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brak 04-15-2009 06:35 PM

AGC fuse ratings
 
Something I always wondered about. Most AGC fuses sold online only show amperage rating, not voltage.
I have a bunch of AGC fuses that say "250V" and then have various amperage ratings and a few 32V fuses.

So, question is - is amperage rating of a fuse dependent on voltage? I would guess it must be (3 amps at 250V would carry a lot more power and require thicker wire than 3 amps at 12V).

If that is true - why doesn't anyone post voltage ratings for AGC fuses online? If it is not true - how do they work?

scottyt 04-15-2009 06:39 PM

no volts dont really matter
a 5 amp fuse will blow at 5 amp at 110 volts or 12 volts. the volt rating you see is a max insulation rating, ie the glass on a 12 volt fuse would not hold up to a 250 volt load shorted out. thats why there are not many ranges, its just to cover them all.

Faster 04-15-2009 06:42 PM

The 250V is the limit where the fuse will not arc across itself (ie voltage over 250 V could get by a blown fuse)

Current is current.... a fuse will blow at its amp rating regardless of how much voltage is involved.

brak 04-15-2009 06:47 PM

Quote:

Originally Posted by Faster (Post 475773)
The 250V is the limit where the fuse will not arc across itself (ie voltage over 250 V could get by a blown fuse)

Current is current.... a fuse will blow at its amp rating regardless of how much voltage is involved.

Hmm. Interesting. I guess thats a good thing - I don't have to throw away my 250V stash. But how does it do that?

sailingdog 04-15-2009 07:21 PM

Brak—

If you think of a circuit as a tube and the electrons as marbles... the amperage is how many marbles pass a given point in a given amount of time... the voltage is how steeply tilted the tube is... 250 volts would be a tube that is at a 80˚ angle from horizontal and the marbles move pretty quickly... 12 VDC would be the same tube at a 20˚ angle from horizontal... the marbles still move, but not as quickly...

If you cut a section out of the tube... that would be a blown fuse. The higher the voltage, the more likely the marbles are to jump the gap. At 20˚ from horizontal a small gap would probably drop any marbles trying to cross. At 80˚ the gap probably has to be a bit larger for the marbles not to jump it. :)

brak 04-15-2009 10:29 PM

Quote:

Originally Posted by sailingdog (Post 475786)
Brak—

If you think of a circuit as a tube and the electrons as marbles... the amperage is how many marbles pass a given point in a given amount of time... the voltage is how steeply tilted the tube is... 250 volts would be a tube that is at a 80˚ angle from horizontal and the marbles move pretty quickly... 12 VDC would be the same tube at a 20˚ angle from horizontal... the marbles still move, but not as quickly...

If you cut a section out of the tube... that would be a blown fuse. The higher the voltage, the more likely the marbles are to jump the gap. At 20˚ from horizontal a small gap would probably drop any marbles trying to cross. At 80˚ the gap probably has to be a bit larger for the marbles not to jump it. :)


:) :) Thanks, I had physics in college all the way up to wave and magnetism (and then some electronics, VLSI and other crap :) ). I have a general idea, though it's been a very long time :)

What I don't get is - the power passing over any given circuit is I*V. So, the same amps (I) and higher voltage (V) would cause more energy to be passing through the fuse and, subsequently, should create more loss in the same resistance, more heat and have it melting faster. Yet apparently it does not matter.

Oh well, may be I should go ask for tuition money back ;)

dacap06 04-16-2009 06:12 AM

Quote:

Originally Posted by brak (Post 475905)
:)
What I don't get is - the power passing over any given circuit is I*V. So, the same amps (I) and higher voltage (V) would cause more energy to be passing through the fuse and, subsequently, should create more loss in the same resistance, more heat and have it melting faster. Yet apparently it does not matter.

Correct. Only current counts. For fuses, the more useful form of the elementary power equation is P = I**2 * R. The resistance of a fuse is very low in comparison to the resistance of other devices in the circuit, so very little of the voltage drop happens across the fuse. However, the current working in the material generates heat in proportion to the square of the current, and fuses are made of a material that melts at a low temperature. If the current is too high, the material melts and the fuse blows.

brak 04-16-2009 08:44 AM

Quote:

Originally Posted by dacap06 (Post 475993)
Correct. Only current counts. For fuses, the more useful form of the elementary power equation is P = I**2 * R. The resistance of a fuse is very low in comparison to the resistance of other devices in the circuit, so very little of the voltage drop happens across the fuse. However, the current working in the material generates heat in proportion to the square of the current, and fuses are made of a material that melts at a low temperature. If the current is too high, the material melts and the fuse blows.


Oh, that's it ;) Current is governed by the rest of the circuit! Thanks - that explains it and makes perfect sense :)


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