2 Parallel connected cables resistance.Going from Diamater (mm) to m^2?

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Homework Help Overview

The discussion revolves around calculating the resistance of aluminum and copper cables connected in parallel, given their diameters, lengths, and resistivities. Participants are exploring how to convert the diameters from millimeters to square meters for use in resistance calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conversion of cable diameters to cross-sectional areas, questioning the correct method to achieve this. There is mention of using the formula for area based on radius and the need to convert units properly.

Discussion Status

Some participants have provided calculations for the cross-sectional areas and resistances, while others are clarifying the conversion process and addressing potential misunderstandings regarding resistivity values. Multiple interpretations of the conversion method are being explored.

Contextual Notes

There is confusion regarding unit conversions, particularly the relationship between square millimeters and square meters. Participants are also discussing the implications of resistivity values for the materials in question.

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A load taking 18.2A is supplied by aluminum and copper cables connected in parallel. The lengths of the copper and aluminium cables are 78m and 64m respectively. The diameter of the aluminium 1.8mm and that of the copper 2.1mm . The resistivity abd copper is 0.018uOHMm (Ohm.meter) and 0.031uOHMm (Ohm.meter) respectively. Calculate the following:

a. The current in each cable
b. The voltage drop across each cable.
c. Power dissipation in each cable.

Now I think if I could get the total resistance of both cables I could just handle them as if they where Resistors and work out all the questions but can't seem to remember how to m^2 of the cables from their current mm reading.

Think if I could do that it would be easy to work out R with R=l.p/a formula were l = meter , a = cross section (m^2) , p = (Ohm.Meter)

Please help.
 
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that is how you go about it. Be careful... they have given diameter and cross section area = pi r^2
 
Ok so I came up with this..

ITS a huge mess

Aluminium Cable = R1
Copper Cable = R2
I =18.2A
R1s = 78m
R2s= 64m
R1a = r^2 π=(0.9)^2 (π)= 2.54mm^2÷1000=0.00254421m^2
R2a = r^2 π=(1.05)^2 (π)= 3.4629525mm^2÷1000=0.0034629525m^2
R1Ωm= 0.018μΩm ×〖10〗^(-6)=0.000000018Ωm
R2Ωm= 0.031μΩm ×〖10〗^(-6)=0.000000031Ωm
R1=78(0.000000018 )/((0.00000254421) )=0.9138=913.8mΩ
R2=64(0.000000031)/((0.0000034629525) )=0.332664=332.66mΩ
 
When converting mm2 to m2 it is important to remember that there are a million square mm in a square meter; a thousand for each meter and a square meter is 1m x 1m.

Also, it looks like you've interchanged the resistivities of the metals. Copper should be the better conductor, so its resistivity should be smaller.
 
Last edited:
Im sorry I might sound stupid but I am confused.
So you can't just go 2mm^2/1000 = 0.002m^2 ?
Should then say 2mm/1000*1000 ?
Wouldnt that just complicated question even further?

Thanks for the replies :)
 
Totter said:
Im sorry I might sound stupid but I am confused.
So you can't just go 2mm^2/1000 = 0.002m^2 ?
Should then say 2mm/1000*1000 ?
Wouldnt that just complicated question even further?
It is what it is :smile:. No matter what it does to the question, there are a million square millimeters in a square meter! By the way, a million can be written as 106.

Scientific notation was invented to clean up all this leading zero nonsense. It's a great tool that you should get used to.
Thanks for the replies :)
Glad to be of help.
 
So 2.54mm^2 would be 2.54mm^2 x10^-6 = 0.00000254m^2
 
Totter said:
So 2.54mm^2 would be 2.54mm^2 x10^-6 = 0.00000254m^2

Unit conversions are often spelled out in the following manner, by multiplying by ratios. Each ratio has a true value of one, being the ratio of a number of units equal in "value" to some other number of units. Thus if there are one million mm2 in one meter2, then:

$$2.54\; mm^2 \times \frac{1\; m^2}{10^6\;mm^2} = 2.54\times10^{-6}\; m^2$$

Notice how all the work ends up being cancelling out the units from numerators and denominators and adding up the exponents. No need to muck around with decimal places.
 

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