Kirchhoff's Voltage Law

Discussion in 'Homework Help' started by RdAdr, Feb 7, 2016.

  1. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    Consider the circuit from minute 6:57:



    So I have an electric field there everywhere in the loop and I can write integral of E*dl as being v and I end up with kirchhoff's rule. (you don't have to watch all the video to respond to this question, just the circuit)

    But what if I had an ideal current source in series with a resistor? How do we write integral of E*dl in a closed loop in this case? Do we even have an electric field inside the current source?

    If not, how do we reconcile the current source with kvl?
     
    Last edited: Feb 7, 2016
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,395
    497
  3. nwvlab

    New Member

    Feb 3, 2016
    4
    1
    When you have an ideal current source, the voltage across it become your unknown.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    First, note that the lecture is about non-conservative electric fields in which KVL doesn't hold.

    Second, if you have a current source then it still has electric fields but you need to determine what the voltage across the supply is via analysis. Just as you need to determine what the current is through a voltage source via analysis.
     
    nwvlab likes this.
  5. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    I am aware of superposition and how to calculate the unknowns in a circuit with any kind of sources.
    I was referring more to how the current source fits with KVL. This picture of going from E*dl to KVL.
    I do not see any integrals in that link. Read my question again.

    Yes, I know. In a non-conservative field the voltage can not be uniquely defined between two points and hence KVL does not hold. If the change of the magnetic flux is too noticeable and created by an internal current (and not an external source) then it is captured inside the lumped element which is the inductor.


    Yes, via analysis. In the case of the voltage source, the current is determined by the rest of the circuit. In the case of the current source, the voltage is determined by the rest of the circuit.

    I thought last night about this. In the case of the voltage source in series with a resistance we apply integral of E*dl and see that V=VR. Then we apply the Ohm's law to get the current:I=U/R. So in this case we do not ask: how do I reconcile the KCL with the voltage source? Because if I choose a node, then what is the current coming from the voltage source? The current is determined by the resistance.

    Similar, in the case of a current source in series with a resistance we apply not integral of E*dl, but integral of J*ds, J-current density and see that I=IR. Then we apply the Ohm's law to get the voltage:V=I/G So in this case we must not ask: how do I reconcile the KVL with the current source? Because if I choose the loop, then what is the voltage across the current source? The voltage is determined by the resistance.

    And that is because the current source is the dual of the voltage source.

    That's my take on it. Thanks for your answer.
     
    Last edited: Feb 8, 2016
Loading...