# Homework Help: Find the diameter of copper wire

1. May 9, 2008

### looi76

[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. May 9, 2008

### Hootenanny

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

3. May 9, 2008

### looi76

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

4. May 9, 2008

### Hootenanny

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

5. May 9, 2008

### looi76

The lengths are also the same im confused!

6. May 9, 2008

### Hootenanny

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. May 9, 2008

### looi76

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

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

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

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

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

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

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

Is my answer correct?

8. May 9, 2008

### Hootenanny

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. May 9, 2008

### 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?

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

### Hootenanny

Staff Emeritus
Your method is correct, but your final answer is of by ~7x10-6 due to rounding errors.

11. May 9, 2008

### looi76

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. May 9, 2008

### Hootenanny

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

13. May 9, 2008

### looi76

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

14. May 9, 2008

### Hootenanny

Staff Emeritus
A pleasure