text: # Homework Help: Temperature coefficient of resistance
  
  1. May 10, 2015
  
  ### DevonZA
  
  1. The problem statement, all variables and given/known data
  
  2. Relevant equations
  
  3. The attempt at a solution
  
  1. A copper rod, 100 mm long and 2.5mm diameter, has a resistance of 340micro ohms at 15degrees Celsius. If the rod is drawn out into a wire of uniform diameter 0.5mm, calculate its resistance at 60 degrees Celsius. Assume the temperature coefficient of resistance to be 0.0043/degree Celsius at 0 degrees Celsius.
  
  2. R1/R2=1+alpha0xt1/1+alpha0xt2
  
  3. (340 uohm)(L2/L1) / (R2/R1)^2
  = (340 uohm)(2500/100)(2.5/0.5)^2 = 0.2125 ohms.
  Then use the temperature equation:
  R = (0.2125 ohm)(1 + 0.0043*45) = 0.25 ohms
  
  ^^ This is an answer that I found online but I don't know where the person got the 2500 value for L2 from?
  
  The answer given in the textbook is R=0.251ohms
  
  Thanks,
  Devon.
  
  2. May 10, 2015
  
  ### Suraj M
  
  when you draw the given wire into another wire of different diameter the length of the wire changes, you can find out the new length of the wire by taking constant volume, i do have my doubts though,( because of the change in temperature).
  
  3. May 10, 2015
  
  ### DevonZA
  
  Could you provide an example?
  
  4. May 11, 2015
  
  ### Suraj M
  
  Oh ok.
  if i take a cylinder of length 4m and diameter 1m, if i now recast the cylinder into a new cylinder of diameter 2m, find the length of the new cylinder..
  do this by considering volume as a constant,
  Do you have a formula for volume of a cylinder!?
  
  Last edited: May 11, 2015
  5. May 12, 2015
  
  ### DevonZA
  
  V=pir^2h
  
  V=pi(0.5)^2(4)
  =pi
  
  pi=pi(1)^2h
  pi/pi=h
  h=1
  
  Last edited by a moderator: May 12, 2015
  6. May 12, 2015
  
  ### Suraj M
  
  excellent! now why don't you apply the same to your question?
  Ignore the resistances for a while, we'll deal with it later!
  also we are assuming that 10°C is quite less and we're ignoring the change in volume..
  PS: Please refrain from using such language
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