# 2 resistance problems,

• feelau
In summary, the conversation discusses two resistance problems, one involving a rod with different resistivities and the other involving a temperature and copper wire. For the first problem, the formula R=p(l/A) is used to find the resistance, but there is a mistake in the calculation. For the second problem, the formula Rf=Ri(1+alpha(Tf-Ti)) is used, with a reference temperature of 20C. The conversation also addresses a formula error in the first problem and clarifies the correct approach for calculating resistance in this scenario.

## 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

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!