Calculate 0.100 ohms Resistance of 5.46m Lead Strip

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Homework Help Overview

The discussion revolves around calculating the resistance of a lead strip and determining the necessary length of a copper strip to achieve a specific effective resistance when placed in parallel. The problem is situated within the context of electrical resistance and material properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the resistance formula R = ρL/A and the method for combining resistances in parallel. Questions arise regarding the calculation of the required length of copper to achieve the desired effective resistance.

Discussion Status

The conversation includes attempts to calculate the resistance of the lead strip and explore the implications for the copper strip. Some participants provide guidance on using the relevant equations, while others express uncertainty about specific calculations and seek clarification on their approach.

Contextual Notes

Participants note challenges with mathematical notation and the need for specific values related to resistivity. There is an indication of confusion regarding the calculations necessary to find the required resistance of the copper strip.

chris85
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Homework Statement

I have a question from an AS level physics paper.Calculate the resistance of a 5.46m length of lead strip,0.450cm wide and 1.25mm thick.What length of similar strip made of copper would have to be put in parallel with the lead strip for the effective resistance of the combination to be 0.100 ohms? (p of Cu and Pb are 1.72x10*-8 and 20.6x10*-800



Homework Equations


I worked the resistance of the lead to be 2 ohms using R+pL/A i know the equation needed is 1/R = 1/r1 + 1/r2 but I am not sure how to find the length of copper needed to get the combine resistance?



The Attempt at a Solution

 
Last edited:
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i would post up my working but i don't know how to get all the sighns/symbols up :redface: i have just joined.

Chris.
 
Welcome to the PF, chris85. It looks like you are doing all the right things. Just keep using the equation for the resistance of a uniform bar:

R = \frac{\rho L}{A}

and the equation for combining parallel resistances. If the lead is 2 Ohms (I didn't check your math), what resistance do you need to put in parallel to get down to 0.1 Ohms total? And then how long would a copper bar of the same cross-sectional area A have to be to make that resistance?

BTW, there is a LaTex tutorial in the Tutorials forum: https://www.physicsforums.com/showthread.php?t=8997
 
Last edited:
ok well for the resistance of the lead i did

20.6X10^-6 X 5.46M
__________________ = 2.00 ohms so R1 = 1/2 correct?
5.62X10^-7

Now the resistivity of copper is 1.72x10^8 but I'm having trouble finding a value of resistance for the copper that when added to the lead will give a reciprocol 0.1 for the parallel resistance can someone give me a hint where I'm going wrong?
 

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