Electricity; Calculating Resistance.

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Homework Help Overview

The discussion revolves around calculating the resistance of a strand of aluminium wire, specifically focusing on a wire with a diameter of 4.0mm and a length of 1.0m, given its resistivity. Participants are tasked with determining the resistance per meter of multiple strands of this wire arranged in parallel.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the cross-sectional area and resistance of a single strand, using the formula for resistivity. There is a focus on the implications of arranging multiple strands in parallel and how that affects total resistance. Some participants question the correctness of the textbook answer and explore the calculations for resistors in parallel.

Discussion Status

There is ongoing exploration of the calculations involved, with participants verifying their results against the textbook answer. Some express doubt about the accuracy of the textbook, while others confirm their calculations align with each other, indicating a productive exchange of ideas and reasoning.

Contextual Notes

Participants are working under the assumption that the textbook answer may contain an error, as their calculations yield significantly different results. The discussion highlights the importance of verifying assumptions and calculations in the context of parallel resistances.

FlyingSpartan
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Homework Statement


Calculate the resistance of a strand of aluminium wire of diameter 4.0mm and length 1.0m, given its resistivity is 2.5*10^-8 Ωm. What would be the resistance per meter of 50 strands of this wire, using the strands in parallel as a cable?

Homework Equations


∏d^2/4 for area
ρ=RA/L,
Where ρ = Given resistivity
R= Resistance
A= Cross sectional Area
L= Length

The Attempt at a Solution



4.00mm Diameter → 4*10^-3 Converted to Meters
Length = 1m
Resistivity = 2.5*10^-8 Ωm

So I found out the Area using ∏d^2/4;∏*(4*10^-3)/4 which = 1.26*10^-5 m^2

Then I used the formula
ρ=RA/L,
rearranging for R,
ρL/A = R

and subbing in the values;

2.5*10^-8 * 1 /1.26*10^-5

which gives an answer of 1.989*10^-3 Ωm^-1

thus i multiplied it by 50 which gives a final answer of 0.09945

However my answer differs from the answer of the book which is given as 4.99Ωm^-1

Thanks for any help!
 
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FlyingSpartan said:
which gives an answer of 1.989*10^-3 Ωm^-1

You've got it right up to here, so this is effectively resistance per length of one strand. And the question says that there are 50 of these strands in parallel, so now you've got to work out the total resistance of the cable due to parallel strands.
 
Last edited:
The answer in the book looks wrong... are you certain you read it correctly.
I also got the resistance/m of 1 strand to be 1.98 x 10^-3
50 of these in PARALLEL will not have a resistance of 4.99 Ωm^-1
Is there a power of 10 missing somewhere?
 
I've tried to calculate the resistors in paralell;

1/r = 50 * 1/1.989*10^-3

R= 3.97*10^-5

which is still far from the answer, maybe the book is wrong?
 
I got 3.97*10^-5 as well, so maybe the book is wrong...
 
I get the same as you (your original textbook answer did not have a 10^-5 in it !)
The textbook is wrong but I can't see an obvious error.
One thing you can be sure of, when resistors are in parallel the combined resistance must be less than any of the single resistors.
Cheers
 
Welcome to PF!

Hi FlyingSpartan! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)
FlyingSpartan said:
Calculate the resistance of a strand of aluminium wire of diameter 4.0mm and length 1.0m, given its resistivity is 2.5*10^-8 Ωm.

However my answer differs from the answer of the book which is given as 4.99Ωm^-1
FlyingSpartan said:
I've tried to calculate the resistors in paralell;

1/r = 50 * 1/1.989*10^-3

R= 3.97*10^-5

hmm :rolleyes: … 4.99/3.97*10-5 = 1.26*104

that's the area! (times 10-to-the-something) :redface:

clearly something's gone very wrong here! :smile:
 

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