Copper wire resistance problem

In summary, when a copper wire is stretched until its thickness is reduced by half, its resistance will increase by a factor of 16 in terms of its initial resistance R. This is because both the length and area of the wire change, but the volume remains constant. Therefore, the new resistance can be calculated using the equation R' = 16^2 R.
  • #1
Krajendren
3
0
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.
 
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  • #2


This should probably go into the homework section, and you will need to show your work so far.
 
  • #3


Krajendren said:
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
 
  • #4


SAINATHAN said:
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 ?
 
  • #5


DaleSpam said:
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:
 
  • #6


What equations do you know that relate resistance to length, and amount of substance?
 
  • #8


Krajendren said:
Thanks, could you please mention by how much the resistance will rise ?

it resistance will be increased by two times.
 
  • #10


Is diameter the only thing changing?
 
  • #11


[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:
  • #12


xcvxcvvc said:
[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?

xcvxcvvc said:
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]
 
  • #13


sjb-2812 said:
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
 

1. What is the cause of copper wire resistance?

Copper wire resistance is caused by the flow of electrons through the wire. As the electrons travel through the wire, they encounter resistance from the atoms in the wire's structure, which slows down their movement and creates heat.

2. How does the thickness of the copper wire affect its resistance?

The thickness, or gauge, of a copper wire does have an impact on its resistance. Generally, a thicker wire will have less resistance than a thinner wire because there is more space for the electrons to flow through. This is why thicker wires are often used for higher voltage and current applications.

3. How does temperature affect the resistance of copper wire?

The resistance of copper wire increases as the temperature increases. This is because as the wire heats up, the atoms in the wire vibrate more, creating more obstacles for the electrons to pass through. Therefore, it is important to consider the temperature when calculating the resistance of a copper wire.

4. Can the length of a copper wire impact its resistance?

Yes, the length of a copper wire does impact its resistance. The longer the wire, the more resistance it will have because the electrons have to travel a longer distance, encountering more obstacles along the way. This is why shorter wires are often used for low resistance applications.

5. How can the resistance of copper wire be reduced?

The resistance of copper wire can be reduced by increasing the thickness of the wire or by using materials with lower resistivity. Additionally, keeping the wire at a lower temperature can also help reduce its resistance. Properly sizing and selecting the appropriate wire for a specific application can also help minimize resistance and ensure efficient operation.

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