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Current flowing through a copper wire

  • Thread starter sweetdion
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  • #1
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Homework Statement


1. The wiring in a house must have low enough resistance so that it does not heat up too much while current is flowing. A particular copper wire needs to carry 20 A of current, and it must not dissipate more than 2 watts of power per meter of length. The resistivity of copper is ρ=1.72 x 10^-8 Ωm

a) What is the minimum diameter that the wire must have so that it doesn't heat up too much?

b) The density of copper metal is 9 g/cm^3, and the atomic mass is 63.5 g/mole. Avagadro's number is 6.022 x 10^23 atoms/mole. Assume one charge carrier per atom and determine the density of charge carries n (number of charge carriers per cubic meter) in copper.

c) A copper wire with circular cross-section with radius r=1mm carriers 1 A of current. Determine the drift velocity of the electrons in the wire.


Homework Equations


A=ρL/R A=π r^2 P = I^2R R=V/I P=V^2/R
v=I/nqA


The Attempt at a Solution



a) from A = pL/R
r=squareroot(A/pi)=squareroot(pL/PiR)=1.046 mm

b) d=9g/cm^3 x 1mol/63.5g x 6.022x10^23 charge carriers/1mol x 100 cm^3/1 m^3 = 8.535 charge carriers/m^3

c) r = .001 m
I = 1 A
v= I/nqA = 1/(8.545)(1.6x10^-19)(1.72x10^-8*1 m/.001m)= 4.257 m/s

I was unsure about the coversion from cm^3 to m^3

Also, I'm unsure about the length I am supposed to use. Should it be 1 m because the power was given to me in watts/meter?

Thanks in advance,
--sweetdion
 

Answers and Replies

  • #2
Redbelly98
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Homework Equations


A=ρL/R A=π r^2 P = I^2R R=V/I P=V^2/R
v=I/nqA


The Attempt at a Solution



a) from A = pL/R
r=squareroot(A/pi)=squareroot(pL/PiR)=1.046 mm
We were not given a value for R; you have left out some steps in how you got to your 1.046 mm answer.

b) d=9g/cm^3 x 1mol/63.5g x 6.022x10^23 charge carriers/1mol x 100 cm^3/1 m^3 = 8.535 charge carriers/m^3
You have neglected the 1023 part of Avogadro's number. Also, your units conversion is wrong:

If (1 m) = (100 cm), then
(1 m)^3 = (100 cm)^3 = ____ cm^3 ?

c) r = .001 m
I = 1 A
v= I/nqA = 1/(8.545)(1.6x10^-19)(1.72x10^-8*1 m/.001m)= 4.257 m/s

I was unsure about the coversion from cm^3 to m^3

Also, I'm unsure about the length I am supposed to use. Should it be 1 m because the power was given to me in watts/meter?

Thanks in advance,
--sweetdion
Yes, it is easiest to use 1 m and 2 W to solve the problem. Alternatively, using 2 m and 4 W, or 3 m and 6 W, etc., should give the same answer, since they all use 2 W per m.
 
  • #3
64
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to find R i did R=V/I=0.1V/20A = .005Ω then I plugged it in to get the right answer

thanks for your help!
 

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