Ohm's law and resistivity, again

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Homework Help Overview

The discussion revolves around a problem involving Ohm's law and resistivity, specifically related to calculating the length and diameter of a copper wire given its resistance, mass, and density. The participants are exploring the relationships between resistance, resistivity, length, area, and volume in the context of this problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the formula R = ρL/A and how to manipulate it to find the length and diameter of the wire. There are attempts to clarify the reasoning behind multiplying both sides of the equation by volume to eliminate variables. Some participants express confusion about the steps taken and seek further understanding of the relationships between the variables involved.

Discussion Status

There is an ongoing exploration of the problem with various participants contributing their interpretations and calculations. Some have provided partial solutions and insights, while others are still grappling with the concepts. There is a recognition of correct answers for length and diameter, but not all participants agree on the methods used to arrive at those answers.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a discussion about the correct values for resistivity, which affects the calculations.

Marbles
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Sorry everyone, I forgot to post my efforts in solving the problem, so here I go again:

The question says: Suppose you wish to fabricate a uniform wire out of 1.00 g of copper. If the wire is to have a resistance of R=0.500 Ohm, and a density of d=8920 kg/m3, and if all the copper is to be used, what will be the:

a)length of the wire? >>ans. is 1.82 m

b) diameter of the wire? >>ans. is 0.28 mm


The equation is: R=p(rho)*l(length)/A(cross sectional area)

I found the area only, I'm lost!
But at least, based on the area, my diameter is supposed to be correct, but it wasn't!
I got 0.001>>lol

Anyways, I will show you what I did to find the area and the diameter:

First I found the volume of the wire: v=m(mass)/d(density)=0.001 kg/(8920 kg/m3)
and I got 1.12e(-7) m3

Then I found height, guess how!

R=p*l/A right?? ok, I canceled the length on top (numerator) with the length in the area equation: L*h and I was left with R=p/h, which means h=p/R=1.72e(-80) (rho or resistivity of copper, got it from our textbook)/0.5= 3.44e(-8) m

Now I substituted both values in the formula: A=v/h=1.12e(-7)/3.44e(-8)=3.25 m2

Now I used it to find the diameter: d=A/pi=3.25/3.14(pi)= 0.001 mm

I know that there are several mistakes, so I'd like you all to help me improve my solution and also help me find the length. I'd really appreciate that!
 
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R= Resistance
p = Resistivity
A = Area
L = Length
V = Volume
m = Mass
D = Density

R = pL/A

Rearrange

R/p = L/A

V = m/D = LA

Now, multiply each side by the volume

V(R/p) = LA(L/A)

VR/p = LAL/A = L2

Sqrt(VR/p) = L

That's a true beauty and I'm proud of it. I'm sure you can work it from here.

*I normally wouldn't give such a direct answer, but this one baffled me and took a lot of effort.
 
Last edited:
wow I get everything, except that, why did we multiply both sides by v??
 
Marbles said:
wow I get everything, except that, why did we multiply both sides by v??

Well we know volume.

So, by multiplying each side by it we can cancel out the area value, leaving only length.

This means instead of having two unknowns (A and L) we only have one (L).

It's like adding a constant to each side.
 
I know that area is going to be canceled...but in what sense are we doing this? why? what makes u know that this is the way it's going to work? I'm not getting it!
 
We're doing it so that instead of there being 2 unknowns, we only have 1.

So long as A and L are unknown, it is impossible to solve. So by introducing volume on each side, all it does is remove the A value so we don't have to worry about it.

It's purely to reduce the equation so we can solve it.

Perhaps you should plug in the values and see if it works?
 
aha I see...so that's how it works! wow! I'd never be able to think this way!
But wait! what about the diameter?
 
What about the diameter?

I'd assume the wire is round:

A=pi*r2.

Now you know volume and you know length. Leaving only r to find.

V = A*L
 
isn't it V/L=pi*r^2??
we find r^2 then square root? then why am I getting different answers?
 
  • #10
Right hang on, I'll work it out myself.
 
  • #11
OK

a) answer I got was 1.825m - note resistivity is 1.68E-8 not 1.72E-80 as you have above.

b) answer I got was 2.8E-4m or 0.28mm - remember, the number you get at the end is the radius, you double it for diameter.

Both agree with the answers you gave in the OP.
 
  • #12
omg how did u solve for r?? I'm not able to get it right no matter what I do!
 
  • #13
V = AL

V/L = A

A=pi(r2)

A/pi = r2

Sqrt(r2) = r

2r = Diameter
 
  • #14
Marbles said:
omg how did u solve for r?? I'm not able to get it right no matter what I do!

Are you getting the right answer for the length?
 
  • #15
yes I am getting the right answer for the length!
 
  • #16
Oh yay! I finally got it right! I'm really glad! Thank you so much for you help and support...and above all, your patience!
 

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