temperature coefficient of resistance

Thread Starter

murari

Joined Feb 6, 2008
3
hi

i managed to thoroughly confuse myself working on a project on the temp coeff. i hope someone can help......

if Resistance has a linear variation with temperature over a temperature range, alpha should be a constant over this range....

if we are to find R of Platinum wire (say) at 50°C in terms of R at 0°C,
alpha = 0.004

R = R(0°C) * (1+ 0.004 * 50)
= 1.2 * R(0°C)

but if we calculate R at 0°C in terms of R at 50°C,

R(0°C) = R *(1+0.004 * (-50))
= 0.8 * R

This gives me R(0°C) = 0.96* R(0°C) :confused:

can anyone tell me where i went wrong??
 

recca02

Joined Apr 2, 2007
1,214
R(0°C) = R *(1+0.004 * (-50))
there should be zero instead of -50.

where u put -50 is the place where temp difference between ref and the desired value with sign is substituted. it is (Temp(at which R is to be calculated)-Temp(taken as reference))

A (-50) simply means resistance at -50°C....Chill out!:cool:
 

Thread Starter

murari

Joined Feb 6, 2008
3
R(0°C) = R *(1+0.004 * (-50))
there should be zero instead of -50
Sorry... i guess i wasn't as clear as i could have been....
R throughout stands for R at 50°C while R(0°C) is... well... R at 0°C

In the second case, the diff in temp is -50 if i use 50°C as ref temp...

That is, if alpha is independent of the ref temp...
 

recca02

Joined Apr 2, 2007
1,214
i think the problem here is u're keeping the R constant when changing the reference.

u see at 50°C the ref will change to that value.

in above case u'r ref is seemingly R at 0°C ....say R(0)
in the second case it must be R(50).
and needless to say, they are different.
 
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