How Should L and A Change to Adjust Power Dissipation and Current in a Wire?

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    Power Resistivity
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Homework Help Overview

The discussion revolves around the relationship between the dimensions of a wire (length L and cross-sectional area A), resistivity, and their effects on power dissipation and current when a potential difference is applied. Participants are exploring how to adjust L and A to achieve specific changes in power dissipation and current.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand the relationship between resistance, power, and the dimensions of the wire, questioning whether L and A can change independently. Some participants raise the concept of volume conservation when stretching the wire and its implications for power dissipation.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of how the dimensions of the wire relate to power and current. Some guidance has been offered regarding the conservation of volume, but no consensus has been reached on the specific relationships between L, A, and power dissipation.

Contextual Notes

Participants note that the volume of the wire remains constant during stretching, which introduces constraints on how L and A can change in relation to each other. There is also an acknowledgment of the complexity in determining the new values of L and A based on the desired changes in power and current.

positron96
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A potential difference V is applied to a wire of cross section A, length L, and resistivity p. You want to change the applied potential difference and stretch the wire so that the energy dissipation rate is multiplied by 35 and the current is multiplied by 3. What should be the new values of L and A [in relationship to the old values]?

I know that R = pL/A and P = IV (etc.), but it doesn't make sense that they'd want specific answers for both L and A. Can't either one change without the other and still affect the power?
 
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HINT: The volume of the metal does not change upon stretching.
 
So then volume V = AL. And if you're stretching it then A gets smaller and L gets bigger. But what does volume have to do with power?
 
Last edited:
Fine. It doesn't matter anymore. Thanks.
 
Resistance can be found from L, A, and rho. Power can be found with P = IV = V^2/R.

- Warren
 

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