# Temperature coefficient of resistance

1. May 19, 2010

### Philip Wong

1. The problem statement, all variables and given/known data
ok we did a temperature coefficient of resistance at class, we need to calculate the coefficient of resistance (a) ourselves.
known variable:
R = resistance
ln R = log of resistance
T = tempreature
a = coefficient of resistance
c = the intercept

2. Relevant equations
ln R = aT+c
i.e. R= k exp(aT), where k=exp(c) is a constant.

3. The attempt at a solution
at first I thought I could calculate a as if it is gradient, and therefore worked out the intercept (c). But since a is the coefficient of resistance it didn't make sense to me to do it this way. So I'm lost
Anyways this is my working:
gradient = delta ln R/ delta T
= (lnR_1-lnR_2/T_1-T_2) = -.912185/44
= -0.02682897
therefore intercept (c) = -.912185 - (-.02682897 * 44)
= 10.8925618

put it back to the equation
lnR = a*T+10.893
given T = 64, its correspond ln R = 4.957 (we measured these ourselves, assume correct)

so:
4.957 = a*64+10.893
-5.936=a*64
a=-5.936/64
a=-0.0927

put everything back to original equation:
ln R = -0.0927 T + 10.893

is this even correct??
thanks

2. May 19, 2010

### Philip Wong

if my above working was correct, here is the second part of my question.
during the experiment we only measure the resistance every 2C drop (from 100C to 30C). I've plot a ln R vs T graph, how can a determine the a at temperature of 65C?

3. May 19, 2010

### Philip Wong

arha! I was right in general, but had made a small mistake! I forgot to convert temperature into kelvin before I calculated the coefficient. when I convert it, everything looks more reasonable and correct!