# Find the diameter of copper wire

1. ### looi76

81
[SOLVED] Find the diameter of copper wire

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?

2. ### Hootenanny

9,678
Staff Emeritus
I'm sure you mean resistivity . Anyway, this is a simple application of resistivity, so what is the equation for resistivity?

3. ### looi76

81
$$R = \frac{Pl}{A}$$
What to do next

4. ### Hootenanny

9,678
Staff Emeritus
Well you're looking for when the two resistances are the same, so...

5. ### looi76

81
The lengths are also the same im confused!

6. ### Hootenanny

9,678
Staff Emeritus
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?

7. ### looi76

81
$$\frac{P_c}{A_c} = \frac{P_a}{A_a}$$

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

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

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

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

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

$$r = 7.36 \times 10^{-5}$$

8. ### Hootenanny

9,678
Staff Emeritus
Nice work! Thanks for LaTeX'ing it up for me
You're good up until this point. Your next line is wrong,
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.

9. ### looi76

81
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?

Last edited: May 9, 2008
10. ### Hootenanny

9,678
Staff Emeritus

11. ### looi76

81
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:

12. ### Hootenanny

9,678
Staff Emeritus
I did indeed.
Now you're two orders of magnitude and a rounding error off. I have 4.85...x10-4m.

13. ### looi76

81
Thanks Hootenanny!! I got the final answer correct...

14. ### Hootenanny

9,678
Staff Emeritus
A pleasure