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Adding resistor in parallel? What happens to circuit. (long read!) -Berkley Review

Discussion in 'MCAT Study Question Q&A' started by manohman, 09.23.14.

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  1. manohman

    manohman 2+ Year Member

    Joined:
    07.27.14
    Messages:
    110
    This question comes from berkley review E&M. Picture attached below.

    It asks: if you add ad resistor in parallel, then what happens to the current through the pre-existing resistors?

    The answer is that the current remains unchanged.

    From one viewpoint, I understand, because no matter how many resistors you add to a circuit, the voltage across each resistor will remain the same (crazy right?), which means that the current, since V = IR , will be the same as well, even if another resistor is added.

    But what I dont understand is, the junction rule also tells us that current going into a split and going out of a split must be equal, so intuitively, if you add another lane for current to flow through, then the current through the other lanes should decrease.

    However we know this is not true, because in adding another resistor in parallel ,the equivalent resistance in the circuit decreases, so the current going into the circuit goes up, just enough to offset the new current flowing into the added resistor.

    My issue here is that, when i considered this problem, i first though of the how current going into a split must be equal to current going out, and by that logic, the current through the pre-existing resistors should decrease. Why doesn't that logic apply?
     

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  3. justadream

    justadream 5+ Year Member

    Joined:
    04.29.11
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    2,165
    Don't overcomplicate this.

    Voltage is set by the battery. Since you have stuff in parallel, it's constant no matter what.
    Resistance is set by the resistors. That doesn't change.

    V = IR
    If V and R are the same, I is the same.


    When the switch is closed, you are adding another lane for current. However, the equivalent resistance of the entire thing is smaller. I believe the "gain" from having a smaller resistance goes into providing current for the new lane.*

    Take this with a grain of salt. I could very well be wrong about this explanation. The stuff I wrote above, however, I'm pretty sure is correct.
     
  4. techfan

    techfan 2+ Year Member

    Joined:
    01.29.12
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    117
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    Pre-Medical
    You seem to understand that the voltage across parallel resistors remains constant (if not check out this link, it has a pretty good explanation http://physics.stackexchange.com/questions/80103/why-does-voltage-remains-same-over-parallel-circuit), what you seem to be having a problem with is understanding that you are measuring the current after the system has gone to steady state. The water pipe analogy is a great one, but here it can lead you astray.

    While yes, when you close the switch at first the current would decrease through R1 (the extra lane provided by R5 sucks up some of the current), but once the particles get back to the battery and go around for the second trip there are more lanes for them to go so the total resistance drops, but the current increases to compensate. Here's a pretty good picture to help get the concept down.

    [​IMG]
     

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