How Does Stretching a Wire Affect Its Resistance Ratio?

[Mdhiggenz]

Homework Statement

Hey guys a problem similar to this is going to be on my exam on Thursday, and I'm a bit stuck.

A wire of length L and cross-sectional area A, has resistance R.

What will be the resistance Rstretched of the wire if it is stretched to twice its original length? Assume that the density and resistivity of the material do not change when the wire is stretched.

However my professor wants us to do this problem as a ration, so he will ask us to find the ratio R2/R1.

So what I did was

L2=2L1

R₁=P*L1/A1

R₂=P*2L1/A2


R2/R1= 2L1P/A2 * A1/L1P

R2/R1= 2A1/A2

and now I am stuck.

Thanks for the help

Homework Equations

The Attempt at a Solution

 

[SammyS]

Homework Statement

Hey guys a problem similar to this is going to be on my exam on Thursday, and I'm a bit stuck.

A wire of length L and cross-sectional area A, has resistance R.

What will be the resistance Rstretched of the wire if it is stretched to twice its original length? Assume that the density and resistivity of the material do not change when the wire is stretched.

However my professor wants us to do this problem as a ration, so he will ask us to find the ratio R2/R1.

So what I did was

L2=2L1

R₁=P*L1/A1

R₂=P*2L1/A2


R2/R1= 2L1P/A2 * A1/L1P

R2/R1= 2A1/A2

and now I am stuck.

Thanks for the help

Is the volume of the wire after it's stretched the same as the volume of the unstretched wire?

 

[Mdhiggenz]

I think it would double as well since it is getting longer, but why would the volume have anything to do with the problem?

 

[SammyS]

I think it would double as well since it is getting longer, but why would the volume have anything to do with the problem?

Yes, the volume is important. It let's you find the ratio of the final cross-sectional area to the initial cross-sectional area .

If the volume doubles, the the density will be reduced to half.

 

[Mdhiggenz]

Oh snap. So wouldn't the Area be reduced to half? and then I can make another relation

A1=1/2A2?

 

[Mdhiggenz]

Ok that works out since that got me the correct answer, but I don't understand how to put that into a ratio form.

 

[CWatters]

The volume is constant as no new copper is being added. If length doubles the area halves.

The resistance will be proportional to length and inversly proportional to area.

Write two equations eg

Rinitial = ?
Rstretched = ?

Then divide one by the other...

Rstretched/Rinitial = ?/?

Some terms on the right (such as the restivity of copper) will cancel.

Over to you to fill in the missing bits.

 

[Mdhiggenz]

Watters thanks for the response, and I was able to get the correct answer, just not sure how to put it in a ratio form.