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Find the diameter of copper wire

by looi76
Tags: copper, diameter, solved, wire
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looi76
#1
May9-08, 05:47 AM
P: 81
1. The problem statement, all variables and given/known data
Find the diameter of copper wire which has the same resistance as an aluminum wire of equal length and diameter 1.2mm. The reactivities of copper and aluminum at room temperature are 1.7x10^-8Ωm and 2.6x10^-8Ωm respectively.


Can someone please explain to me the way of solving this question?
Thnx in advance
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Hootenanny
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May9-08, 05:57 AM
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Quote Quote by looi76 View Post
1. The problem statement, all variables and given/known data
Find the diameter of copper wire which has the same resistance as an aluminum wire of equal length and diameter 1.2mm. The reactivities of copper and aluminum at room temperature are 1.7x10^-8Ωm and 2.6x10^-8Ωm respectively.


Can someone please explain to me the way of solving this question?
Thnx in advance
I'm sure you mean resistivity . Anyway, this is a simple application of resistivity, so what is the equation for resistivity?
looi76
#3
May9-08, 05:59 AM
P: 81
Quote Quote by Hootenanny View Post
I'm sure you mean resistivity . Anyway, this is a simple application of resistivity, so what is the equation for resistivity?
[tex]R = \frac{Pl}{A}[/tex]
What to do next

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May9-08, 06:05 AM
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Find the diameter of copper wire

Quote Quote by looi76 View Post
[tex]R = \frac{Pl}{A}[/tex]
What to do next
Well you're looking for when the two resistances are the same, so...
looi76
#5
May9-08, 06:14 AM
P: 81
Quote Quote by Hootenanny View Post
Well you're looking for when the two resistances are the same, so...
The lengths are also the same im confused!
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#6
May9-08, 06:18 AM
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Quote Quote by looi76 View Post
The lengths are also the same im confused!
You have two equations,

[tex]R = \frac{\rho_c \ell}{A_c}[/tex]

And

[tex]R = \frac{\rho_a \ell}{A_a}[/tex]

For the copper and aluminium wire respectively. Can you see what to do next?
looi76
#7
May9-08, 06:33 AM
P: 81
Quote Quote by Hootenanny View Post
You have two equations,

[tex]R = \frac{\rho_c \ell}{A_c}[/tex]

And

[tex]R = \frac{\rho_a \ell}{A_a}[/tex]

For the copper and aluminium wire respectively. Can you see what to do next?
[tex]\frac{P_c}{A_c} = \frac{P_a}{A_a}[/tex]

[tex]P_c = 1.7 \times 10^{-8}\Omega{m}[/tex]
[tex]P_a = 2.6 \times 10^{-8}\Omega{m}[/tex]

[tex]A_a = \pi{r}^2[/tex]
[tex]A_a = \pi \times \left(\frac{1.2 \times 10^{-3}}{2}\right)^2[/tex]
[tex]A_a = 1.1 \times 10^{-6} m^2[/tex]

[tex]\frac{P_c}{A_c} = \frac{P_a}{A_a}[/tex]

[tex]\frac{1.7 \times 10^{-8}}{\pi{r}^2} = \frac{2.6 \times 10^{-8}}{1.1 \times 10^{-6}}[/tex]

[tex]r = \sqrt{\frac{1.7 \times 10^{-8} \times 2.6 \times 10^{-8}}{2.6 \times 10^{-8}\pi}}[/tex]

[tex]r = 7.36 \times 10^{-5}[/tex]

Is my answer correct?
Hootenanny
#8
May9-08, 06:40 AM
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Nice work! Thanks for LaTeX'ing it up for me
Quote Quote by looi76 View Post
[tex]\frac{P_c}{A_c} = \frac{P_a}{A_a}[/tex]

[tex]P_c = 1.7 \times 10^{-8}\Omega{m}[/tex]
[tex]P_a = 2.6 \times 10^{-8}\Omega{m}[/tex]

[tex]A_a = \pi{r}^2[/tex]
[tex]A_a = \pi \times \left(\frac{1.2 \times 10^{-3}}{2}\right)^2[/tex]
[tex]A_a = 1.1 \times 10^{-6} m^2[/tex]

