
#1
Oct212, 08:41 PM

P: 248

1. The problem statement, all variables and given/known data
What diameter must a copper wire have if its resistance is to be the same as that of an equal length of aluminum wire with diameter 3.94mm? Give your answer in mm 2. Relevant equations R = ρL/A where A is the cross sectional area, L is the length and ρ resistance of the material ρ_{Copper} = 1.72x10^{8} Ωm ρ_{Aluminum} = 2.82x10^{8} Ωm 3. The attempt at a solution So I knew that R_{copper} = R_{Aluminum} (1.72x10^{8}Ωm)(L)/A_{copper} = (2.82x10^{8}Ωm)(L)/(0.00394m) A = 0.00240 m A = πr^{2} r = 0.027657 m diameter = 55.31499 mm But the system says its wrong and I don't see where I've gone wrong. 



#2
Oct212, 09:39 PM

HW Helper
PF Gold
P: 1,848

[Edit: On a much lesser note, and I'm not sure if this matters, according to some sources I've looked at on the Internet, 1.72x10^{8} Ωm is the resistivity of annealed copper. Plain old copper has a very similar resistivity, but slightly different. Use whatever your course material advises. And again, the difference is small and I'm not sure if copper vs. annealed copper matters.] 



#3
Oct212, 09:57 PM

P: 248

I see what you're saying.
The cross section area od aluminum is π(0.00197m)^{2} 3.8809x10^{6} m^{2} Then solving for the cross sectional area of copper I get A_{copper} = 7.4364x10^{6} m^{2} r_{copper wire} = sqrt(7.4363x10^{2}/π) Diameter of copper wire = 2r = 3.077059 mm Thanks for your help 


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