1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

2nd Kirchhoff's law

  1. Jun 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Apply the 2nd Kirchhoff's law in:

    attachment.php?attachmentid=70441&stc=1&d=1402248289.png

    2. Relevant equations
    3. The attempt at a solution

    ##V_1 = V_3 + V_5 + V_2##

    ##V_7 = V_4 + V_5##

    ##0 = V_2 + V_4 + V_6##

    ##V_1 + V_7 = V_3 + V_6##

    All right!?
     

    Attached Files:

    • asd.PNG
      asd.PNG
      File size:
      3.2 KB
      Views:
      186
  2. jcsd
  3. Jun 8, 2014 #2
    No, not all right. You have to chose a direction for the Voltage drops V2, V3, etc. and then be consistent with that choice through out the equations. Some of the terms will inevitably end up with a minus sign attached to them.
     
  4. Jun 8, 2014 #3
    Also, Application of Kirchhoff's 2nd to that circuit produces only three independent equations. You wrote down four equations. Even after fixing the signs one of those equations will be useless. Correct but useless.
     
  5. Jun 8, 2014 #4
    hmm... if v1 > v7 so this scheme is correct:
    attachment.php?attachmentid=70442&stc=1&d=1402253759.png

    and if v1 < v7 so this scheme is valid:
    attachment.php?attachmentid=70443&stc=1&d=1402253759.png
     

    Attached Files:

    • 1.PNG
      1.PNG
      File size:
      4 KB
      Views:
      117
    • 2.PNG
      2.PNG
      File size:
      4 KB
      Views:
      107
  6. Jun 8, 2014 #5
    OR

    for v1 > v7 the correct is:
    attachment.php?attachmentid=70448&stc=1&d=1402257086.png

    and v1 < v7 is:
    attachment.php?attachmentid=70449&stc=1&d=1402257086.png

    ?
     

    Attached Files:

    • 01.PNG
      01.PNG
      File size:
      3.6 KB
      Views:
      91
    • 02.PNG
      02.PNG
      File size:
      3.6 KB
      Views:
      86
  7. Jun 8, 2014 #6
    Don't waste time trying to guess the direction of the currents. Just chose one that seems reasonable. If you make a mistake somewhere, it is a self correcting mistake because the solution for those currents that were chosen in the wrong direction will turn out to be negative numbers indicating that the current is in the opposite direction.
     
  8. Jun 8, 2014 #7
    Ok... so why the my equations in the 1st post are wrong?
     
  9. Jun 9, 2014 #8
    Dear Jhenrique
    for writeing KVL equation at first you should define the direction of current or voltage of each element.it depends on you how you can assume the element voltage or current sign or direction.for example i set the voltage sign in this form
    5433016700_1402290090.png
    now for first loop we write KVL
    -V1+V2+V5-V3=0
    and for third loop we have
    -V7-V4+V5=0

    above equations were an example of setting voltage signs.after set the voltage sing and write KVL equation you should solve them.i assume that we find V5=2 and V3=-1 so the circuit could be same as this
    7415916400_1402290457.png
    or combine the sign of voltage that specified and the sign of voltage that you obtain. redraw the circuit like this
    2507908400_1402290458.jpg
     
    Last edited: Jun 9, 2014
  10. Jun 9, 2014 #9
    Your approach is very interesting!
     
  11. Jun 9, 2014 #10
    I said don't waste time guessing the direction of the currents. Just chose one. Your 1st post is wrong because you didn't chose a direction
     
  12. Jun 9, 2014 #11
    I found another way more easy of solve it (my ideia isn't solve the problem, is understand the dynamical of the electrical circuit)

    if I apply the hydraulic analogy like this video:


    and the superpostion theorem:
    http://en.wikipedia.org/wiki/Superposition_theorem

    together with KVL
    ##\oint \vec{E} \cdot d\vec{s} = 0##

    in this case:
    attachment.php?attachmentid=70470&stc=1&d=1402348364.png

    and choose a mesh and a sense of circulation, so the "ramps" (source or resistor) would have positive voltage and the declivities correspond to negative voltage!
     
    Last edited by a moderator: Sep 25, 2014
  13. Jun 9, 2014 #12
    Yes, that's the way to solve the problem. Guess what you will get out of it. Yes, you get Kirchhoffs' laws. No kidding...
     
    Last edited by a moderator: Sep 25, 2014
  14. Jun 10, 2014 #13
    The 2nd Kirchhoff's law is quite elaborate! I never would understand the ideia, the principle, the fundament, the dynamical of the thing if I hadn't watched the caltech's video...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: 2nd Kirchhoff's law
  1. Kirchhoff's Laws (Replies: 2)

  2. Kirchhoff's law (Replies: 22)

  3. Kirchhoff laws (Replies: 9)

  4. Kirchhoff's Laws (Replies: 4)

  5. 2nd Kirchhoff's law (Replies: 4)

Loading...