Change in Resistance due to Thermal Expansion

Click For Summary
SUMMARY

The discussion centers on calculating the temperature of a tungsten filament in a lightbulb, which has a resistance of R0=12 Ohms at 20 degrees Celsius and 140 Ohms when hot. The filament's temperature was determined to be 2400 degrees Celsius, considering thermal expansion with a coefficient of 5.5 x 10^-6 degrees Celsius^-1 and a temperature coefficient of resistivity of 0.0045 degrees Celsius^-1. Participants sought assistance in calculating the percentage change in resistance due to thermal expansion and changes in resistivity, indicating a need for clarity on the relevant equations.

PREREQUISITES
  • Understanding of thermal expansion and its effects on materials
  • Familiarity with the temperature coefficient of resistivity
  • Knowledge of resistance calculations in electrical circuits
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the derivation of the equation for resistivity change: rho(T)=rho(0)(1+alpha(T-T(0)))
  • Learn how to calculate percentage changes in resistance using formulas: R(expansion) - R(0)/R(0) and R(change in rho) - R(0)/R(0)
  • Explore the properties of tungsten and its thermal expansion characteristics
  • Investigate the impact of temperature on electrical resistance in various materials
USEFUL FOR

Students in physics or electrical engineering, educators teaching thermodynamics or materials science, and professionals involved in the design of electrical components that experience thermal changes.

munchy35
Messages
25
Reaction score
0

Homework Statement



The filament of a lightbulb has a resistance of R0=12 Omegas at 20 degrees Celsius and 140 Omegas when hot.


a) Calculate the temperature of the filament when it is hot, and take into account the change in length and area of the filament due to thermal expansion (assume tungsten for which the thermal expansion coefficient is approx 5.5 * 10^ - 6 degrees Celsius^-1.

Assuming an average temperature coefficient of resistivity alpha = 0.0045 degrees celsius^-1.

b.) In this temperature range, what is the percentage change in resistance due to thermal expansion?

c.) What is the percentage change in resistance due solely to the change in rho(p)?

Homework Equations




rho(T)=rho(0)(1+alpha(T-T(0))

part b. ) R(expansion) - R(0)/R(0) = %

part c.) R(change in rho) - R(0)/R(0) = %

The Attempt at a Solution



Well part a, I figured out...it's 2400 degrees Celsius.

I just don't know how to do the other parts. Any direction.equations would be helpful. =/
 
Physics news on Phys.org
nevermind. figured it out!
 
i have the same problem, how did you do parts b and c?
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
Replies
3
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 13 ·
Replies
13
Views
21K