Resistivity equations

[Gary Tollison]

Hi folks.
I'm getting ready for some upcoming exams and I find myself tripping up over resistivity equations. I've been working through example questions, but this one confuses me.

Example:
Find the CSA of a single copper cable 500 metres long, which carries a current of 2.5A, and has voltage drop of 3.58V, resistivity of copper being 1.72x10(-8) ohm/m.

any help would be great!
Thanks

Gary

 

[The Ghost]

 

[Gary Tollison]

thanks for the reply.

I know how to get the resistivity with a known CSA, my problem is I cant work out how to calculate the CSA of the cable, when the CSA is unknown.

 

[The Ghost]

You are normally given the CSA in a question, or as you say you could not work out the answer. Ok sorry just seen its a reverse question. Silly me. So reverse engineer the figures.

 

[Gary Tollison]

You are normally given the CSA in a question, or as you say you could not work out the answer. Ok sorry just seen its a reverse question. Silly me. So reverse engineer the figures.

 

[Simonslimline]

use ohms law to work out Ohms. V/I

I then used the formula and put a Value for CSA in that is not known yet.

I will, give you a clue, It is less than 10mm2.

I expect there is any easier way though.

 

[marconi]

You are given the formula R = rho x l/csa.

One has to re-arrange this formula to put your unknown, the csa = ...

The trick is to do the same thing to both sides of the equality (=).

So, csa x R = csa x rho x l/csa which can be re-arranged on the right hand side to:

rho x l x csa/csa. But csa/csa = 1.

One can now write csa x R = rho x l

Use the same trick again but this time divide both sides by R to arrive at csa = ...

I leave you to finish it off.

 

[Wilko]

Did they stuff up the units for resistivity in the question(?) - should be rho in Ohm.meter, as per @Vortigern post, iirc .

 

[happysteve]

Did they stuff up the units for resistivity in the question(?) - should be rho in Ohm.meter, as per @Vortigern post, iirc .

They usually do. I keep sending corrections to C&G...

 

[Lucien Nunes]

I don't know what that page is an extract from but it trips over this a couple of times. The resistance of a one-metre cube of copper isn't any number of Ω⋅m, then when they do mean Ω⋅m they call it ohms per metre. Doesn't anybody check this stuff before printing (could that be another 'replica' like the dodgy BS7671s?). People don't make these mistakes in daily life ('I'll have half a minute of potatoes and two dozen per-egg-squared please') so it's a bit poor that they crop up so often in supposedly technical stuff.

Marconi gives the right strategy - rearrange the formulas you have, and/or substitute into one another until you have your unknown in terms of your known values. You can also substitute V=IR into it so that you have only one equation to solve, that takes in all the data given in the question and spits out the CSA. But, if you find that too abstract, you can calculate in stages. i.e. you know the voltage drop and current, so you can calculate the resistance, and once you know that you can use it and the resistivity and length to calculate the CSA. Sometimes knowing the solution of the intermediate steps is helpful because it can show up mistakes, or feed other parts of a multi-part question.

 

[Drew34]

Have a look at transposition of formulas mate. It's evil but necessary

 

[Loki]

I find using the diagonal method the easiest.

r = pl
-----
a

So in this case you want to find a (csa)

The pl would move diagonally across to become:

r = a
----
pl



If you wanted l then you would use the same technique:

r = pl
-----
a


Move p below r

r = l
---- ----
p a



Now a move across diagonally to r leaving l on its own

ra = l
-----
p



Hope this posts keeping the format as its done on my mobile

 

[Loki]

Nope it didn't lol

So here's a couple pics

 

[AJshep]

Quite a long time since I did this so i could be way off but are you sure it's not

a= pl/r

If I were to break It down..

Step 1. (Move "a" over the equals sign to the top of the equation) so you end up with.

ra=pl

Step 2 (move "r" over the equals sign so it ends up at the bottom of the equation e.g. divide instead of multiply)

a=pl/r

 

[telectrix]

transposition of formulae is easy if you just bear in mind that you can do anything like add, subtract,multiply, or divide .just ensure that wtatever you do, do it to both sides of the equation. e.g. if you divide the right side by x, then divide the left side by x. both are then still equal.