Talk:Resistivity

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1 resistivity of Al
2 References
3 Volume?
4 combining resistivity and conductivity tables
5 Resistivity equation
- 5.1 The general equation is incorrect

[edit] resistivity of Al

After searching other locations, I'm finding very different versions of resistivity for Al. Something closer to 2.6 or 2.8 rather than 2.282... --Hobit 14:19, 11 October 2005 (UTC)

–Fixed. I used the value from aluminium, which was taken at 20°C. This is equivilent to 293K, as used in this article). —deanos {_p^t_a^a_*^l_g^k_e} 15:18, 15 October 2005 (UTC)

[edit] References

The table of resistivity values looks exactly like the one in my physics textbook (Physics for Scientists and Engineers, with Modern Physics by Serway and Jewett). Is there a missing citation here or am I missing something? Mahsmanj

I've added references for most of the materials, and for those where I couldn't I've moved them from the article to here. I also redirected Table of resistivities here, because the table here was exactly the same.

Material	Resistivity (ohm metres)	Temperature coefficient per kelvin
Chromium	1.8 × 10^-7	.0000059
Tin	1.15 × 10^-5	.0042
Silver, German	3.3 × 10^-5	.0004
Seawater	2.0 × 10^-1 [1]	?
Pure water	2.5 × 10⁵	?
Human skin	approximately 5.0 × 10⁵	?

Kevin 09:30, 4 May 2006 (UTC)

[edit] Volume?

Does anybody know how resistivity relates to the volume of an object? That is, if I had something that looked more like a sphere than a thin wire, and wanted to calculate its resistance using its resistivity, how would I do it?

Its wikipedia custom to sign your name with four tildes ~~~~, welcome to wikipedia. As for your question, I think you could calculate it with an integral if you really want to : ) . Start by approximating it as a series of circles with some thickness dx, where the thickness of circles in the series goes from 0 to the radius (of the sphere). Add all the resistivities of those circles up, and take the limit as the distance between different sized circles goes to 0.

That would probably be a difficult integral to figure out how to set up, but you could easily approximate it without taking such a limit. Simply approximate the sphere as say 5 different circles, and see what the resistivity comes out to. Fresheneesz 22:50, 27 May 2006 (UTC)

It probably involves integrals, though I don't know if the equations are straightforward. Remember that this isn't a straight volume or surface area type calculation. You're calculating the resistance of the object, but the resistance will be different for the same object depending on how the electrodes are arranged. Would you have to calculate the current flow through each differential cross section of the entire object? This gets into bulk resistance calculations that I am not familiar with (but would like to be). — Omegatron 19:00, 24 February 2007 (UTC)

The total resistance of a body is not just determined by its volume, but the area of the contacts (contact points). If you know the volume of the body, and the area of the contacts, then you can calculate the bodies "relative resistivity length" by simply dividing the volume by the area of the contacts. This will give you the value for "l". ZoftWhere 08:48, 15 March 2007 (UTC)

[edit] combining resistivity and conductivity tables

I think it would be better if we combined the tables of resistivity and conductivity since they are so fundamentally linked. Please discuss this at Talk:Electrical conductivity. Fresheneesz 22:52, 27 May 2006 (UTC)

[edit] Resistivity equation

somebody has just defaced the general equation: it should be rho=R multiplied by A divided by length

[edit] The general equation is incorrect

I have changed it twice but someone keeps changing it back. It should be

${\rho={R \left. \frac{l}{A} \right.}}$

and not

${\rho={R \left. \frac{A}{l} \right.}}$

I do not make out that I understand everything that is involved with resistivity as I am only a year 11 pupil studying it for my GCSE but about this I am quite certain and I have also verified it with my teacher. Some one told me to check this up on an External sitewhich I did. There equation of

${\ {R=\left. \frac{\rho l}{A} \right.}}$ is the same as ${\ {R=\rho \left. \frac{l}{A} \right.}}$

which can be rearranged to form

${\rho={R \left. \frac{l}{A} \right.}}$

If anyone can explain why it should be the other way please try. Mrpowers999 16:16, 24 February 2007 (UTC)

If someone keeps reverting your edits, there's probably a good reason for it! Don't keep making the same edits without discussing it!

Your math is wrong. You say:

${\ {R=\rho \left. \frac{l}{A} \right.}}$ which can be rearranged to form ${\rho={R \left. \frac{l}{A} \right.}}$

but this is incorrect. Think of it this way. Divide both sides by L and multiply both sides by A:

${\ {R=\rho \left. \frac{l}{A} \right.}}$ which can be rearranged to form ${R \left. \frac{A}{l} \right. = \rho.}$

Also you should know that your equation is wrong because the units don't come out right. The correct units for resistivity are Ω·m. If you divided length by area you would get Ω/m.

Please be more careful when editing if you aren't sure of something. — Omegatron 18:32, 24 February 2007 (UTC)

I got out the old pencil and paper and realised you were right :( This will mean a few ammendments to my physics coursework 'sigh' Mrpowers999 00:50, 25 February 2007 (UTC)

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