Why does a re-formed copper rod have a resistance of 4R instead of 2R?

AI Thread Summary
When a cylindrical copper rod is re-formed to twice its original length while maintaining the same volume, its cross-sectional area decreases to half of the original size. The resistance of a conductor is determined by the formula R = ρ(L/A), where ρ is resistivity, L is length, and A is cross-sectional area. With the new length being 2L and the new area being A/2, the resistance becomes R' = ρ(2L/(A/2)), simplifying to R' = 4R. Thus, the resistance increases to four times the original value due to the combined effect of increased length and decreased area. Understanding this relationship is crucial for solving similar problems in physics.
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Homework Statement


A cylindrical copper rod of length L and cross-sectional area A is re-formed to
twice its original length with no change in volume. If the resistance between its
ends was originally R, what is it now?


Homework Equations


R=p(L/A)


The Attempt at a Solution


Using the above equation the answer would be 2R, however the correct answer is 4R. Why is this? I am studying for a test tomorrow and need to understand why this is so.
 
Physics news on Phys.org
If it has the same volume but twice the length, what will happen to the cross-sectional area?
 
alex3 said:
If it has the same volume but twice the length, what will happen to the cross-sectional area?

ahhh, it will be 1/2 of the original, thanks.
 

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