[tex]\frac{P_c}{A_c} = \frac{P_a}{A_a}[/tex]

[tex]\frac{1.7 \times 10^{-8}}{\pi{r}^2} = \frac{2.6 \times 10^{-8}}{1.1 \times 10^{-6}}[/tex]
You're good up until this point. Your next line is wrong,
Quote Quote by looi76 View Post
[tex]r = \sqrt{\frac{1.7 \times 10^{-8} \times 2.6 \times 10^{-8}}{2.6 \times 10^{-8}\pi}}[/tex]
A further word of caution: be careful with rounding errors, try not to round too early in your calculation, either leave all your calculations to the end, or store the intermediate answers in your calculator.
looi76
#9
May9-08, 07:14 AM
P: 81
Welcome Hootenanny! and thanks for the help

[tex]r = \sqrt{\frac{1.7 \times 10^{-8} \times 1.1 \times 10^{-6}}{2.6 \times 10^{-8}\pi}}[/tex]

[tex]r = 4.78 \times 10^{-4}m[/tex]

Is this right?
Hootenanny
#10
May9-08, 07:20 AM
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Quote Quote by looi76 View Post
Welcome Hootenanny! and thanks for the help

[tex]r = \sqrt{\frac{1.7 \times 10^{-8} \times 1.1 \times 10^{-6}}{2.6 \times 10^{-8}\pi}}[/tex]

[tex]r = 4.78 \times 10^{-4}m[/tex]

Is this right?
Your method is correct, but your final answer is of by ~7x10-6 due to rounding errors.
looi76
#11
May9-08, 07:38 AM
P: 81
Quote Quote by looi76 View Post
[tex]\frac{P_c}{A_c} = \frac{P_a}{A_a}[/tex]

[tex]P_c = 1.7 \times 10^{-8}\Omega{m}[/tex]
[tex]P_a = 2.6 \times 10^{-8}\Omega{m}[/tex]

[tex]A_a = \pi{r}^2[/tex]
[tex]A_a = \pi \times \left(\frac{1.2 \times 10^{-3}}{2}\right)^2[/tex]
[tex]{\color{red}A_a = 1.1 \times 10^{-6} m^2}[/tex]

[tex]\frac{P_c}{A_c} = \frac{P_a}{A_a}[/tex]

[tex]\frac{1.7 \times 10^{-8}}{\pi{r}^2} = \frac{2.6 \times 10^{-8}}{1.1 \times 10^{-6}}[/tex]

[tex]r = \sqrt{\frac{1.7 \times 10^{-8} \times 2.6 \times 10^{-8}}{2.6 \times 10^{-8}\pi}}[/tex]

[tex]r = 7.36 \times 10^{-5}[/tex]

Is my answer correct?
Quote Quote by Hootenanny View Post
Your method is correct, but your final answer is of by ~7x10-6 due to rounding errors.
By rounding errors I think you meant the area of Aluminum [tex]P_a[/tex] right?

[tex]A_a = \pi{r}^2[/tex]
[tex]A_a = \pi \times \left(\frac{1.2 \times 10^{-3}}{2}\right)^2[/tex]
[tex]A_a = 1.1309734 \times 10^{-6} m^2[/tex]

[tex]r = \sqrt{\frac{1.7 \times 10^{-8} \times 1.1309734 \times 10^{-6}}{2.6 \times 10^{-8}\pi}}[/tex]

[tex]r = 4.8 \times 10^{-6}[/tex]

Is this right?
Hootenanny
#12
May9-08, 07:40 AM
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Quote Quote by looi76 View Post
By rounding errors I think you meant the area of Aluminum [tex]P_a[/tex] right?
I did indeed.
Quote Quote by looi76 View Post
[tex]r = 4.8 \times 10^{-6}[/tex]

Is this right?
Now you're two orders of magnitude and a rounding error off. I have 4.85...x10-4m.
looi76
#13
May9-08, 07:50 AM
P: 81
Thanks Hootenanny!! I got the final answer correct...
Hootenanny
#14
May9-08, 07:55 AM
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Quote Quote by looi76 View Post
Thanks Hootenanny!! I got the final answer correct...
A pleasure


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