Circuits - Calculating resistance/conductance etc

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SUMMARY

The discussion focuses on calculating the resistance, conductance, resistivity, and conductivity of a cylindrical copper conductor subjected to a voltage of 4.25mV and a current of 5A. The resistance was calculated as 0.00085Ω, leading to a conductance of 1176.47Ω-1. The resistivity was determined to be 2.07 × 10-5 Ω·m. The conductivity, which is the reciprocal of resistivity, is confirmed as σ = 1/ρ, providing clarity on the relationship between these electrical properties.

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Homework Statement


Source with voltage 4.25mV is ran along a 10cm long cylindrical copper conductor which has a cross sectional radius of 0.8mm. If the current was measured at 5A flowing through the conductor calculate: the resistance and conductance of the conductor and the resistivity and conductivity of the material the conductor is made from.

Homework Equations



The only equations we were given last week which is supposed to refer to this problem type are:
<br /> I=\frac{Q}{t} \\<br /> Q=ne \\<br /> I=\frac{ne}{t} \\<br /> I=frac{neAl}{t}=neAv_d \\<br /> v_d=\frac{I}{nAe} \\<br /> J=\frac{I}{A}=nv_de \\<br /> \rho=\frac{E}{J} \\<br /> \rho=(\frac{V}{I})(\frac{A}{l}) \\<br /> R=\frac{V}{I}=\frac{\rho l}{A} \\<br />

The Attempt at a Solution



For the resistance i have done:
<br /> R=\frac{V}{I}=\frac{0.00425}{5}=0.00085Ω <br />
For the conductance I looked it up that its the inverse so I just did
<br /> \frac{1}{0.00085}=1176.47Ω^{-1}\\<br />
for the resistivity i have done
<br /> \rho=\frac{RA}{l} \\<br /> \rho=\frac{(0.00085)(\pi 0.0088^2)}{0.1}=2.07\times 10^{-5} \\<br />
It is the conductivty that is confusing the hell out of me, don't even know where to start.

Any help is really appreciated. Thanks.
 
Last edited:
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Oh right OK thank you :)

Is what I have already done correct? Apart from the little typo mistake where I used 0.88mm for the radius as it should be just 0.8mm.
 
It looks correct.

ehild
 

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