- #1
redsealelectron
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Homework Statement
A piece of aluminum wire has a resistance of 20C, 15.5 [tex]\Omega[/tex]. If this wire is melted down and used to produce a second wire having a length four times the original length, what will the resistance of the new wire be at 20C? (Hint: The volume remains the same)
Homework Equations
R = [tex]\rho[/tex][tex]\stackrel{l}{a}[/tex]
or
R2 = R1(1 + [tex]\alpha[/tex]1[t2 - t1])
The Attempt at a Solution
R2 = R1(1 + 0)
R2 = R1 * 4
R2 = 62.0
I have plugged in the information into equation number two. As the delta t = 0, I didn't use the temperature coefficient. My answer was to multiply the initial resistance by 4, which gave me 62.0 [tex]\Omega[/tex].
Some guys in my class are saying that you have to multiply by 16, and the correct answer is 248 [tex]\Omega[/tex].
I don't see why they would multiply by 16, sure its four times longer, but does that mean the diameter of the wire went down? The volume of the wire remained constant, so I assume yes, it did get longer increasing the resistance by four, but the volume remained the same so that is all that is needed.
Can anyone see if I have missed something?