Node analysis

Discussion in 'Homework Help' started by kidi3, Oct 16, 2012.

  1. kidi3

    Thread Starter New Member

    Oct 16, 2012
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    0
    I currently trying to train my skills in analyzing a Electric circuit using the node analysis and mesh method.

    But the answers I get isn't the same as the one giving.

    http://personal.georgiasouthern.edu...43-lec/exercises-quiz3-mesh-nodal-w-ans.pdf

    I tried calculating Exercise 1 and 5 using Node analysis.

    these are my calculations for exercise 1
    http://snag.gy/Rfcp7.jpg

    And these are my calculations for excersice 5
    Har også lige prøvet opgave 5

    For V1 = 12v

    V2 = -2V


    I1 =6+(-3) = 3A

    V1 = 12V

    I2 = -1A

    -1 * 2 = -2V...


    And here is another one http://snag.gy/Dnuqf.jpg

    http://snag.gy/Rm2fs.jpg
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    You have here two unknown nodal voltages Vo and where we have 1Ω resistor

    Are you sure that this is a solution for exercise 5
    For node V1

    3A + V1/4Ω = 6A + (V2-V1)/1Ω

    And for V2

    5A = V2/2Ω + 6A + (V2-V1)/1Ω

    And solution
    V1 = 4V l V2 = 2V

    http://www.wolframalpha.com/input/?i=3+%2B+V1%2F4+%3D+6+%2B+%28V2-V1%29%2F1%2C+5+%3D+V2%2F2+%2B+6+%2B+%28V2-V1%29%2F1+


    You made a error in calculations X = - 12/11
    http://www.wolframalpha.com/input/?i=(12+-+X)/6+++(0+-+X)/4+-+(X+-+(-6))/2+==+0
     
    Last edited: Oct 16, 2012
  3. kidi3

    Thread Starter New Member

    Oct 16, 2012
    4
    0
    About exercise 1.

    I see there is 2 resistors for the voltage source at 40V.
    Are those 2 in series or parallel.

    The equation I wrote gave me V_0 to become 25,0714V. But thats incorrect?..

    So should i change the resistors in parallel?

    About Exercise 5
    I didn't know how i should use the node equation here, because i seemed to be simple.
    I shoudn't have thought so..

    But using my knowlegde i get these equations.
    Are they correctly written?
    http://snag.gy/LWdjY.jpg


    And about the last one..
    Yeah I see my mistake.. - to +..

    Thanks for your reply.. I hope you help me with my other questions.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
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    You can not do that because you have a 5A current source

    [​IMG]

    So we need to nodal equations for two unknown voltages V1 and Vo.

    No, you forget about current sources.

    [​IMG]

    For V1 node we can write

    I1 + I2 = I3 + I4


    So we can write

    3A + V1/4Ω = (V2-V1)/1Ω + 6A
     
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  5. kidi3

    Thread Starter New Member

    Oct 16, 2012
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    Ok.. I think i understand it now..

    what about..

    is it the direction of I0?
     
  6. kidi3

    Thread Starter New Member

    Oct 16, 2012
    4
    0
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Hmm, This current source has 1mA or 1A current ??

    [​IMG]

    First I assume that all current flow into the Va node.
    And all current which entering the node are positive ( give them "+").
    So I write

    1mA + (-Va)/R3 + (Vo - Va)/R2 = 0

    1mA - Va/510Ω + (0.5V - Va)/330Ω = 0


    And the solution

    Va = 503.929mV

    As for your equations, you made slight error.

    (0.5 - V)/330 - (-V)/510 - 1 = 0


    You also assume that all current entering into a Va node.
    but you assume that all entering currant are negative.
    So the correct nodal equations look like this:

    -(0.5 - V)/330 - (-V)/510 - 1 = 0


    -(0.5 - V)/330 +(V)/510 - 1 = 0

    Va = 200.661V

    Next time you should use this convention:
    Assume that all current flow out from the node and give them "+" if they flow out from the node.

    So we can write

    Va/R3 + (Va - Vo)/R2 + (-1A) = 0

    Va/510 + (Va - 0.5)/330 -1A = 0

    and solution

    Va = 200.661V
     
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