How Does Resistivity Affect Current Flow in Unconventional Conductors?

AI Thread Summary
The discussion focuses on calculating the resistance of a rubber band and a penny using the formula R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. For the rubber band, the resistivity is given as 1E13 ohm-m, and its dimensions are 19 cm long and 2.2 mm in diameter. For the penny, which is made of copper, users are advised to look up the specific resistivity value for copper and use its dimensions of 1 mm thick and 2 cm in diameter. Participants emphasize the importance of unit conversion and accuracy in calculations. The thread concludes with a warning against attempting these experiments at home.
jen0519
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Homework Statement


(a) Assuming a resistivity of 1E13 (horseshoe thing)m, calculate the resistance between the ends of a rubber band that is 19 cm long and 2.2 mm in diameter.


(b) Calculate the resistance between the 'heads' and 'tails' sides of a penny that is 1 mm thick and 2 cm in diameter.


(c) WARNING! Do not try this at home! What is the expected current that each would carry if it were connected across a 108-V power supply?
rubber band (A)
penny (A)

Homework Equations



R=pL/a

The Attempt at a Solution



So I thought for the first one that I would use R =pL/A but I think you have to change some of the numbers and I'm not sure how.
 
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jen0519 said:

Homework Statement


(a) Assuming a resistivity of 1E13 (horseshoe thing)m, calculate the resistance between the ends of a rubber band that is 19 cm long and 2.2 mm in diameter.


(b) Calculate the resistance between the 'heads' and 'tails' sides of a penny that is 1 mm thick and 2 cm in diameter.


(c) WARNING! Do not try this at home! What is the expected current that each would carry if it were connected across a 108-V power supply?
rubber band (A)
penny (A)


Homework Equations



R=pL/a


The Attempt at a Solution



So I thought for the first one that I would use R =pL/A but I think you have to change some of the numbers and I'm not sure how.

Welcome to the PF. Yes, that is the correct equation to use. You are given the resistivity rho, and just need to use the dimensions you are given to get the length L and the cross-sectional area A. Be careful with your units...

And for the penny, you will need to look up the resistivity rho number of copper.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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