# Calculate Resistance Area: P x L/A

• suppo
In summary, the conversation was about finding the resistance of a copper conductor with given dimensions using the formula R = PL/A. There was some confusion about the dimensions and calculations, but in the end the correct area was determined to be 1.8x10^-4 m^2.
suppo
Im a first year electrical apprentice and I can't work out the area for resistance. The formula is R = PL/A :eg 20mm by 9mm conductor the answer has to be in meters squared can you show me the formula please?

Is that a rectangle,or an ellipse?I think you mean

$$R=\rho\frac{l}{S}$$

Daniel.

Are you just asking how to find out how many meters-squared an area 20 mm by 9 mm is?

If so, just convert each length into meters (0.020 m x 0.009 m) and multiply them.

- Warren

A question I have for homework is find the resistance of a 500 meter length of copper conductor having a cross sectional area of 20mm by 9mm? I know the resistivity of copper is 1.72 x10-8 and L is given 500, the A I think is 1.8x10-4. So R=PL/A !.72x10-8 x 500 / 1.8x10-4. The answer I get is 4.7 but the answer I'm supposed to get is 0.133 . Can you tell mr where I've gone wrong?

Again,check your arithmetics.The surface is okay.

Daniel.

I get 0.0477 ohms (this number is also within 10% of a number generated from the american wire gauge tables, so I believe it is correct).

Are you sure you have written down the dimensions correctly ?

The actual question is. A copper conductor is 200 meters long and has a rectangular cross section of 12mm by 8mm. It's resistance is 0.01 ohm. Find the resistance of 500 meters of a copper conductor having a cross section of 20mm by 9mm?

That's something else.

$$R=\rho\frac{l}{S}$$

Use this for the first conductor to find $\rho$ and then use this resistivity to find the resistance of the 2-nd conductor.

Daniel.

I did try that Daniel but I still can't arrive at the answer the course notes gives . I think my calculations for area isn't right. Is 20mm by 9mm 20+9=29 x 10-3squared =2.9x10-5 or is it 20x9=180x10-3squared 1.8x10-4. Or is there another way?

$$20 \mbox{mm}\cdot 9 \mbox{mm}=180 (\mbox{mm})^{2}=1.8\cdot 10^{-4}\mbox{m}^{2}$$

That's the area.

Daniel.

Thanks mate!

## 1. What does "P x L/A" represent in the formula for calculating resistance area?

The term "P x L/A" represents the perimeter (P) multiplied by the length (L) divided by the cross-sectional area (A). This is used to calculate the resistance area, which is a measure of the resistance of a material to the flow of electricity.

## 2. How is the resistance area calculated using the formula "P x L/A"?

The resistance area is calculated by first determining the perimeter and length of the material, and then dividing the product of these values by the cross-sectional area. This will give the resistance area in units of ohms.

## 3. Why is the resistance area important in scientific research?

The resistance area is important in scientific research because it helps determine the conductivity of a material, which is a crucial factor in understanding its electrical properties. This information is necessary for a wide range of applications, from designing electronic circuits to studying the behavior of materials in various environments.

## 4. How does the resistance area affect the flow of electricity?

The resistance area directly affects the flow of electricity. The higher the resistance area, the more difficult it is for electricity to flow through a material. This is because a larger resistance area means there is more resistance to the movement of electrons, leading to a decrease in current flow.

## 5. Can the resistance area be changed?

Yes, the resistance area can be changed by altering the dimensions of the material or by changing the material itself. For example, increasing the length or cross-sectional area of a wire will increase its resistance area, while using a different material with a higher or lower conductivity will also affect the resistance area. Additionally, external factors such as temperature can also impact the resistance area of a material.

• Electrical Engineering
Replies
6
Views
1K
• Electrical Engineering
Replies
18
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Engineering and Comp Sci Homework Help
Replies
6
Views
1K
• Electrical Engineering
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Electrical Engineering
Replies
4
Views
1K
• Electrical Engineering
Replies
1
Views
1K
• Electrical Engineering
Replies
7
Views
4K
• Electrical Engineering
Replies
15
Views
1K