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Finding temperature coefficient of resistivity of the alloy

  1. Feb 18, 2016 #1
    Suppose a wire made from an unknown alloy and having a temperature of 20.0°C carries a current of 0.529 A. At 52.4°C the current is 0.378 A for the same potential difference. Find the temperature coefficient of resistivity of the alloy.

    tempinital = 20C
    tempfinal = 52.4C
    currentInital = .529A
    currentFinal = .378A
    Voltage = constant but unknown


    how do you go about this problem i have tried manipulating ohms law ( I= V/R ) to remove voltage but that always leaves me with IR= IR.

    TCOR = temperature coefficient of resistivity;

    because the resistance isn't a constant i cant factor it out of RF = RI (1 - TCOR(TF - TI))

    i cant use either form because i don't have resistance to solve for resistivity.

    if someone could point me in the right direction that would be great.
     
  2. jcsd
  3. Feb 18, 2016 #2

    berkeman

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    Staff: Mentor

    Since the two measurement voltages were the same, you can write that equation, but you should distinguish between them...

    [tex]I_1 R_1 = I_2 R_2[/tex]

    So you can find the ratio of the two resistances. Then use the ratio of the two temperatures (be sure to use absolute temperature, not degrees Celsius) to help you find the coefficient of resistivity...
     
    Last edited: Feb 18, 2016
  4. Feb 18, 2016 #3
    The resistance is constant in each of the tests. I don't see why you can't use the equation. Also, resistivity is proportional to resistance for the same wire. Is it that you don't know how to solve the equation for TCOR?

    Also, RF = V/IF and RI=V/II
     
  5. Feb 18, 2016 #4
    how can the resistance be constant if the voltage is constant and the current changes?
     
  6. Feb 18, 2016 #5
    It's constant with location along the wire.

    ##R_1=V/I_1## and ##R_2=V/I_2##. So,
    $$\frac{R_2}{R_1}=\frac{I_1}{I_2}$$
     
  7. Feb 18, 2016 #6
    right so you can changed the formula to be
    I2/I1 = (1 - TCOR(TF - TI))
     
  8. Feb 18, 2016 #7
    Excellent!!!
     
  9. Feb 18, 2016 #8
    so then plugging everything in you would get

    .378/.529 = (1- x(52.4-20))
    meaning x = .0088100072
     
  10. Feb 18, 2016 #9

    berkeman

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    Always remember to include units in your calculations and answers... :smile:
     
  11. Feb 18, 2016 #10
    well that's my problem it says my answers units should be C-1 but by using the formula wouldn't the units just be C
     
  12. Feb 18, 2016 #11

    berkeman

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    Include units in your calculation above at each step for each quantity. What do you get now?
     
  13. Feb 18, 2016 #12
    .7145557656 (A/A cancels) = (1 - (32.4xC))
    - (32.4xC) = .2854442344
    - xC = .2854442344/32.4
    x = .0088100072 C
     
  14. Feb 18, 2016 #13

    berkeman

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    No. Keep the units with the quantity that has them. You don't know x's units until you have solved the equations...

    0.378[A]/0.529[A] = (1- x(52.4[C]-20[C]))

    Now can you keep going with that?
     
  15. Feb 18, 2016 #14
    well that would mean the Amps cancel leaving that as just a number and the only unit left is C which x picks up when its distributed across those numbers. correct
     
  16. Feb 18, 2016 #15

    berkeman

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    Nope.

    Since the lefthand side (LHS) of the equation is unitless, the RHS has to also be unitless. On the RHS, "1" is unitless, so x(52.4[C]-20[C]) has to be unitless. The temperatures have units of [C], so what units must x have? :smile:
     
  17. Feb 18, 2016 #16
    oh it must have 1/C
     
  18. Feb 18, 2016 #17
    It Makes perfect sense to me but the homework program told me i got the wrong answer. is there some mistake in my calculation?
     
  19. Feb 18, 2016 #18

    berkeman

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    Looking back at this equation, resistance goes up with temperature. If the final temperature is higher than the initial temperature, should you be subtracting or adding the delta resistance? I don't know if that's the error, but it jumps out at me...
     
    Last edited: Feb 18, 2016
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