Physics Help Forum The cable resistence constant with the temperature

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Jul 2nd 2018, 02:18 AM   #11
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 Originally Posted by studiot |The words steamroller and nuts come to mind, can you take it to its conclusion to show the maths works?
Sure.

From this:

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

Expand brackets and move $\displaystyle l_t$ and $\displaystyle l_c$ to LHS and RHS respectively

$\displaystyle \rho l_t + \rho_{0,t} l_t \alpha_t T_0 - \rho_{0,t} l_t = -\rho l_c - \rho_{0,c} l_c \alpha_c T_0 + \rho_{0,c} l_c$

Factorise

$\displaystyle l_t \left(\rho + \rho_{0,t}\alpha_t T_0 - \rho_{0,t} \right) = - l_c \left(\rho - \rho_{0,c} \alpha_c T_0 + \rho_{0,c}\right)$

Then

$\displaystyle \frac{l_t}{l_c} = -\frac{\rho - \rho_{0,c} + \rho_{0,c} \alpha_c T_0}{\rho - \rho_{0,t} + \rho_{0,t} \alpha_t T_0}$

Jul 2nd 2018, 04:47 AM   #12
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And this?

 And where was To given in the original question.

Jul 2nd 2018, 06:44 AM   #13
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 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.

Last edited by benit13; Jul 2nd 2018 at 06:46 AM.

 Tags cable, constant, resistence, temperature

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