# Resistance problem (Simple electrical problem)

1. Mar 10, 2012

### Rito3d03

1. The problem statement, all variables and given/known data
This question comes from one of my previous exam
and I still couldn't solve it after days of trying
hope someone can guide me through this thanks

A copper wire has a resistance of 10Ω at 10°C.
Determine the resistance of the wire at 80°C.
Given that the temperature coefficient of the wire is 0.00393C-1 at 0°C.

2. Relevant equations
R2 = R1(1+$\alpha$1*ΔT)

3. The attempt at a solution
I don't know where to start when I don't have the temperature coefficient of R1

I know this may look silly because it maybe very easy to all of you

2. Mar 10, 2012

### cepheid

Staff Emeritus
We're not talking about two different components R2 and R1.

We're talking about the same component, which has two different resistances at two different temperatures.

R1 is the resistance at temperature T1

R2 is the resistance at temperature T2.

You can confirm this by noticing that if T2 = T1, then delta T = 0, and hence R2 = R1.

In this case, delta T = T2 - T1 = 80 C - 10 C = 70 C, and you already know R1, which is the resistance at T1 = 10 C. So all you have to do is use the equation to solve for R2.

3. Mar 10, 2012

### Rito3d03

but how about the $\alpha$1
I can use the coefficient at 0 C to solve different temperature?

4. Mar 10, 2012

### cepheid

Staff Emeritus
I think you can assume alpha is constant. It's not strictly true, but the linearity holds within a certain temperature range around the temperature at which the coefficient was measured. 80 degrees may seem like a fairly large range, but unless you've been given additional information about how the alpha coefficient itself varies, I don't see any other choice other than using the given value and assuming it to be constant with temperature.

5. Mar 10, 2012

### Rito3d03

I finally find out the way to solve this question
first find out R0 which is the resistance at 0°C
10 = R0 (1+0.00393*(10-0))
then I can use the equation to solve the rest of the problem
and figure out R80
R80= R0 (1+0.00393*(80-0))

I guess i was stuck on some logic problem
glad i can finally finish this and move on
thanks for the help