Circuits - Calculating resistance/conductance etc

[FaraDazed]

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:
$$I=\frac{Q}{t} \\ Q=ne \\ I=\frac{ne}{t} \\ I=frac{neAl}{t}=neAv_d \\ v_d=\frac{I}{nAe} \\ J=\frac{I}{A}=nv_de \\ \rho=\frac{E}{J} \\ \rho=(\frac{V}{I})(\frac{A}{l}) \\ R=\frac{V}{I}=\frac{\rho l}{A} \\$$

The Attempt at a Solution

For the resistance i have done:
$$R=\frac{V}{I}=\frac{0.00425}{5}=0.00085Ω$$
For the conductance I looked it up that its the inverse so I just did
$$\frac{1}{0.00085}=1176.47Ω^{-1}\\$$
for the resistivity i have done
$$\rho=\frac{RA}{l} \\ \rho=\frac{(0.00085)(\pi 0.0088^2)}{0.1}=2.07\times 10^{-5} \\$$
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.

 

[ehild]

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.

It is the reciprocal of resistivity σ=1/ρ

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/resis.html

ehild

 

[FaraDazed]

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.

 

[ehild]

It looks correct.

ehild

 

[HalfLostAndSearching]

First of all, great job on correctly calculating the resistance and conductance of the conductor! To calculate the resistivity and conductivity of the material, we can use the following equations:

\rho=\frac{RA}{l} \\
\sigma=\frac{1}{\rho}=\frac{l}{RA}

Where \rho is the resistivity, \sigma is the conductivity, R is the resistance, A is the cross-sectional area of the conductor, and l is the length of the conductor.

Plugging in the values we have, we get:

\rho=\frac{(0.00085)(\pi 0.0088^2)}{0.1}=2.07\times 10^{-5} \Omega\cdot m \\
\sigma=\frac{1}{2.07\times 10^{-5}}=4.83\times 10^4\, \Omega^{-1}\cdot m^{-1}

This means that the material the conductor is made from has a resistivity of 2.07\times 10^{-5}\, \Omega\cdot m and a conductivity of 4.83\times 10^4\, \Omega^{-1}\cdot m^{-1}.

Hope this helps! Keep up the good work.