A deeper look at voltage drops?..

Discussion in 'General Electronics Chat' started by Hesamu, Apr 22, 2011.

  1. Hesamu

    Thread Starter New Member

    Apr 22, 2011
    Consider the attached circuit. I know that no current passes throw R(2) given the 9V drop on R(1) which consumes all the voltage V(2). Please consider my following question as more of a physics type; why are electrons which are being driven toward the V(2) source have to spend the same amount of work as those being driven towards the V(1) source that is 9v. I know that KVL has to be valid in this circuit. but I want to know the reason if there is any. This came to my mind as to the fact that if there was a simple voltage divider circuit with one source and two or more resistors, voltage drops would be divided among the resistors and would sum up to the voltage supplied by KVL; in other words.... the voltage source did whatver ever it takes inorder to drive electrons form its negative to its positive terminal. Why wouldn't electrons do the same in this circuit?

  2. Adjuster

    Well-Known Member

    Dec 26, 2010
    :confused:I do not understand quite what you are asking, but I will try to help.
    The work done on an electron travelling from the common point to either of the two voltage sources would be the same because their potentials are equal, but I do not think this is the answer you are looking for.

    Are you wondering why no electrons are going to be driven towards the V2 source? This is simply because there is no potential difference to drive any current through R2. If you were to increase V2 slightly some electron current would flow towards it, and conversely if V2 were made less than V1 a current would flow in the opposite direction.

    Is it easier for you to see this if R1 is removed (made open-circuit)? This does not change the voltages, since we assume perfect voltage sources. The amount of current in R2 depends on the difference between the two potentials, and when this is zero no current flows.

    If you assume a conventional current I flowing from V2 to V1, I=(V2-V1)/R2.
  3. wmodavis

    Well-Known Member

    Oct 23, 2010
    For electrons to be 'driven' there must be a voltage difference to do the driving. Since you show two 9volt opposing voltage sources connected across r2 the actual forcing voltage is zero. So no electrons are being driven toward v2 caused by v1. It does not even matter that r1=r2. Whatever their values are the result would be the same.
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
    Both the circuit arrangements and the circuit elements presented at elementary level are greatly simplified versions of reality.

    Teachers do this to highlight the important points about the theory.
    These points are the real lessons at this stage.

    There are always a few students who try to be 'clever' by arranging these simplified models in impossible configurations and then saying. It doesn't work or it leads to contradictions or whatever.

    In electronics this often involves asking what happens when two ideal voltage sources are put in opposition.

    Don't do this, it will hinder your progress.

    Real voltage sources have additional characteristics that modify the model.
  5. Hesamu

    Thread Starter New Member

    Apr 22, 2011
    yes, i know that V1 wouldn't be able to push any electrons through V2 cause of no potential difference, but then...why aren't there any electrons being pushed in the V2 circuit by V2's driving force. To elaborate, why wouldn't V2's loop act on its own such that its driving force would drop some voltage across R1 that is not equal to V1's drop across R1 and the remaining voltage from V2 drops across R2. I know that my statement sounds absurd because it completely ignores KVL. Well....you could consider my question to, as annoying as it sounds, to explain the physics of electrons in this circuit.
    Anyways, I have reached some kind of conclusion. I think that electrons would just prefer to choose the path with less resistance that is V1's loop and go through R1 and not through both R1 and R2 which is V2's loop; this would only hold if V1 and V2 are both of same potential. Thoughts?
  6. Adjuster

    Well-Known Member

    Dec 26, 2010
    The point is that V1 and V2 ARE equal in this case. If they were not equal, current would indeed flow in R2.

    The existence of R1 actually makes no difference to the situation, since the voltage sources are ideal and have no internal resistance.

    You really have to accept that the current flowing through any resistor is defined by the difference of potential across it.

    Ohm's Law tells us I=V/R. If V= 0, I=0. That's all there is to say.
  7. JMac3108

    Active Member

    Aug 16, 2010
    You seem to be unwilling to accept the fact that it takes a voltage difference to create a current, or in your terminology "push electrons". V2 will not push any electrons through R2 because the voltage is the same on both sides of R2.

    Remove R1 from your circuit and look at it for a while and you'll see the point.