How Does Stretching a Wire Affect Its Resistance?

AI Thread Summary
Stretching a wire to twice its original length while maintaining constant volume results in a significant change in resistance. The resistance formula R = pL/A must account for both the increased length and decreased cross-sectional area. When the length doubles, the area is halved, leading to a quadrupling of resistance. Therefore, if the original resistance is 0.010 ohms, the new resistance would be 0.040 ohms. Understanding these relationships is crucial for accurate calculations in physics.
Mitchtwitchita
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Homework Statement



A wire has a resistance of 0.010 ohms. What will the wire's resistance be if it is stretched to twice its original length without changing the volume?


Homework Equations



R = pL/A


The Attempt at a Solution



R = p(2L)/A
pL/A = R/2
=0.010/2
=0.0050 ohms

I think I this is wrong. Can anybody let me know? And, if so, can you please steer me in the right direction?
 
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It's wrong for two reasons:
- You only considered the change in length. How does the area change?
- You didn't calculate the change in resistance due to the doubled length correctly. Assuming everything else remains the same, what happens to the resistance if the length doubles?

You must account for both the changing length and cross sectional area.
 
Mitchtwitchita said:

Homework Statement



A wire has a resistance of 0.010 ohms. What will the wire's resistance be if it is stretched to twice its original length without changing the volume?


Homework Equations



R = pL/A


The Attempt at a Solution



R = p(2L)/A
pL/A = R/2
=0.010/2
=0.0050 ohms

I think I this is wrong. Can anybody let me know? And, if so, can you please steer me in the right direction?

You started with the correct equation, but there are two things wrong with your work. First, if you stretch the wire and the volume stays constant, what else changes in addition to its length?

Second, the resistance has to be going up, not down. Your algebra gets a bit off in your calculation. Remember, you are solving for R, not pL/A.
 
So, if the length doubles, the area would have to be halved?

R = p(2L)/(1/2)A?
 
Mitchtwitchita said:
So, if the length doubles, the area would have to be halved?

R = p(2L)/(1/2)A?

Correct-a-mundo. So by inspection, what happens to the resistance?
 
R = (0.010) x 4 = 0.040?
 
Mitchtwitchita said:
R = (0.010) x 4 = 0.040?

Correct. The resistance doubles because the length doubles, and it doubles again because the area is halved. Good job!
 
Thanks for the for the help!
 
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