Temperature Coefficient of Resistance of Two Wires in Series

[chap126]

Homework Statement

Two wires A and B are connected in series at 0^oC and resistance of B is 3.5 times that of A. The resistance temperature coefficient of A is 0.04% and that of the combination is 0.1%. Find the resistance temperature coefficient of B.

Homework Equations

R_t=R₀(1+α₀t)
α_t=α₀/1+α₀t

The Attempt at a Solution

This is an example problem form A Textbook of Electrical Technology Vol. I. So the solution comes from that book:
It is seen that R_B/R_A = 3.5 = 0.003/(0.001-α) --> α = 0.000143^oC^-1

My problem is that this relationship is not clear to me. From the looks of the given solution it is saying that:
R_B/R_A=α_A-α_AB/α_AB-α_B and then solving for α_B. I'm confused if this is the correct relationship and if so where it came from? How do you relate them without at least one other measurement at some other temperature?

 

[chap126]

Ok, I think I figured it out. The given solution was way too simplified, and I'm still not sure how the author worked it out, but here's what I got:

From the given equation we know

R_t=R₀(1+α₀t) --> R_1,A=R_0,A(1+α_0,A*1)

and because A and B are in series we also know that

-->R_1,A+R_1,B=(R_0,A+R_0,B)(1+α_0,AB)

since the problem tells us that R_0,B = 3.5R_0,A, we can solve for R_1,B in terms of R_0,A

R_1,B=(R_0,A+3.5R_0,A)(1+α_0,AB)-R_0,A(1+α_0,A)

we also know that R_1,B=R_0,B(1+α_0,B)= 3.5R_0,A(1+α_0,B)

Then set the two equations equal and solve for α_0,B as the problem asks

3.5R_0,A(1+α_0,B) = (R_0,A+3.5R_0,A)(1+α_0,AB)-R_0,A(1+α_0,A)

α_0,B=((1+3.5)(1+α_0,AB)-(1+α_0,A)-3.5)/3.5
α_0,B= 1.42x10^-4

If anyone can explain how the author solved it, I'd be happy to see it. His way seemed much faster.

 

[rude man]

The given answer can't be right. α_B, the coefficient of the higher resistor, must be greater than the coefficient of the entire resistance α since α_A, the coefficient of the smaller resistor is less than α. But your teacher "sees" α = 0.000143 ^oC^-1 which is < 0.001 = α. In fact, he/she has α_B < α_A!

Do it like this: assume R = T = 1 without loss of generality.
Then 1(1+α_A) + 3.5(1+α_B) = 4.5(1+α).

 

[chap126]

The given answer can't be right. αB, the coefficient of the higher resistor, must be greater than the coefficient of the entire resistance α since αA, the coefficient of the smaller resistor is less than α. But your teacher "sees" α = 0.000143 oC-1 which is < 0.001 = α. In fact, he/she has αB < αA!

Do it like this: assume R = T = 1 without loss of generality.
Then 1(1+αA) + 3.5(1+αB) = 4.5(1+α).

Now I'm really confused. The answer I got matches the one in the book, but this way gives a different answer. Where did I go wrong? Any hints?

 

[rude man]

Now I'm really confused. The answer I got matches the one in the book, but this way gives a different answer. Where did I go wrong? Any hints?

Problem is, it's often not easy to work through posters' math when a lot of it is posted.

I think my equation is straight-forward enough that it should be understandable. It's just

R(1+α_AT) + 3.5R(1+α_BT) = (R+3.5R)(1+αT)

with R = 1 ohm and T = 1 deg. C.

It should also be intuitively obvious that α_B has to be > α since α_A < α.

 

[Engine Grace]

Hello, this type is problem has a general solution (quite latent tho).

@_a×R_a/(Total resistance) + @_b×R_b/(total resistance) = @ _TOTAL.
NOW JUST SUBSTITUTE...