Why do wires melt?

[i.am.lost]

Hi all,

I have been set an experiment of relating to the current needed to melt a wire (predominantly dealing with R = ρL/A). This sounds like a really dumb question (because it is :redface:) but why does the wire melt when too much current is passed through?

Also, what is the correlation between the current that would melt a wire and the wire's resistance? And what the heck is the difference between resistance and resistivity (let me guess, resistivity is a standard measure, whereas resistance is length-dependent??)

Any help would be great. Many thanks you guys, and hope your day is going good.

[dst]

The wire melts due to generated heat (given by P = I^2/R and the heat capacity of the wire) which of course comes from resistance.

Resistivity is just a sort of "resistance density" if you will. In classical physics you have a lot of these "density" type quantities like pressure, density (who'd have thought), etc. It's resistance per unit length, and nothing more - ohms per metre, Ω/m, Ωm^-1, etc, etc. Resistance is length dependant as you say.

A little bit of dimensional analysis goes a long way here.

[kamerling]

Note that the unit of resistivity is Ohm metre Ωm and not Ω/m.

[Google_Spider]

Resistivity is just a sort of "resistance density" if you will. In classical physics you have a lot of these "density" type quantities like pressure, density (who'd have thought), etc. It's resistance per unit length, and nothing more - ohms per metre, Ω/m, Ωm^-1, etc, etc. Resistance is length dependant as you say.



It's resistance per unit length, and nothing more - ohms per metre, Ω/m, Ωm^-1, etc, etc.

Not exactly. :smile:

If the resistivity of a wire 2m long and having area of cross-section 1 cm^2 is 5 Ωm, then the resistance is not 5*2 Ω. Resistance will be \frac{5*2}{0.0001}Ω

[Google_Spider]

Due to resistance there is a drop in potential. This potential energy is converted into heat energy and if the T reaches the melting point of the wire, it melts!