How Does Resistance Change with Different Metals and Dimensions?

AI Thread Summary
The discussion centers on calculating the resistance of an aluminum bar and finding the equivalent length of a copper wire with the same resistance. The formula for resistance, R = pL/A, is emphasized, with a focus on correctly determining the area of the rectangular cross-section. Errors in calculating the area and misunderstanding the formulas are highlighted, particularly the incorrect use of pi and the confusion between resistance and area. The conversation also touches on the capacity of different wire gauges to safely carry current, specifically comparing aluminum and copper wires. Ultimately, the thread illustrates the importance of accurate calculations and understanding fundamental concepts in physics.
sonrie
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An aluminum bar 3.74 m long has a rectangular cross section 1.11 cm by 5.30 cm.

What is the resistance ?

R = ____ ohms

What is the length of a copper wire 1.53 mm in diameter having the same resistance?

L= ____ m.
 
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The things you need are:
A formula for finding the resistance when you know the resistivity of a metal and its geometric shape.
A table that lists the resistivity values for aluminum and other metals.
 
please! answer me! i have 1 problem.
A No.10 copper wire will sofly carry up to 30 Amperesof current.If the copper wire is to be replaced by an aluminium wire and the only availiable guages of aluminium wire in stock are Nos; 8,10,12,and 14 which one would be able to carry 30 A safely? Explain how you made your choice and which data you referred to.
 
well the formula is R= p L/A, so i plugged in the numbers which were p = 2.82*10^-8, L was 3.74 and the Area is pi r^2 my answer was 9.71 *10^-6, which is incorrect. What did i do wrong? Help!
 
For area i did 1.11*5.30= 5.88^2 =.034, then i multiplied by pi =1.08*^2.
 
sonrie, your formula Area = pi r^2 looks odd. Do you know what the "r" stands for? What is the area of a rectangle?

p.s. also, watch those units. I am seeing m, cm, and mm all being used, so we need to be careful with that.
 
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the r stands for the resistance, p is the resistivity, L is the length of the wire and A is the Area.
 
r does not mean resistance in the "Area = pi r^2" formula. But that is irrelevant to getting the resistance of the rectangular wire, so let's drop it and try to get back on track.

We have the formula:
R= p L/A

Earlier you gave values for p and L which look good. We just need the correct value for area A.

Do you know how to find the area of a rectangle? Earlier you said
For area i did 1.11*5.30= 5.88^2 =.034, then i multiplied by pi =1.08*^2
This statement was wrought with errors. The area of a rectangle is a simple multiplication, and it does not involve pi.
 
Area = .5 B*H, which is .5*1.11*5.30 = 2.94 Is that correct ?

So then my final equation should look like

R= 2.82*10^-8 3.74/2.94 = 3.59*10^-7 Is that what you got?
 
  • #10
I'm going to give up now, as it appears you are deliberately doing this wrong in order to engage a response from people.

My apologies if I am mistaken. Good bye.
 
  • #11
Hmm, I don't think you can jump to that conclusion..
 
  • #12
astrorob said:
Hmm, I don't think you can jump to that conclusion..

Here is the evidence. These are all things sonrie has claimed in this thread:

1. The area of a rectangular cross section is given by the formula Area = pi r^2

2. In the above formula, the "r" is resistance.

3. 1.11*5.30= 5.88^2 =.034

4. The area of a rectangle is given by 0.5 B H

People can draw their own conclusions, just as I have done.
 
  • #13
I've dealt with people who struggle quite badly with physics and mathematics and you'd be surprised what people come out with.. They just throw formulas that they've heard around until they hit the correct one without understanding the mechanism behind them.

What you may think as unimaginably easy might not be so to someone less gifted.
 
  • #14
So my answer is wrong!
 
  • #15
I'll acknowledge that I am not 100% certain, and will not be "reporting" him. But I am certain enough to not spend more of my time trying to help this person.

He is still welcome to enlist the help of other members if he is in earnest.
 
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