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Will any charge flow through the circuit?

  1. Dec 16, 2016 #1
    https://drive.google.com/open?id=0B3IenBvKWb4PZW9rRWg4cXptT2c
    I had to find the resistance of this circuit between A and B. My teacher said that since the blank wires have zero resistance, the potential difference across the ends of the blank wire is zero(according to v = ir) and thus 1,4 are equipotent and 2,5 are equipotent and the circuit is thus simplified.The equivalent resistance comes out to be 2r/5.

    MY QUESTION:Will charge flow only through the circuit having resistance or will it flow through the circuit having no resistance or will it flow though both the branches.

    MY REASONING:Actually i am confused.
    One way in which i can reason is:the potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow thorough 1-4 and 2-5.
    Second way i can reason is that current always chooses the least resistant path and hence all the current will flow through 1-4 and 2-5.
     
  2. jcsd
  3. Dec 16, 2016 #2

    vanhees71

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    Somehow the figure got lost, and this kind of questions should be posted to the homework forum!
     
  4. Dec 16, 2016 #3
    @vanhees71 should i shift this to homework forum?
     
  5. Dec 16, 2016 #4

    cnh1995

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    It would be better if you did. Also, please upload the image in the thread itself instead of providing its link.
    Look up the terms 'open circuit' and 'short circuit'. In an open circuit, voltage is (or can be) nonzero but current is zero while in a short circuit, voltage is zero but current is (or can be) non zero. Ideal wires are kind of a short circuit.
     
  6. Dec 16, 2016 #5
    1. The problem statement, all variables and given/known data

    https://drive.google.com/open?id=0B3IenBvKWb4PZW9rRWg4cXptT2c
    upload_2016-12-16_12-28-36.png

    I had to find the resistance of this circuit between A and B. My teacher said that since the blank wires have zero resistance, the potential difference across the ends of the blank wire is zero(according to v = ir) and thus 1,4 are equipotent and 2,5 are equipotent and the circuit is thus simplified.The equivalent resistance comes out to be 2r/5.
    Will charge flow only through the circuit having resistance or will it flow through the circuit having no resistance or will it flow though both the branches.

    2. Relevant equations

    V=IR


    3. The attempt at a solution
    Actually i am confused.
    One way in which i can reason is:the potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow thorough 1-4 and 2-5.
    Second way i can reason is that current always chooses the least resistant path and hence all the current will flow through 1-4 and 2-5.
     
    Last edited by a moderator: Dec 16, 2016
  7. Dec 16, 2016 #6

    BvU

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    The drawing is intentionally misleading. Manipulate it a little bit to make it clearer:

    1) Put 2-4 upright by pulling 2 down and 4 up. You get a diamond shape with 1-2 on the lower left and 4-5 on the upper right; 2-3-4 is the vertical axis.

    2) shorten 1-4 and 2-5 until grokking is (*) :smile:

    (*) insight breaks through -- borrowed from Heinlein
     
  8. Dec 16, 2016 #7

    cnh1995

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    Instead of doing such reasoning, I'd siggest you redraw the circuit in such a way that you could easily simplify it. Replace the equipotentials with a single node.

    Edit: I see BvU suggested the same while I was typing...
     
  9. Dec 16, 2016 #8

    Merlin3189

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    I don't know the answer to that, though I understand that very large currents can flow in superconductors, which people seem to say have zero resistance.
    I think the "effort" with superconductors goes into changing the current. If there is no current, you'll need something, like a PD, to start current flowing. Once you have a current, no PD is required to maintain it. But another emf is required to reduce the current again.

    In fact, for your circuit the wire links 1-4 and 2-5 will have some resistance (supposedly very small), so you can assume current flows through them in any case.
    If you assumed a very small resistance, say r/10^6, you could work out the more complex circuit and get the right answer to 5sf.
    I think we just assume connecting wires are zero resistance to simplify the calculation for all practical purposes. And then we allow it to conduct any current without any PD (even though, as with a superconductor, it would need a transient PD to initiate current flow.).

    I don't understand this. I agree about equal PD across parallel circuits, but see no reason why you then assume it is zero here nor why you assume no current flows.
    Well your premise is simply wrong. Current does not choose the path of least resistance nor any path. Current flows along any and all paths available to it.

    I think the notion of current flow along the path of least resistance may come from currents in insulators? In an ideal insulator no current can flow anywhere. But if the field strength is great enough, you get dielectric breakdown and the insulator can conduct.
    Since field strength is PD/distance (volts/metre) then a homogeneous insulator breaks down along the shortest path between the sources of the PD. If there are variations in dielectric strength, then some parts will breakdown sooner (at lower field strength) and, by becoming conductive allow the field to increase elsewhere causing breakdown in higher dielectric strength areas. This can lead to a very irregular shaped current path - the path of least aggregate dielectric strength?
     
  10. Dec 16, 2016 #9

    rude man

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    The potential across an equipotential (like an ideal wire) is zero but so is its resistance, giving i = V/R = 0/0 which is indeterminate so you can't associate zero potential with zero current.
     
  11. Dec 16, 2016 #10

    gneill

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    This is actually a bit of folklore that is not true. It's based upon lay observations of how water flows through terrain, or how lightning seems to find a single shortest route to ground once it strikes. Neither is strictly true, and the false generalization is based on observing the end result of a "carved path" left in the wake of the largest current.

    In a circuit, current will take every path available to it provided there is a potential difference to drive it. Parallel paths are no problem, and the current will divide amongst the available paths according to the relative conductivities of the paths. That is to say, while a lower resistance path will take more current than a higher resistance one, every path still conducts some current. The exception to this is when one of the parallel paths happens to be a perfect conductor (such as an ideal wire in a circuit diagram). Then the zero resistance path will clamp the potential difference across all the paths to zero. There being no potential difference across the other paths, there will be no potential to drive current through those path's components, while the zero resistance path allows all the current to flow through itself unimpeded.
     
  12. Dec 16, 2016 #11
    @BvU could you please provide a diagram. I am finding it hard to imagine the changes.
     
  13. Dec 16, 2016 #12

    phinds

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    No, this is EXACTLY what the problem is supposed to teach you --- how to change circuit diagrams so as to show clearly what is serial and what is parallel. If we redraw the diagram we are spoon feeding you the answer to this question and that is against forum rules.

    And by the way, this specific example is a classic for teaching circuit redrawing. We've seen it here before several times. You need to figure this one out on your own now.
     
  14. Dec 16, 2016 #13

    cnh1995

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    Well, go step-by-step.
    Name the resistor terminals as a,b,c and so on and connect the equipotential terminals together.
    Edit: I see the terminals are already numbered. Nvm, you can use the numbers directly.
     
  15. Dec 16, 2016 #14

    vanhees71

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    Just to let you know, I can't do anything of this, and I also don't see your figures or links.
     
  16. Dec 16, 2016 #15

    davenn

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    start a new thread in homework, use the homework template and upload any images to the thread (as one other person said)

    I will ask a mentor to delete this thread once you have that sorted out :smile:


    Dave
     
  17. Dec 16, 2016 #16

    berkeman

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    Two threads merged after the Mentors struggled with a server issue all day. Yay!
     
  18. Dec 17, 2016 #17
    @BvU @phinds , I did it. After struggling for about half an hour i was able to imagine pulling and reshaping the circuit into the diamond shaped structure. I also discovered another approach in which the nodes are taken at convenient positions and then the resistors between the nodes are drawn by tallying it with the original circuit. What do you think of this?
     
  19. Dec 17, 2016 #18

    cnh1995

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    Did you get the answer as 2r/5? Or do you need some more help?
    Is there any circuit diagram you wanted to show us with this post? Because I can't see any image attached.
     
  20. Dec 17, 2016 #19
    @cnh1995 , No. I am not referring to any circuit. I want to know what others think of the second approach which i used.
    By the way i think that this approach is more convenient than pulling the nodes and reshaping the circuit.
     
  21. Dec 17, 2016 #20

    ehild

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    As the currents have to reach the resistors, current should flow through the wires. The ends of the wires are at the same potential, but their resistance is zero, so any current can flow resulting zero potential across them. Simplify the circuit, determine the current (magnitude and direction) through each resistor, and determine the current flowing through the wires in the original set-up using Kirchhoff's nodal law.
     
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