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Another KVL-KCL

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Find Vx, Is and power on the dependent source.
    Capture_1.jpg


    3. The attempt at a solution
    http://s17.postimg.org/3x4hszzkf/002.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 23, 2013 #2

    gneill

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    Staff: Mentor

    Your determination of vx is fine. However, Is is not the only current flowing through the 4 Ω resistor; there's some current coming from the 29 A source, flowing through the 4 Ω + Is combination, then to the components on the right.

    You should be in a position to know how much current is entering and leaving that middle bit.
     
  4. Oct 23, 2013 #3
    I did KCL on node a and got 30 A. Also calculated the current through the 4 Ω resistor -1.5 A.
    Then KCL on node b, Is = 31.5 A
     
  5. Oct 23, 2013 #4

    gneill

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    Staff: Mentor

    Draw the directions of the currents through the resistors. Then do KCL again :wink:
     
  6. Oct 23, 2013 #5
    Oh ok. 28 not 30. Then Is = 26.5.
     
    Last edited: Oct 23, 2013
  7. Oct 23, 2013 #6

    gneill

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    All of them. You have the voltages across them all.
     
    Last edited: Oct 23, 2013
  8. Oct 23, 2013 #7
    ....
     
  9. Oct 23, 2013 #8

    gneill

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    Staff: Mentor

    No, not 26.5 A. I think you've got a current direction wrong. Sketch them on your diagram and pay attention to polarities -- remember, Vx is negative...
     
  10. Oct 23, 2013 #9
    Here's the new sketch. If i'm not wrong then 29.5 A.
     
  11. Oct 23, 2013 #10

    gneill

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    You're not wrong :smile:
     
  12. Oct 23, 2013 #11
  13. Oct 24, 2013 #12
    Another day, Another KVL-KCL problem.

    Find V0 using KVL-KCL and Ohm's law.

    Capture.jpg

    Here's my solution
    http://s21.postimg.org/usa92r31z/001.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  14. Oct 24, 2013 #13

    gneill

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    You really should place new questions in new threads, if for no other reason than more people are likely to see it.

    For this problem it looks like you've got the right ideas, but you're losing way too much accuracy by not keeping enough decimal places in your intermediate results. Keep intermediate values in fraction form or keep several more decimal places ("guard digits").

    For example, knowing that I1 is 4 mA and it's flowing through a 4 kΩ resistor you can tell that the potential drop across that resistor should be 16 V exactly. That means the sum of the 42V supply and the drop across the 6 kΩ resistor must add to 16V, so that the drop is 26 V across that resistor. But in your determination of i2 you rounded the result to 4.3 mA, and then when you calculated the potential drop across that resistor you arrived at 25.8 V.

    The situation gets worse when you round small values, as the resulting percentage change is more drastic. You determined the current through the 18 kΩ resistor to be 1.63 mA, but it should have ended up as 1.889 mA. So the early roundings of values has landed you with a 14% error so far...
     
  15. Oct 24, 2013 #14
    Is it enough if i write three decimal digits?
     
  16. Oct 24, 2013 #15

    gneill

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    Staff: Mentor

    Three is better, yes. If your calculator has enough memory locations, write three digits but use the stored full-accuracy values for calculations. As always, round final results to the appropriate number of significant figures.
     
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