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Homework Help: Linear algebra and circuits

  1. Aug 29, 2010 #1
    1. The problem statement, all variables and given/known data
    its about setting a system of equations. . .about "nodes" and how in = out
    I have the following circuit
    [PLAIN]http://img820.imageshack.us/img820/8508/circuit.png [Broken]



    2. Relevant equations

    I= v/r

    3. The attempt at a solution

    I get for the first equation in node 1 (the middle circle in the leftmost column)

    I also know that I = V/R so plugging in values I get:

    4 + 2 = I_2

    then for the second node (the rightmost) I would have

    I_2 = I_3 + I_1 = 4 + 2


    so does that mean that I_2 is just 6? (the addition of I_1 and I_3)

    . . .if so then was this problem just a trick problem or something? cause I dont need any system or anything do I??
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 29, 2010 #2

    lanedance

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    Homework Helper

    check you solutions are consistent (its a little hard to follow)

    all the voltage drops around a closed loop must sum to zero
     
  4. Aug 29, 2010 #3
    Your equations are not correct.


    How did you get [tex] I_{2} = 4 +2 [/tex] ?


    You are not applying nodal analysis properly. More importantly, when using nodal analysis you are solving for voltages not currents.

    First, pick the node on the right as ground or your voltage preference.

    Second, write the equation for the sum of the total current entering and leaving a single node in terms of voltages and resistance.


    Show us what you have done so far.
     
  5. Aug 29, 2010 #4
    I think I got it. . .
    R_1 I_1 + R_2 I_2 = 16V
    R_2 I_2 + R_3 I_3 = 8V
    and
    I_1 +I_3 = I_2

    which if I plug in values of R and use I's as my variables I can get 3 equations

    [1 -1 1 0]
    [4 1 0 16]
    [0 1 4 8]

    I can use calculator at this point so I get 3, 4 and 1 for the currents respectively.
    correct?
     
  6. Aug 29, 2010 #5
    Your equation is still not correct.

    In your first loop your equation should be
    [tex] R_{1} I_{1} + R_{2}( I_{2}-I_{3})= 16[/tex]

    Do you understand why ?

    Can you fix the second equation too ?
     
  7. Aug 29, 2010 #6
    Mmmm I thought you could treat both loops separately (use the loop 1 of I_1 and I_2) and set their voltages togheter. . .
    so from your equation R_2 is multiplied by both currents that are passing trough it. . .but shouldnt you consider R_3 also then??
     
  8. Aug 29, 2010 #7
    No.

    This is mesh analysis. The currents flowing through a resistor which is on the boarder of two different current loops is the sum of both currents.

    Do you know how to setup mesh currents ?


    Think about it this way. Suppose you had a pipe and water was flowing to the right side of the pipe at 4m/s and also a pump was pumping water to the left side of the pipeat 3m/s. What would the speed of the water following through the pipe be ?

    A. 4m/s
    B. 3m/s
    C. 1m/s

    The answer is obvious, it will be C since the opposite flows are opposing each other. The same is true here.
    http://en.wikipedia.org/wiki/Mesh_analysis

    Can you look up mesh analysis in your book ?


    Alternatively, you can use the method I initially suggested. You can also use nodal analysis to do this question. I believe that was the orginal intenstions of the author ( considering the two nodes in the circuit).
     
  9. Aug 29, 2010 #8
    No mesh analysis in the book. as you mentioned the purpose is to use node analysis. . .
    thats why my original equations were set that way. . .

    in node 1 I can see that current 1 is going in, current 3 is going in and current 2 is comming out. . .but then current 1 and 2 are known (since you know the voltage and the resistance) so thats why I asked if it was 6. . .(which didnt make sense so I put a question mark) cause I think it was too simple if it actually was 6.

    ----- that was wrong I agree

    then the second thing I remembered that in a closed loop the voltage remains the same, so I did a loop around 16V source, which then I used it as a node came to the conclusion that the voltage of current 2 and curren 1 should be 16V and samething with the voltage of current 2 and 3 to 8V. . .
    now that Im explaining it it doesnt sound coherent :P . . .
    -------------

    so lets go into the nodes only.

    in node_1 we have I1R1 + I3R3 = I2R2
    and in node_2 we have I2R2 = . . . I1R3 +I3R3 ??<-- here Im not sure since the resistor is after the PSU not before. . .so maybe its I2R2 = 16V + 8V ?


    *edit*
    I was just checking your numbers. . .why would it be I_2 - I_3 and not +
    Im not sure If you're reading the diagram right or im doing it wrong but what I see is too loops that come togheter at the middle, and at the middle (node 1) they combine (not cancel)
     
  10. Aug 29, 2010 #9
    Your equation for node 1 is correct but you need to solve for the currents in terms of node voltages. Make node 2 to be zero and then use the fact that current in = current out.
     
  11. Aug 29, 2010 #10
    then if the node has 0V then that means that I2R2 = 0? and if it splits into the currents then I_3 + I_1 = 0 ?
     
  12. Aug 29, 2010 #11
    No.

    Okay to make this simple. I_2=6.

    The voltage on node 1 is V.
    V= I_2*1
    V= I_2

    (16-V)/4 = I_1

    (8-V)/4= I_3.

    I_2= I_1+I_3;



    Solving for V results in V =6.
    Okay i believe you had the right idea all along but you were not clear while explaining your question.

    I wasn't sure were you were pulling numbers from that is why i asked you to write out the equations
     
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