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The current of a nichrome tube

  1. Mar 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A 60.0 hollow nichrome tube of inner diameter 1.60 , outer diameter 4.20 is connected to a 4.00 battery.What is the current in the tube?
    [tex]\rho[/tex] = 1.5*10^-6 ohm for nichrome

    2. Relevant equations
    I=dletaV/R
    R=[tex]\rho[/tex]L/A -a should be the crossectional area

    3. The attempt at a solution
    need R so use R=pL/A
    A=(4.2E-3)^2(pi)-(1.6E-3)^2(pi)=4.7375E-5m
    plug into R
    R=((1.5E-6ohm)(0.6m))/(4.7375E-5m)=0.0189972 ohms
    then use the resistance to solve for current of the wire
    I=4V/0.0189972ohm=210.6A
    This isn't the right answer and I don't know what I did wrong! Could someone help me out here, please!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 1, 2010 #2

    berkeman

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    Staff: Mentor

    Your technique and calculations look okay to me, but I could be missing something.

    One thing, though, is your units are incorrect in a couple places. The units of rho are not Ohms (what are they?), and units of area are most certainly not meters.

    It would also be helpful if the problem statement had units on all numbers. I had to make assumptions about your calculations later using the numbers you stated in the problem.
     
  4. Mar 2, 2010 #3
    Sorry about the missing units! hee they are!
    A 60.0 cm long hollow nichrome tube of inner diameter 1.60mm , outer diameter 4.20mm is connected to a 4.00 V battery. The units for P is ohm m, and area should have been m^2. My bad! Oh and by the way the p value wasn't given in the problem, thought I should mention that.
     
  5. Mar 2, 2010 #4

    berkeman

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    Staff: Mentor

    That's much clearer, thanks. But are you still getting the wrong answer? Maybe the value of rho is off?
     
  6. Mar 2, 2010 #5
    I looked it up and it goes from 1.0E−6 to 1.5E−6ohm m, so I kinda had the right value. I guess i should try the first value as well, however I doubt if its right.
     
  7. Mar 3, 2010 #6
    okay so I tried 1E-6, that didn't work so i looked again at what I did for the cross section, found out you can get 2 different values, if you use 4.2E-3 or just keep it in mm then convert you get 2 different numbers, the conversion to mm being the largest. So i tried all possible combinations, and none are right. Heres what i got for 1E-6 as the p value , 315.8 A and if you do it antoher way 3.158E5 A, for the new way with the original value i had its 2.106E5 A. none of these answers are right, or mastering physics is not accepting them.
    I really want to know what I am doing wrong!
     
  8. Mar 3, 2010 #7

    ideasrule

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    Homework Helper

    You forgot to convert diameter to radius.
     
  9. Mar 4, 2010 #8
    Wow, that would deffinetly help. Can't belive I forgot something so symple... okay I drew the picture wrong so that messed it up. Thankyou, now let me see if what I get is right now!
     
  10. Mar 4, 2010 #9
    Yay it did help, but you have to convert the radius to m first and then you get the right answer of 52.6 A. Thankyou, to the both of ya!
     
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