How Does Changing Length and Area Affect the Resistance of Copper Wires?

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Homework Help Overview

The discussion revolves around the resistance of copper wires, specifically comparing a wire of length L and cross-sectional area A to another wire of length 2L and area 1/2 A. Participants are exploring how these changes affect resistance using the formula R = ρL/A.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss substituting values into the resistance formula to find the new resistance. Questions arise regarding the value of resistivity (ρ) and its relevance to the problem. There is also speculation about the expected outcome of the resistance calculation.

Discussion Status

The discussion is active, with participants attempting to clarify the relationship between length, area, and resistance. Some guidance has been offered regarding the use of the resistance formula, but there is no consensus on the final answer or the value of resistivity.

Contextual Notes

Participants are considering the resistivity of copper and its implications for the problem, but there is uncertainty about how to apply this information correctly. The specific values for resistivity are being debated, and assumptions about the conditions of the wires are also under discussion.

zelda1850
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Homework Statement



a copper wire of length of length L and cross sectional area A has resistance R. A second copper wire at the same temperature has a length of 2L and a cross sectional area of 1/2 A. what is the resistance of the second copper wire?

Homework Equations



R = pL/A

L = 2
A = 1/2
p = how do i find p

The Attempt at a Solution



p stand for resistivty but how do i know it in this problem
 
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In the relevant equation replace L by 2L and A by A/2 to get new resistance.
 


oh should the answer come out to be 1R
 


yes.
 


zelda1850 said:
oh should the answer come out to be 1R
No.
New resistance R' = ρ*2L/A/2 = ...?
 


i think p would be 1.72 x 10 exponent -8 since it is on copper
so i do 1.72x10 exponent -8 * 2 m/ 2 a is that the right equation?
 

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