Stupid question of the day....

AllTel - Jim Hubbard · 2005-07-30T03:15:20+00:00

I am curious about what would happen to an electrical current in 2 situations.....Assume that you have 2 wires that, when joined, complete a closed electrical DC circuit with electrons flowing thusly.....------------ ============eeeeeeeeee eeeeeeeeeeeeeee------------ ============If you flattened out the end of each wire where they connect , would the resulting electron paths be more like figure A or Figure B?Figure A --- === --- === --- ===--- ===eeeeeeeee eeeeeeeeeeeeeeee--- === --- === --- === --- ===Figure B --- e e === --- eee eeeeee === --- eeeee eeeeeeeeee ===--- eeeeeee eeeeeeeeeeee ===eeeeeeeeeee eeeeeeeeeeeeeeeee--- eeeeeee eeeeeeeeeeeee === --- eeeee eeeeeeeee === --- eee eeeee === --- e e ===(Please note that the vast # of "e"lectrons shown in Figure B is simply to show the path's of electrons. )The second portion of my question is....If the flattened portions were increases in mass (if each wire were connected to a metal cube and the cubes were brought together to complete the circuit) how would it effect electron flow where the cubes touch?Thanks for your help.

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TokaMundo 21 years ago

On Tue, 02 Aug 2005 22:01:40 GMT, "daestrom" Gave us:

Since a thin round copper wire has a very low emissivity it wont give up its heat all that fast. This will mean that your thermal gradient won't be as prevalent as you suggest. The proof is when one takes a copper wire and places it across a battery's terminals. Notice how the entire wire turns a nice cherry red quite evenly, all the way up to where it is attached to any form of sinking element.

The current throughout the wire will be even, and it is that current which generates the heat, or more precisely, the resistance to said current flow.

If the wire were giving up its heat real fast, like that of a finned heat sink with air passing over it, I might agree. In the case of bare copper, however, the temperature throughout the wire is going to be very even. Your gradient will be nearly undetectable.

For a very large diameter copper bus, it MIGHT have a slight gradient between the center and the outer surface, but not much. For wire, it is as even as even gets.

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TokaMundo 21 years ago

On Tue, 02 Aug 2005 22:05:54 GMT, "daestrom" Gave us:

Actually, it is an oxide inhibitor, and its function is to seal away oxygen.

It too is conductive.

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TokaMundo 21 years ago

On Tue, 02 Aug 2005 22:58:07 GMT, "Bob Myers" Gave us:

Hey Bobby... f*ck you. Filter me like you have already claimed to do, you petty asswipe.

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Repeating Rifle 21 years ago

Some people say that there is no such thing as a stupid question. Obviously there seems to be no shortage of stupid answers.

Bill

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TokaMundo 21 years ago

On Tue, 2 Aug 2005 22:58:33 +0000 (UTC), snipped-for-privacy@manx.misty.com (Don Klipstein) Gave us:

The table I saw shows the AC and DC resistance as being exactly the same for both.

Your flaw is where you failed to note the topic given in the CRC handbook.

60 Hz is NOT high frequency... at all.

Try some calculations at 100 kHz and you'll see that those frequencies down near zero (ie 60Hz) yield very nearly nil difference.

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TokaMundo 21 years ago

On Tue, 02 Aug 2005 23:05:04 GMT, "Bob Myers" Gave us:

As if declaring that someone has "no practical knowledge" isn't a personal attack.

F*ck off retard. You have social problems.

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John Popelish 21 years ago

The Navy also deals with lots of salt water. ;-)

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John Larkin 21 years ago

There are no stupid questions, only stupid people.

John

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Don Klipstein 21 years ago

Maybe your table rounds? Please cite source as well as I did.

I did note this, as you quoted above.

The formulas are functions of wire diameter, wire resistivity and frequency, and do not lose validity merely because a thick wire has AC resistance greater than DC resistance at a frequency that is easy to label "NOT high".

And as you asked... Ratio of AC resistance to DC resistance of 17 mm diameter copper wire at 100 KHz is about 21.5. This does not invalidate the calculation for 60 Hz.

You would have been better off claiming that resistance 2% higher at 60 Hz than at DC is a negligible increase.

- Don Klipstein ( snipped-for-privacy@misty.com)

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Autymn D. C. 21 years ago

I'm not a son, you illiterate and uncouth obscurantist troll. I already used reason in the proof, which you wilfully ignore. It uses maths too, which puts a limiting case on the Lorentz corrections, remembering that GR and QM are incompatible. It looks like my other replier didn't read it too or he would be forced to agree with me. The people here are retards.

-Aut

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Autymn D. C. 21 years ago

There are stupid questions, those that could be easily found on one's own.

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Autymn D. C. 21 years ago

You can't prove anything. You're wrong in everything.

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TokaMundo 21 years ago

On Tue, 02 Aug 2005 22:12:37 -0400, John Popelish Gave us:

Exactly! :-]

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TokaMundo 21 years ago

On Tue, 02 Aug 2005 19:50:04 -0700, John Larkin Gave us:

Sometimes I find it hard to believe that we actually call ourselves sentient beings with the way some of you act.

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TokaMundo 21 years ago

On Wed, 3 Aug 2005 04:02:04 +0000 (UTC), snipped-for-privacy@manx.misty.com (Don Klipstein) Gave us:

The whole point is that it does NOT list the resistances as being different at that frequency, for that wire, AND it had several digits shown after the last significant figure given.

That and pumping the numbers through the formula myself with a 20+ mantissa calculator show the same result.

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