Find The Power Supplied by The Ideal Current Sources Using KVL and KCL

Discussion in 'Homework Help' started by Mohamed Hesham Mohammed, Oct 8, 2016.

  1. Mohamed Hesham Mohammed

    Thread Starter Member

    Oct 8, 2016
    33
    0
    I tried To figure All Kcl in Every Node And KVL for Each CLosed Path But I ended Up with Huge Number Of Equations I tried posting this Question on Diffferent Forums but No Hope this is Actually a Tough One.I did't Post any Steps Because It is about 20 Equations.
     
    • Cap.PNG
      Cap.PNG
      File size:
      72.9 KB
      Views:
      26
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,164
    post the steps anyway. that will give the clues for the members to help you. Was there a specific question the professor wanted you to answer?
     
  3. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,286
    331
    Why do you have 20 equations? Do you know how to do mesh or nodal analysis?

    The first thing you should have done is to eliminate the 5Ω resistor in parallel with the 4Io source, and also eliminate the 3Ω resistor in parallel with the 5Vy source. Those two resistors have no effect on the rest of the circuit because they're in parallel with voltage sources.

    If you do this you will only have 5 meshes and that means 5 equations to solve.
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492

    Hi,

    I could help you with this but i have to know for sure if that 5 ohm resistor is connected to that 4*Ia dependent current source. It's hard to tell because there is no dot there, but there are dots in some places and not in others. We can do it both ways i suppose, but knowing will save time.
    So are all crossing lines dotted or not?

    Yes a number of pre-simplifications look promising.
    But also, i only see about 12 nodes. Where are the "20" equations coming from or was that just an enlightening exaggeration?
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,788
    4,808
    So you don't want to post your work but yet you expect people to somehow be able to tell you where you are going wrong.

    We are NOT mind readers.
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hello again,

    While waiting for your reply here is a cleaned up drawing. I assume those two crossing lines have a dot too so i added a dot, but if that is not correct then you'll have to say so. The stuff i added is shown in blue with all the original still in black. I also assumed that one current is 'I' sub 'a' but it may be 'I' sub 'alpha', although the analysis will work out the same either way.
    Also, since this looked like an interesting network i decided to solve it and i show one node solution but only that one, and i did not double check that solution yet so it may be wrong, so you have to do it yourself too. If you have any problems after this you'll just have to state what is not working for you. You might also show how you are going about doing this, as there are some shortcuts possible due to a number of independent substructures.
    In any case you have to solve systems of simultaneous equations where there could be a number of equations not just 2 or 3 as in many simpler problems. That means you have to have a way to solve more than 3 equations. You can do it by hand or use automated software if that is allowed in your course. I used automated software to save time and energy :)

    When checking however be sure to note my choice for the ground location which was really arbitrary. Another choice would be on the other side of that lower 3 ohm resistor but that changes the solution at the node shown with the displayed voltage 670/499 volts.
     
    Last edited: Oct 9, 2016
  7. DGElder

    Member

    Apr 3, 2016
    345
    85
    Using KVL:

    You can ignore the the two meshes with the dependent voltage sources and the 5 and 3 ohm resistors in parallel since those resistors have no effect on the other mesh currents - as the Electrician already noted above.

    This leaves you 5 meshes and 5 mesh currents to find. However you can not write a KVL equation for two of them because they have current sources in the mesh - which have unknown voltages. However one of those meshes has a self defined current of 5A due to the included unshared independent current source. The other mesh with a current source shares that current source with the 5A mesh so you can write an equation relating the two mesh currents through the shared 2A source.

    This gives you five equations in 5 unknown mesh currents - if you write the dependent voltage source voltages in terms of mesh currents.
    Due to the known 5A mesh this easily reduces to 3 equations in 3 unknowns to solve simultaneously.

    I solved the circuit and got a different result than Mr. Al. I checked a few currents and voltages for self consistency, but not everywhere so I could have made an error as well. In any case for the 4Ia voltage source I got 12V.
     
    Last edited: Oct 9, 2016
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,286
    331
    He did the same thing in the other forum he mentioned; posts an initial post and then no further response.

    Perhaps the policy for homework help could be expressed as a slight variation on an old saying: "AAC helps those who help themselves".
     
  9. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,286
    331
    It's interesting that all 5 mesh currents are small integer values; the node voltages are also all integer values. Whoever created this circuit most likely went to the trouble of making that happen. Usually a network this complicated wouldn't have an all integer solution. Doing so makes for easier grading of students' answers! As soon as I saw my result came out as all integers, I was 90% sure I had the right result. I then solved it using nodal analysis and got the same answers. I find that solving with both methods is a good way to verify answers.
     
  10. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hi,

    I'll have to go over my result then, i didnt get to do that yet. Electrician says he got all integer values so if you got that too then you might have the right result too.
    Yeah it looks like the OP may not be coming back.

    LATER:
    Ok, verified the all-integer solutions. All integer voltages ranging from -15 to 50 volts.
     
    Last edited: Oct 9, 2016
  11. RBR1317

    Active Member

    Nov 13, 2010
    232
    48
    While the thrust here has been mesh analysis, it may be simpler to approach the solution via node analysis. To this end I have annotated the circuit diagram in preparation for writing the node equations. #1-Color code the super-nodes. #2-Designate the voltage nodes & ground node. #3-Mark the super-node offset voltages on the diagram. #4-For controlled sources, express the control variable in terms of the node voltages. Note that the purpose for marking-up the circuit diagram is to make it easy to write the node equations. The particular choice of the ground node can also affect the complexity or simplicity of the node equations.
    Screenshot from 2016-10-09 18-52-28.png
    Following image is an extract from a Maple worksheet showing how simple the node equations can be. And yes, the node voltages compute to whole integer values. Note: Result has not been verified.
    Screenshot from 2016-10-09 18-53-17.png
     
  12. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hi again,

    Without going over your equations and just going by the three results, they look correct.
    Where your Vb is i get 20v with the ground placed in the previous location.
     
Loading...