A wire with a resistance R is lengthened to 1.25

  • #1
138
1
My answer for this problem doesn't agree with the answer in the back of my textbook book:

A wire with a resistance R is lengthened to 1.25 times its original length by being pulled through a small hole. Find the resistance of the wire after it has been stretched.

R = pL/A

p = resistivity
L = length
A = area of a cross section perpendicular to the length
r = radius

R1 = pL1/A1

L2 = 1.25L1 so r1/r2 should equal 1.25 right.

The decrease in radius should be proportional to the increase in length.

So r2 = r1/1.25

Therefore R2 = p(1.25L1)/[pi](r1/1.25)^2 = R1(1.25*1.56) = 1.95R1

The answer in the back of my textbook is R2 = 1.56R1

Where have I gone wrong?

Thanks
 
Last edited by a moderator:

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,950
19
The decrease in radius should be proportional to the increase in length.

Should it? How would you go about proving this?
 
  • #3
138
1
I'm assuming it's proportional. If it's not then I'm lost
 
  • #4
enigma
Staff Emeritus
Science Advisor
Gold Member
1,754
15


Originally posted by discoverer02

L2 = 1.25L1 so r1/r2 should equal 1.25 right.


Volume is constant, and volume is L*A, not L*r
 
  • #5
138
1
You're right. Stupid me!!

The area would have to shrink proportionally: A2 = A1/1.25 so R2 = 1.25*1.25R1 = 1.56R1.

Thank you very much for the help. :smile:
 

Related Threads on A wire with a resistance R is lengthened to 1.25

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
4K
Replies
8
Views
7K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
0
Views
7K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
379
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
7
Views
7K
  • Last Post
Replies
3
Views
3K
Top