Capacitor decay?

Question

I think that capacitor decay is a random process since it is very similar to radioactive decay but one person is trying to prove me that it is not I can’t find an answer in search engine may be you know it?

Answer 1

At a microscopic level, of course, it is a random process just like everything else is. At the macroscopic level, though, you can treat the draining of the capacitor (or a lump of radioactive stuff decaying) as determined by set laws of physics.

The similarity of the draining capacitor to radioactive decay you notice is because the charge on the capacitor and the amount of undecayed material in a radioactive source obey the same differential equation, so their solutions have the same form. I wouldn't read too much physical significance into it.

Good question though. Don't let some jerk tell you otherwise.

Answer 2

Capacitor decay and radioactive decay are two quite different processes which happen to obey the same equation. However, radioactive decay is a random process, at the atomic level, and capacitor decay is not.

Just because the description of two things is similar (eg same equation) doesn't mean the mechanism is the same.

Answer 3

the first person who answered this question is a ****, as of course radioactive decay is a random process. i think u could well argue that all decays are a random process as there is not property that would favour one molecule for example to "decay" before another

Answer 4

Capacitive decay is not random in the sense that it can be plotted and it follows an exponential curve.

The actual orientation of the curve is determined by whether you are measuring the charge decay in coulombs, voltage decay in volts or current decay in amps.

What's wrong with going to Wiki?

Answer 5

OK RC series circuit. i=V/R ohms law (i=i(t))
C=Q/V defn of capacitance
i= Q' defn of current (decreasing - sign)
>Q'=-Q/RC or solving,Q= Qoexp(-t/RC) (Qo=initial charge) Integral of 1/Q= LnQ so LnQ=-t/RC +const etc
For random radioative decay obv decay rate depends on population of undecayed atoms
N'=-kN Solving N=Noexp(-kt) (No= initial atom pop) and int 1/N =LnN as b4
It's just it's the same mathematical model for both processes..

Answer 6

Show me your evidence for random discharge.

Its very predictable - get a voltmeter and a cap and record it accurately. a few 10,000uF at say 20V to start with. check every 10 secs. Plot it. Try with various large bleed resistors.

Answer 7

It decays exponentially, and it very predictable:

(The link is to an experiment that will demonstrate this)

Answer 8

Radioactive decay is not random either.
Learn what you are taught and don't question it till you know what you're talking about.