Can you give me some help,suggestions about these circuits ?

Discussion in 'General Electronics Chat' started by slavidbg, May 25, 2007.

  1. slavidbg

    Thread Starter New Member

    May 25, 2007
    1
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    These are the tasks:
    specify I=?
    simplify the circuits
    Make Power balance
    V=?
    this is for the upper circuit,with C
    for the second without simplifying the circuit,W=?,without V,and to specify I=?
    For the upper circuit: (simplifying)
    we are removing from the branches L1 and L2,also the whole branch with C,and remove the V from the branch.
    After this we are defining the direction of each I,how ? choosing own direction or we are looking the shown directions left and down ?
    then on Kirhoff's first law - sum of all I equals to 0
    and the second - sum of all E equals to sum of all U=R.I
    how to specify whether it is + or - ?
    if it is coming out of the node is +,if it is coming in the node is - ?
    then we have to make the balance of all powers ?
    the sum of P of all consumators equals to the sum of P of all sources
    P=U.I ? how to determine this ? I've heard something about using I11,I22,I33 and also G11,G22,G33 ? How ?
    And about the second one any suggestions ?
    Please give some directions:)
     
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  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
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    No values are provided for current, voltage, or components.
     
  3. martincho8002

    Member

    May 26, 2007
    12
    0
    I=3A
    R1=6Ω , R2=4Ω , R3=5Ω , R4=4Ω , R5=5Ω , R6=8Ω , R7=3Ω , R8=2Ω
    E1=30V , E2=40V , E3=30V , E4=20V , E5=50V , E6=20V
    L1=45mИ , L2=40mИ
    C=8μF

    for the second one ( lower one)

    E1=100V , E2=120V
    R1=6Ω , R2=4Ω , R3=4Ω , R4=10Ω , R5=10Ω
    L1=30mИ , L2=50mИ , L3=20mИ , L4=40mИ
    C1=200μF , C2=300μF, C3=300μF, C4=220μF
    К12=0,5, К13=0,4, К23=0,6
    f=50Hz
    φ1=45° ,φ2=60°, φ3=30°

    don't pay attention if any of the data is additional and not used...
     
  4. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    have u read about network analysis.
    mesh and nodal analysis?
    do u wish to know about the the polarities and direction with signs?
    summation of power= power supplied by battery this a theorem in network analysis dont know what it is called but isnt that obvious?
    try solving by writing kvl for each loop and then express it in matrix form
    but this might be too lengthy.
     
  5. martincho8002

    Member

    May 26, 2007
    12
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    I want someone to tell me the equations based on the Kirchhoff's Voltage Law in the loops which I have selected as follows(after ofcourse the circuit is simplified ... I think that the V , C and the bobbins are off):
    1st loop. the upper left triangle.
    3rd loop.the lower left triangle.
    2nd loop.the right triangle that is left.
     
  6. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    can u replace inductors cap and voltage (seems like current source) with short circuits here?
    then using star delta transformations there wud be no loops.try it.(or m i missing something?)
    try posting the loops after simplification u are talking about.
    in kvl u can trace a circuit in any direction and use any convenient sign convention u like.
    try this one.trace a loop across each resistance element drop voltage such that current enters positive and leaves at negative signs. voltages from +ve to -ve taken as postive or drop and -ve to +ve as negative. summation will then be zero.
     
  7. martincho8002

    Member

    May 26, 2007
    12
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    what i am asking is all theoritical... i want it for a theorithical problem example .. this is not a real circuit
     
  8. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    I believe that you can define the currents and voltages with any directions and signs you want to use. Then, using those, you can write the sets of equations for Kirchoff's Voltage Law (KVL) (sum of all voltage drops around any loop = 0) and Kirchoff's Current Law (KCL) (sum of all current into OR out of any node = 0). The currents' directions and voltages' signs will be "automatically" taken care of as you write the equations, if you pick either one of "into" or "out of" for each node, for KCL, and pick one direction to proceed in, around each loop, for KVL. The solutions will be in terms of the directions and signs as you have defined them.

    - Tom Gootee

    http://www.fullnet.com/~tomg/index.html

    -
     
  9. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    what exactly made u feel my answer was practical.
    quote that sentence. i wud like to clarify my point.
     
