temperature coefficients

• Sep 30th 2010, 08:41 PM
bigwave
temperature coefficients
Temperature coefficients of resistance for copper and platinum are respectively
$0.00427^oC^{-1}$, and $0.00357^oC^{-1}$
Resisters made of copper and platinum have respective resistences $17\Omega$ and $18\Omega$ at $20^oC.$
Calculate the temperature at which the two resistances will be equal. Ans. $165.4^oC$

I am very new to this but knowing that

$R_t=R_o\Left(1+ \alpha T \Right)$

using this simultaneously is the way to solve it. however apparently it assumes some things that I don't know to set this up....

• Sep 30th 2010, 09:46 PM
Gusbob
Couldn't you just substitute your values for copper and platinum individually to find R0 for both metals? (i.e. for copper sub Rt = 17, a = 0.00427, T = 20)

Then you could just eqaute R0(copper)(1+a(copper)T) = R0(platinum)(1+a(platinum)T) and solve for T.
• Oct 6th 2010, 09:19 PM
bigwave
here is an example problem
Quote:

Originally Posted by Gusbob
Couldn't you just substitute your values for copper and platinum individually to find R0 for both metals? (i.e. for copper sub Rt = 17, a = 0.00427, T = 20)

Then you could just eqaute R0(copper)(1+a(copper)T) = R0(platinum)(1+a(platinum)T) and solve for T.

not sure if I follow just how this is plugged in. but here is an example problem to show somewhat how this is supposed to be solved... however I don't know how they got $\alpha$ and $R_0$ from this

Electrical conductivity, and hence resistence, is a slowly varying function of temperature. To a good approximation, the Resistance $R_T$ at temperature $T(^o C)$ is related to the resistance $R_o$ at $0^o$ by:

$R_T=R_o(1+\alpha T)$

where $\alpha$ is a constant known as temperature coefficient of resistence of the material. If a coil has a resistence of $4.0\Omega$ at $20^o C$ and $4.52 \Omega$at $80^o C$ calculate (a) the temperature coefficient of resistance of the material; (b)the resistance of the coil at $100^o C$
(a)
$R_{20}=R_o(1+20\alpha)=4.00\Omega$
$R_{80}=R_o(1+80\alpha)=4.52\Omega$

Solving simultaneously, $\alpha = 2.27 \times 10^{-3}^o C^{-1}$ and $R_o = 3.38\Omega$

(b)
$R_{100}=(3.83)[1+10092.27\times10^-3] = 4.69\Omega$

thanks ahead for help explaining this.(Cool)