# Find the diameter of copper wire

• looi76
In summary, to find the diameter of a copper wire with the same resistance as an aluminum wire of diameter 1.2mm, you can use the equation R = (ρl)/A and set the resistivity of copper and aluminum equal to each other. Solving for r, you get a diameter of 4.85x10^-4m.

#### looi76

[SOLVED] Find the diameter of copper wire

## Homework Statement

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?

looi76 said:

## Homework Statement

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

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

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

looi76 said:
$$R = \frac{Pl}{A}$$
What to do next
Well you're looking for when the two resistances are the same, so...

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

The lengths are also the same I am confused!

looi76 said:
The lengths are also the same I am 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?

Hootenanny said:
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?

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

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

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

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

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

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

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

Nice work! Thanks for LaTeX'ing it up for me
looi76 said:
$$\frac{P_c}{A_c} = \frac{P_a}{A_a}$$

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

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

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

$$\frac{1.7 \times 10^{-8}}{\pi{r}^2} = \frac{2.6 \times 10^{-8}}{1.1 \times 10^{-6}}$$
You're good up until this point. Your next line is wrong,
looi76 said:
$$r = \sqrt{\frac{1.7 \times 10^{-8} \times 2.6 \times 10^{-8}}{2.6 \times 10^{-8}\pi}}$$
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.

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:
looi76 said:
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?

looi76 said:
$$\frac{P_c}{A_c} = \frac{P_a}{A_a}$$

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

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

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

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

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

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

Hootenanny said:

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:

looi76 said:
By rounding errors I think you meant the area of Aluminum $$P_a$$ right?
I did indeed.
looi76 said:
$$r = 4.8 \times 10^{-6}$$

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

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

looi76 said:
Thanks Hootenanny! I got the final answer correct...
A pleasure

## 1. What is the standard measurement unit for the diameter of copper wire?

The standard measurement unit for the diameter of copper wire is millimeters (mm).

## 2. How do I measure the diameter of copper wire?

The diameter of copper wire is measured using a micrometer or a caliper. Place the wire between the two jaws of the measuring tool and read the measurement on the scale.

## 3. What is the average diameter of copper wire used in electrical wiring?

The average diameter of copper wire used in electrical wiring is 1.63 mm or 0.064 inches.

## 4. Can the diameter of copper wire vary?

Yes, the diameter of copper wire can vary depending on its gauge. The higher the gauge number, the thinner the wire will be.

## 5. Why is it important to know the diameter of copper wire?

Knowing the diameter of copper wire is important because it helps determine the wire's current-carrying capacity and its resistance. This information is crucial in designing and installing electrical systems.