Current change after inserting thin conductive foil into resistor

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SUMMARY

The discussion focuses on the effect of inserting a thin conductive foil into a system of two square parallel metal plates filled with a conductive liquid. The original current through the plates is calculated using Ohm's Law (V=IR), and it is concluded that the current will increase to (9/8) times the original current when the foil is introduced. Participants emphasize the importance of considering the conductive foil as an equipotential zero-ohm link across the resistors, which significantly alters the current distribution in the circuit.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Familiarity with series and parallel resistor configurations
  • Knowledge of electrical conductivity and resistivity concepts
  • Basic principles of circuit analysis involving equipotential surfaces
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  • Study the effects of inserting conductive materials in parallel plate capacitors
  • Learn about the implications of equipotential surfaces in electrical circuits
  • Explore advanced circuit analysis techniques for complex resistor networks
  • Investigate the properties of conductive liquids and their impact on current flow
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Electrical engineers, physics students, and anyone interested in circuit design and analysis, particularly in scenarios involving conductive materials and their effects on current flow.

student_man
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Homework Statement
Given are two square parallel metal plates, with dimensions a×a whose distance between them is d where d≪a. The space between the plates is filled with a conductive liquid with resistivity p. The plates are connected to a constant voltage source U. How much will the current through the plates change when a conductive foil is inserted into the space between the plates and bent in the middle as shown in the figure? The current at any point between the plates flows in the direction perpendicular to the plane of the upper plate.
Relevant Equations
Circuit attached
I tried splitting it up into 4 individual resistors, adding them up (each side serially and then both sides with each other as parallel) and then using V=IR but it ended up being the same current as the original.
(The answer should end up being (9/8)* the original current)
Screenshot 2023-10-05 185408.png
 
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student_man said:
Homework Statement: Given are two square parallel metal plates, with dimensions a×a whose distance between them is d where d≪a. The space between the plates is filled with a conductive liquid with resistivity p. The plates are connected to a constant voltage source U. How much will the current through the plates change when a conductive foil is inserted into the space between the plates and bent in the middle as shown in the figure? The current at any point between the plates flows in the direction perpendicular to the plane of the upper plate.
Relevant Equations: Circuit attached

I tried splitting it up into 4 individual resistors, adding them up (each side serially and then both sides with each other as parallel) and then using V=IR but it ended up being the same current as the original.
(The answer should end up being (9/8)* the original current)View attachment 333071
Welcome to PF.

Interesting problem! Can you show your math where you ended up getting the same overall current?
 
berkeman said:
Welcome to PF.

Interesting problem! Can you show your math where you ended up getting the same overall current?
Here is my math
SmartSelect_20231005_194723_Samsung Notes.jpg

I hope it's understandable : )
 
I believe you left out the horizontal short-circuiting wire between the for 4 resistors... :wink:
 
berkeman said:
I believe you left out the horizontal short-circuiting wire between the for 4 resistors... :wink:
I'm not really sure how to take it into consideration in the equations😅
The only information I'm given about the piece of foil is that it's conductive.
 
The foil is an equipotential. It is a zero ohms link across a bridge of four resistors. Solve parallel then series, not series the parallel.
 
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Baluncore said:
The foil is an equipotential. It is a zero ohms link across a bridge of four resistors. Solve parallel then series, not series the parallel.
Thank you! :biggrin:
 
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