R through a wire: What's the max error in diameter if your R has 1% uncertainty?

AI Thread Summary
The discussion centers on determining the maximum error in measuring the diameter of a wire when its resistance has a 1% uncertainty. The formula used is R = ρl/(πr²), where the diameter D is derived from resistance R. Participants suggest solving the problem symbolically by considering the resistance variations of 1.01R and 0.99R to find corresponding diameters. Clarifications are made regarding the squaring of D in the formula, leading to the need for square root calculations in the final results. The conversation emphasizes the importance of precision in calculations and understanding the relationships between resistance and diameter.
anban
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Homework Statement



I have a wire of some resistance, resistivity, and length.

What is the maximum error in measuring the diameter that you can have if the resistance is to have 1% uncertainty?


Homework Equations



R = \frac{ρ*l}{\pi*r^{2}} for which I substituted the area as \pi(.5 D) ^{2}.


The Attempt at a Solution



I picked an arbitrary R and found the corresponding D. Then, I took 1% of that arbitrary R and found the corresponding D. The issue is that D's were basically the same number. I'm not sure if I could just go out to more decimal places or if I'm just wrong...
 
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anban said:

Homework Statement



I have a wire of some resistance, resistivity, and length.

What is the maximum error in measuring the diameter that you can have if the resistance is to have 1% uncertainty?

Homework Equations



R = \frac{Rρl}{\pi*r^{2}} .

The Attempt at a Solution



I picked an arbitrary R and found the corresponding D. Then, I took 1% of that arbitrary R and found the corresponding D. The issue is that D's were basically the same number. I'm not sure if I could just go out to more decimal places or if I'm just wrong...

You meant R = \frac{ρl}{\pi*r^{2}} , didn't you?

Solve the problem symbolically. If the resistance can differ from R by one percent it can be 1.01 R or 0.99 R. What are the corresponding diameters and how are they related to the original D? By what percent are they different from it? ehild
 
ehild said:
You meant R = \frac{ρl}{\pi*r^{2}} , didn't you?

Typo, thank you!

Solve the problem symbolically. If the resistance can differ from R by one percent it can be 1.01 R or 0.99 R. What are the corresponding diameters and how are they related to the original D? By what percent are they different from it?

Big help! Turns out I was off by a decimal point. My answer makes a lot more sense now. Thank you very, very much.
 
Another clarification-- the D in the denominator is squared. Does that mean I need to take the square root of my result?
 
What is your result at all?

ehild
 
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