Homework Help: What is the resistance of copper wire?

1. Nov 7, 2008

subopolois

1. The problem statement, all variables and given/known data
what is the resistance of copper wire in this situation? 200 meters long strand of wire, with a resistivity of 0.625x10^-6

2. Relevant equations
im not sure if this is the right equation but:
resistance=resistivity x length/area

3. The attempt at a solution
this is the part im stuck at, i have the resistivity and the length of the wire, but what do i use for the area? this is what i have so far:
resistance= 0.625x10^-6 x 200/area
as you can see i have no idea what area to use?

2. Nov 7, 2008

mgb_phys

You can't without the area. does it give you a wire gauge number, like SWG or AWG ?
Unless it's the resistance per unit length.
What units are given - that number doesn't look familair for copper.

3. Nov 7, 2008

subopolois

well it a problem im trying to solve, i have to find the radius of the copper wire given the resistivity and length, here is the exact question
what is the radius of 200m of copper wire with the same resistance of sea water? (are sea water=0.25, copper=0.625x10^-6 resisivites

4. Nov 7, 2008

subopolois

it also gives the dimensions of sea water as 200 m long and a cross section of 1 m^2

5. Nov 7, 2008

mgb_phys

So the question is what is the diameter of a copper wire with the same resitivity of a 'wire' of seawater 200m long and 1m^2 CSA?

You simply need the relative resistivity of sea water and copper then work out how many times lower the resistance of the copper is.

6. Nov 7, 2008

subopolois

yes, how i read the question was: what is the radius of a copper wire with the same resistance of seawater and 200m long, given that the cross section of water is 1 meter square.

i think both of our interpretations are correct. here is the work ive done so far on the question, does it make sense to you?

since the question wants it in terms of same RESISTANCE of sea water i have to find the resistance of sea water, rearranging the equation roe=RS/L (roe= resisivity, R=resistance, S= cross sectional area and L= length) for resistance i get:
R=roe x L/S
= 0.25 ohm.m x 200m/ 1 m^2
= 50 ohm

now i find since i have the resisivity of copper, i rearrange the original equation to solve for the cross sectional area, S:
roe= RS/L
0.625x10^-6= (50)(s)/200
1.25x10^4= 50s
s=2.5x10^-6

since this is the cross sectional area, and i assume the cross section of copper wire is circular, i can use the area of a circle to solve for r:
A=pi r^2
2.5x10^-6= pi r^2
7.96x10^-6= r^2
8.92x10^-4 m= r

so for a radius of copper wire i get 8.92x10^-4 meters or 0.892 millimeters.

does this seem right to you?

7. Nov 7, 2008

mgb_phys

Resistance =$$\rho$$ * S / A

So if the resistance is equal

$$\rho$$w * L / A w=$$\rho$$cu * L / A cu

so given that Lenght is the same.

Acu / Aw = $$\rho$$cu/$$\rho$$w

$$\rho$$w = 0.2 ohm m
$$\rho$$cu = 1.7x10-8 ohm m, where did you get 0.625x10^-6?

Acu = 1.7x10-8 / 0.2 = 8.5 -8 m2
A = pi r2, so r = 0.1mm

8. Nov 7, 2008

gabbagabbahey

His problem statement gives $\rho_{cu}=0.625 \times 10^{-6} \Omega \text{cm}$; resistivity is a function of temperature, so this this value is perfectly reasonable.

9. Nov 7, 2008

subopolois

sorry, what are you trying to say?

10. Nov 7, 2008

gabbagabbahey

Just that you should use whichever value of the resistivity is given to you in the question....mgb gave a different value (1.7 *10^-8 ohm m) for the resistivity of copper, but that value is for copper at about 25C.

11. Nov 7, 2008

subopolois

the 0.625x10^-6 came right from my question. what are your S, A, and L values represinging ie. roe= resisivity

12. Nov 7, 2008

mgb_phys

A = 0.625E-6 / 0.25 = 2.5 × 10-6 m^2
r = sqrt(A/pi) = 0.89mm using your values

13. Nov 7, 2008

subopolois

alright! so using both your method and mine, either way we get the same answer. thank you to all

14. Nov 7, 2008

mgb_phys

Yes - I was confused by the 50 in your calcs - so I wanted to work it through fully to make sure it was correct.

15. Nov 7, 2008

subopolois

its alright, thank you for your help