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 $\displaystyle \alpha $ and $\displaystyle R_0$ from this

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

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

where $\displaystyle \alpha $ is a constant known as *temperature coefficient of resistence* of the material. If a coil has a resistence of $\displaystyle 4.0\Omega $ at $\displaystyle 20^o C$ and $\displaystyle 4.52 \Omega $at $\displaystyle 80^o C$ calculate *(a)* the *temperature coefficient of resistance *of the material; *(b)*the resistance of the coil at $\displaystyle 100^o C$

*(a)*

$\displaystyle R_{20}=R_o(1+20\alpha)=4.00\Omega$

$\displaystyle R_{80}=R_o(1+80\alpha)=4.52\Omega$

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

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

thanks ahead for help explaining this.(Cool)