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):confused:

    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
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    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
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    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
    2,810
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    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
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    % is implied but good engineering practice means it should have been explicitly stated.
     
    djsfantasi likes this.
  8. #12

    Expert

    Nov 30, 2010
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    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.

    Thank you for your patience.
     
  9. evilclem

    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
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    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. :rolleyes:
     
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