Changes in Resistance due to Temperature

Click For Summary
SUMMARY

The discussion focuses on the changes in resistance, length, and cross-sectional area of a copper conductor when subjected to a temperature change of 5.0 °C. The relevant equations include the linear expansion formula, delta L = L initial (delta T) (alpha), and the resistance formula R = ρL/A. The percentage changes calculated for length and area are 0.0085% and 0.017%, respectively. The correct approach requires determining the resistivity (ρ) of copper at different temperatures to accurately compute the resistance change.

PREREQUISITES
  • Understanding of linear thermal expansion and its coefficient.
  • Familiarity with the relationship between resistance, length, area, and resistivity.
  • Knowledge of the properties of copper as a conductor.
  • Basic algebra for manipulating equations involving resistance and physical dimensions.
NEXT STEPS
  • Research the temperature dependence of resistivity for copper.
  • Learn how to apply the formula R = ρL/A in practical scenarios.
  • Explore the concept of thermal expansion in different materials.
  • Study the effects of temperature on electrical conductivity in metals.
USEFUL FOR

Students studying physics or engineering, electrical engineers, and anyone interested in the thermal properties of materials and their impact on electrical resistance.

dari09
Messages
2
Reaction score
0

Homework Statement



When a metal rod is heated, not only its resistance but also its length and its cross-sectional area change. The relation R = L/A suggests that all three factors should be taken into account in measuring at various temperatures.
(a) If the temperature changes by 5.0 C°, what percentage changes in R, L, and A occur for a copper conductor? The coefficient of linear expansion is 1.7 10-5/K.


Homework Equations



delta L = L initial (delta T) (alpha)

R = density L/A

The Attempt at a Solution



I got the answer to the percentage changes of L and A using the equation
L = L initial (delta T) alpha.
I tried to plug those answers into the equation R = density L/A which gives the answer.5
Unfortunately that answer is wrong and and now I am currently lost.

The percent changes of L and A are:
delta L / L initial = .0085%
delta A / A initial = .017%
 
Physics news on Phys.org
Hi dari09, welcome to PF. Unfortunately, [itex]\rho[/itex] is used to mean multiple things in engineering, including density and (in this case) resistivity, which is a material property with units [itex]\Omega\cdot m[/itex]. You can see how that correctly makes ohms the unit of resistance.

You need to find the resistivity [itex]\rho[/itex] of copper at the two different temperatures.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K