Posted by rick on Wednesday, January 30, 2008 at 9:16pm.
The 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
where are you getting 22.95% and what is delta T
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R = 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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Whoa - I forgot the definition. dr wls did it right !
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In 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
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drwls is correct. To get the percentage subtract 1.22 from 1.5, then divide by 1.22
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