Current through a conductor with varying cross-sectional area

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SUMMARY

The discussion centers on the behavior of electric current through a conductor with varying cross-sectional areas, specifically addressing the constancy of current as described by Kirchhoff's Current Law (KCL). Participants clarify that the current remains constant across different sections of the conductor, despite variations in area, due to the conservation of charge. The conversation also touches on the role of resistance and voltage drop in this context, and how these factors interact within a series circuit. Additionally, the flow of charge through capacitors is explained, emphasizing the role of electrons as charge carriers.

PREREQUISITES
  • Understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with electric current and charge carriers
  • Basic knowledge of series circuits and resistance
  • Concept of conservation of charge in electrical systems
NEXT STEPS
  • Study the implications of Kirchhoff's Current Law in complex circuits
  • Explore the relationship between resistance and cross-sectional area in conductors
  • Learn about the behavior of capacitors in DC circuits
  • Investigate the concept of electric fields and charge induction in conductors
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Electrical engineering students, physics learners, and professionals interested in understanding current flow in conductors and the principles governing electrical circuits.

WWCY
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Homework Statement


Screen Shot 2017-02-08 at 3.25.34 PM.png

Homework Equations



I = nqVA
J = E/ρ
J = I/A

The Attempt at a Solution


The underlying assumption was that current was constant across all 3 bits of the conductor, and thus answer is b.

The concept I can't grasp is this: Why is current constant? Shouldn't a smaller A mean smaller current as well?

Is there a 'force' shoving more q through a smaller A to compensate and is there a mathematical way to show that current is indeed independent of area?

Help is greatly appreciated!
 
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Current is freely moving electrons in a conductor sum total of all these electrons takes while conducting electricity.
Same number of electrons pass thru all the cross section of a conductor the
Only the resistance across the varies depending on the cross sectional area, and voltage drop
 
malemdk said:
Current is freely moving electrons in a conductor sum total of all these electrons takes while conducting electricity.
Same number of electrons pass thru all the cross section of a conductor the
Only the resistance across the varies depending on the cross sectional area, and voltage drop
I don't quite get what you mean, could you elaborate?
 
Consider Kirchhoff's Current Law.
 
gneill said:
Consider Kirchhoff's Current Law.
Im not sure how to apply it in this scenario, my concept of it is that it applies to junctions?
 
WWCY said:
Im not sure how to apply it in this scenario, my concept of it is that it applies to junctions?
Yes. The machined cylinder might be roughly modeled as having three distinct regions with their own resistance. Draw the equivalent circuit.
 
gneill said:
Yes. The machined cylinder might be roughly modeled as having three distinct regions with their own resistance. Draw the equivalent circuit.

Apologies for not drawing the circuit as I'm hanging about outside. However if i were to draw it, i'd draw 3 distinct resistors connected via a wire in a series connection. Is this right?

If so, how does Kirchoff's Law apply? It says in my text that a junction has to be a connecting point between > 3 separate conductors.
 
An analogy would help you,
Imagine water flowing thru a pipe-with various cross sections -the rate of flow across all the cross sections of the pipe same, irrespective of the changes in area, similarly the current across this conductor same across all sections
Google conservation of charge
 
malemdk said:
An analogy would help you,
Imagine water flowing thru a pipe-with various cross sections -the rate of flow across all the cross sections of the pipe same, irrespective of the changes in area, similarly the current across this conductor same across all sections
Google conservation of charge

Thank you, but I'm still not seeing how the variables in the I equation i quoted will behave across cross-sectional areas, do you mind elaborating?
 
  • #10
It is series circuit I1 =I2=I3 =I
(R1+R2+R3) I =V
 
  • #11
You might consider with one branch
 
  • #12
Sorry, you might consider this a junction with one branch only
 
  • #13
WWCY said:
If so, how does Kirchoff's Law apply? It says in my text that a junction has to be a connecting point between > 3 separate conductors.
I suppose they must have some reason for defining a junction in that manner, but it doesn't matter. How do they define a node?
 
  • #14
Node is a point where the incoming and outgoing currents are equal
 
  • #15
malemdk said:
Node is a point where the incoming and outgoing currents are equal
That's a pretty unhelpful definition as it presumes KCL holds before proving it. Everywhere else I've seen a node defined it's a point where two or more components connect.

