HELP A problem on resistivity (1 Viewer)

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HELP!!!A problem on resistivity

1. The problem statement, all variables and given/known data
For some application it is important that the value of a resistance not change with temperature. For example, suppose you made a 3.70 KΩ resistor from a carbon resistor and a Nichrome wire-wound resistor connected together so the total resistance is the sum of their separate resistances. What value should each of these resistors have (at zero degree celsius) so that the combination is temperature independent


3. The attempt at a solution
Resistivity of carbon resistor, ρ: 3~60*10^-5
Resistivity of Nichrome resistor, ρ: 100*10^-8
ρ=ρo[1+α(T-To)]
But resistance is R=ρ*(l/A)
I can write the equation like
R=Ro[1+α(T-To)]
However, the question doesn’t have length and area of the resistor, otherwise I can use that to solve for Ro, so I don’t really know how to continue on with the problem.

Any help would be great. Thank you very much
 

Delphi51

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Re: HELP!!!A problem on resistivity

I don't see any information about how the resistivity varies with temperature for carbon and nichrome. If one increases while the other decreases, it should be possible to find a solution. You need the constant that multiplies (T-To). I don't think you need the length or area. The resistance will be proportional to [1+α(T-To)].
 
Re: HELP!!!A problem on resistivity

I don't see any information about how the resistivity varies with temperature for carbon and nichrome. If one increases while the other decreases, it should be possible to find a solution. You need the constant that multiplies (T-To). I don't think you need the length or area. The resistance will be proportional to [1+α(T-To)].
Do you mean that R=[1+α(T-To)] for both carbon and nichrome resistor?
α for carbon resistor is -0.0005 and α for nichrome resistor is 0.0004
If R=[1+α(T-To)], then R for carbon would be 1.01 and R for nichrome would be 0.992
where T=0 and To=20 since we are trying to find the resistance at zero degree celsius
but the sum of both resistance would only be 2.002
However, the question says that the resistance for both is 3700
Or, should I use the R I got for both resistor and find the ratio between them to solve the problem?
 

Delphi51

Homework Helper
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Re: HELP!!!A problem on resistivity

I'm thinking that
Rn[1+.0004*DT] + Rc[1-.0005*DT] = 3700
where DT is short for the change in temperature from 0, Rn and Rc are the resistances at zero degrees. Perhaps you could find Rn and Rc by using a couple of different values for DT to create two equations in the two unknowns.
 
Re: HELP!!!A problem on resistivity

I'm thinking that
Rn[1+.0004*DT] + Rc[1-.0005*DT] = 3700
where DT is short for the change in temperature from 0, Rn and Rc are the resistances at zero degrees. Perhaps you could find Rn and Rc by using a couple of different values for DT to create two equations in the two unknowns.
Sorry, but I still think that DT should be 0-20=-20, because the values of α for both copper and nichrome are at 20 degree celsius.
Or do you mean radomly select numbers and plug into the equation to solve the problem? However, I don't really know how to solve the problem in this way...

I have tried to solve the problem in another way:
Rn=1+.0004*(0-20)=0.992
Rc=1+(-.0005)*(0-20)=1.01
0.992/1.01=0.982
0.982Rc=Rn
3700=Rc+0.982Rc
Rc=1866.8 and Rn=1833.2

I am not sure whether it is the correct way to approach the problem
 
Last edited:

Delphi51

Homework Helper
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Re: HELP!!!A problem on resistivity

The .0004 and -.0005 must be accurate over some temperature range, or the formula is useless. If you can't find the values closer to 0 degrees, you pretty much have to assume those are pretty good over the range you are interested in.

If you take Rn[1+.0004*DT] + Rc[1-.0005*DT] = 3700 at DT = 0, you have
Rn + Rc = 3700
At DT=20, Rn*1.008 + Rc*.99 = 3700.
This solves to Rn = 2056 and Rc = 1644 ohms.
This answer makes sense to me. I don't understand the .992/1.01 method, which seems to have two entirely different values for Rn and Rc (Rc=1866.8 AND Rc=1.01).
 

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