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Jun 28th 2018, 01:39 AM

#1  Junior Member
Join Date: May 2017 Location: Sri Lanka
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 The cable resistence constant with the temperature
I am confused with this question, difficult to find steps to solve it.
A carbon cable and a tungsten cables are connected each and made a cable. The resistivity of tungsten is 5.6*10^8Ωm and carbon is 3.5*10^5Ωm. The temperature coefficient of resistivity ,tungsten=0.0045°C1 and carbon=0.0005°C1. The resistance of resultant cable is not change with the temperature. What is the ratio of the two lengths of the carbon and tungsten cables.(Assume that the two cables have same cross sectional area).
Last edited by osalselaka; Jun 28th 2018 at 02:12 AM.
Reason: corrected the content

 
Jun 28th 2018, 02:07 AM

#2  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 1,035

Originally Posted by osalselaka I am confused with this question, difficult to find steps to solve it.
A carbon cable and a tungsten cables are connected each and made a cable. The resistivity of tungsten is 5.6*10^8Ωm and carbon is 3.5*10^5Ωm. The temperature coefficient of resistivity ,tungsten=0.0045°C1 and carbon=0.0005°C1. The resistance of resultant cable is not change with the temperature. What is the ratio of the two lengths of the carbon and tungsten cables.(Assume that the two cables have same surface area). 
First step to solving this is to get your facts right.
The temperature coefficient for one of these is (must be) negative.
and further the temperature coefficient refers to resistance, not resistivity.
Can you find out which one?
I assume you mean cross sectional area not surface area.
Can you proceed with this information.
Post again if you need further help.

 
Jun 28th 2018, 02:13 AM

#3  Junior Member
Join Date: May 2017 Location: Sri Lanka
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Thanks for showing the mistake. Now see the edited content.
Last edited by osalselaka; Jun 28th 2018 at 02:20 AM.

 
Jun 28th 2018, 03:53 AM

#4  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 1,035

Originally Posted by osalselaka Thanks for showing the mistake. Now see the edited content. 
Fine, but it is not clear if you can now complete it by yourself, which would be best.
So now for some Physics.
The temperature coefficients tell us that the resistance of the tungsten section increases with temperature and the resistance of the carbon section decreases with temperature.
So the change in the one must balance out the change in the other, per degree C.
So can you write an equation equating these changes?

 
Jun 28th 2018, 06:26 PM

#5  Junior Member
Join Date: May 2017 Location: Sri Lanka
Posts: 27

By using the equation, ρ= ρ°{1+ α(TT°) , As you said,I consider that ΔT is same for the both.(without considering positive or negative signs at the calculations) Then I got ratio of resistivity of two cable materials.(answer:1/70) After that I can't see a way to continue. I don't no the way I started would necessary for solve it. Am I correct at the path of start, and if it is correct can you give me the guidelines for continue.
Last edited by osalselaka; Jun 28th 2018 at 09:54 PM.

 
Jun 29th 2018, 01:54 AM

#6  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 1,035

You are thinking too hard.
But not about what I said, (is that because of translation?)
Following on from what I said the temperature coefficient refers to resistance.
Note the difference.
Let us consider a change of 1degreeC for each part of the cable.
Resistance is resistivity times length divided by cross sectional area by definition.
So
$\displaystyle {R_t} = \frac{{{\rho _t}{L_t}}}{A}$
and
$\displaystyle {R_c} = \frac{{{\rho _c}{L_c}}}{A}$
Noting that we are given equal cross sectional areas, A
So changes of resistance for 1 degreeC are
$\displaystyle \Delta {R_t} = {\alpha _t}\frac{{{\rho _t}{L_t}}}{A}$
and
$\displaystyle \Delta {R_c} = {\alpha _c}\frac{{{\rho _c}{L_c}}}{A}$
But these must sum to zero since there is no overall change.
$\displaystyle \Delta {R_c} + \Delta {R_t} = 0$
rearranging
$\displaystyle \Delta {R_c} =  \Delta {R_t}$
Thus
$\displaystyle \frac{{\Delta {R_c}}}{{\Delta {R_t}}} =  1$
Substituting for the delta R s
$\displaystyle  1 = \frac{{{\alpha _c}}}{{{\alpha _t}}}*\frac{{{\rho _c}}}{{{\rho _t}}}*\frac{{{L_c}}}{{{L_t}}}*\frac{A}{A}$
Can you find the ratio we want now and finish it?
Last edited by studiot; Jun 29th 2018 at 02:01 AM.

