Holy Moly -- Residual Electricity????

Is there some such concept as "residual electricity"???
There was a problem with a computer at work...it's a new Dell running
WinXP Pro...everything's fine, I go to lunch and come back to a blank screen that won't wake up from power-save/sleep mode!! I do the obvious and check connections, making sure they're secure and whatnot. I turn off the computer and turn it back on a few times, to no effect!
Tech Support suspects some kind of a "power management" issue -- whatever that is -- and suggests that I leave the system off for a few minutes, literally, to let things "clear"...whatever that means. Sure enough, however: it works!
So now I'm here asking, because Tech hasn't the time to puzzle over it with me, WHAT HAPPENED??? And how come shutting off power for a few seconds isn't comparable to leaving power off for a few minutes????????????
TIA!
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Capacitors.
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Jim Pennino

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capacitors usually discharge when the current is switched off - their main job is to smooth current, not to store volts....
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They discharge, but it can take some time depending on the capacitors. A few minutes is not at all unusual.
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| Jupiter Jones wrote: | They discharge, but it can take some time depending on the capacitors. | A few minutes is not at all unusual.
Yes, indeedy. I have a Heathkit AR-1500 (amp receiver) that, when powered off playing music, it can take about 15 ---> 20 seconds before it stops playing, and this is with decent bass. I usually get up the stairs and into the living room before the sound cuts out. The capacitors are one of the biggest put into home entertainment electrons. Also, the transformers are huge. ______________________________________________________Gerard S.
|> Gordon wrote: |> capacitors usually discharge when the current is switched off - |> their main job is to smooth current, not to store volts....
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On Sat, 17 Nov 2007 09:59:55 -0800 (PST), Saucerhunter

Or a huge, charged pulse cap.
Wouldn't "reducing voltage fluctuations" also cause the current to be "smoothed" as well? Simple Ohm's law would say yes. If voltage peaks or "fluctuations" are reduced, then so too would be the instantaneous currents associated with the circuit that said "reduced fluctuation" voltages are passing though.
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In sci.physics GoldIntermetallicEmbrittlement

Yes, but you are confusing cause and effect.
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On Mon, 19 Nov 2007 02:45:02 GMT, snipped-for-privacy@specsol.spam.sux.com wrote:

No. YOU are trivial, bitch.
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Yet another foul mouthed little snot...
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snipped-for-privacy@youdontknowjack.org says...

No. The reason the voltage is fluctuating is because the current is fluctuating. The capacitor supplies the "surge" current when the load demands it (but the supply, can't). It effectively lowers the impedance of the supply but doesn't reduce current fluctuations, Indeed may increase them.
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Keith

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----------------------------
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An easy proof is a low power HV supply. One designs them ONLY with enough stored energy to do the customer spec.
We use the same supply for a high precision, 0.00006% ripple X-Ray supply as we do for the higher ripple, higher power unit. The only difference is that the high precision unit has an RC network on the output, which makes for very low ripple, but cannot handle much loading over a couple uA, yet the same supply with "storage caps" across the output will push several uA, but have a slightly greater ripple.
One is to make for a very clean X-Ray flux emission where the cleaner the DC rail is, the cleaner the electron beam striking the target is, and the cleaner the X-Ray flux will be, and, and the other is for pushing a Geiger counter where a fair deal of ripple in the output rail is perfectly tolerable for proper , precise circuit operation. Same supply, different output stage, different results.
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Not at all. Think about a load like a microprocessor. The current of the load may fluctuate a couple of orders of magnitude. The current varies all over the place but the voltage can't.
Of course the capacitor may increase the current in the rectifier considerably, as well.
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Keith

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On Mon, 05 Nov 2007 12:23:37 -0800, Prisoner at War
There is a such thing as an overtly crosspoting Usenet retard!
You qualify!
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In sci.physics snipped-for-privacy@gmail.com wrote:

And the charge on a capacitor is measured in...
Yep, volts.
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Coulombs actually.
Q(charge)=CV
The product of capacitance and voltage
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Stuart Winsor

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Where do you buy a coulombmeter to MEASURE the charge on a capacitor?
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On Sun, 18 Nov 2007 17:25:03 GMT, snipped-for-privacy@specsol.spam.sux.com wrote:

Read voltage, read stated capacitance, apply formula, achieve result.
Get clue.
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This is a measurment.

These are not.

Yes, do and learn to read the words that are written and not the words you think are there.
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Jim Pennino

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On Sun, 18 Nov 2007 23:55:02 GMT, snipped-for-privacy@specsol.spam.sux.com wrote:

Read voltage, read capacitor, apply formula... That IS a coulombmeter, idiot.
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