## Calculating heat loss and conduction through a wire

I'm creating a project where I am attempting to melt solder paste using the current running through two copper wires, in a kind of lap joint. I'm trying to figure out how much current would actually be going through the wires if I use a 9V battery. I'm not sure where to take into account heat losses and the conduction of heat through the wire.
I'm new to circuits and all things electrical, so any help would be great! thanks!
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 First read this http://en.wikipedia.org/wiki/Resistivity http://phet.colorado.edu/sims/resist...a-wire_en.html and find wire resistance, next use Ohm's law to find the current. I = V/R
 Thank you for the post. I understand that part. My question is about the current. The resistivity of solder paste is very low and so the current I calculate is very high (over 1000 A). This isn't actually a realistic number right? Is there some kind of loss in practice?

Recognitions:

## Calculating heat loss and conduction through a wire

The highest resistance in your circuit will be the internal resistance of the battery. Small 9V batteries won't deliver more than a few amps of current and only for a short time.

On the other hand a 12V car battery could deliver 500A or more, but DON'T TRY THAT EXPERMENT. You run a serious risk of getting burned, and also the battery may release explosive gas which could be ignited by a spark when you try to disconnect the circuit. (You willl quite likely wreck the car battery as well, but that's a minor problem compared with the safety issues).
 The 9V battery short circuit current is close to 5A at the beginning. The 9V battery internal resistance is between 1.7Ω to 2.8Ω.
 So, in calculating the current, I take into account the resistance of the battery? and not the solder paste or the wire? Is that just because it is negligible in comparison with the battery?

 Quote by electricnov So, in calculating the current, I take into account the resistance of the battery? and not the solder paste or the wire? Is that just because it is negligible in comparison with the battery?
Yes, the battery internal resistance is the main factor limiting the short circuit current.
At least for 9V battery.