Question for the resistance of the wire

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SUMMARY

The discussion centers on calculating the resistivity of copper and the resistance of a wire drawn from a copper rod. The resistivity of copper at 20℃ is established as 0.0173 x 10^-6 ohm m. When the copper rod, originally 0.4m long and 4.0mm in diameter, is drawn into a wire with a diameter of 0.8mm, the resistance at 60℃ is calculated to be 0.398 ohm. The key formula used is R = ρL/A, where adjustments for length due to diameter change are critical.

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bckcookie
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Member advised to use the homework template for posts in the homework sections of PF.
I think about many times for few days, but I still can't find out the correct answer! Maybe my steps are somethings wrong, please advise me! Thanks!

A copper rod, 0.4m long and 4.0mm in diameter, has a resistance of 550μΩ at 20℃. Calculate the resistivity of copper at that temperature. If the rod is drawn out into a wire having a uniform diameter of 0.8mm, calculate the resistance of the wire when its temperature is 60℃. Assume the resistivity to be unchanged and the temperature coefficient of resistance of copper to be 0.00426/℃.Correct answer : 0.0173 x 10^-6 ohm m, 0.398ohm

My steps as follows:

the resistivity of copper : 0.0173 x 10^-6 ohm m ( I've got it, no problem!)

the resistance of the wire :

Area : π x [(0.8/2)x10^-3]^2

R0 = Pl/A =[ 0.0173 x 10^-6 x 0.4] / 3.14 x 1.6 x 10^-7 =0.01377ohm

R1 /R2= (1+0.00426x20) / (1+0.00426x60)

R1/0.01377ohm = 0.8652

R1 = 0.1184ohm (wrong answer!)
 
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bckcookie said:
R0 = Pl/A =[ 0.0173 x 10^-6 x 0.4] / 3.14 x 1.6 x 10^-7 =0.01377ohm

The equation R0 = ρL/A is correct but you forgot about L changing when you substituted the numbers.

L is no longer 0.4m
 
I don't know what is the value of changing L and how to calculate it, please advise me!
 
Last edited:
If the diameter is reduced the length will increase. Assume the volume of copper is constant.
 
I got it, thanks so much!
 

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