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 Electricity and Magnetism Electricity and Magnetism Physics Help Forum 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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post MaxBit General Physics 1 Nov 8th 2012 06:42 PM moodylurker08 Kinematics and Dynamics 5 Sep 19th 2009 01:58 AM MEC Kinematics and Dynamics 2 Apr 22nd 2009 10:20 AM SPEEDTRONIC General Physics 1 Feb 26th 2009 09:53 PM