Win Hill: Inverse Marx Generator ??

Sure

Of course not, but Larkins original statement was "eventually", not "after one second". You are the only one to add this condition, obviously if you stop discharging the cap early you will not get all the charge out!

One ampere-second does not have to mean that "one amp flowed for one second", It means an amount of charge *equal to* that transferred by one amp flowing for one second. It could be 2 amps for half a second, 10 amps for 0.1s, 1000 amps for 1 millisecond.

Or an exponential decay.

Well, no. The man said:

"John Larkin" wrote in = message news: snipped-for-privacy@4ax.com...

^ ^ ^ ^ ^ ^ The mathematical definition of "eventually" is, "as t approaches = infinity".

Larkin never stated that the capacitor was to discharge for only one = second. I don't know where you got that from.

Tim

--=20 Deep Friar: a very philosophical monk. Website:

Do they still make CRTs?

John

Priceless! Thanks for the laugh! Isaac Newton had nothing on you.

John

Quoting myself,

"Slap a 1-ohm resistor across the 1F/1v cap and discharge it. You'll get 1 ampere-second out of it eventually."

How did you miss the words "discharge" and "eventually"? I worded the situation as carefully as I could, figuring some whiney dork or another would get pretend-lawyer pickey. Sigh.

How many ampere-seconds would you get if it was discharged by a 10 ohm resistor?

John

Seven miles? Downhill?

John

Obviously.

John

Hilarious! I guess if you put a lighted flashlight in an insulated box, the temperature inside can't go up.

John

"Side" with whomever you like. But both "laws" apply simultaneously. ...Jim Thompson

Am I to suppose that half an amp for two seconds is not a coulomb? Or that an exponentially decaying current can never integrate to one coulomb?

Yup. Energy is conserved in the restacking-the-caps case, but charge isn't. That was my simple point all along. Thanks for the confirmation.

John

The photons entered the game from "outside the box" as someone opined.

As for your "explanation" above... :-(

If I'm so wrong and Larkin is so right, WHY don't Hill and Hobbs jump to his defense? ...Jim Thompson

Right. This being an electronics design group, and specifically a discussion about capacitors on schematics and not electrostatics, I assume that when we talk about the charge on a capacitor, it's zero if there's no voltage across its terminals, and any subsequent charge stored by the cap is equal to the integral of applied current.

Q = CV = integral(I.dt)

where V is the voltage across the cap C in farads, V volts, t seconds

or in the rectangular case,

Q = CV = IT

starting from some initial point of Q=0, V=0.

Q is measured in coulombs, which are dimensionally ampere-seconds.

Given this definition of charge on/in a capacitor, one could further define the "total charge" in a circuit somehow. Carefully.

Does anybody want to argue about that?

John

I think the highest voltage Tek CRT was around 28KV, in the 519. Relativistic effects are still small at 30KV, about 5% on mass, 1.5% on velocity, probably small enough to ignore.

John

sage

Energy is always conserved.

Charge gets moved around. Whether you find it to be "conserved" or not depends on where you look for it.

Both ways...

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Nowhere did I say that charge cannot be removed (via the resistor, or whatever) from a capacitor... you MORONIC OBFUSCATOR!

Certainly! Do you? ...Jim Thompson

-- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at

But they don't carry charge. They do carry energy - so are you saying you can convert energy into "charge"? I agree - this is true for "Q=CV" "charge-as-used-in-electronics", but false for "number of electrons - number of protons" physicists charge.

I have yet to see yours...

Something about wrestling with pigs? :)