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9 comments
Member

Join Date: Jun 2009
Posts: 5

Voltage Drop Calculations

12/11/2009 12:59 PM

Just a quick question for the experts.

In this volt drop calculations where does the 0.0164 come from

if E = (0.0164 x I x L) / S

then S = (0.0164 x I x L) / E

so S = (0.0164 x 1 x 200)/0.5 = 6.56 squ.mm

regards

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Anonymous Poster
#1

Re: Assistance please,

12/11/2009 1:58 PM

Too blure but it looks like copper's resistance coefficient. Homework done. 53 64 69 6E 6D 69 70 4D 6C 65.

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Anonymous Poster
#5
In reply to #1

Re: Assistance please,

12/12/2009 2:04 AM

Don't feel bad , he does not look like a student, look at his earlier posts.

The resistivity is that of copper at room temperature

http://hypertextbook.com/facts/2004/BridgetRitter.shtml

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#6
In reply to #1

Re: Assistance please,

12/12/2009 9:25 AM

53 64 69 6E 6D 69 70 4D 6C 65

S d i n m i p M l e

S i m p l e M i n d ????

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Anonymous Poster
#7
In reply to #6

Re: Assistance please,

12/12/2009 8:51 PM

Grouch! Next time I'll make it difficult.

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#2

Re: Assistance please,

12/11/2009 2:08 PM

I think , its k factor regarding insulation. Reply me if it is correct

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#3
In reply to #2

Re: Assistance please,

12/11/2009 3:53 PM

It's not insulation factor -- it resistivity of conductor.

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#4

Re: Voltage Drop Calculations

12/11/2009 10:53 PM

Ohm's Law states E = IR. R can also be expressed as pl/A where p(rho) = resistivity in ohm-meters, l = length, A = cross-sectional area of conductor. so E = I.(pl/A) You have solved for A (cross sectional area, S in your nomenclature) and the 0.0164 appears to be corresponding to resistivity. I have no idea what material this is as you have not provided that information to us. There's a list of resistivity values for common metals on the internet http://metals.about.com/gi/dynamic/offsite.htm?site=http://www.reade.com/Particle_Briefings/elec_res.html

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#8

Re: Voltage Drop Calculations

12/14/2009 3:26 AM

British Standard 7671?

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#9

Re: Voltage Drop Calculations

12/15/2009 9:32 AM

Thankyou all for replying, and thank you Ron for an explicit answer, all is understood now!

regards

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Anonymous Poster (3); Bill (1); indel (1); interengineer (1); jhonson (1); PWSlack (1); Ron George (1)

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