what's wrong with this circuit?

Sorry -- I think I was confusing coulombs and joules here before.

Elsewhere, I agree with your simplified for fried brains version ;)

Grant.

My brain hurts from parsing that paragraph ;) To me it was easier to see the inductor loading up flux from input and automagically voltage adjusting on the flyback part of the cycle into the battery, minus various bits taking a loss.

Grant.

Okay- that's better, this circuit gives Qbatt/Qsolar=3DVsolar/(Vbatt- Vsolar).

So, you admit that all this time, you were insulting people when you use this "term" on them?

You really stepped right on that asshole trap, eh? How's that ankle?

Duh! Obviously. You'd have to be awfully dumb to ask a question like that. But giant rats aren't noted for their IQs anyhow.

John

(See, it's not hard to make up original insults. Or original circuits. No need to copy other peoples'.)

That's the beauty of CoE. If it violates CoE, you don't have to analyze all that math; it's wrong.

John

Correct that to Qbatt/Qsolar=3DVsolar/Vbatt, so twice as good as the ED circuit.

Ah! Okay, I understand why we were confused.

Cheers, John

On the other hand, one must be very careful in applying CoE. Remember the old discharging a C into an equal-valued C and winding up with only half of the original energy?

John

So, rather than me going through all this, please tell me your conclusion.

Thanks, John

Careful? No, just remember to account for all of the 'E'. Remember CNF?

Of course, but WTF is CNF?

Fred -

I'm having a bit of trouble accepting your analysis. Are you analyzing the greatly simplified example I supplied above or are you analyzing the circuit I posted?

Thanks, John

I guess not. ;-) (Cold Nuclear Fusion)

Oh. (snicker) You got me.

Really? So where is this "original insult", idiot? Nothing in your response is original in any way, shape or form.

You are truly stupid.

Somehow, I have serious doubts about you. I'd bet that you have.

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Yes, I remember that one. I think the lost energy is in the resistance of the wires, but if you use larger wires, the current is greater, time is less, and loss is the same. Must be someway around it?

A 1 farad cap charged to 1 volt will be Q=3DCE, or 1 coulomb of charge, and energy of 1/2 CE^2 =3D 1/2 Joule. But when connected to another 1F cap of zero charge, the charges equalize so each cap has 1/2 coulomb of charge. So, from E=3DQ/C, E is only .5/1 =3D 500 millivolts for each cap, and the cap energy is now 0.5 * 1 * 0.5^2 =3D 1/8 joule, or 1/4 joule for both caps combined.. That's a loss of 50%

-Bill

I started using "word salad" here, to refer to your posts. Ditto "old hens" and "clucking", whose targets have taken to echoing.

And I invented Always Wrong, as a play on Massive Prong.

How about Tiny Wombat? Archimedes' Boy-Toy? Ultimate Janitor?

Try it!

John

As I mentioned in another post, the loss comes from radiation in the case of a circuit with zero resistance (a thought experiment, for example).

John