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Circuit Problems : Series and Parallel Combination

  1. Aug 4, 2011 #1
    1. The problem statement, all variables and given/known data
    As Shown in the Figure

    http://img713.imageshack.us/img713/3342/unledcdq.jpg [Broken]



    Find the i1 and the power dissipated in 10 resistor ohms


    2. Relevant equations

    P=I^2 x R
    P= I x E
    Ohms law
    E= I x R

    3. The attempt at a solution

    @node A I=-i2 +Io
    @node B i2=-i1-i3
    @node C Io=i4-i3
    @node D I=i1 + i4

    Im trying to solve for the R total, to find the I total and find the individual Current in the resistor.. but im having trouble in determining which in the circuit is in parallel and series

    i came with this answer when i try my work

    P=66.04

    but my prof gave me a range of 570 - 650 ... my calculation is wrong

    anyone who can help?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 4, 2011 #2

    SammyS

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    This is not a circuit that can be analyzed by series - parallel combinations.

    Either use Kirchhoff's rules, or do a http://en.wikipedia.org/wiki/Y-delta_transform" [Broken].

    After a Y - Delta Transform you could then do a series - parallel analysis.
     
    Last edited by a moderator: May 5, 2017
  4. Aug 4, 2011 #3

    PeterO

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    Curious that current flows out from point B along all 3 wires. I wonder where is is supposed to be coming in from?

    Perhaps you have just drawn that arrow in the wrong direction.
     
    Last edited by a moderator: May 5, 2017
  5. Aug 4, 2011 #4
    Is my idea right about what you said that i need to solve the R1 R3 and R4 must be solve using that Y- Delta transform then having the result to be series in R2 then parallel in Ro ? is that correct?
     
    Last edited by a moderator: May 5, 2017
  6. Aug 4, 2011 #5
    i dont know... but the figure is right.. the three current is going out of point B
     
  7. Aug 4, 2011 #6

    gneill

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    It's fine. It just means that one or more of the currents will calculate out to have a negative value.

    You haven't shown a value for the voltage source E1.

    You should show more of your solution calculations so that we can determine how we can help you. What's your basic approach? Are you working with KVL loop equations, KCL node equations, mesh equations? Something else?
     
    Last edited: Aug 4, 2011
  8. Aug 4, 2011 #7
    aw.. i forgot to put the value in E1 its 220V..
    so itried using that Y-Delta Transformation,, then came to answer Rt= 15Ohms and the It= 14.67A..

    so im stuck again.. i dont know what should i use next to find i1.. should i use KCL by node
    or should i use the KCL in series and parallel ?
     
  9. Aug 4, 2011 #8

    gneill

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    Why not try a nodal analysis approach? There are four nodes labelled in the diagram, and it looks like node D would make a natural choice for reference (ground) node. Node A already has its node voltage fixed by the voltage source, so you really only need concern yourself with writing equations for nodes B and C.

    Can you write the KCL equations for nodes B and C?
     
  10. Aug 4, 2011 #9
    At node B
    i2+i3+i1=0
    at node C
    i3+Io=i4

    now tryring what you said, ill inform you of my answer and correct me if im wrong.
     
  11. Aug 4, 2011 #10
    so i got this answers.

    i1= 2.44A
    i3= 7.304A
    then my P= 533.48

    is this right?
     
  12. Aug 4, 2011 #11

    gneill

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    It looks very close to what I see as answers. Perhaps you're rounding some intermediate results thus introducing roundoff error into the results. It's hard to tell unless you show your actual computations.
     
  13. Aug 4, 2011 #12

    SammyS

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    It looks like gneill has you well on the way using nodal analysis - not my forte. He (gneill) answers lots of circuit questions.

    To answer your questions in the above quoted text:
    R2 is in series with one leg of the Y, R0 is in series with another leg of the Y, (Those two combinations are in parallel with each other.) and the third leg is in series with that combination.

    From these you can get I0 & I2. Then go back to the original circuit to get the other currents, etc.​
     
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