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Why does this bulb get brighter after the switch is closed

  1. Apr 18, 2017 #1
    1. The problem statement, all variables and given/known data
    pKYgD0X.jpg

    2. Relevant equations
    None

    3. The attempt at a solution
    The answer is B, bulb B will be brighter than before.

    My thought was that initially, the current gets split so that 2/3 I0 goes to the branch with 2 bulbs, since it has twice the resistance of the first branch, and that 1/3 I0 goes to the first branch. And when the switch is closed, that initial junction doesn't change, so I picked D.

    Can anyone explain where I went wrong?
     
  2. jcsd
  3. Apr 18, 2017 #2

    lewando

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    Gold Member

    Your thought on how current was distributed was not correct (edit: study "current division").

    "the initial junction doesn't change, so I picked D" requires additional clarification.

    Ask yourself what is the voltage across A before and after the switch closure.

    Ask yourself what is the voltage across B before and after the switch closure.
     
    Last edited: Apr 18, 2017
  4. Apr 18, 2017 #3
    I still don't really understand.

    So the voltage across A initially would be equal to emf of the source, right? And the voltage across B is a fraction of the emf,with the other fraction shared with the other lightbulb?
     
  5. Apr 18, 2017 #4

    gneill

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    Staff: Mentor

    Right, for when the switch is open. The "fraction" would be 1/2 if the bulbs are identical.

    How about when the switch is closed? Let's put some labels on various points of the circuit when the switch is closed:

    upload_2017-4-18_21-23-28.png

    Can you identify groups of labels that share the same potential? (Let's say the potentials are with respect to the location labeled "e").
     
  6. Apr 19, 2017 #5
    Would e, c, d and f have the same potential? And then a and b would also have the same potential?
     
  7. Apr 19, 2017 #6

    gneill

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    Staff: Mentor

    Yes, that's correct.

    So now can you compare the potential differences that appear across the bulbs A and B?
     
  8. Apr 19, 2017 #7
    If lightbulbs A and B have the same resistance, wouldn't the potential across ac be equal to the potential across bd?
     
  9. Apr 19, 2017 #8

    gneill

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    Staff: Mentor

    Yes, but not just because they have the same resistance. You've already ascertained that the potential at a and b is the same, as is the potential at c and d. So regardless of the resistance of the bulbs the potential differences across them must be the same. This is true of any components connected in parallel -- they share the same potential difference if they share the same connection points.
     
  10. Apr 22, 2017 #9

    Hey, sorry I haven't answered in a while.

    So the potential from a to f will be the same as the potential from b to f, right? So if this is the case then why does bulb B shine brighter than bulb A if both paths have the same potential shared between two bulbs?
     
  11. Apr 22, 2017 #10

    gneill

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    Staff: Mentor

    Where does it say that bulb B will shine brighter than bulb A?
     
  12. Apr 22, 2017 #11
    Oh I guess I assumed they both shined at the same level initially.
     
  13. Apr 22, 2017 #12

    gneill

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    Staff: Mentor

    And what are your thoughts on that now?
     
  14. Apr 22, 2017 #13
    I don't really know how to talk about it in terms of electric potential, but when the switch is closed, some of the current that was going through path ae now goes through path af instead. So now bulb B gets a greater share of the current going through bf, so it shines brighter.

    Is this correct?
     
  15. Apr 22, 2017 #14

    gneill

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    Staff: Mentor

    Actually, the current flowing through bulb A (path ae) remains the same in both cases. The potential difference is the same in each case: it is the potential difference supplied by the battery since it connects to nodes a and e.

    It is the current through bulb B that changes when the switch closes. When the switch is open the path from b to e involves two bulbs in series (the path is bdfe). Two bulbs in series presents more resistance to current flow than one, so the current through that path is less than the current though the bulb A path. When the switch is closed, both paths have just a single bulb across the battery.

    upload_2017-4-22_12-39-30.png
     
  16. Apr 22, 2017 #15
    Ah okay, I understand now.

    So, since path bdce has a lower potential, there will no current going through the third bulb?
     
  17. Apr 22, 2017 #16

    gneill

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    Staff: Mentor

    Locations have potential. Potential differences drive currents. As you noted previously, when the switch is closed locations c,d,e, and f all have the same potential. So locations d and f have zero potential difference, so there's no potential difference to drive current through the third bulb.
     
  18. Apr 22, 2017 #17
    Okay, thank you.
     
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