A smaller cross-sectional area increases the resistance of a conductor?

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SUMMARY

The resistance of a conductor increases with a smaller cross-sectional area due to Ohm's Law, as demonstrated by the relationship \( R = \frac{1}{\sigma} \frac{l}{S} \), where \( R \) is resistance, \( \sigma \) is conductivity, \( l \) is length, and \( S \) is the cross-sectional area. This principle is further explained through the Drude model, which describes current flow in conductors. Additionally, at high frequencies, the skin effect causes current to flow only in a thin surface layer, effectively increasing resistance. Understanding these concepts is crucial for analyzing electrical properties in conductors.

PREREQUISITES
  • Ohm's Law in differential form
  • Drude model of electrical conduction
  • Concept of skin effect in conductors
  • Basic understanding of electrical resistance and conductivity
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  • Study the implications of Ohm's Law in various conductor geometries
  • Explore the Drude model and its applications in solid-state physics
  • Investigate the skin effect and its impact on high-frequency AC circuits
  • Learn about the relationship between cross-sectional area and resistance in practical applications
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Electrical engineers, physics students, and anyone interested in understanding the fundamental principles of electrical resistance and conduction in materials.

Viona
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why smaller cross-sectional area increases the resistance of a conductor?
 
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The answer to that question is basically Ohm's law, with some other things(Drude model of current in conductors).
Hold on while I expand my reply with more details.
 
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@Viona, just fyi, a simple Google search would have given you the answer immediately
 
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Well basically we can start with Ohm's law in differential form ##J=\sigma E##, then if ##S## is the cross section and ##l## the length of the conductor we will have have that the current is ##I=JS=\sigma ES##, while the voltage in the two points of the conductor is ##V=El##, and hence the resistance $$R=\frac{V}{I}=\frac{El}{\sigma ES}=\frac{1}{\sigma}\frac{l}{S}$$.

I don't know if you were lookin for something like an intuitive explanation, like with the water through a pipe analogy, if the pipe has smaller cross section then we need to apply bigger pressure difference (analogy Voltage) in its two ends in order to achieve the same volumetric flow rate (analogy current).
 
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phinds said:
@Viona, just fyi, a simple Google search would have given you the answer immediately
I tried but the results did not convince me!
 
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Viona said:
I tried but the results did not convince me!
What sort of explanation you would find convincing? What is your intuition behind this?
 
What I said in post #3 hold for the DC or Quasi DC (low frequency AC) case, where we can consider the electric field ##E## to be uniform through out the conductor. If the wavelength is small enough (or equivalently frequency high enough) in comparison to the dimensions of the conductor, then the electric field ##E## varies in space and time as a wave and what I said in post #3 don't hold. More specifically in high frequencies what we have is the skin effect where the ohmic resistance is effectively increased because the current flows only in a thin surface layer (i.e. the "skin") of the conductor.
 
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Viona said:
I tried but the results did not convince me!
Then you should SAY so in your post. Since you gave no indication of any research, it was a natural assumption that you didn't DO any. When you know something about a subject, tell us and then tell us what you don't understand about it. That gives us better focus on presenting an answer for you.
 
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