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I was discussing Ohm's Law with my teacher and asked about what would happen if the resistance of a circuit was zero, to which he replied that the wire would melt. This got me wondering about plasma/ionised gas and whether or not it would have zero/negative resistance.

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  • $\begingroup$ To be clear, a wire with zero resistance would not melt as resistance is required to produce heat. I am assuming the "zero resistance" comment was implying a short circuit (i.e., no resistance other than the intrinsic resistivity of the wire and power supply), in which case, yes the wire would have issues. It may not melt though, e.g., aluminum sublimes if super heated like this... $\endgroup$ – honeste_vivere Sep 23 '16 at 21:05
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The conductivity of plasmas is very high, though not infinite.

In a metal wire the applied voltage accelerates the conduction electrons, but the electrons collide with and scatter off the atoms that make up the crystal lattice of the metal. This transfers energy from the electrons to the metal and the metal heats up as a result. The energy lost as heat is what leads to resistance.

In a plasma applying a voltage causes the electrons to accelerate, but because a plasma has a much lower density than a metal the scattering and consequent energy loss of those electrons is very low. There will still be some scattering because electrons will scatter off the positive ions and any unionised gas molecules and indeed each other. So although the resistance is low it won't be zero.

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  • $\begingroup$ I agree with the plasma answer, which is the main point here, but have some concerns with the metal. In solid state physics the scattering in a crystal is not rally on each and every atom, as one might get the impression here; that would lead to a mean free path of 2 Angstrom only. Scattering takes place at defects like dislocations; or dynamically with lattice excitations such as phonons. (sorry for nitpicking) $\endgroup$ – mikuszefski Apr 10 '15 at 10:15
  • $\begingroup$ Yet another one, what about superconductors? $\endgroup$ – mikuszefski Apr 10 '15 at 10:16
  • $\begingroup$ @mikuszefski: yes, I've given a simplified description because I thought trying to explain electron-phonon scattering would be too complicated given that the OP is clearly a school student. I also think introducing superconductivity is an unnecessary complication, though of course the student is welcome to look at my answer to Superconductivity reasons (Intutitive) :-) $\endgroup$ – John Rennie Apr 10 '15 at 10:19
  • $\begingroup$ I get the point, just wanted to avoid conceptual misunderstandings. I agree that this is rather difficult as the theory of conductivity in metals is very different to that in plasmas. So I agree it would be enough to provide this as further reading ;). Superconductivity is a different animal again, not even talking about high Tc. I just wondered about the teachers conclusion that the wire would melt. Simply speaking, infinite conductivity means no scattering, so why should it melt? $\endgroup$ – mikuszefski Apr 10 '15 at 10:25
  • $\begingroup$ @mikuszefski: for a zero resistance wire and an ideal voltage source you get an infinite current, so the power dissipated in the wire in $\infty^2\times 0$ and this is undefined. However the power dissipated $\rightarrow \infty$ as $R \rightarrow 0$, so the teacher is correct in this sense. Of course for a real voltage source you need to include the internal resistance of the PSU. In this case the power dissipated in the wire would go though through a maxium then go to zero as $R \rightarrow 0$. Again, this is tangential to the OP's question. $\endgroup$ – John Rennie Apr 10 '15 at 10:42

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