Calculating heat loss and conduction through a wire

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Discussion Overview

The discussion revolves around calculating heat loss and current conduction through copper wires in a project involving melting solder paste with a 9V battery. Participants explore the implications of wire resistance, battery internal resistance, and practical limitations in circuit design.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant is attempting to determine the current through copper wires when using a 9V battery to melt solder paste, expressing uncertainty about heat losses and conduction.
  • Another participant suggests calculating wire resistance and using Ohm's law to find the current, providing links for further reading.
  • A participant questions the realism of a calculated current over 1000 A due to the low resistivity of solder paste, implying that practical losses must be considered.
  • It is noted that the internal resistance of the battery is the highest resistance in the circuit, which limits the current to a few amps, especially for small 9V batteries.
  • Participants discuss the short circuit current of a 9V battery, which is stated to be around 5A, with an internal resistance between 1.7Ω to 2.8Ω.
  • There is a clarification about whether to consider the resistance of the battery versus the solder paste or wire, with an emphasis on the battery's internal resistance being the main limiting factor for a 9V battery.

Areas of Agreement / Disagreement

Participants generally agree that the internal resistance of the battery is a significant factor in determining current, but there is uncertainty regarding the impact of solder paste and wire resistance in practical scenarios.

Contextual Notes

Participants have not fully resolved the implications of heat loss and the actual resistance values in the circuit, leaving some assumptions about the solder paste and wire resistance unaddressed.

Who May Find This Useful

This discussion may be useful for individuals interested in electrical circuits, particularly those exploring the practical aspects of current flow and resistance in projects involving soldering and battery use.

electricnov
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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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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?
 
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?
 
electricnov said:
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.
 

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