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Resistivity Coursework problem

  1. Mar 5, 2005 #1
    OK doing some physics coursework at the moment where I have to measure the resistivity of Constantan. Anyway, for those who don't know, the resistivity of constantan is 49x10^-8.

    Anyway the formula for those who don't know for resistivity is...

    RA/l

    So anyway at length of 100cm, with a diameter of 0.464mm, I got these readings in a circuit. Current = 0.33, Voltage = 2.13, which gives the resistance as 6.45. (I did repeat readings to get an average but this is just one reading as an example.)

    Anyway after finding the Cross-sectional area of the constantan by doing A=pi*r^2. Now to do this, I halve the diameter to get 0.232mm. This is the same as 0.000232m (metres). So after squaring this I get, 5.3824^-8. After multiplying this by pi, I get the cross-sectional area to be 1.69...^-7. Multiply the cross-sectional area by the resistance (6.45) which gives 1.09...^-6. Finally dividing by the length (1metre), which leaves the same answer.

    I know this is a lot to read, and I'm asking a lot for help, but please, can you find what I've done wrong, or anything?

    I really appreciate all help :smile:
     
  2. jcsd
  3. Mar 5, 2005 #2
    What are the units? check the units
     
  4. Mar 5, 2005 #3
    Not quite sure now. Using multimeters at college, I think the ampmeter was set to "10A" and the voltmeter on "20V"

    How would this affect the readings?

    Would this mean that I'd have to halve my readings of p.d. before putting them into the formula R=V/I to work out Resistance. If yes, then that would make sense, as I did a quick test and using a halved p.d. gives a final value of resistivity very close to the real value.

    Thanks in advance.
     
    Last edited: Mar 5, 2005
  5. Mar 5, 2005 #4

    Integral

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    If you use the given value of [itex] \rho [/itex] and the length and area of your wire to compute an expected resistance you will find a signifiant difference. You need to verify your current and voltage data.
     
  6. Mar 5, 2005 #5
    So you're guessing that it's my current/voltage readings which are incorrect somehow?
     
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