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Lab: Plot graph of resistance, R (in Ohms) versus 1/d^2

  1. Nov 17, 2016 #1
    1. The problem statement, all variables and given/known data

    We were given five samples of nichrome wire, each with a different diameter but the same length. A micrometer was used to measure the diameters. We then measured and recorded the electrical resistance, R in Ohms of each nichrome sample with a DMM. I will try to summarize all of this as clearly as possible below:

    Sample #, Diameter (± 0.01 mm), Resistance
    20, 0.80 mm, 23.6 ohms
    22, 0.62 mm, 36.5 ohms
    24, 0.56 mm, 57.4 ohms
    28, 0.31 mm, 159.0 ohms
    30, 0.26 mm, 226.8 ohms

    Length of wire: 10.00 ± 0.01 m


    We are required to plot a graph of resistance, R (in Ohms) versus the inverse square of the diameter (in m^-2), i.e.; "R" vs. 1/d^2. We then have to calculate the slope of the graph, and from the slope determine the resistivity p of nichrome wire. We then compare our value of p with the published value of p = 1.18 x 10^-6 ohm*m.

    2. Relevant equations

    The electrical resistance R (Ohm) of a particular cylindrical sample of material is related to the resistivity by:

    R = pL/A = 4pL/pi^2 = (4pL/pi)(1/d^2)
    R = 1/d^2

    where L is the length of the wire, A = (pi*d^2)4 is the cross-sectional area, p is the resistivity, and d is the diameter of the wire.

    3. The attempt at a solution

    My problem lies in calculating the inverse squared diameter itself, and plotting the values. My first instinct was to convert the diameters to m, so instead of 0.80 mm we have 0.80 x 10^-3 m. I then square the value, then invert it to get 1562500 m^-2. HOW does this make any sense? It also looks off when i try to graph the values on excel; with 1/d^2 on the x-axis and R on the y-axis, the slope should be 4pL/pi, right? I've written out all of the equations relevant to this lab right out of the lab manual -- so why are the 1/d^2 values so large? Did I miss something, what could I be doing wrong?

    Any help would be immensely appreciated!!
     
  2. jcsd
  3. Nov 17, 2016 #2

    TSny

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    Why do you say that this looks wrong?
    Yes.
     
  4. Nov 17, 2016 #3

    Student100

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    First, why would the cross sectional area of the wires be ##A=(\pi*d^2)4##? I'm assuming you meant: ##A=\frac{(\pi*d^2)}{4}##.

    For the inverse diameter equals R, what would the slope of the best fit be? ##y=mx+b## so ##R=m(1/d^2)## I don't explicitly see anything wrong with your attempt, and the values should be quite large when converted to meters. So I think you're on the right track.

    I dunno why they have you do this way, seems kind of roundabout to me, unless I'm reading the post wrong.

    Whoops, didn't see that TSny had already replied.
     
  5. Nov 17, 2016 #4
    Basically my graph is throwing me off since it's not linear. Would it be incorrect if I take the most linear portion of the graph and calculate the slope manually from there, instead of adding a trendline and having excel do it?
     

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  6. Nov 17, 2016 #5
    Sorry, you're right, it is over 4, not *4
     
  7. Nov 17, 2016 #6

    Student100

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    Check calculations and add the zero point. plot.jpg

    Unless I mucked it up myself, should be nice and linearish. Also, adjust your x scale, and do a best fit, instead of whatever that is. You should be okay, I think you just need to check the scale really and one number might off.
     
  8. Nov 17, 2016 #7

    TSny

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    I agree with Student100. Your data looks good, but it appears that you didn't plot the data correctly.
    Check the 4th data point in particular.
     
  9. Nov 17, 2016 #8

    Untitled.png
    That looks more like it... thank you so much for your help!
     
  10. Nov 17, 2016 #9

    Student100

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    Now it's looking good. Remember to label your axis and give a title. Also, not sure if you're required to plot the error bars or not, but if so don't forget them! :smile:
     
  11. Nov 17, 2016 #10
    Perfect. Thank you for all your help!
     
  12. Nov 17, 2016 #11
    Does this look ok?

    Untitled.png
     
  13. Nov 17, 2016 #12

    Student100

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  14. Nov 17, 2016 #13
    I don't think we are required to add error bars but is it better to include them? Thanks again! :smile:
     
  15. Nov 17, 2016 #14

    Student100

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    I would remove them then, I didn't check if they were right, and no sense to do so if not required.
     
  16. Nov 17, 2016 #15
    OK, will do, thank you. I am now comparing my experimental value with the theoretical value. The slope is 2x10^-5, so I worked out the following:

    |[(1.18 x 10^-6 ohm*m) - (2 x 10^-5 ohm*m)]/(1.18 x 10^-6 ohm*m)| x 100 = 1595 %

    That is huge! I worked it out a couple of times and I can't seem to figure it out. Is it possible that my slope is off?
     
  17. Nov 17, 2016 #16
    EDIT -- Wait, the slope does not give me p! I need to solve for p, since that is 4pL/pi and sub in the given value for L. Oops.
     
  18. Nov 17, 2016 #17

    TSny

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    Your graph is giving the slope to only one significant figure. How many sig figs for the slope do you think you should have?
     
  19. Nov 17, 2016 #18
    Considering the diameter is measured to 2 sig figs, the slope should have 2 as well, right?
     
  20. Nov 17, 2016 #19

    TSny

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    Sounds good to me.
     
  21. Nov 17, 2016 #20
    Thank you! One last question -- if the uncertainty for measuring the diameter is ± 0.00001 m, how would that change in the inverse square of the diameter? I would think to convert the absolute uncertainty to a relative uncertainty; i.e. 0.0008 ± 0.00001 m would be 0.0008 m ± 1.25 %, and then multiply that by 2 since it is being squared, and then happens if I take the inverse? Do I just divide 1 by 0.0008 and leave the relative uncertainty alone?
     
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