# Temp. Coeff. Resistance

Discussion in 'Homework Help' started by Steve1992, May 11, 2012.

1. ### Steve1992 Thread Starter Senior Member

Apr 7, 2006
100
0
Hi,
From a text book:
A platinum resistance temperature sensor has:
Resistance = 100Ω at 0°C
Temp. coefficient = +0.385 per °C

Therefore at 50°C = 119.25Ω
and at 150°C = 157.75Ω

When I used the data in the temperature coefficient of resistance equation I get a resistance of 2k at 50°C: 2k = 100Ω(1 + 0.385x50°C)

Caveat
I may have misunderstood the question, so if a mistake isnt obvious dont waste your time.

Last edited: May 11, 2012
2. ### BSomer Member

Dec 28, 2011
433
106
It was a simple math error that I can see. The resistance increases 0.385 Ohms per degree C. Go from there and you should see the error.

3. ### #12 Expert

Nov 30, 2010
17,897
9,316
IIRC it doesn't add, it multiplys, as in +.385%/C
The lack of units in the text book is the problem.

Please post the "temperature coefficient of resistance equasion".

4. ### mlog Member

Feb 11, 2012
276
36
$R = 100 (1 + \alpha T )$

where $\alpha=0.00385$/°C

5. ### #12 Expert

Nov 30, 2010
17,897
9,316
That makes sense.
R = resistance at 0 C times [1 plus (alpha times difference in temperature)].
alpha isn't .385, it is .385 PERCENT per centigrade.

at 50 C, R = 100 ohms times (1+ .385% times 50)
R = 100 (1+ .1925)
R = 119.25 ohms

Got it?

Steve1992 likes this.
6. ### djsfantasi AAC Fanatic!

Apr 11, 2010
3,496
1,253
I am totally confused by the responses. I think #12 hit it on the head. In the e-book on this site, it appears as if the temperature coefficient of resistance are all in the range of .00xxx, which is consistent with #12's calculations. Same with Wikipedia and EngineeringToolbox... All sources only state that α is in units/°C and there is no reference to percentage.
Or is that implied, #12?

7. ### crutschow Expert

Mar 14, 2008
16,545
4,459
% is implied but good engineering practice means it should have been explicitly stated.

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8. ### #12 Expert

Nov 30, 2010
17,897
9,316
If I may indulge in a bit of a rant, the only place for implied labels is on the scratch pad next to your soldering iron. Proper math is imperative in electronics. It's hard enough even if you're good at it. To throw student into a math problem with implied labels is inexcusable.

9. ### evilclem Active Member

Dec 20, 2011
118
16
It has nothing at all to do with percentages.

It is a positive temperature coefficient (hence the + sign) of 0.385 ohms per deg C.

The resistance at 0 deg C is 100 ohms as given.

Therefore for 50 deg C it is: 100 + (50 * 0.385) = 119.25 Ohms. That is, the initial value plus 0.385 ohms for every degree over the initial values given temperature.

The percentage theory will catch you out with a device whose resistance at 0 deg C is not 100 Ohms.

10. ### crutschow Expert

Mar 14, 2008
16,545
4,459
That's still sloppy notation. If it is indeed +0.385 ohms per deg C then it should be stated as +0.385Ω per °C, not +0.385 per °C.