Resistance and Resistivity

In summary, the problem involves finding the resistance of a longer wire that is drawn out through a die, assuming that the resistivity and density of the material are unchanged. Two methods were used to solve the problem, with both resulting in the same answer of 54. The first method involved integration, while the second method was simpler and more intuitive, demonstrating the inverse relationship between length and area. Ultimately, the problem can be solved by recognizing that the new resistance is proportional to the square of the increase in length.
  • #1
Ajwrighter
42
0
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?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
mer584 said:
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.
 
  • #4
Ajwrighter said:
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.
 
  • #5
berkeman said:
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.
 
  • #6
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.
 
  • #7
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
 
  • #8
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.
 

1. What is resistance?

Resistance is the measure of an object's ability to impede the flow of electric current. It is represented by the symbol "R" and is measured in ohms (Ω).

2. What factors affect resistance?

The factors that affect resistance include the material of the object, its length and cross-sectional area, and its temperature. Different materials have different inherent resistances, longer objects have more resistance, and thinner objects have less resistance. Additionally, as temperature increases, resistance also increases.

3. What is resistivity?

Resistivity is the measure of a material's resistance per unit length and cross-sectional area. It is represented by the Greek letter "ρ" (rho) and is measured in ohm-meters (Ωm). It is a property of the material and is used to calculate resistance in different objects made of the same material.

4. How is resistance calculated?

Resistance can be calculated using Ohm's law, which states that resistance is equal to the voltage (V) divided by the current (I). This can be represented by the equation R = V/I. Alternatively, resistance can also be calculated using the formula R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area.

5. How does temperature affect resistance?

As mentioned before, temperature has a direct effect on resistance. As the temperature of a material increases, the resistance also increases. This is because at higher temperatures, the atoms in the material vibrate more, causing more collisions with electrons and hindering the flow of current. Some materials, however, have a negative temperature coefficient, meaning their resistance decreases as temperature increases.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
502
  • Introductory Physics Homework Help
Replies
7
Views
654
  • Introductory Physics Homework Help
Replies
2
Views
664
  • Introductory Physics Homework Help
Replies
5
Views
807
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
27
Views
3K
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
346
Back
Top