Discuss Cables in parallel formula in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

Jay Sparks

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Hi Guys

Quick one for you.
A cable that is 25M long and has a resistance of 2.3 m ohms/M ( milli ohms per metre ). A second Identical cable is to installed in parallel with this conductor. What is the total resistance of the two conductors connected in parallel?

1st conductor: 25 x 2.3 = 57.5
57.5 divided by 1000 = 0.0575

2nd conductor: 25 x 2.3 = 57.5
57.5 divided by 1000 = 0.0575

1 divided by 0.0575 + 1 divided by 0.0575 = 34.782

1 divided by 34.782 = 0.028

So 0.028 is my answer. Is this correct?:confused:

Sorry but I have no divide symbol on my computer.

Any help would be great. Got my 2391 in Jan so I am craming like mad.

Cheers

Jay
 
Looks good to me mate.


Sent from my iPhone using Tapatalk
 
hi mate
2.3 x 25 x 2 divide by 1000= 0.115ohm
is the 2.3 m/ohms per metre at 20 c if so then have to include a multiplier if 70c cable T/E multicore ie a core incorporated in a cable or bunched max temp 70c then
2.3 x 25 x 2 x 1.2 divide by 1000 = 0.138ohm or rounded up to 0.14ohm
but i may be wrong and mixing up what you are asking and someone elso more clued up may know better
 
Last edited:
hi mate
2.3 x 25 x 2 divide by 1000= 0.115ohm

Thanks for that but I think the way you have done it is if the cables where in series. Because there in parallel you have do: 1 divided by answer then 1 divided by total answer. Or so I was told when working with resistors. There was no mention of temp in the question so I guess they dont need it. Its on a past 2391 paper and I just wanted to know if I was working it out ok.

Cheers for the replys guys

Jay
 
R = (R1xR20)/(R1+R2) for resistors in parallel
 
RT = R1XR2
.......R1+ R2

as telectrix stated or when two resistances of equal resistance in parallel total resistance is half the resistance of 1 resistance.
example
R1=5 OHM R2=5 OHM
5/2 RT=2.5 OHM
 
Last edited by a moderator:
and the same rule for capacitors in series. Ct= (C1 x C2)/(C1 + C2)
 
but what about inductances in parallel, king?
 
lots of Ls and pies if i remember ( i only left colege in 2007 ) i have a terible memory, especialy formulas i would never use. If i ever need a formula i have a handy ap for my ipod, or google works just as well or if realy, realy bored i pull out my old college folders and remember how intelegent i used to be ,before jibs grading system knocked me on my bottom.
 
To remember resistances in parrallel "Divide the product by the sum"...........but for only 2 x resistances. If you have 3 x resistances use the formula for 2 x resistances then using that answer using the formula again for the 3rd resistance and so on.
 
To remember resistances in parrallel "Divide the product by the sum"...........but for only 2 x resistances. If you have 3 x resistances use the formula for 2 x resistances then using that answer using the formula again for the 3rd resistance and so on.
better way is................. 1/R = 1/R1 + 1/R2 + 1/R3 ....

eg. 3 10ohm in parallel

1/Rt = 1/10+1/10+1/10 = 3/10 : Rt = 10/3 = 3.33
 
Last edited:
hi mate
2.3 x 25 x 2 divide by 1000= 0.115ohm

Thanks for that but I think the way you have done it is if the cables where in series. Because there in parallel you have do: 1 divided by answer then 1 divided by total answer. Or so I was told when working with resistors. There was no mention of temp in the question so I guess they dont need it. Its on a past 2391 paper and I just wanted to know if I was working it out ok.

Cheers for the replys guys

Jay

Hi Jay you are correct i have worked it out for a series circuit sorry mate
I have looked into it and should have done it the way you said lol thanks for pointing my error learn something new every day mate cheers
 

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