[SOLVED] Find the diameter of copper wire

[looi76]

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

[Hootenanny]

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 :wink:. Anyway, this is a simple application of resistivity, so what is the equation for resistivity?

[looi76]

I'm sure you mean resistivity :wink:. Anyway, this is a simple application of resistivity, so what is the equation for resistivity?

R = \frac{Pl}{A}
What to do next :smile:

[Hootenanny]

R = \frac{Pl}{A}
What to do next :smile:
Well you're looking for when the two resistances are the same, so...

[looi76]

Well you're looking for when the two resistances are the same, so...

The lengths are also the same :frown: im confused!

[Hootenanny]

The lengths are also the same :frown: im confused!
You have two equations,

R = \frac{\rho_c \ell}{A_c}

And

R = \frac{\rho_a \ell}{A_a}

For the copper and aluminium wire respectively. Can you see what to do next?

[looi76]

You have two equations,

R = \frac{\rho_c \ell}{A_c}

And

R = \frac{\rho_a \ell}{A_a}

For the copper and aluminium wire respectively. Can you see what to do next?

\frac{P_c}{A_c} = \frac{P_a}{A_a}

P_c = 1.7 \times 10^{-8}\Omega{m}
P_a = 2.6 \times 10^{-8}\Omega{m}

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

\frac{P_c}{A_c} = \frac{P_a}{A_a}

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

r = \sqrt{\frac{1.7 \times 10^{-8} \times 2.6 \times 10^{-8}}{2.6 \times 10^{-8}\pi}}

r = 7.36 \times 10^{-5}

Is my answer correct?

[Hootenanny]

Nice work! Thanks for LaTeX'ing it up for me :approve:
\frac{P_c}{A_c} = \frac{P_a}{A_a}

P_c = 1.7 \times 10^{-8}\Omega{m}
P_a = 2.6 \times 10^{-8}\Omega{m}

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

\frac{P_c}{A_c} = \frac{P_a}{A_a}

\frac{1.7 \times 10^{-8}}{\pi{r}^2} = \frac{2.6 \times 10^{-8}}{1.1 \times 10^{-6}}
You're good up until this point. Your next line is wrong,

r = \sqrt{\frac{1.7 \times 10^{-8} \times 2.6 \times 10^{-8}}{2.6 \times 10^{-8}\pi}}

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]

Welcome Hootenanny! and thanks for the help

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

r = 4.78 \times 10^{-4}m

Is this right?

[Hootenanny]

Welcome Hootenanny! and thanks for the help

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

r = 4.78 \times 10^{-4}m

Is this right?
Your method is correct, but your final answer is of by ~7x10-6 due to rounding errors.

[looi76]

\frac{P_c}{A_c} = \frac{P_a}{A_a}

P_c = 1.7 \times 10^{-8}\Omega{m}
P_a = 2.6 \times 10^{-8}\Omega{m}

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

\frac{P_c}{A_c} = \frac{P_a}{A_a}

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

r = \sqrt{\frac{1.7 \times 10^{-8} \times 2.6 \times 10^{-8}}{2.6 \times 10^{-8}\pi}}

r = 7.36 \times 10^{-5}

Is my answer correct?

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 P_a right?

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

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

r = 4.8 \times 10^{-6}

Is this right?:uhh:

[Hootenanny]

By rounding errors I think you meant the area of Aluminum P_a right?

I did indeed.

r = 4.8 \times 10^{-6}

Is this right?:uhh:
Now you're two orders of magnitude and a rounding error off. I have 4.85...x10-4m.

[looi76]

Thanks Hootenanny!! I got the final answer correct...

[Hootenanny]

Thanks Hootenanny!! I got the final answer correct...
A pleasure :smile: