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Current through a conductor with varying cross-sectional area

  1. Feb 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2017-02-08 at 3.25.34 PM.png
    2. Relevant equations

    I = nqVA
    J = E/ρ
    J = I/A

    3. 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!
     
  2. jcsd
  3. Feb 8, 2017 #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
     
  4. Feb 8, 2017 #3
    I don't quite get what you mean, could you elaborate?
     
  5. Feb 8, 2017 #4

    gneill

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    Staff: Mentor

    Consider Kirchhoff's Current Law.
     
  6. Feb 8, 2017 #5

    Im not sure how to apply it in this scenario, my concept of it is that it applies to junctions?
     
  7. Feb 8, 2017 #6

    gneill

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    Staff: Mentor

    Yes. The machined cylinder might be roughly modeled as having three distinct regions with their own resistance. Draw the equivalent circuit.
     
  8. Feb 8, 2017 #7
    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.
     
  9. Feb 8, 2017 #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
     
  10. Feb 8, 2017 #9
    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?
     
  11. Feb 8, 2017 #10
    It is series circuit I1 =I2=I3 =I
    (R1+R2+R3) I =V
     
  12. Feb 8, 2017 #11
    You might consider with one branch
     
  13. Feb 8, 2017 #12
    Sorry, you might consider this a junction with one branch only
     
  14. Feb 8, 2017 #13

    gneill

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    Staff: Mentor

    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?
     
  15. Feb 8, 2017 #14
    Node is a point where the incoming and outgoing currents are equal
     
  16. Feb 8, 2017 #15

    gneill

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    Staff: Mentor

    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.
     
  17. Feb 8, 2017 #16
    It's not a definition,
     
  18. Feb 9, 2017 #17
    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?
     
  19. Feb 9, 2017 #18
    Thanks!
     
  20. Feb 9, 2017 #19

    gneill

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    Staff: Mentor

    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.
     
  21. Feb 9, 2017 #20
    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?
     
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