Coefficient of resistance formula

[Weightofananvil]

Hi all,
I've been a long time reader of this forum but haven't signed up for whatever reason.
So thanks for the wealth of information and solutions to my headaches.

I'm trying to figure out a coefficient of temperature equation and I'm having a hard time with the manipulation
for some reason.

I know the formula is:
a=alpha

R2=R1[1+a(T2-T1)]

However, the variable I'm solving for is T2.
I'm given R1,T1,R2,Coefficient of metal @ 20 degrees Celsius.
I'm a little ashamed because I know this isn't hard at all but its been a long time
since I've been at this stuff.

Thanks.

 

[olaney]

R2/R1=1+aT2 -aT1
R2/R1 -1 +aT1 = aT2
1/a * [R2/R1 -1 +aT1] = T2

 

[Weightofananvil]

R2/R1=1+aT2 -aT1
R2/R1 -1 +aT1 = aT2
1/a * [R2/R1 -1 +aT1] = T2

Thanks olaney. I actually only had 1 little thing wrong..
Can you elaborate on why [r2/r1 -1 +aT1] end up in brackets?

I ended up with 1+ r2/r1 -1 +aT1 / a
I realize you just devided by alpha to remove it from T2.
Just trying to understand why it just goes 1/a instead of 1 over the whole works?

Is it because of the brackets in the initial formula?
could your second line be written [r2/r1 -1 +at1] = aT2 ?

 

[Weightofananvil]

Hey,
still struggling with this question

formula is: R2=R1[1+a(T2-T1)]

T2= ?
T1 = 20 Celcius
R1 = 1000 Ohms
R2 = 880 Ohms
Alpha = at 20 celcius is 0.006C^-1

Any help would be greatly appreciated.

 

[meBigGuy]

Can you elaborate on why [r2/r1 -1 +aT1] end up in brackets?

Because, given R2/R1 -1 +aT1 = aT2, you need to multiply every term on the left by 1/a.

You could just say (R2/aR1) -(1/a) + (T1) = T2.

 

[sophiecentaur]

It sounds very trite to say this but, as my old Maths master (and every other one else, probably) used to tell us "do the same thing to both sides of an equation and you still have equality. That's the basis of all 're-arrangements' in algebra to bring the wanted variable to one side. People get very good at this and can often miss out steps without going wrong. What they write down in the short form can be confusing. If you go just one step at a time for each line of the process, you can't (well you know what I mean) go wrong. Practice, practice, I'm afraid.

 

[Weightofananvil]

Because, given R2/R1 -1 +aT1 = aT2, you need to multiply every term on the left by 1/a.

You could just say (R2/aR1) -(1/a) + (T1) = T2.

It sounds very trite to say this but, as my old Maths master (and every other one else, probably) used to tell us "do the same thing to both sides of an equation and you still have equality. That's the basis of all 're-arrangements' in algebra to bring the wanted variable to one side. People get very good at this and can often miss out steps without going wrong. What they write down in the short form can be confusing. If you go just one step at a time for each line of the process, you can't (well you know what I mean) go wrong. Practice, practice, I'm afraid.

Thanks both for you help.
I found out that it wasn't the formula rearrangement that was my error. I goofed in reading the measurement for temperature coefficient. It being in ^-1Celcius made me believe it was the coefficient multiplied by ^-1. This led to very high abstract numbers. Silly mistake... In the end me big guy and me both had the same formulas just written a little different.

sophiecentaur - Thanks for the comment. Coming back into doing math after a few years it is very true its always the small things your missing and not some huge aspect of it. As my mistake was here, I just need to slow down a bit.

 

[sophiecentaur]

Stick at it.

 

[jim hardy]

Mr Anvil
the guys are right it's an arithmetic exercise not a physics one.

I know the formula is:
a=alpha

R2=R1[1+a(T2-T1)]

expand that
R2 = R1 + aR1 (T2 - T1)
re-arrange
R2= aR1(T2-T1) + R1
which looks a lot like the y=mx + b we learned to work in grade school.
aR1 is the slope , (T2-T1) is x, and R1 the intercept "b" ? Do i remember the names right

There's a reason they use the first format for temperature measurements
and that is every metal alloy has its own particular coefficient of temperature "a"
but no two sensors will have exactly the same R1, in fact resistive sensors come in 10 ohm, 25 ohm, 100 ohm, 200 ohm , 1000 ohm, & so on.
So even if they are made from the same metal they'll all have different slopes. Same "a" X different R1's = different slopes.
So, they give you R1 and "a" for a sensor, and let you figure out value of aR1.

Here's some a's from hyperphysics
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/rstiv.html

looks like you might have a 1000 ohm iron-ish wire sensor ?you might read up on the "Callendar Van Dusen" equation.
It is used a lot in temperature measurement. It includes a square term to allow a closer fit to the slightly non-linear R vs T curve Mother Nature gave metals

 

[Weightofananvil]

Mr Anvil
the guys are right it's an arithmetic exercise not a physics one.

expand that
R2 = R1 + aR1 (T2 - T1)
re-arrange
R2= aR1(T2-T1) + R1
which looks a lot like the y=mx + b we learned to work in grade school.
aR1 is the slope , (T2-T1) is x, and R1 the intercept "b" ? Do i remember the names right

There's a reason they use the first format for temperature measurements
and that is every metal alloy has its own particular coefficient of temperature "a"
but no two sensors will have exactly the same R1, in fact resistive sensors come in 10 ohm, 25 ohm, 100 ohm, 200 ohm , 1000 ohm, & so on.
So even if they are made from the same metal they'll all have different slopes. Same "a" X different R1's = different slopes.
So, they give you R1 and "a" for a sensor, and let you figure out value of aR1.

Here's some a's from hyperphysics
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/rstiv.html
View attachment 89452

looks like you might have a 1000 ohm iron-ish wire sensor ?you might read up on the "Callendar Van Dusen" equation.
It is used a lot in temperature measurement. It includes a square term to allow a closer fit to the slightly non-linear R vs T curve Mother Nature gave metals

Thanks for the reply Jim! That was helpful