How Does Resistivity Affect Current Flow in Unconventional Conductors?

[jen0519]

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.

 

[berkeman]

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.

 

[kerimek]

Dear student,

Thank you for your question. Let's first review the basics of resistance and Ohm's law. Resistance is a measure of how much a material opposes the flow of electric current. It is represented by the symbol R and is measured in ohms (Ω). Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them. Mathematically, this can be represented as I=V/R, where I is the current in amperes (A), V is the voltage in volts (V), and R is the resistance in ohms.

Now, let's apply this knowledge to the given problems. For part (a), we are given the length and diameter of a rubber band, as well as its resistivity (ρ). The first step is to convert the diameter from millimeters to meters, which gives us a value of 0.0022 m. Next, we can calculate the cross-sectional area of the rubber band using the formula A=πr^2, where r is the radius (half of the diameter). Plugging in the values, we get A=3.14*(0.0011)^2=3.8E-6 m^2. Finally, we can use the formula R=pL/A to calculate the resistance. Plugging in the values, we get R=(1E13)*(0.19)/(3.8E-6)=5E8 Ω. This means that if the rubber band were connected across a 108-V power supply, it would carry a current of approximately 0.2 μA (using I=V/R).

For part (b), we are given the thickness and diameter of a penny. Similarly, we can convert the diameter from millimeters to meters, which gives us a value of 0.02 m. Then, we can calculate the cross-sectional area using the formula A=πr^2, where r is the radius (half of the diameter). Plugging in the values, we get A=3.14*(0.01)^2=3.1E-4 m^2. Finally, we can use the formula R=pL/A to calculate the resistance. Plugging in the values, we get R=(1E13)*(0.001)/(3.1E-4)=3.2E10 Ω.