Current through a conductor with varying cross-sectional area

In summary, the conversation discusses the concept of current being constant across a conductor and the application of Kirchhoff's Current Law in understanding this concept. The conversation also touches upon the definition of a node and how KCL applies to all points in a circuit. It also briefly mentions how charge flows through a capacitor and how current-in equals current-out in this component.
  • #1
WWCY
479
12

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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  • #2
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
 
  • #3
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?
 
  • #4
Consider Kirchhoff's Current Law.
 
  • #5
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?
 
  • #6
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.
 
  • #7
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.
 
  • #8
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
 
  • #9
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!
 

Related to Current through a conductor with varying cross-sectional area

1. What is the relationship between current and cross-sectional area of a conductor?

The current through a conductor is directly proportional to its cross-sectional area. This means that as the cross-sectional area increases, the current also increases, and vice versa.

2. How does varying the cross-sectional area affect the current through a conductor?

Changing the cross-sectional area of a conductor will cause a change in the resistance of the conductor, which will in turn affect the current. If the area decreases, the resistance increases and the current decreases, and vice versa.

3. What is the formula for calculating current with varying cross-sectional area?

The formula for calculating current with varying cross-sectional area is I = A * v, where I is the current, A is the cross-sectional area, and v is the velocity of the electrons flowing through the conductor.

4. How does the shape of a conductor's cross-sectional area affect the current?

The shape of a conductor's cross-sectional area can affect the current by changing the resistance of the conductor. For example, a circular cross-section will have a lower resistance compared to a rectangular cross-section of the same area, resulting in a higher current.

5. Can current flow through a conductor with varying cross-sectional area at a constant rate?

No, current cannot flow through a conductor with varying cross-sectional area at a constant rate. This is because the change in cross-sectional area will cause a change in resistance, which will affect the flow of electrons and therefore, the current.

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