Resistance and Resistivity


by Ajwrighter
Tags: resistance, resistivity
Ajwrighter
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#1
Mar14-08, 01:23 PM
P: 42
1. A wire with a resistance of 6ohms is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.

R = p (L/A)

Working out this problem I came across two different methods. When I initially looked at the problem my first reaction was to use integration which lead me to do this
R`= 6
(6/3) = 2. then R =(6^2) *(3/2) =54

the other method led me to do this 3/(1/3) = 9 then 9*6 = 54

Are they both not correct ways of doing it? My friend says I just pulled Numbers out of my hat and got lucky. But I can get the same answer both ways on any problem. Would the first way not be a correct way of integrating?
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mer584
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#2
Mar14-08, 01:51 PM
P: 39
I'm not exactly sure what you did there.
But the trick to this is seeing that as L becomes 3L you are going to see the same amount of decrease in your area as it stretches and flattens out.
Your new area will therefore be 1/3A.
this means you have 3L/(1/3A) = 9(L/a).
SO what we see is 9 times the old resistivity which was 6.
So 54 should be the correct answer.
Ajwrighter
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#3
Mar14-08, 01:57 PM
P: 42
Quote Quote by mer584 View Post
I'm not exactly sure what you did there.
But the trick to this is seeing that as L becomes 3L you are going to see the same amount of decrease in your area as it stretches and flattens out.
Your new area will therefore be 1/3A.
this means you have 3L/(1/3A) = 9(L/a).
SO what we see is 9 times the old resistivity which was 6.
So 54 should be the correct answer.
yea i got that. its a easy problem. I'm talking about the math. two completely deferent methods and both will be exactly correct every single time. I need an explanation for it. Particularly method 1.

berkeman
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#4
Mar14-08, 02:20 PM
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Resistance and Resistivity


Quote Quote by Ajwrighter View Post
yea i got that. its a easy problem. I'm talking about the math. two completely deferent methods and both will be exactly correct every single time. I need an explanation for it. Particularly method 1.
Your second method and mer's explanation are the simpler and more intuitive way to do the problem. To keep it simple, start by convincing yourself that L and A vary inversely when the volume is kept constant (like in this problem). Once you convince yourself of that, there is no need to break the area A down into an r^2 component. The first method you used just appears to be the next more complicated way of doing it, compared to method #2. Not different, just one step more complicated than necessary.
Ajwrighter
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#5
Mar14-08, 03:01 PM
P: 42
Quote Quote by berkeman View Post
Your second method and mer's explanation are the simpler and more intuitive way to do the problem. To keep it simple, start by convincing yourself that L and A vary inversely when the volume is kept constant (like in this problem). Once you convince yourself of that, there is no need to break the area A down into an r^2 component. The first method you used just appears to be the next more complicated way of doing it, compared to method #2. Not different, just one step more complicated than necessary.

I have no doubt that the second method was faster for most, but we did a quiz similar to this problem and I regurgitated the first method in less than 30 seconds and was beat only by one other person in time. (thats only because he was sitting in front of the professors desk). I didn't even think about the problem I just did it, and came to here as to why my method was different from everyone else's.
Mapes
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#6
Mar14-08, 04:34 PM
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What "method" are you talking about? Can you list the steps? All I see is

[tex]R^\prime=R^2 \left(\frac{L^\prime}{L}\right)\left(\frac{L^\prime}{LR}\right)[/tex]

Why did you decide to multiply these terms together? What did you integrate? Sorry if I'm not seeing it, please explain.
Ajwrighter
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#7
Mar14-08, 10:31 PM
P: 42
not integrated, just broken down is all as Berkeman has said. for instance if the resistance was 8 and the length was 4 times the original length. One method of solving out what the resistance would be is to take the length 4 and divide by 1/4 = 16 and then multiply 16 time 8 which = 128 . the other way is to take the 8 and divide by 4 = 2 then take the 8 and square it then multiply the length (4/2) * 8^2 = 128. so in part everything looks like this in its final :
Method 1: 8 *(4/(1/4)) = 128
Method 2: (8^2)(4/2) = 128
rl.bhat
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#8
Mar14-08, 10:47 PM
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The resistance of a wire is given by R = rho*L/A. When you stetch the wire the volume remains constant. So we can rewright the formula as R = rho*L*L/A.L. AL = volume of the wire which remains constant. So the new resistance is proportional to the square of the increase in the length. So R = 6*3^2 = 6x9 = 54.


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