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Adding resistors in parallel - Current Change through resistors

  1. Sep 23, 2014 #1
    1. The problem statement, all variables and given/known data
    A resistor is added to a circuit in parallel.. How does the current through the pre-existing resistor change?


    2. Relevant equations
    V=IR


    3. The attempt at a solution


    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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    Last edited: Sep 23, 2014
  2. jcsd
  3. Sep 23, 2014 #2

    gneill

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    Staff: Mentor

    If you have an ideal voltage source then it will produce any amount of current in order to satisfy its mission of providing that fixed potential difference. It doesn't matter what the load is, it will produce enough current to maintain that potential difference. So if you add resistors in parallel, the source will produce more current to supply every one of the added resistors with the current it needs to show that potential difference.

    The situation would be different if the voltage source were replaced by a current source. Then the total current would be fixed and would divide through the parallel paths, so the potential difference across the parallel resistors would drop as more parallel resistors are added.
     
  4. Sep 23, 2014 #3
    ah okay that makes sense. So in the latter case with a current source, the voltage across each battery would still be the same, but the current would decrease across each?

    And so when we consider such problems with batteries, its under the assumption that the battery has infinite stores of chemical energy with which to convert into the potential energy (or is it electric energy?) to sustain the potential difference that drives the charge.

    Sorry I know this yet another question, and i can post it in a new thread if that would be more appropriate, but what conceptually is electric potential? I know it's what drives charges to flow, and what causes a current, and the resulting current is determined by this driving force and the resistance to the resulting flow, but it's different than just a push like with an electric force between two charges. So what is it exactly?
     
  5. Sep 23, 2014 #4

    gneill

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    Staff: Mentor

    Um, sorry? If you replace the battery (voltage source) with a current source, then there's no battery in the circuit. A current source is not the same thing as a voltage source (although they both deal with voltage and current!). The voltage source produces a fixed potential difference no matter how much current it needs to produce or consume in order to do so. The current source produces a fixed current no matter what potential difference it has to create in order for that to happen.

    If you have parallel resistors supplied by a single current source, then adding more resistors in parallel will not change the amount of current that the source produces. What will happen is that the given amount of current is shared out to all the resistors. More resistors means each individual resistor gets less of the total. Less current through a resistor means a lower potential drop across it (Ohm's Law).
    That is true for ideal batteries 9ideal voltage supplies).
    Potential difference is the result of an electric field. The push is there from the field acting on the charges, and the potential difference represent the work that will be done by that field on a charge moving from one location to another. Note that the unit of potential, the Volt, is a composite unit: 1 Volt = 1 Joule / Coulomb.
     
  6. Sep 23, 2014 #5
    Ah I see. Thank you. Sorry I butchered my wording regarding current source (and i misunderstood) but that's a lot clearer now.

    So the difference between a current source and a voltage source (battery) is that if you add resistors in parallel with a voltage source, the current through each resistor does not change, though the current leaving the battery does (while the voltage of course remains constant as well). With a current source, the current leaving the source is constant, but if you add resistors in series, the current going through each resistor will decrease accordingly per resistor added.

    So electric field tells us about the force (push or pull) that would be felt by a charge at a given distance, it is Felectric/q because it is for any given charge. So does the potential difference simulate the work that it would take to move a charge from one point to another? V = J/C so the energy it takes to move a single charge. Where does distance factor in? Thanks this is helping me clarify my foundational understanding a lot!
     
  7. Sep 23, 2014 #6

    gneill

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    Staff: Mentor

    You mean adding resistors in parallel with a current source, right? Adding resistors in series with a current source doesn't change the current through the resistors. Resistors in series all conduct the same current.
    Don't think of it as distance but as differences in location. The field strength at a given location tells you the force acting on a charge at that location. Yes, the potential difference between locations tells you how much work (Joules) has to be done on a charge (Coulombs) to move that charge from one location to the other. The path or distance doesn't matter -- The electric field is a conservative field.
     
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