand that they operate in a fundamentally different way than capacitors and do not have an "equivalent capacitance" or anything like that. I also understand that they are generally more appropriate for large-scale energy storage. What would be the best way to measure the amount of energy stored in a battery bank fed by an arbitrary voltage source?

EDIT

I understand the solution to capacitor issue #1. Since capacitors have a charge time equal to about 5*tau=5*R*C (that gets it to about 98% charge), the amount energy that builds up in the capacitor is dependent on the amount of time the voltage source supplies a voltage. I haven't done the math, I would guess that the amount of energy that has built up in the capacitor over a given time will never exceed the amount of energy expended by the voltage source over that same time.

I'm still not sure what to do about issue #2, though. Do the same rules surrounding time constants apply when the voltage is rapidly changing? That is to say, if the voltage supplied by the voltage source momentarily exceeds the voltage built up across the capacitor, do the same rules tau=R*C and Vc=Vs(1-exp(t/tau)) apply? Could I use a diode to avoid capacitor discharge if the voltage supplied by the voltage source is lower than the voltage built up across the capacitor?

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  • $\begingroup$ -1 Not clear what you are asking. Are you asking us to calculate the amount of energy which your apparatus would produce? And why is there no maximum value of C? Don't you set the value of C by the number, size and configuration of capacitors which you use? $\endgroup$ – sammy gerbil Jan 5 '18 at 15:19
  • $\begingroup$ In a nutshell, yes, that's what I'm asking. The reason why C has no maximum value is because (in theory) you could have a capacitor with infinite capacitance, thereby making the harvested energy infinity. $\endgroup$ – Michael Aronson Jan 5 '18 at 19:04
  • $\begingroup$ If (in practice) you use a capacitor with finite capacitance, then you store a finite amount of energy, and there's no problem using $E=\frac12CV^2$. Doesn't that deal with issue #1? $\endgroup$ – sammy gerbil Jan 5 '18 at 21:51
  • $\begingroup$ The thing is, though, there's no reason why you couldn't choose a capacitor with a capacitance that makes the value of E higher than the energy expended by the voltage source over a given time, violating the Law of Conservation of Energy. I think I've figured it out, though—since capacitors have a charge time, a sufficiently large capacitor won't charge all the way up over the duration of time the voltage source is supplying a voltage. I'll edit my question to reflect this. $\endgroup$ – Michael Aronson Jan 6 '18 at 22:41
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I will phrase it in civil engineering terms.

a capacitor is a reservoir, the voltage across the capacitor is equal to the depth of water behind the dam; the reservoir integrates water flow over time, the capacitor integrates charge flow over time.

the capacitance of the capacitor can be though of as the amount of charge flow required to raise its voltage by a volt; the capacitance of the reservoir is the water flow into it required to raise its level by one foot. a "large" capacitor is one in which the amount of charge flow into it needed to raise its voltage by one volt is large; a "large" reservoir is one in which the amount of water flow into it required to raise its level by one foot is large. the surface area of the reservoir is hence related by analogy to the capacitance rating of the capacitor.

the energy stored in the capacitor e = (1/2) cv^2 where c is the capacitance of the capacitor and v is the voltage. You know how to calculate the energy stored in the reservoir, yes?

Model the shaken coil generator as a pump with a piston and some check valves to prevent backflow. Now you can see that one stroke of the pump is analogous to one current surge from the generator and the voltage rise of the capacitor with pump strokes scales exactly as the amount of water level rise in the reservoir as the pump fills it.

This analogy should allow you to do the algebra required to solve your problem. let us know it if isn't. best regards, niels

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  • $\begingroup$ Thanks, this analogy helps a lot. I'm actually a structural engineer—I don't know a ton about modeling pipe systems, but the idea of a capacitor being a reservoir that takes time to charge/fill up makes a lot of sense. $\endgroup$ – Michael Aronson Jan 6 '18 at 22:52
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Voltmeter..? You could also use a oscilloscope

The maximum energy stored is equal to the voltage in the capacitance formula: C=Q/V, given that the electric field is uniform, V=Ed.. there is a fundamental unit charge so use C~(1.6x10^-19)/V...

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html#c2

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  • $\begingroup$ I'm not looking for the voltage output by the coil—I already know that. I want to know the total amount of energy that can be stored by placing this voltage across a battery or capacitor. $\endgroup$ – Michael Aronson Jan 5 '18 at 2:07
  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html#c2 his might help $\endgroup$ – Randomrok and petcaveman Jan 5 '18 at 2:49
  • $\begingroup$ The maximum energy stored is equal to the voltage in the capacitance formula: C=Q/V, given that the electric field is uniform, V=Ed.. there is a fundamental unit charge so use C~(1.6x10^-19)/V... $\endgroup$ – Randomrok and petcaveman Jan 5 '18 at 2:58

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