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circuit breaker power overflow

phileaton

Oct 11, 2010
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When a break occurs in a conductive material, the electricity can't get past. A common analogy to this is when a water pipe is missing a section, the water doesn't continue on, it falls out of the pipe circuit. What I was wondering was, if like in the water energy, depending on the water pressure and the size of the gap, some water would still make it past the gap, is this the same for an electrical circuit? With the variables being the electrical pressure, Amps I assume, and the physical size of the breaker? Is this a visible effect? Thanks!
 

Militoy

Aug 24, 2010
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Yes - the analogy holds up - and it is a visible effect. If the electrical pressure (voltage) is high enough, it will jump the gap. The visible effect is called an arc.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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There are other ways too.

You can convert the electrical energy to another form (electromagnetic radiation -- e.g. light), and convert that back to electrical energy at another point.

In your analogy, it's a bit like boiling water at one point and capturing the steam somewhere else.

And just like you could see that steam would probably escape, the same thing happens when you try to do it with electricity. The amount you capture is a fraction of what was sent. That's not a problem if you're not distributing power (or, by analogy, water). It works fine for radio and television for example.
 

phileaton

Oct 11, 2010
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That's really cool. Thanks! Is there a maximum range that the energy can be successfully transferred, and is there a method of finding the maximum range? (i.e. a formula) Also, what variables are there to increase the range? Power output I assume and a better way to guide with less interference and a good receiver?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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How far away is the furthest star you can see in the sky? The range of electromagnetic radiation is essentially limited only by the sensitivity of your receiver.

The most common mathematical relationship is the inverse square law. If you double the distance between the source and the target, the intensity reduces by a factor of 4 (i.e. relative intensity = 1/(relative distance)^2).

As for an arc -- have you seen lightning? An arc can be pretty long too.
 
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