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Question related to resistance, please help?
1.00 g of copper is used to make a wire that has a resistance of 0.500 ohms. Find a) the length of the wire, and b) the diameter of the wire.
I'm not really sure where to start on this one. I think i have to use
R = (rho)*(L / A) where rho is the resistivity, L the length, and A the cross-section diameter. The resistivity of copper is 1.72x10^(-8) ohm-meters.
Thanks for your answers.
Thank you so much, can't believe I didn't think of that. The answers are in the back of the book, so i know it is correct. I'll choose Best Answer when I can.
2 Answers
- Paschal HLv 61 decade agoFavorite Answer
You have a good start. You also need the density of copper so that you can calculate how much wire can be made with 1.00 g of copper. I don't have that number available to me, I would guess you do somewhere.
With that number and your R = (rho)*(L / A) equation you should be able to come up with a second equation that will have L, A, Density, and 1.00g.
Then you will have two equations and two unknowns.
- BonyLv 51 decade ago
Continuing Paschals work
From R = (Rho) L/A, -----L in meters, A in m^2
A = (Rho)L/R, ____ A = 1.72 x10^-8 x L/0.5 ------Eq1
Density of copper is 8.96g/cm^3.
So the volume of 1g of wire is (1/8.96)cm^3.
Convert to m^3 =(1/8.96) x 10^-3
Think of wire as very long thin cylinder.
Volume = A x L
A = Vol/L _____ A = (1/8.96) x 10^-3 / L -------- Eq2
Eq1 = Eq2
1.72 x10^-8 x L/0.5 = (1/8.96) x 10^-3 / L
L^2 = (0.5 x (1/8.96) x 10^-3)/ 1.72 x10^-8
L = 56.96m = Answer
