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Circuit analysis is this correct?

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data
    We have the problem in the figure where E1= 9 V,E3= 5 V,E8=-5 V,J5= 3 A,J7=-1 A. R1= 8 Ohm,R2= 6 Ohm,R3= 6 Ohm,R4=7 Ohm, R6= 5 Ohm, R8= 1 ohm.
    HA0DfSN.jpg


    2. Relevant equations

    I have to find the currents and the potential in the elements using KVL and KCL.

    3. The attempt at a solution
    I know how to solve this,but the problem is the arrow in the left,near E8.Does that mean that the current should be always directed from the right to the left? WITHOUT taking this in consideration I have found the current first in R8.
    I8= E8/R8=-5/1= -5 A
    In the nod 1
    I6+I8=J7
    -5+I6=-1 and so I6= 4 A
    In the nod 2
    I1=E1/R1=9/8 A

    I4+I6=I1 so I4+4=9/8 so I4=-23/8 A.

    To find R2 I apply KVL in the upper left part of the circuit and I have
    R4I4- R2I2+R1I1+E1=0

    Remember,I havent taken in consideration the fact that the direction of the current should always be from right to lef.
     
  2. jcsd
  3. May 27, 2013 #2

    gneill

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    Staff: Mentor

    I don't see an arrow to the left of E8. I see a vertically oriented arrow above and to the right of E8. Is this the one you mean? This arrow is not labelled in any way so its meaning is unknown. Does the original problem statement refer to it?

    If the problem doesn't specify assumed current directions and potential polarities for the individual resistors and current supplies then perhaps they just want magnitudes for results. Or are you to answer by filling in the values on the diagram?
    How do you justify the above expression?
     
  4. May 27, 2013 #3
    They just want the magnitude.:)
    I have found the current running through the resistor R8, by using the Ohm Law..
     
  5. May 27, 2013 #4

    gneill

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    Okay then, how do you justify using E8 as the potential across R8?
     
  6. May 27, 2013 #5

    phinds

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    Uh ... say what???
     
  7. May 27, 2013 #6
    you are right gneill I cant justify it :/
     
  8. May 27, 2013 #7

    gneill

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    So, it looks as though you'll have to write and solve a set of equations for the circuit. Nodal analysis looks promising...
     
  9. May 27, 2013 #8
    No I have to solve this with KVL and KCL...
     
  10. May 27, 2013 #9
    but still,give me a hint for nodal analysis?
     
  11. May 27, 2013 #10
  12. May 27, 2013 #11

    gneill

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    Nodal analysis is an application of KVL and KCL. It's just a formalized way of applying them. Have you learned nodal analysis yet? If not, fall back to direct application of KVL and KCL; Label the currents and write KCL at the nodes and KVL around the loops. You could also think about using superposition, solving for the voltages and currents for each source independently, then summing the results.
     
  13. May 27, 2013 #12
    I have learnt nodal analysis yes,and thank you..
     
  14. May 27, 2013 #13
    I solved this with KCL and KVL and I got I2 = 1 A,I4=3 A and so on..am I right?
     
  15. May 27, 2013 #14

    gneill

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    It's very difficult to evaluate "and so on", but your values for I2 and I4 don't look right. You'll have to show your work.
     
  16. May 27, 2013 #15
    Let's call the nod on the left 4 and the one in the middle 5. I write KCL for each node.So I have for the node 5 : I4+J7+I2=I5 and so I4+I2= 4 A. For the node1 I have J7+I6=I8.So I8-I6=-1 A. and for the node 4 I have J5+I3=I8 and so 3+ I3=I8. For the node 3 I have I1=I2+I3. And for the node 2 I have I1+I4=I6.

    For the loop in the left,on the top I have the KVL
    R4I4-R2I2-R1I1+9=0
    R4(4-I2)-R2I2-R1(I2+I3)=-9
    NOTE : I noticed that during my calculations I forgot to put 9 and I just wrote R4(4-I2)-R2I2-R1(I2+I3)=0..
    anyway I will continue so I can know if my way is correct.
    Here I find I3 in terms of I2 and also I1 in terms of I2.
    Then I have I4=4-I2 but I4 also =I6-I1 . so I have 4-I2=I6-I1.(1)
    I also have I8=I6-1 and I8=3 +I3.Here I have I6=I1+I3+3. I replace this In (1) and so I find I2 and on...
     
  17. May 27, 2013 #16

    gneill

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    You should take your diagram and pencil-in all the component values, label all the nodes (some are already there), and pencil-in all the currents with their assumed directions. It's important to do this in order to be consistent with the signs of currents when you write the KCL equations; a given current will flow OUT of one node and INTO another. As it stands I can't tell what directions are assumed for your currents.

    If you apply the classic nodal analysis technique you get to choose one node as a reference node and only have to write equations for the remaining nodes (in this case there will be 4 such nodes, so 4 equations will be required to solve for the 4 unknown node voltages).
     
  18. May 27, 2013 #17
  19. May 27, 2013 #18

    gneill

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    Okay, so it looks like you've got a method and a plan that should work for you, if you can sort out all the equations!
     
  20. May 27, 2013 #19
    I have already solved this but I still get weird numbers :/ Anyway thank you!
     
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