Circuits - Calculating resistance/conductance etc

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Homework Help Overview

The problem involves calculating resistance, conductance, resistivity, and conductivity for a cylindrical copper conductor given specific parameters such as voltage, length, and current. The subject area is electrical circuits and material properties.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate resistance and conductance using provided equations and expresses confusion regarding the calculation of conductivity. Some participants clarify that conductivity is the reciprocal of resistivity.

Discussion Status

The discussion includes attempts to verify calculations and clarify concepts. Some guidance has been offered regarding the relationship between resistivity and conductivity, but the original poster is still seeking further understanding.

Contextual Notes

The original poster notes a typo in their calculations regarding the radius of the conductor, which may affect their results. There is an emphasis on ensuring accuracy in the parameters used for calculations.

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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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