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Help with some Resistance Questions

  1. Nov 7, 2004 #1
    Okay we just start electricity in class and almost instantly we got a small lab. I did part of it, but I am having trouble with some of the questions and have no idea where to start from.

    The first question is Determine the R value of a resistor that uses 4 m. of copper wire with a thickness of 1/3x the one used in the lab.
    Now I have to find the ratio, but how? Also the resistor board used is 1 meter. I'm not sure, but would it be
    R2= Resisitivity x 4 m. / Pie (4r)^2

    R1= Resistivity x 1 meter / Pie r^2

    How do I find the ratio?

    2nd question was "Determine how much wider the diameter of the thick nichrome wire is in comparison to the thin nichrome wire".
    How would I start such a question?

    3rd question was "If the resistivity of nichrome is 100x10^-8 ohms/meter, determine the resistivity of copper"
    How would I start to solve this question also?

    Thanks (I'm not looking for an answer, Just how to start solving it)
     
  2. jcsd
  3. Nov 8, 2004 #2
    How do you find the ratio? I'm sorry but I don't quite understand your problem. You seem to know all the parameters. What is the problem? If the thickness is an issue then the thickness refers to the diameter (twice the radius). And you have used

    [tex]R = \rho\frac{L}{A}[/tex]

    so where are you stuck?

    Cheers
    Vivek
     
  4. Nov 8, 2004 #3
    Doesn't matter, I figured it out and I got it right ^^
     
  5. Nov 8, 2004 #4
    ummm can u please temme how u did it?? havent done electricity for a while now and cant seem to remember how to do it.. what is the value of RHO ( the P symbol ) in this case??

    thanks
     
    Last edited: Nov 8, 2004
  6. Nov 8, 2004 #5
    okay so we have the resistance of the copper from the lab, which we use

    R1=P x L / A
    Then we have the changes, which is 4x length and 1/3 the thickness. So, since we want the value of the resistor, we do:

    R2=P x 4L / pie (1/3r^)2 <--- this is pie r^2 because the wire is circular

    Since the resistance of the wire was 0.4 ohms (I'm doing this all off by memory BTW because I handed in my lab) then all we have to do is find the ratio. Since we have 1/3 in the brackets, we square to get 1/9, then the ratio is 4 / 1/9, which is 1/36. So 0.4 x 1/36 = 1.1 x 10^-2 ohms.
     
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