Resistance from Microscopic Ohm's law

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SUMMARY

The resistance of a cylindrical resistor made from a conductive material with conductivity \(\sigma\) is calculated using the formula \(R = \frac{L}{\sigma A}\). This derivation begins with Ohm's Law, where the current density \(J\) is defined as \(J = \sigma E\), and the relationship between voltage \(V\), electric field \(E\), and length \(L\) is utilized. The final expression for resistance eliminates the electric field terms, confirming that resistance is directly proportional to length and inversely proportional to the product of conductivity and cross-sectional area.

PREREQUISITES
  • Understanding of Ohm's Law and its applications
  • Familiarity with electrical conductivity and its significance
  • Knowledge of cylindrical geometry and related formulas
  • Basic principles of electric fields and current density
NEXT STEPS
  • Study the derivation of resistance in different geometries, such as spherical and planar resistors
  • Explore the impact of temperature on conductivity and resistance
  • Learn about the applications of carbon composition resistors in electronic circuits
  • Investigate advanced topics in materials science related to resistive materials
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Students studying electrical engineering, physics enthusiasts, and professionals involved in circuit design and analysis will benefit from this discussion.

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Homework Statement



Your task is to calculate the resistance of a simple cylindrical resistor with wires connected to the ends, such as the carbon composition resistors that are used on electronic circuit boards. Imagine that the resistor is made by squirting material whose conductivity is \sigma into a cylindrical mold with length L and cross-sectional area A. Assume that this material satisfies Ohm's law. (It should if the resistor is operated within its power dissipation limits.)

What is the resistance R of this resistor?
Express the resistance in terms of variables given in the introduction. Do not use V or I in your answer.

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Homework Equations



J=\SigmaE
V=EL
I=JA
R = V/I

The Attempt at a Solution



I start out with Ohm's Law and get resistance R to be R = V/I and I know V = EL and I = JA so I get resistance R to be R = EL/JA. I also know J = \SigmaE. So I get resistance R to be R = EL/\SigmaEA

Would the two E's cancel out?
 
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I will continue from your R = EL/JA expression.
J = \sigmaE
therefore,
R = EL/\sigmaEA
the E's cancel out and your answer should be R = L/\sigmaA

Hope this helps.
 
Ah, wrong sigma. Got it, thanks.
 

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