  10. martincho8002

    Member

    May 26, 2007
    12
    0
    I did not pay enough attention to your answer, I'm sorry for that.Now I've posted the simplified circuit.I think that the branch in red is useless... but anyway... now that i've posted the simplified circuit can someone write to me the equations for each loop and find the unknown I's so I can check If my answers are correct
    [​IMG]
     
  11. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    was that v a voltmeter?

    ok i m writng down the kvl
    LOOP ORDER
    1----2
    :
    3
    lets say all currents are in clockwise direction, since we have three loops.
    first loop (upper left)
    E4+I1.R4+(I1-I2).R3 - E3+E1+(I1-I3)R1 =0 -------------(1)
    -E5+E6+I2.R6+E2+E3+(I2-I1)R3+I2.R5 =0---------------(2)
    -E1-E2+I3.R2+(I3-I1)R1 =0-----------------------------(3)
    according to ur circuit (red part suggests its removed right or is it short circuited in that case there will be four loops)
    if u dont understand any part or think i have erred somewhere post it.
    (since i was having tea and listening to jrock at the same time :) )
    try for other circuit incase u have difficulty feel free to post.
     
  12. martincho8002

    Member

    May 26, 2007
    12
    0
    Is it probable that I receive non exact values for the currents for example I1=3,06.... amp
    May be the red part must not be removed to get a correct answer ? If a voltmeter is removed does the branch stay ?

    I think you have erred on row two because Currents I1 and I2 don't go trough R3
     
  13. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    I1 AND I2 are current in left and right hand loops sure they do go thru r3
    which current do u feel goes thru r3 then?and I2-I1 because their directions are oppsite.
    (try this on paper in each loop assume direction of current clockwise.these currents are independent of other currents and flow independently in each wire.the current in a element with more than one currents is equal to algebraic sum of the individual currents considered with direction.)

    if its a voltmeter then it makes no difference to remove it completely.
    in case of ammeter u wud have had to replace with short circuit.

    solve it and u may even land up with negative values of current. 3.06 is nothing compared to what values u'll get in cases of more complex circuits.

    i m saying again if u have any doubts ask it, there is nothing like a bad question.

    (also a general advice to everybody try clicking the new post button every now and then its a good idea to get the latest post many dont even notice new posts which are replaced by newer posts in other threads :))
     
  14. martincho8002

    Member

    May 26, 2007
    12
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    I meant Currents I1 and I3 dont go trough R3 ...
     
  15. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    check the equation i made the correction.
    only I3 does not go thru r3
     
  16. martincho8002

    Member

    May 26, 2007
    12
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    this would be too much to ask but me and my friend slavidbg .. cannot get the correct answers for I... it seems we are doing mistake somewhere on the way..
     
  17. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    it might just be the case,
    that capacitors and inductors shud not have been disconnected,
    is it an assignment question?
    have you people been taught nodal analysis,
    are the sources ac or dc,
    is it a question based on initial conditions,
    from what i gather form the op,
    you will have to solve with inductance, capacitance kept intact,
    something that is annoying me is why have they given meters in an analytical question.
    have you solved problems using impedance ,matrix because without having a thorough knowledge about network analysis the above example is a little hard to solve.
    the equations i posted were for the simplifications that u came up with,
    if i were to solve it i wud have just solved using kvl for each loop witout simplification and solved using matrices.
    say why dont u ask the person who gave u the assignment on how does he
    expects u to solve it.
     
  18. martincho8002

    Member

    May 26, 2007
    12
    0
    can you solve it for me without simplification ?
     
  19. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    ok i'll try to soplve it,
    it might require some time,
    so shud i consider them to be ac sources, v as voltmeter?
    the other question deals with mutual inductance,
    i'll try to solve it by tomorrow,
    there is one more problem,
    i need to do some brushing up on some concepts myself,and i also lost my favorite book for network analysis but i will manage somehow.


    edit:
    i wud like u people also to work hard to get to the answer the aim of the question is for u to learn not me,
    i have already cleared my subjects with good marks ^_^ v . the more u struggle the more u learn,
    that said i wud help as much i can or shud.
     
  20. martincho8002

    Member

    May 26, 2007
    12
    0
    V is voltmeter yes... i asked again and some friend of mine said that the proper way is to solve after simplification which includes removing the L1 , L2 , C , and V...

    can you at least solve me what the Voltmeter would show ?
    I am struggling yes but except the "silent book" there is no one else to explain and im trying to do the balance of the power Rk.ik=Rp.ip^2
     
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