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Kirchhoff's Junction Rule in the AC case...

  1. Oct 2, 2015 #1
    Kirchhoff's laws are valid for DC circuit and also for AC circuits (up to point. I think they become only approximations at high frequencies).

    Let's consider a node (junction) in a AC circuit. Three AC currents with the same magnitude but different relative phases enter the node. It is possible for zero current to exit the node, correct? It seems strange though that 3 currents enter and no current leaves the node. That cannot happen in DC: no charge accumulation can happen at any point in the circuit.

    But what does it physically mean that in the AC case the net current exiting the node is zero? That there is some charge accumulation and that the charge does not have an instantaneous velocity at that point?


    thanks!
     
  2. jcsd
  3. Oct 2, 2015 #2

    anorlunda

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    There is no mystery. The currents (in complex vector form) simply add up to zero.

    Consider a DC wire. We have a 1 amp current traveling to the left and 1 amp traveling to the right. They add up to zero; no current. Why is that mysterious?
     
  4. Oct 2, 2015 #3

    donpacino

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    Lets



    For simplicity lets assume the AC current is a sine wave with a magnitude of 1V and a frequency of 1 Hz.

    The voltage waveform will start at zero, increase to 1, decrease to 0, decrease to -1, etc...

    so if our AC signal is going into an element, the actual instantaneous current going into the element is going to switch between positive and negative.

    also don't forget that sin(x)= -sin(ft+180)

    if we have two signals waves A and B

    A=sin(ft)
    B=sin(ft+180)

    you can say that both signals go into an element from different directions.


    Does that make sense?
     
  5. Oct 2, 2015 #4
  6. Oct 2, 2015 #5

    donpacino

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    When you're dealing with circuitry like this, thinking of the velocity of charge is going to confuse the hell out of you. Think of it at a higher level unless for some reason you need to look into it with that level of complexity.
     
  7. Oct 2, 2015 #6

    anorlunda

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    Also a third way of saying the same thing, for three phase systems.

    What is the numerical value of COS(0)+COS(2*pi/3)+cos(-2*pi/3)? All angles are in radians.
     
  8. Oct 7, 2015 #7
    Hello,

    sure, COS(0)+COS(2*pi/3)+cos(-2*pi/3)=0. It is as if three mono-phase generators where connected together Superposition)
     
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