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calculating area

 
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Apr12-05, 06:25 PM   #1
 

calculating area


Im a first year electrical apprentice and I cant 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?
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Apr12-05, 06:26 PM   #2
 
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Is that a rectangle,or an ellipse???I think you mean

[tex] R=\rho\frac{l}{S} [/tex]


Daniel.
Apr12-05, 06:27 PM   #3
 
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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
Apr13-05, 06:22 AM   #4
 

calculating area


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 Ive gone wrong?
Apr13-05, 08:58 AM   #5
 
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Again,check your arithmetics.The surface is okay.

Daniel.
Apr13-05, 09:36 AM   #6
 
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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 ?
Apr14-05, 03:31 AM   #7
 
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?
Apr14-05, 10:40 AM   #8
 
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That's something else.

[tex] R=\rho\frac{l}{S} [/tex]

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


Daniel.
Apr16-05, 07:42 PM   #9
 
I did try that Daniel but I still cant 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?
Apr16-05, 10:04 PM   #10
 
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[tex] 20 \mbox{mm}\cdot 9 \mbox{mm}=180 (\mbox{mm})^{2}=1.8\cdot 10^{-4}\mbox{m}^{2} [/tex]


That's the area.


Daniel.
Apr16-05, 10:16 PM   #11
 
Thanks mate!
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