How Does Resistance Change with Different Metals and Dimensions?

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Homework Help Overview

The discussion revolves around calculating the electrical resistance of an aluminum bar and comparing it to a copper wire. Participants explore the relationship between resistance, resistivity, and geometric dimensions, particularly focusing on the implications of using different metals and wire gauges.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for resistance, R = p L/A, and the need for resistivity values. There are attempts to calculate the area of the cross-section of the wire, with some confusion regarding the correct formula for a rectangle versus a circle. Questions arise about the implications of using different wire gauges and the ability of aluminum to carry the same current as copper.

Discussion Status

The conversation is ongoing, with some participants providing guidance on the correct approach to calculating area and resistance. There is a mix of interpretations regarding the formulas used, and some participants express frustration over misunderstandings. No consensus has been reached, but there is an active effort to clarify concepts and assist the original poster.

Contextual Notes

Participants note the importance of consistent units (meters, centimeters, millimeters) in calculations. There is also a mention of specific wire gauges available and their current-carrying capacities, which adds complexity to the discussion.

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.
 
Last edited:
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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