Charge Conservation - Hint of the Day

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I heard that before... about burnt rosin. Maybe I'll get out the big soldering gun and burn some....

George

He may be a snarky ol' git, but he's SLOW.

John

I announced yesterday that I had real ($) work.

I'll get there.

Do you still maintain what you said in...

"From: John Larkin Newsgroups: sci.electronics.design Subject: Re: Inverse Marx generator Date: Wed, 07 Jul 2010 08:50:50 -0700 Message-ID: [snip]

Right. If you dump all the energy from one charged cap into another, discharged, cap of a different value, and do it efficiently, charge is not conserved.

John"

is true ?:-)

You can still recant, admit error, and be viewed a gentleman.

However I won't hold my breath.

Of course every expert in the world disagrees with you... why do you think it's called a LAW? ...Jim Thompson

I was expecting an actual answer to this in the recently-revealed paper. I guess I was misled.

You want an answer already? Go to Radio Shack and make them perform to their ad (and if they don't, sue).

Richard Henry proves my point that the "youngsters" here (can't use "young bucks" anymore, John "The Bloviator" Larkin gets sexually aroused :) don't know any fundamentals and can't think for themselves.

All they want is "actual answer(s) ®" :-(

Another "hint": Charge is the time-integral of current.

Not that I think that will help. It's been said multiple times in the thread, though certain obfuscators simply say, "ampere-seconds", to keep it unclear, and make it better fit their loose and hare-brained explanations. ...Jim Thompson

I didn't really expect an answer, especially after reading your "duh!"- level bowl of porridge.

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

A somewhat non-sensical question, designed I suspect to illicit responses that try to answer if definitively so Jim can lambast them for their ignorance.

But I'm going to attempt to give an answer, with some assumptions:

An inductor cannot store net charge. The same current flows in as flows out (as for a capacitor in a circuit, which also cannot sore net charge, for the same reason). Assuming the 1A is steady state there is no voltage across the inductor. We can say the inductor has stored energy, given by (L * I^2)/2.

As an example, we could connect a capacitor across it in parallel, as the voltage across the capacitor rises the current through the circuit falls, until we reach a state where no current is flowing. Assuming no resistive losses, the energy that was in the inductor would then be in the capacitor.

A few definitions: Iinit is the initial current flowing through the circuit, in this case 1A. Q is the absolute value of the charge present on the plates of the capacitor at the time the current reduced to zero (+Q on one plate, -Q on the other).

This means at the point of zero current, (L * Iinit^2)/2 = Q^2/(2 * C), or Q = sqr(C * Init^2 * L) = Iinit * sqr(L*C).

So, the total charge (i.e. the time integral of the current) that flowed around that circuit until the current drops to zero is actually proportional to the square root of the capacitance we placed across the inductor. The inductor has essentially pumped charge around the circuit by voltage given by -L* dI/dt. Note that same amount of charge has come back into the inductor via the circuit.

This is a parallel resonant circuit, the frequency of which is 1/(2 * pi * sqr(L*C)), and will continue to oscillate forever if there are no energy losses.

So, the inductor cannot 'deliver charge' as such, but it can transfer its stored energy by pumping charge around the complete circuit in the same way a water pump pumps water. The total amount of charge it can pump until the current reaches zero is dependent on what is connected to it.

Technically you can place a piece of wire across the inductor and, assuming no resistive losses, the current would continue to flow forever.

Mark.

Malapropism of the day, but a beautiful summation.

Try doing the math, folks. It isn't hard, even if it isn't as much fun as slagging each other off in public, and you learn a lot more per hour.

Cheers

Phil Hobbs

Er, yipe. Solicit!! :-)

It was intended to solicit engineering thought, but it was wasted on the (illicit) illiterate. ;-) ...Jim Thompson

Or elicit? :)

(Not afraid to admit he made me look up exactly what a "malapropism" is, and only saw the error after that)

A double malapropism (if there is such a thing)! In fact elicit is the better one :)

Mark.

Well, ok, but how do you know he doesn't want illicit responses? :-)

Ed

As Phil says, Do the math. I doubt that "clueless-markp" can even come close. ...Jim Thompson

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=A0 =A0 ...Jim Thompson

Break the law, get arrested, jump bail, run to another country that doesn't have an extradition treaty!

Michael