Nodal Analysis

Discussion in 'Homework Help' started by tim_3491, Apr 1, 2008.

  1. tim_3491

    Thread Starter New Member

    Apr 1, 2008
    7
    0
    [​IMG]Hi

    I have been doing this problem with mesh analysis and using matrices to solve the currents etc. However i get a different answer everytime. I understand nodal analysis but this question has an ammeter in it so now i don't no what to do.

    I have the following equations for the attached document
    A: I8=I2+I1
    B: I8=I3
    C: I2+0.5A=I4
    D: I3=I6+I7
    E:
    F:

    And then for the currents

    I1=(Va-Ve)/10
    I2=(Va-Vc)/20
    I3=(Vb-Vd)/30
    I4=(Vc-Ve)/100
    I5=(Vf-Ve)/100
    I6=(Vd-Vf)/100 => Vd/100
    I7=0.5A
    I8=Vb-Va+5V

    But i don't know what to do with the equations and whether or not i can leave 0.5A in there and use it.

    Any help would be appreciated
    [​IMG]
     
  2. RmACK

    Active Member

    Nov 23, 2007
    54
    0
    Sounds really familiar. I went through the same thing 3 years ago...

    Still, here's my two cents.
    I8=I3 is wrong. I8=-I3 as the arrows point opposite ways.
    How about adding to your initial list the following
    E: I5=I1+I4
    F: I6=I5
    Also your statement I8=Vb-Va+5 is wrong, where is the resistance to relate voltages Vb & Va to current I8?? Just use the I3=-I8 here.

    The next step is to substitute your Ix=(Vy+Vz)/n equations into your Ia=3Ib+Ic style equations. Check your answer with a simulation.
    Hope this helps a little, I really need to go to sleep or I'd help you more. :)
     
  3. RmACK

    Active Member

    Nov 23, 2007
    54
    0
    Here is what I did:
    I know that all the currents out of a node sum to zero. I also will be calling the voltages at points on the circuit A B C etc instead of Va, Vb, etc for clarity.
    Remember F=0

    Point E
    E/100+(E-C)/100+(E-A)/10=0
    E+E-C+10E-10A=0
    -10A-C+12E=0 (1)

    Point C
    (C-E)/100+(C-A)/20-0.5=0
    C-E+5C-5A-50=0
    -5A+6C-E-50=0 (2)

    Point A&B lumped together as a supernode
    (A-E)/100+(A-E)/20+(B-d)/30=0
    6A-6E+3A-3C-2B-2D=0
    9A+2B-3C-2D-6E=0 (3)

    POINT D
    (D-B)/30+D/100+0.5=0
    10D-10B+3D+150=0
    -10B+13D+150=0 (4)

    Oh and indidentally, A=B+5 so B=A-5 (5)

    POINT F
    F-E/100+F-D/100=0 etc so F=0 and you get
    D=-E or E=-D (6)

    Now I use (5) to convert all B's to A's and (6) to convert all E's to D's by substitution getting the following set of equations:

    From (1)
    -10A-C-12D=0 (7)
    From (2)
    -5A+6C+D-50=0 (8)
    FROM (3)
    11A-3C+4D-10=0 (9)
    From (4)
    -10A+13D+200=0
    D=(10A-200)/13 (10)
    NOW WE TAKE (7)-2(8):
    -10A-C-12D+10A-12C-2D+100=0
    -13C-14D+100=0
    C=(100-14D)/13 (11)

    Now I substitute (11) and into (7) to convert all C's into D's
    -10A-C-12D=0
    -10A-((100-14D)/13)-12D=0
    NOW SUBSTITUTE IN (10) TO CONVERT ALL D'S TO A'S)
    -10A-(100/13)+((14/13)-12)((10A-200)/13)=0
    Multiply by 13
    -130A-100+((14/13)-12)(10A-200)=0
    Multiply by 13 again
    -1690A-1300+(-142)(10A-200)=0
    -1690A-1300-1420A+28400=0
    3110A=27100
    A=27100/3110=8.714V YAY! :D
    THEN B=A-5=3.714V
    D=(10A-200)/13=-8.682V
    etc etc you can work out voltages and currents from there.
    Attached is a pspice simulation showing that my answers are correct. Hope this helps you.
     
Loading...