Regardless, you can assume that KCL applies to every point in a circuit, including anywhere along a conducting path.
 
  • #16
It's not a definition,
 
  • #17
gneill said:
I suppose they must have some reason for defining a junction in that manner, but it doesn't matter. How do they define a node?

The word "node" has somehow not appeared yet in my text but thanks for providing me with its definition.

So this actually means that Kirchoff's Junction Rule is actually something like Kirchoff's everywhere-around-the-circuit rule?
 
  • #18
malemdk said:
Sorry, you might consider this a junction with one branch only

Thanks!
 
  • #19
WWCY said:
So this actually means that Kirchoff's Junction Rule is actually something like Kirchoff's everywhere-around-the-circuit rule?
Yes. If you think about it you can take any conductor and split it in two and you end up with two "components", one on either side of the split. So KCL applies there.

One component to note which appears to contradict KCL is the capacitor which consists of plates with a non-conducting gap between them through which charge carriers cannot flow. We know that charge builds up on a capacitor plate, so current-in doesn't equal current-out at that point. However, taking the capacitor as whole, an equal current leaves the opposite plate so that overall the current-in equals the current out, satisfying KCL for the component.
 
  • #20
gneill said:
Yes. If you think about it you can take any conductor and split it in two and you end up with two "components", one on either side of the split. So KCL applies there.

One component to note which appears to contradict KCL is the capacitor which consists of plates with a non-conducting gap between them through which charge carriers cannot flow. We know that charge builds up on a capacitor plate, so current-in doesn't equal current-out at that point. However, taking the capacitor as whole, an equal current leaves the opposite plate so that overall the current-in equals the current out, satisfying KCL for the component.

Thank you! Now that you mentioned it, how does charge flow through a capacitor?

Im guessing + charge flows from the + terminal to the connected plate, and the the flow of the - charge from the - terminal to its corresponding plate helps the + charge "continue" on its way to the - terminal. Is this the right concept?
 
  • #21
WWCY said:
Thank you! Now that you mentioned it, how does charge flow through a capacitor?

Im guessing + charge flows from the + terminal to the connected plate, and the the flow of the - charge from the - terminal to its corresponding plate helps the + charge "continue" on its way to the - terminal. Is this the right concept?
Sort of. In reality, in common materials the only charge carrier is the electron. If we abandon conventional current for a moment and consider electron flow and anchored positive atomic nuclei, the build up of electrons on one plate "pushes" electrons off the opposite plate and out its connecting wire (it's the basic static electricity concept of charge induction). Similarly, if you pull electrons off of a plate, electrons will be pulled onto the opposite plate, attracted by the potential of the exposed + charges of the atoms that gave up their electrons. It's the electric field produced by the charges that reaches across the gap to push or pull the charges on the opposite side.
 
  • #22
WWCY said:
Thank you! Now that you mentioned it, how does charge flow through a capacitor?

Im guessing + charge flows from the + terminal to the connected plate, and the the flow of the - charge from the - terminal to its corresponding plate helps the + charge "continue" on its way to the - terminal. Is this the right concept?
Sort of. In reality, in common materials the only charge carrier is the electron. If we abandon conventional current for a moment and consider electron flow and anchored positive atomic nuclei, the build up of electrons on one plate "pushes" electrons off the opposite plate and out its connecting wire (it's the basic static electricity concept of charge induction). Similarly, if you pull electrons off of a plate, electrons will be pulled onto the opposite plate, attracted by the potential of the exposed + charges of the atoms that gave up their electrons. It's the electric field produced by the charges that reaches across the gap to push or pull the charges on the opposite side.
 
  • #23
gneill said:
Sort of. In reality, in common materials the only charge carrier is the electron. If we abandon conventional current for a moment and consider electron flow and anchored positive atomic nuclei, the build up of electrons on one plate "pushes" electrons off the opposite plate and out its connecting wire (it's the basic static electricity concept of charge induction). Similarly, if you pull electrons off of a plate, electrons will be pulled onto the opposite plate, attracted by the potential of the exposed + charges of the atoms that gave up their electrons. It's the electric field produced by the charges that reaches across the gap to push or pull the charges on the opposite side.

Thank you, really appreciate the help!
 

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