Resistance change under temperature

1. Apr 25, 2005

Pengwuino

Ok so we got an aluminum rod that has a resistance of 1.234 ohms at 20 degrees Celsius. I need to calculate the resistance of the rod at 120 degrees by accounting for hte changes in both the resistivity and the dimensions of the rod.

I calculated the resistance would be 1.238 ohms.

Can someone verify this for me?

2. Apr 25, 2005

OlderDan

There is a chance someone would verify your approach if you stated how you did the problem and provided the needed constants such as the thermal expansion coefficients and the resistivity dependence on temperature.

3. Apr 25, 2005

Pengwuino

ok here we go, at 20 degrees C...

expansion coeffecient: 24 * 10-6 C^-1 (yah my book says 24, not 2.4)
resistivity coefficient: 3.9 * 10-3 C^-1

Is there a simpler way to use that latex programming? I could show my work but it seems like it would take an hour to type up the formulas correctly.

4. Apr 25, 2005

OlderDan

So the new length and cross sections of the rod for a temperature change of 100 degrees C are

$$L = L_0(1+24*10^{-4})$$

$$A = A_0(1+24*10^{-4})^2$$

and the new resistivity is

$$\rho = \rho_0(1+3.9*10^{-1})$$

The new resistance is

$$R = \frac{\rho_0(1+3.9*10^{-1})*L_0}{A_0(1+24*10^{-4})}$$

$$R = R_0\frac{(1+3.9*10^{-1})}{(1+24*10^{-4})}$$

$$R = 1.238\Omega*1.387 = 1.717\Omega$$

One of us did it wrong. I think some people have tools for converting to LaTex. I don't have one, and it is a bit tedious.

5. Apr 25, 2005

Pengwuino

We did it in radically different ways too.... someone throw me something that will allow me to do it in less then 5 hours and ill show you lol.

I was looking at it conceptually and thought ok, put a arbitrary number in for A and you can figure out how long a piece of wire would be (I used a wire of 1mm each side so its 1mm^2) and figured the wire would be some 45 meters long or so. I then plugged the length into the length expansion formula and got like a .1 meter expansion and then i plugged in the length into resistance formula using the new p from the resistance-temperature dependance and the resistance change was very small so i figured 2 very small changes should result in a much smaller change and thought my answer could very well be right.

If i just plugged in A = 0.00001m^2 and did all teh calculations correctly, would i still get the right answer (kind of a cheating way of checking your answer?)?

Added: ah crap, i just plugged in 0.001m as the side value (so 1x10^-6m^2) and got 1.717 ohms.... crap, wonder where i went wrong.

Ah double crap! I didnt account for the expansion in both directions. Crap crap crap, i hate my life.... good thing i did this a day in advance :D

Last edited: Apr 25, 2005
6. Apr 25, 2005

OlderDan

Any combination of area, length, and resistivity that gives you the original resistance will work. You can make up any starting numbers you want as long as rho*L/A is 1.238 ohms. My calculation shows that all those numbers are replaced in the end by the original resistance.

7. Apr 25, 2005

Pengwuino

Yah i think leaving A alone screwed me up because it along with the resistivity coefficient just dissappeared in my equation.