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What happens in a short circuit situation (parallel)?

  1. Jun 25, 2012 #1

    JJBladester

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    Gold Member

    1. The problem statement, all variables and given/known data

    What will be the current flowing theoretically through each resistor this short-circuit case?

    short_circuit.png


    2. Relevant equations

    [tex]V=IR[/tex]
    [tex]I=V/R[/tex]
    [tex]R_{T}=\frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+...\frac{1}{R_{N}}}[/tex]
    For a closed loop, [itex]\sum I_{entering}=\sum I_{leaving}[/itex]

    3. The attempt at a solution

    My answer: The current through each resistor will be effectively zero. This is because a practically resistance-free path exists (the short circuit) through which essentially all of the current will flow.

    The professor's response: Theoretically, the current through the 4-ohm resistor is infinite.

    I disagree. How can the current through any of the resistors (4Ω, 6Ω, or 12Ω) be anything other than ~0 A when the current will seek the path of least resistance (the short circuit)?
     
  2. jcsd
  3. Jun 25, 2012 #2
    The 4-ohm resistor short circuits, thus making its resistance zero, from ohm's law.

    I = v/r, since r is 0, I is infinity.
     
  4. Jun 26, 2012 #3

    NascentOxygen

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    Staff: Mentor

    I doubt that your professor would have said precisely that. He probably said that the current through branch 1 would theoretically be infinite.

    There will be zero current through all resistors, R1, R2, and R3. There is no current through them because all of the current is going through the short circuit which I understand has been placed in parallel to R1. Although all of the current that the source can supply will go through the short circuit, it of course won't be infinite because no practical source will be able to maintain its output at 2 volts regardless of the load, not right down to short-circuit conditions.

    Perhaps your professor meant that all of the current flowed through R1because R1 was accidently set to 0 ohms? If that's the case, then it is no longer correct to refer to R1 as the "4 ohm resistor". :smile:
     
  5. Jun 26, 2012 #4

    JJBladester

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    Oh, but he did!
     
  6. Jun 26, 2012 #5

    HallsofIvy

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    I presume he was talking about the "four ohm resistor" after the short circuit when would no longer be "four ohm". Still calling it a "four ohm resistor" is misleading but the concept is correct. (And "theoretically" is important- there will be some resistance in the wires which would keep the currant from being "infinite".)
     
  7. Jun 26, 2012 #6

    JJBladester

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    Also, the current cannot be infinite because the battery cannot push out an infinite number of electrons.

    But yes, HallsofIvy, I agree that you cannot call the 4 Ω resistor as such after the short circuit. That's what really confused me.

    I responded to the professor as follows:

    Perhaps I am misinterpreting your drawing. That is the only explanation I can come up with that would make sense. If the short-circuit branch in Case C is not in addition to, but rather a replacement for the normally-operating 4 Ω resistor, then your explanation makes sense.
     
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