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Kirchoffs Current Law to find voltage?

  1. May 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Screen_Shot_2015_05_14_at_5_09_25_pm.png
    Am having trouble solving the initial part of this problem, I think I am being confused by the dual voltage sources.

    2. Relevant equations

    I=V/R ΣI at a node =0

    3. The attempt at a solution
    V3/R3 - (V1/R1+V2/R2) = 0
    V3=12-VA
    V1=4-VA
    V2=VA-4

    However this must be incorrect somewhere as It is impossible to isolate Va from the above equations. Am normally very strong mathematically but find basic electronics very confusing for some reason!
     
  2. jcsd
  3. May 14, 2015 #2

    ehild

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    Apply KCL for the node A: that means you express the currents through the resistors R1, R2, R3 in terms of the voltages VA, V1, V2, and then apply the nodal law. Remember, the current flowing through a resistor is proportional to the voltage difference between its terminals.
     
  4. May 14, 2015 #3
    I understand that part, however I thought that the equations I had put in my attempted solution were in terms of voltage differences and the three voltages? I thought that the way the circuit was laid out, all three voltage differences would be dependant on Va?
     
  5. May 14, 2015 #4

    ehild

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    Yes. And KCL provides an equation for VA. Do as it is said in the problem.
    So what is I1, the current through R1?
    What is I2, the current through R2?
    What is I, the current through R3?
     
  6. May 14, 2015 #5

    CWatters

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    You didn't mark up the circuit so it's very hard to work out what you are doing. It doesn't look right. You have terms like V2/R2 which doesn't make sense. V2 is the 12V source on the right hand branch. R2 is a resistor in the middle branch.

    Do what others have suggested. Start by marking the circuit with the currents I1, I2, I3 and any voltages you need.

    Apply KCL but don't skip the very first step which is to sum the currents. There should be an equation of the form

    Ia + Ib + Ic...... = 0

    Don't go straight to equations with V/R in them.
     
  7. May 14, 2015 #6

    CWatters

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    Just to add...

    It's extremely easy to make trivial sign errors when applying KVC/KCL. It's essential to mark up the diagram because you need to refer to it at the end. For example if a voltage or current turns out to be negative you need to know which direction you defined as positive.
     
  8. May 15, 2015 #7
    Thanks for your replies everyone, just at work atm but will mark up the diagram and post my current calculations tonight
     
  9. May 16, 2015 #8
    Marked_circuit.png
    This shows what I have attempted, I think where I am going wrong is the way I am doing the voltage differences but I am really struggling to understand specifically where I am going wrong
     
  10. May 16, 2015 #9

    ehild

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    The currents are right , write up the Nodal Law: them current entering the node is equal to the current leaving it.
     
  11. May 16, 2015 #10
    Okay cool, so if i1+i2+i=0

    Then (-Va/8)+((Va-4)/6)+((Va-12)/4)=0

    But I can't work out how to solve for Va from here, is there another equation I am meant to be using to solve with substitution?
     
  12. May 16, 2015 #11

    ehild

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    Va is the only unknown. Just collect the terms with Va.
     
  13. May 16, 2015 #12
    Oh of course! brain fade..

    I calculated Va as 12.57V so therefore i has a value of (12.57-12)/4 = .143A

    Thanks for your help!

    I will attempt to solve the thevenins problem by myself but will post later if I can't get there
     
  14. May 16, 2015 #13

    ehild

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    Check it once more. I1 goes into the node, but the other two leave it.
     
  15. May 16, 2015 #14
    Okay so now I have (-Va/8)-(Va/6)-(Va/4)=(-4/6)-(12/4)

    Which gives Va as 6.77V and a corresponding i of -1.31A

    The negative current means i is flowing the opposite way to what I have marked (towards top of diagram)
     
  16. May 16, 2015 #15

    CWatters

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    You didn't define if +ve is into or out of the node.

    Is that equation consistent with your diagram (which defines some as +ve going into the Va node and some as leaving)?
     
  17. May 16, 2015 #16

    CWatters

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    If you define +ve as current going into the node then I make it...

    I1 + (-I2) + (-I) = 0
    or
    I1 - I2 - I =0
     
  18. May 16, 2015 #17

    ehild

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    Correct! Well done!
     
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