Current change after inserting thin conductive foil into resistor

In summary, inserting thin conductive foil into a resistor alters its electrical properties by reducing resistance and potentially increasing current flow. This modification can lead to enhanced performance in electronic circuits but may also raise concerns regarding heat dissipation and component reliability.
  • #1
student_man
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1
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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  • #2
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?
 
  • #3
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 : )
 
  • #4
I believe you left out the horizontal short-circuiting wire between the for 4 resistors... :wink:
 
  • #5
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.
 
  • #6
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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  • #7
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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FAQ: Current change after inserting thin conductive foil into resistor

What happens to the current when a thin conductive foil is inserted into a resistor?

Inserting a thin conductive foil into a resistor typically decreases the overall resistance of the circuit. This is because the foil provides an additional path for the current to flow through, effectively creating a parallel resistance. According to Ohm's Law (V=IR), if the resistance decreases and the voltage remains constant, the current will increase.

How does the thickness and material of the conductive foil affect the current change?

The thickness and material of the conductive foil significantly influence the extent of the current change. Thicker foils and materials with higher conductivity (such as copper or silver) will further reduce the resistance, resulting in a larger increase in current. Conversely, thinner foils or those made from less conductive materials will have a smaller effect.

Will the insertion of conductive foil impact the power dissipation in the resistor?

Yes, the insertion of conductive foil will impact the power dissipation in the resistor. Since power dissipation in a resistor is given by P = I²R, an increase in current (I) due to decreased resistance (R) will generally lead to higher power dissipation. However, the exact change depends on the specific configuration and the resultant resistance of the combined system.

Is it possible for the thin conductive foil to cause overheating in the circuit?

Yes, it is possible for the thin conductive foil to cause overheating in the circuit. The increase in current can lead to higher power dissipation, which manifests as heat. If the circuit components are not designed to handle the increased power, this can result in overheating and potential damage to the resistor or other circuit elements.

How can the insertion of conductive foil be modeled in circuit analysis?

The insertion of conductive foil can be modeled as adding a parallel resistor to the original resistor. The total resistance of the circuit can be calculated using the formula for parallel resistances: 1/R_total = 1/R1 + 1/R2, where R1 is the resistance of the original resistor and R2 is the resistance of the conductive foil. This model allows for a more precise analysis of the changes in current and voltage within the circuit.

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