Resistance of a Wire at varying temperatures

In summary: It's hard to say definitively without knowing the specific material properties and geometry, but your approach seems sound.
  • #1
Enharmonics
29
2

Homework Statement


Determine the resistance of wire at room temperature and at the melting point of the given material.

GIVEN:
Material = Aluminum
Diameter of wire ##= 0.5 mm = 5.0 * 10^{-4} m##
Length of wire ## = 0.5 m = 5.0 * 10^{-1} m##
Temperature Coefficient: ##\alpha = 3.9 * 10^{-3}C^{-1}##
Base Resistivity: ##\rho_{0} = 2.82 * 10^{-8} \Omega * m##
Melting Point of Aluminum: ##T = 660.3^{\circ} C##
Room Temperature: ## T_{0} = 20^{\circ} C##

Homework Equations



Variation of Resistivity w/ Temperature: ##\rho = \rho_{0}(1 + \alpha(\Delta T)##

Pouillet's Law: ##R = \rho * \frac{L}{A}##

The Attempt at a Solution



I've already finished working the problem out, but in doing so, I had to make a few assumptions I'm not sure are correct. Actually, I'm not even sure if my methodology is correct to begin with.

So, for the Resistance at room temperature, I assumed that the Resistivity given for Aluminum in my textbook is its resistivity at room temperature (not sure if that's right).

The second assumption I made is that the cross-sectional area A of a wire would be given by

$$ A = \pi r^2 = \pi *(\frac{5.0 * 10^{-4} m}{2})^{2}) = 2.0 * 10^{-7} m^{2}$$

Which is based purely on me picturing what each "slice" would look like if I were to cut across a wire an arbitrary number of times. Again, not sure if this is valid, but a circle seemed like the closest geometric shape to what I pictured.

If that's the case, then applying Pouillet's Law yields

$$ R_{room} = \rho_{0} * L/A = (2.82 * 10^{-8} \Omega * m)(\frac{5.0 * 10^{-1} m}{2.0 * 10^{-7} m^{2}}) = 7.1 * 10^{-2} \Omega$$

Then calculating the resistivity of the wire at the melting point I've got

$$ \rho_{mp} = (2.82 * 10^{-8} \Omega * m) + (2.82 * 10^{-8} \Omega * m)(3.9 * 10^{-3}C^{-1})(660.3^{\circ} C - 20^{\circ} C) = 9.9 * 10{-8} \Omega * m$$

So the Resistance of the wire at its melting point is

$$R_{mp} = \rho_{mp} * L/A = (2.82 * 10^{-8} \Omega * m)(\frac{5.0 * 10^{-1} m}{2.0 * 10^{-7} m^{2}}) = 2.5 * 10^{-1} \Omega$$

But is this correct? I can't help but feel like the Resistance of the wire should be higher at the melting point than at room temperature for some reason.
 
Physics news on Phys.org
  • #2
In your last equation line you seem to have copied in the wrong resistivity (ρ0 instead of ρmp), but I take it that's just a typo.
The final answer looks reasonable (it is more than the room temperature resistance!)
 

Related to Resistance of a Wire at varying temperatures

1. How does the resistance of a wire change with temperature?

The resistance of a wire generally increases with temperature. This is because as the temperature increases, the atoms in the wire vibrate more, causing more collisions with the moving electrons and impeding their flow. This increase in collisions results in a higher resistance.

2. Why does the resistance of a wire increase with temperature?

The increase in resistance with temperature is due to the change in the material's resistivity. As the temperature increases, the resistivity of most materials also increases, making it more difficult for electrons to flow through the wire and resulting in a higher resistance.

3. How does the length of the wire affect its resistance at different temperatures?

The length of a wire is directly proportional to its resistance. This means that as the length of the wire increases, its resistance also increases. This relationship remains consistent at different temperatures, meaning that a longer wire will have a higher resistance at both high and low temperatures.

4. How does the material of the wire influence its resistance at varying temperatures?

Different materials have different resistivity values, which affect their resistance at varying temperatures. For example, a copper wire will have a lower resistance compared to an iron wire of the same length and thickness, as copper has a lower resistivity. This difference in resistivity also means that the resistance of a copper wire will increase less with temperature compared to an iron wire.

5. Can the resistance of a wire be accurately predicted at different temperatures?

While the resistance of a wire can be estimated at different temperatures using mathematical models, it is not always accurate due to other factors that may affect resistance, such as impurities in the wire or external factors like humidity. Experimentation is the most accurate way to determine the resistance of a wire at varying temperatures.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
809
  • Introductory Physics Homework Help
Replies
2
Views
879
  • Introductory Physics Homework Help
Replies
3
Views
188
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
17
Views
594
  • Introductory Physics Homework Help
Replies
8
Views
1K
Back
Top