Originally Posted by **studiot** And this? |

Well, $\displaystyle T_0$ is usually 20 degrees Celsius and quick look online verifies that the resistivities are valid for that temperature, but now that you've asked me to look at it, I think made a mistake because the result shouldn't depend on whether Celsius or Kelvin systems are used, yet the result I obtained does. The mistake I made was in the removal of the temperature-dependence. The result should have included the $\displaystyle T_0$ terms because it is the change in resistance with temperature difference that should be removed. That would mean that

$\displaystyle \rho_{0,t} l_c \alpha_c (T-T_0) + \rho_{0,c} l_c \alpha_c (T - T_0) = 0$

Which, following through the same mathematics, would yield

$\displaystyle \frac{l_t}{l_c} = \frac{\rho - \rho_{0,c}}{\rho - \rho_{0,t}}$

Which is not very useful because we don't know the compound resistivity.

I think your solution is better. After all, if you differentiate the RHS of the equation with the resistances in series, then

$\displaystyle \frac{dR}{dT} = \rho_{0,t}l_t\alpha_t + \rho_{0,c}l_c \alpha_c$

Setting this to zero (because the resistance shouldn't change with temperature in the compound wire) gives

$\displaystyle \rho_{0,t}l_t\alpha_t + \rho_{0,c}l_c \alpha_c = 0$

$\displaystyle \frac{l_t}{l_c} = - \frac{\rho_{0,c} \alpha_c}{\rho_{0,t} \alpha_t}$

which is the same result.