1. Nov 30, 2012

### V0ODO0CH1LD

I was wondering about some basic concepts of electromagnetism.

Firstly, resistance is kind of like the measure of obstruction in a conducting material; right? So maybe the atomic properties of the conducting material are such that the current of electrons gets obstructed and ends up colliding with stuff in the way and as a result it generates all sorts of forms of energy. Is that right?

Something else I was wondering is how does the dimensions of a conducting wire relate to how much heat it can dissipate. Before it starts melting, of course. I tried searching for it but I didn't have any luck. I found an article on joule heating which talks about the amount of heat generated by a certain current going through a conductor with a certain resistivity (P = RI2). I don't even know if what I understood from the article is correct.

Anyway, if I double the external area of a piece of conducting wire. Will it be able to dissipate twice the amount of heat?

2. Nov 30, 2012

### Staff: Mentor

Right.

Usually, heat is dissipated via the surface of a resistor, and it depends on the surface (is it in air? maybe even air flow? Water-cooled? ...).

That is the power which heats the resistor.

Maybe. Depends on the environment.

3. Nov 30, 2012

### V0ODO0CH1LD

A conducting wire will melt when the heat generated by the current is bigger than the heat that can be dissipated by the surface of the wire, right? At least it makes sense..

My question is if something like a 1A current is the least necessary so that a wire of a certain material will start to melt when it goes through it. What would be the least amount of current necessary to melt a wire of twice the surface area? Twice the current?

4. Nov 30, 2012

### haruspex

You'd want twice the power per unit length, but if it has twice the surface then presumably it has 4 times the cross section, so 4 times the conductance. P = I2R ~ I2/r2; As ~ r; so P~As implies I2~r3.

5. Dec 1, 2012

### V0ODO0CH1LD

So if I make the wire twice as long I could have twice the amount of power going through it. But what is "r" in your post? Are you saying that I2R is proportional to I2/r2? And that the surface area is proportional to r? How do you get that I2 is proportional to r3? And is r the radius of the wire?

6. Dec 1, 2012

### Staff: Mentor

If you want to use a certain current to melt a wire, use a thin wire. Its resistance will help to generate more heat, and it has a smaller surface at the same time.

r is the radius of the wire.
I2 ~ r3 means that you need to increase current by a factor of 8 if you increase the radius by a factor of 4 (as 82 = 43).