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When is resistance large enough to be measurable?

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  1. Sep 13, 2016 #1
    1. The problem statement, all variables and given/known data
    We conducted experiments using copper wires to observe the effect length has on resistance. We measured lengths from 5 - 30 cm and used a multimeter to measure the voltage while supplying a constant current from a power supply package. The multimeter measures in mV. Plotting the graph of length against power and got a strait line graph which indicated power increased with length, which I thought was great because that's what's supposed to happen according to theory. Our teacher said he expected random results and not the strait line we observed because resistance shouldn't be noticeable in the lengths of wire we used.

    2. Relevant equations
    I'm perplexed, I thought what we had recorded was right.

    3. The attempt at a solution
    No idea what's going on, help understanding this will be very much appreciated.
     
  2. jcsd
  3. Sep 13, 2016 #2

    cnh1995

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    Homework Helper

    Sounds correct.
    The resistance of the wires is very small and usually, the meter shows random values very close to zero. But if you have set the meter in mV range and the voltage drops are indeed in mV, the meter readings should be correct.
    Practically, if you got increasing voltage readings (in mV) with increase in length of the wire, I think what you've recorded is correct.
     
  4. Sep 13, 2016 #3

    Merlin3189

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    Gold Member

    I can see both sides of this argument. I too would expect it to be difficult to measure such low resistances. But you used a good method and could conceivably have measured them.
    You don't say what thickness of wire you used, what current you used, nor what the slope of your line was. So I have no idea what you should have been able to measure.
    1mm diameter or about 20 swg would have a resistance of about 20mΩ/m. So if you passed 1A of current, you could have measured 1mV every 5cm, which might be possible with a good multimeter. If you used thinner wire, like 26 swg about 0.5mm diameter, the resistances and voltages would be 4x as big - even more possible.

    I haven't tried it out, but I think my power supply could give a constant current up to 2 A and my meter could measure these voltages within 0.1mV, so I think it is possible that you got a reasonable straight line. It would be interesting to know the details - what wire you used (diameter), what current you used and how you knew it stayed constant, what meter you used and, of course, your actual results.

    I don't know what your teacher had in mind. The resistances of a few milliOhms are small in comparison to a light bulb for example, and my multimeter couldn't read less than 1Ω with any accuracy at all. So just using it on the resistance range and putting it across a short length of wire would be hopeless.
     
  5. Sep 13, 2016 #4
    Thank you so much for these responses. We forgot to measure the diameter of the wire but I'd estimate it to be 1mm or less. We supplied a current of 3A which we controlled directly from the power supply, and the equation of the straight line we plotted for length against power was,
    y = 1.8691x - 1.66
    We've identified a systematic error from the power supply which actually supplied 2.93A.
    I'll try to get the actual measure of the diameter of the wire as well as the type of multimeter.
     
  6. Sep 13, 2016 #5

    Merlin3189

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    Gold Member

    Ok. That equation looks like a line of best fit for a spreadsheet table. If you'd drawn it, there would be no constant term, as (0,0) is definitely on the line. (If both probes of the multimeter go to the same point, the pd must be 0.)

    The question then becomes, what values has the spreadsheet used for x and y? I'm guessing you entered your data as millivolts, so y could be in mV. The x readings could be in cm or 5cm or whatever steps you chose to measure. If I guess for a start that you took a reading every cm and proceed from there.

    (Your top reading at 30 cm is supposed to be 1.8691 x 30 - 1.66 = 54mV I'd fudge this to remove the -1.66 and guess that the best line would be about y= 1.83x )

    But if we just take the slope of your line as 1.8691 mV/cm and divide by the current to give 1.8691mV/2.93A = 0.6379 mΩ/cm = 0.06379 Ω/m , (which amazingly enough is plausible for a thin copper wire!) we can now try to estimate the thickness. (I'll assume circular cross section and ignore kinks & dents.)

    R = ρ L / A where resistivity ρ is about 1.7 x 10-8 Ωm
    So A = 1.724 x 10-8 x 1 / 0.06379 = 0.27 x 10-6 m2 = 0.27 mm2
    So r2 = 0.27/π = 0.086 mm2 and r = 0.29 mm or diameter = 0.58 mm call it 24 swg. (or 0.6mm if that's how they sell wire now.)

    If the x values were really every 5cm, then your resistance per m is 1/5 of what we calculated, about 0.0127 Ω/m and the wire comes out nearer 1.3mm (multiply by √5) around 18swg.

    Either is possible, but 24swg sounds a bit fine. I think people would pick thicker wire, just for ease of handling and lower risk of breakage, so 18 swg is my bet!

    If I try to correct for the impossible -1.66, I should get a slightly less steep line, implying a lower resistance and therefore slightly thicker wire, but it doesn't make enough difference to jump 2 swg steps, so I'll stick with 18swg. (or 16 awg )

    My feeling is that a spreadsheet might overestimate the steepness of the line, leading to overestimating the resistance and underestimating the thickness. But I'll wait to see how far adrift I am, before thinking about further "corrections".
     
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