Understanding Resistance and Temperature in Electrical Circuits

  • Thread starter Thread starter feelau
  • Start date Start date
  • Tags Tags
    Resistance
Click For Summary
SUMMARY

This discussion focuses on solving two resistance problems in electrical circuits. The first problem involves calculating the resistance of a rod composed of two materials with different resistivities, where the total resistance is the sum of individual resistances calculated using the formula R = ρ(l/A). The second problem addresses how temperature affects current in a copper wire, utilizing the formula Rf = Ri(1 + α(Tf - Ti)), where α is the temperature coefficient of resistivity. Participants emphasize the importance of correctly identifying reference temperatures and the need to adjust calculations accordingly.

PREREQUISITES
  • Understanding of electrical resistance and resistivity
  • Familiarity with the formula R = ρ(l/A) for calculating resistance
  • Knowledge of temperature coefficients of resistivity
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the concept of resistivity and its dependence on material properties
  • Learn how to apply the formula R = ρ(l/A) in complex scenarios
  • Research the effects of temperature on resistance in conductors
  • Explore the significance of reference temperatures in resistivity calculations
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of resistance and temperature effects in electrical circuits.

feelau
Messages
61
Reaction score
0
2 resistance problems, please help

Homework Statement


Um so I have two problems, they should be fairly easy after explanation. So the first problem is that there's a rod, 65 cm long. 25 cm of it is a material with resistivity a and the rest of it is another material with resistivity b. cross section is same throughout and I'm suppose to find resistance.
The second problem is a temperature problem, so we have an initial temp of 60C and the measured current on copper wire is 1.3A the same voltage is applied when it is -88C with the same wire. We're suppose to find the new current.

The Attempt at a Solution


So for the first problem I don't understand how to do this at all. I think integration might be involved but I'm not sure. For the second problem I've set up an equation with the formula they provided in textbook: Rf=Ri(1+alpha(Tf-Ti)) alpha is temp coeficient of resistivity which is in book. So my equation is V/If=(V/1.3A)(1+.0039(-88-60)) because same voltage is applied voltage cancels out. I then solved for I final but the answer is not correct. I don't know what I'm doing wrong. Any advice will help. Thank you
 
Physics news on Phys.org
Question 2: Check your formula for resistance.

The constant .0039 will only apply when Ti is a fixed reference temperature. From your number 0.0039, I would guess the reference temp is either 20 degC or 300 degK - check in your book.

Assuming the reference temperature was 20C: if the resistance is R at 20C, then it is R(1 + .0039(60-20)) at 60C and R(1 + .0039(-88-20)) at -88C. The value of R wll cancel out, like the voltage did.

Question 1: You should have a formula for the resistance of a rod in terms of its resistivity, length, and cross section area. The two parts of the rod act like two resistors in series.
 
Last edited:
oh! i didn't realize that there's a reference temperature thing. Thanks. But for question 1, I tried that approach and its not correct :( k so the resistivity is 6.00x10^-3, one side is 25 cm long, then other is 40 cm long. the cross section is 3mm. I figured that since both sides have same resistivity i can just use R= p(l/A) to get resistance but apparently that's not the right way. So I have no idea how to approach this
 
feelau said:
I figured that since both sides have same resistivity i can just use R= p(l/A) to get resistance but apparently that's not the right way.
According to your first post, the resistivities are different because the materials are different. The total resistance would then be the sum of the individual resistances
 
yes, you see, this homework system is on the internet and numbers are randomized. I however, ended up with same resistivity for both material so the equation should just be resistivity multiply by total length of rod and divided by cross section area correct? But when I enter this answer into the homework system, it says my answer is not correct. I also tried doing the resistance separately and I end up with the same answer which is 1.3 ohms so I'm not really sure what I should do now
 
nvm i got it, i misread something, the supposedly cross section area is only one side, I had to square it. thanks for all the help!
 

Similar threads

Replies
9
Views
1K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
1K
Replies
44
Views
4K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 10 ·
Replies
10
Views
917
Replies
14
Views
2K