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October 25, 2014

October 25, 2014

Posted by **rick** on Wednesday, January 30, 2008 at 9:16pm.

- physics -
**drwls**, Wednesday, January 30, 2008 at 9:25pmThe relative change in resistivity as the wire heats up is 1.50/1.22 = 1.2295. That is a 22.95% INCREASE.

(delta T)* coefficient of resistivity = 0.2295

delta T = 0.2295/4.5*10^-4 = 510 C

T = 510 + 23 = 533 C

- physics -
**rick**, Wednesday, January 30, 2008 at 9:34pmwhere are you getting 22.95% and what is delta T

- physics -
- physics -
**Damon**, Wednesday, January 30, 2008 at 9:37pmR = V/i , V is constant here

V/i2 = V/i1 (T2-T1)k

i1 = i2 (T2 - 20) k

1.5 = 1.22 (T2 - 20)(4.5 * 10^-4)

15000 = 5.49 T2 - 110

You can get it from there

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**Damon**, Wednesday, January 30, 2008 at 9:43pmWhoa - I forgot the definition. dr wls did it right !

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**drwls**, Wednesday, January 30, 2008 at 11:47pmIn answer to your followup questions: delta T is the increase in temperature. The 22.95% resistance increase was explained in my first calculation. 1.2295 is the inverse of the current decrease factor, 1.22/1.50. Since I = V/R, the current varies inversely with the resistance at constant voltage

- physics -
**anonymous**, Tuesday, July 17, 2012 at 12:02pmdrwls is correct. To get the percentage subtract 1.22 from 1.5, then divide by 1.22

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