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Copper wire resistance problem

  • Thread starter Krajendren
  • Start date
A copper wire is in the form of a cylinder and has a resistance R. it is stretched till its thickness reduces by half of its initial size. Find its new resistance in terms of R.
 
28,310
4,659
Re: resistance

This should probably go into the homework section, and you will need to show your work so far.
 
13
0
Re: resistance

A copper wire is in the form of a cylinder and has a resistance R. it is stretched till its thickness reduces by half of its initial size. Find its new resistance in terms of R.
its resistance will be increased.because R is always inversely proportional to length
 
Re: resistance

its resistance will be increased.because R is always inversely proportional to length
Thanks, could you please mention by how much the resistance will rise ?
 
Re: resistance

This should probably go into the homework section, and you will need to show your work so far.
I am really sorry to confuse you but this question appeared for my board exams recently . Could you please help me out . :smile:
 
445
5
Re: resistance

What equations do you know that relate resistance to length, and amount of substance?
 
13
0
Re: resistance

Thanks, could you please mention by how much the resistance will rise ?
it resistance will be increased by two times.
 
445
5
Re: resistance

Is diameter the only thing changing?
 
393
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Re: resistance

[tex] R = \rho \frac{h}{A}[/tex]

We must first notice that both area and length change so that the volume remains constant. We assume that the stretching results in a prefect cylinder that we began with. Therefore:
[tex] V = Ah = \pi r^2 h[/tex] and [tex] V[/tex] is a constant.

Thickness refers to the diameter of the cylinder, so to decrease its thickness by half, you must decrease its diameter by half (or its radius by half). So the new area is:
[tex] A' = \pi (\frac{r}{2})^2 = \pi \frac{r}{4}[/tex]

Therefore, to destroy no matter and to keep volume constant, we must multiply [tex]h'[/tex] by a constant to keep [tex]V'[/tex] equal to [tex]V[/tex]:
[tex] V = V' = \pi r^2 h = (\pi \frac{r^2}{4})(Kh)[/tex]
Seen above, K must equal four to cancel out that fractional four, thus bringing both left and right equations equal and thus keeping volume constant.

To sum:
[tex] h' = 4h[/tex]
[tex] A' = \pi \frac{r^2}{4} = (\pi r^2) \frac{1}{4} = \frac{A}{4}[/tex]

We now plug in our new height and our new area into the equation for resistance:
[tex] R' = \rho \frac{4^2h}{A}= 4^2(\rho \frac{h}{A}) = 4^2 R = 16R[/tex]
 
Last edited:
445
5
Re: resistance

[tex] R = \rho \frac{h}{A}[/tex]

We must first notice that both area and length change so that the volume remains constant. We assume that the stretching results in a prefect cylinder that we began with. Therefore:
[tex] V = Ah = \pi r^2 h[/tex] and [tex] V[/tex] is a constant.

Thickness refers to the diameter of the cylinder, so to decrease its thickness by half, you must decrease its diameter by half (or its radius by a fourth). So the new area is:
[tex] A' = \pi (\frac{r}{4})^2 = \pi \frac{r}{16}[/tex]
Are you sure? Aren't radius and diameter directly proportional, so that if you decrease the diameter by a half, you also decrease the radius by a half?

Therefore, to destroy no matter and to keep volume constant, we must multiply [tex]h'[/tex] by a constant to keep [tex]V'[/tex] equal to [tex]V[/tex]:
[tex] V = V' = \pi r^2 h = (\pi \frac{r^2}{16})(Kh)[/tex]
Seen above, K must equal sixteen to cancel out that fractional sixteen, thus bringing both left and right equations equal and thus keeping volume constant.

To sum:
[tex] h' = 16h[/tex]
[tex] A' = \pi \frac{r^2}{16} = (\pi r^2) \frac{1}{16} = \frac{A}{16}[/tex]

We now plug in our new height and our new area into the equation for resistance:
[tex] R' = \rho \frac{16^2h}{A}= 16^2(\rho \frac{h}{A}) = 16^2 R[/tex]
 
393
0
Re: resistance

Are you sure? Aren't radius and diameter directly proportional, so that if you decrease the diameter by a half, you also decrease the radius by a half?
you're right
 

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