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So I know that $V=IR$ works for circuits, but for the case of an arc-before the arc jumps, there is a potential difference, but no current, but there isn't infinite resistance is there? I don't understand how to compute a finite resistance for an arc that would come out as infinite in some other cases.

Nic
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    Before the arc, there is a very very high resistance, and a very very small current. That's why it is approximated as an infinite resistance and a zero current. The resistance changes when the arc is created because the air is ionized and provides more conducting electrons. Calculating the arc resistance is not a straightforward task. What is your goal? – fffred Jun 10 '13 at 05:51
  • I was just curious, there is a current though? So there are tiny currents all around us? –  Jun 10 '13 at 06:29
  • For a sufficiently tiny definition of tiny yes, there are tiny currents all around us. :) – Michael Jun 10 '13 at 06:36
  • You can estimate the tiny current from a standard value of air's resistivity, a given voltage, and a given apparatus (surface and distance between two electrodes). – fffred Jun 10 '13 at 06:43
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    Just for completeness (random knowledge is always nice), Ohm's law is indeed incorrect. In superconductors, for example, it gives wrong answers and we need to use the London equations. – Wouter Jun 10 '13 at 10:31
  • Ohm's law is a grand lesson in learning about ideal vs 'real' in physics. – Nic Jun 10 '13 at 11:32
  • Also, why are there arcs? If the current is always there, then why doesn't it just gradually become a bigger stream, as opposed to arcing? –  Jun 10 '13 at 21:26

2 Answers2

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I don't understand how to compute a finite resistance for an arc that would come out as infinite in some other cases.

Arc formation is a sufficiently non-stationary and nonlinear process. So, one has to use dynamic circuit theory, where the resistivity in Ohm's law is a complex number and contains both active and reactive components depending on the applied voltage.

That is in general. In practice, modeling the arc's resistivity in both static and dynamic (transient) regimes is very hard problem which was attempted to be solved by many groups. Searching in Google you can find several approaches based on the equivalent circuit method, where conducting chanel is approximated by a set of resistors, capacitors and inductors. Understanding of this phenomena is strongly related to microscopic nature of electron transport in conducting channel.

freude
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  • Can you maybe link one or two approaches that you think are relevant? – Bernhard Jun 10 '13 at 11:37
  • There are many different regimes of electric arc AC/DC, high/low-power etc. Each requires own method. Here is an example of modeling http://link.springer.com/article/10.1007%2FBF01574384 – freude Jun 10 '13 at 13:16
  • Also, I have found some relevant info in this book http://www.amazon.ca/Electric-Fuses-A-Wright/dp/086341379X – freude Jun 10 '13 at 13:24
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Ohm's law in the circuitry sense can be derived from the electromagnetic sense from the equation

$$\vec J=\sigma\vec E$$

That is, current density is the conductivity times the electric field [with current density as the analog to current, conductivity the analog to the inverse of the resistance, and the electric field as the analog to voltage]. But this equation is only true for certain materials, in particular those that have sufficient free electrons. Air isn't loaded up with them so air is particularly non-ohmic [that is, it doesn't follow $V=IR$]. As an example, we know that non-zero electric field often leads to no current [when a charge is built up before the current starts flowing]. This graphic from wiki is useful [the two on the left are ohmic, the two on the right aren't]. http://en.wikipedia.org/wiki/File:FourIVcurves.svg

jazzwhiz
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  • You take Ohm's low in a very narrow sense. It has dozens generalizations. For me, as a solid-state physicist, it is just a wrapper formula which can be anything, since the conductivity is not necessary a constant, but a complex-valued function of many arguments. – freude Jun 10 '13 at 14:14