Temperature Coefficient of Resistivity

[piratebill]

Hi

I was wondering if anyone could help. My teacher isn't very good at explaining things and he's given us a question without going through the reasoning behind the equation.

Aluminium wire has a resistance of .250Ω at 0°C. Give that the temperature coefficient of resistivity at 0°C for aluminium is 0.004 find the resistance of the wire at 200°C

My answer is Rt = .250 x (1+.004x200) = 0.45Ω

What I don't get however is why isn't it just .250Ω + (0.004 x 200) which would equal 1.05Ω

I thought the temperature coefficient was how much the resistance increases per increased degree, which you then add to the original resistance.

If anyone could help explain ild be very grateful :)

 

[berkeman]

Hi

I was wondering if anyone could help. My teacher isn't very good at explaining things and he's given us a question without going through the reasoning behind the equation.

Aluminium wire has a resistance of .250Ω at 0°C. Give that the temperature coefficient of resistivity at 0°C for aluminium is 0.004 find the resistance of the wire at 200°C

My answer is Rt = .250 x (1+.004x200) = 0.45Ω

What I don't get however is why isn't it just .250Ω + (0.004 x 200) which would equal 1.05Ω

I thought the temperature coefficient was how much the resistance increases per increased degree, which you then add to the original resistance.

If anyone could help explain ild be very grateful :)

Welcome to the PF.

The only error that you've made is that the coefficient of resistivity applies to differences in temperature. So if you start at 25C and go to 200C, what is the delta temperature?

Oh, and it's best to put units on everything in your calculations. What are the units on the coefficient of resistivity?

 

[berkeman]

Actually, there may be one more error, but I'm not sure. Please write out the full corrected equation and show the answer so we can check it. Thanks.

 

[piratebill]

hi, thanks for the welcome :)

The question I have is:

A long thin aluminium wire has a resistance of 0.250Ω at 0°C. Given that the temperature coefficient of resistivity at 0°C for aluminium is 4x10-3/0°C find the resistance of the wire at 200°C

Ive simply transported my values into this equation:

RT = Rref [1 + ɑ(T-Tref)]

R200°C = R0°C [1 + ɑ(200-0)]

R200°C = 0.250Ω (1 + 4x10-3 x 200)

R200°C = 0.45Ω

But that's just blindly following the formula. What i don't get is that i thought the temperature coefficient was just the measure of change in resistance as the temperature increases.

So an increase of 200°C, would simply mean you multiply the coefficient (4x10-3) by 200 and add that to your original resistance (0.250Ω)

Basically what I am asking is, how come the formula is:

R200°C = 0.250Ω (1 + 4x10-3 x 200)

And not simply

R200°C = 0.250Ω + (4x10-3 x 200)

I mean why do you have to multiply everything in the brackets by 0.250Ω?

 

[Emilyjoint]

The temp coeff of 0.004 means that the resistance increases (changes) by 0.004 of the resistance at 0C for each change of 1C in temp.
That is what the equation R = R0(1 + αt) means
R = R0 + R0αt

 

[piratebill]

sorry i still don't understand why you have to multiply it by R0 :/

 

[berkeman]

sorry i still don't understand why you have to multiply it by R0 :/

Since the increase is just a percentage, you need to multiply that percentage increase by the original resistance to find out what the full change in resistance is. If you have a 1% increase in resistance due to temperature, that will result in twice as big a change in resistance for a 200 Ohm resistor versus a 100 Ohm resistor.

Does that help?

 

[piratebill]

arh, the coefficient is a percentage of the original resistance at 0°C

So to find the actual resistance increase per degree, I need to find out what 0.004% of .250 is

To do that I have to times my original resistance of .250 x the coefficient of 0.004 = 0.001Ω

this means that an increase in each degree causes a change of 0.001Ω in the material

Because I have a 200°C increase i need to multiply 0.001Ω by 200 = 0.2Ω

Add that to my original resistance 0.25Ω + 0.2Ω = 0.45Ω

Many thanks everyone, I very much appreciate the help youve given me :)