 
Jun 29th 2018, 03:14 AM

#7  Senior Member
Join Date: Oct 2017 Location: Glasgow
Posts: 482

Originally Posted by osalselaka I am confused with this question, difficult to find steps to solve it.
A carbon cable and a tungsten cables are connected each and made a cable. The resistivity of tungsten is 5.6*10^8Ωm and carbon is 3.5*10^5Ωm. The temperature coefficient of resistivity ,tungsten=0.0045°C1 and carbon=0.0005°C1. The resistance of resultant cable is not change with the temperature. What is the ratio of the two lengths of the carbon and tungsten cables.(Assume that the two cables have same cross sectional area). 
Pouillet's law is
$\displaystyle R = \rho \frac{l}{A}$
where R is resistance, $\displaystyle \rho$ is the resistivity, l is the length of the cable and A is the crosssectional area of the cable. Furthermore, the resistivity can be temperaturedependent, which is usually expressed as a linear relationship relative to some reference resistivity, $\displaystyle \rho_0$, at a reference temperature, $\displaystyle T_0$.
$\displaystyle \rho = \rho_0 \left[1 + \alpha(TT_0)\right]$
where T is temperature and $\displaystyle \alpha$ is the temperature coefficient of resistivity.
Using these laws we can build a relationship for the resistance of the whole cable. We know that two cables attached end to end add up in series, so
$\displaystyle R = R_c + R_t$
$\displaystyle \rho \frac{(l_t + l_c)}{A}= \rho_{0,t}\frac{l_t}{A} \left[1 + \alpha_t(TT_0)\right] + \rho_{0,c}\frac{l_c}{A} \left[1 + \alpha_c(TT_0)\right]$
$\displaystyle \rho (l_t + l_c)= \rho_{0,t} l_t \left[1 + \alpha_t(TT_0)\right] + \rho_{0,c} l_c \left[1 + \alpha_c(TT_0)\right]$
If we expand the brackets and then rearrange this so that all of the terms not dependent on T are on the LHS and all the ones that are are on the RHS, we get
$\displaystyle \rho (l_t + l_c)  \rho_{0,t} l_t + \rho_{0,t} l_t \alpha_t T_0  \rho_{0,c} l_c + \rho_{0,c} l_c \alpha_c T_0 = \rho_{0,t} l_t \alpha_t T + \rho_{0,c} l_c \alpha_c T$
Since we know there is no temperature dependence, we can say that
$\displaystyle \rho_{0,t} l_t \alpha_t T + \rho_{0,c} l_c \alpha_c T = 0$
Which can occur because $\displaystyle \alpha_c$ is negative. that leaves
$\displaystyle \rho (l_t + l_c)  \rho_{0,t} l_t + \rho_{0,t} l_t \alpha_t T_0  \rho_{0,c} l_c + \rho_{0,c} l_c \alpha_c T_0 = 0$
I think it should be fairly straightforward to rearrange this to get the ratio of the lengths of the cables.
Last edited by benit13; Jun 29th 2018 at 03:21 AM.

 
Jun 29th 2018, 04:01 AM

#8  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 1,035

Originally Posted by benit13 Pouillet's law is
$\displaystyle R = \rho \frac{l}{A}$
where R is resistance, $\displaystyle \rho$ is the resistivity, l is the length of the cable and A is the crosssectional area of the cable. Furthermore, the resistivity can be temperaturedependent, which is usually expressed as a linear relationship relative to some reference resistivity, $\displaystyle \rho_0$, at a reference temperature, $\displaystyle T_0$.
$\displaystyle \rho = \rho_0 \left[1 + \alpha(TT_0)\right]$
where T is temperature and $\displaystyle \alpha$ is the temperature coefficient of resistivity.
Using these laws we can build a relationship for the resistance of the whole cable. We know that two cables attached end to end add up in series, so
$\displaystyle R = R_c + R_t$
$\displaystyle \rho \frac{(l_t + l_c)}{A}= \rho_{0,t}\frac{l_t}{A} \left[1 + \alpha_t(TT_0)\right] + \rho_{0,c}\frac{l_c}{A} \left[1 + \alpha_c(TT_0)\right]$
$\displaystyle \rho (l_t + l_c)= \rho_{0,t} l_t \left[1 + \alpha_t(TT_0)\right] + \rho_{0,c} l_c \left[1 + \alpha_c(TT_0)\right]$
If we expand the brackets and then rearrange this so that all of the terms not dependent on T are on the LHS and all the ones that are are on the RHS, we get
$\displaystyle \rho (l_t + l_c)  \rho_{0,t} l_t + \rho_{0,t} l_t \alpha_t T_0  \rho_{0,c} l_c + \rho_{0,c} l_c \alpha_c T_0 = \rho_{0,t} l_t \alpha_t T + \rho_{0,c} l_c \alpha_c T$
Since we know there is no temperature dependence, we can say that
$\displaystyle \rho_{0,t} l_t \alpha_t T + \rho_{0,c} l_c \alpha_c T = 0$
Which can occur because $\displaystyle \alpha_c$ is negative. that leaves
$\displaystyle \rho (l_t + l_c)  \rho_{0,t} l_t + \rho_{0,t} l_t \alpha_t T_0  \rho_{0,c} l_c + \rho_{0,c} l_c \alpha_c T_0 = 0$
I think it should be fairly straightforward to rearrange this to get the ratio of the lengths of the cables. 
Does that work?
The resistivity of carbon is not equal to the resistivity of tungsten so part of the cable have different resistivities, but you seem to have shown them to be the same.

 
Jun 29th 2018, 05:53 AM

#9  Senior Member
Join Date: Oct 2017 Location: Glasgow
Posts: 482

Originally Posted by studiot Does that work?
The resistivity of carbon is not equal to the resistivity of tungsten so part of the cable have different resistivities, but you seem to have shown them to be the same. 
In the above formulas, $\displaystyle \rho_{0,t}$ is the reference resistivity of tungsten and $\displaystyle \rho_{0,c}$ is the reference resistivity of carbon. They are different. Only the crosssectional areas for both cables are the same, which was indicated in the question.
Last edited by benit13; Jun 29th 2018 at 05:57 AM.

 
Jun 29th 2018, 08:22 AM

#10  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 1,035

Originally Posted by benit13 In the above formulas, $\displaystyle \rho_{0,t}$ is the reference resistivity of tungsten and $\displaystyle \rho_{0,c}$ is the reference resistivity of carbon. They are different. Only the crosssectional areas for both cables are the same, which was indicated in the question. 
Sorry I missed that, with all those subscripts flying around.
And where was To given in the original question.
The words steamroller and nuts come to mind, can you take it to its conclusion to show the maths works?

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