1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

You have a fine wire with a cross sectional area of 9.4×10−10 m2

  1. Mar 17, 2012 #1
    You have a fine wire with a cross sectional area of 9.4×10−10 m2 and a length of 1.1 m. You wish to determine what it is made out of, so you connect it to a standard AA battery (1.5 V) and measure a current of 0.048 A. What material is the wire most likely made out of?
    The options for the resistivities are as follows:
    Iron (but iron is wrong, so discard)
    Aluminum: 2.7E-8
    Silver: 1.6E-8
    Copper: 1.7E-8

    I found the resistance by multiplying the current (.048 A) and voltage (1.5). I got .072 ohms.
    I'm sort of confused about the area though. Would it be A=∏(9.4×10−10)2, or what?

    Then when I solve I get .072=ρ(1.1/2.95E-9)
    ρ= 1.9E-10 ohm meters.

    I only have one shot of getting the answer. I'm confused, do it right, and in that case is it closest to aluminum or silver? My friend had aluminum, and we might have the same numbers. PLEASE HELP!
     
  2. jcsd
  3. Mar 17, 2012 #2

    PeterO

    User Avatar
    Homework Helper

    Re: Resistivty

    Read those two bits in red and see if you are still confused.
     
  4. Mar 18, 2012 #3
    Re: Resistivty

    Yes, I am, because my friend did the same thing as me and it's wrong. When I used the area like that I got 6.0E-11 ohm meters. Neiter one of those options is remotely close to that number. Or do I put the area into the Area equation? Divide it by pi to get the radius and then solve?
     
    Last edited: Mar 18, 2012
  5. Mar 18, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Re: Resistivty

    Check what Ohm's Law states. Your resistance calculation is not correct.
     
  6. Mar 18, 2012 #5
    I'm sorry, I meant that I divided the voltage by the current.
     
  7. Mar 18, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    Re: Resistivty

    Okay, so what resistance does that give you?
     
  8. Mar 18, 2012 #7
    Re: Resistivty

    the resistance was .072 ohms.
     
  9. Mar 18, 2012 #8

    gneill

    User Avatar

    Staff: Mentor

    Re: Resistivty

    Recheck that value!
     
  10. Mar 18, 2012 #9
    Re: Resistivty

    31.25 ohms...oh I see what I did with the resistance. I had the equation as V/I but I still multiplied them. Silly me. But what about the area? Is that actually the number I use?
     
  11. Mar 18, 2012 #10
    Re: Resistivty

    When I do the calculations I get 2.67E-8 ohm meters. So it's aluminum?
     
  12. Mar 18, 2012 #11

    gneill

    User Avatar

    Staff: Mentor

    Re: Resistivty

    :confused: You're given the cross sectional area. What other number do you have in mind?
     
  13. Mar 18, 2012 #12
    Re: Resistivty

    I don't know, my friend said he did it a different way, but he got the right answer. He confused me, because I thought it was the area. I guess it is. Just wondering.
     
  14. Mar 18, 2012 #13

    gneill

    User Avatar

    Staff: Mentor

    Re: Resistivty

    If it's closest to the corresponding value given for aluminum, then yes :smile:
     
  15. Mar 18, 2012 #14
    Re: Resistivty

    Well, aluminum is Aluminum: 2.7E-8 ohm meters, so I'm gonna go with yes. Thanks!
     
  16. Mar 18, 2012 #15

    PeterO

    User Avatar
    Homework Helper

    Re: Resistivty

    Resistance is given by R = ρL/A using resistivity, length and cross-section area

    R = V/I from Ohms law

    Combining you have ρL/A = V/I

    so ρ = VA/LI

    since you were given the values of V, A, L and I it should just have been a matter of plugging in the values and see what you get.

    Note that the actual Resistance value does not ever have to be calculated.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: You have a fine wire with a cross sectional area of 9.4×10−10 m2
  1. Cross sectional area (Replies: 2)

Loading...