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philaxgt500

i'm in year 3 and i'm stuck on a year 1 question, is there any hope for me??
i was wondering if someone could jog my memory on this 301 question please.

question:

A copper conductor has a rho value of 17.5 x 10-9. calculate the resistance of the copper wire if it's length is 20 meters and it's CSA is 2.5mm2

cheers phil
 
Hi Phil.

Ok so resistance (R) = (Rho x l)/s where l = length and s = CSA.

Are you sure you've got your units right?? as the resistivity of copper is 1.72x10-8 and they usually give the diameter of the wire not the CSA. So you would also need to work out the CSA = pie x r squared

Or is this a hypothetical question??

Let me know.

Cheers:)
 
resistance r = pl / a (ohms) p (the greek letter rho) is the resistiivity value for the material l is the length and a is the cross sectional area got the book next to me lol
 
its been 9 years since I was in college and I aint got no books by me either. The old grey cells aint what they were:(.

I think I was right thought just with different letters. R= Rho (P) x length (L) / CSA (S).

Where L = length and S = CSA. Your book says CSA = A though.

Makes the same sum, thought doesn't it????:confused:
 
I don't remember any of this, Cheers for the help.
And now for your next question,
A conductor 600mm long carries a current of 300 milli amps and is at 90 degrees to a magnectic field with density or 1.5T. determine the force exerted on the conductor.
 
erm dunno lol

Trust me.;)

I don't remember any of this, Cheers for the help.
And now for your next question,
A conductor 600mm long carries a current of 300 milli amps and is at 90 degrees to a magnectic field with density or 1.5T. determine the force exerted on the conductor.

F=BIL but as your current is at 90 degrees to the field then the formula is F = BIL cosine theta. As the angle is 90 degrees then cosine theta = 1.

Where B = flux density, I = current, and L = length.

So - 1.5 x 0.003 x 600 = 2.7N.

I think. It's been a while since I did this stuff and my swede is aching a bit now.:)
 
Last edited by a moderator:
Thanks alot everyone. i should hopefully be able to do the next 50 questions, But if i do get stuck i know where to come :D unless of course you want me to scan the page and put it up so you all can have a go at the 52 questions?? ;)
 

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