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Help figuring out nodal analysis

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Lpa6j.png

    2. Relevant equations

    Equivalent resistance?
    Nodal analysis?
    V=IR

    3. The attempt at a solution

    2V0lf.png
    I was thinking of using nodal analysis at the points designated by the arrows. The bottom one would be connected to ground. My question is if I picked out the nodes correctly. Is the branch with the current source supposed to be a node?

    Also, is the total current of the circuit .4444444 ?
     
  2. jcsd
  3. Sep 25, 2011 #2

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    Welcome to PF, naivy! :smile:

    You're nodes look about right.
    But your leftmost 2 arrows are connected with a wire without resistor, so their voltages must be the same.

    I'm not sure what you mean by total current though.
    What do you mean by total current?

    As far as I'm concerned, as far as you can talk about a total current that would just be 5 mA.
     
  4. Sep 25, 2011 #3
    Could I also just do something like finding the equivalent resistance?

    I calculated it to be 3.6 kΩ. Then I multiplied this by 5 mA to get Vo to be 18V.

    Is this answer correct? Thanks. If not, could you please help me some more with nodal analysis?
     
  5. Sep 25, 2011 #4

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    Yes, you can also find the equivalent resistance.
    But I don't think that makes your problem easier.

    And no, your calculated values are not correct.

    If you try for the equivalent resistance, you should rearrange the circuit in such a way that the top of the current source is on one side, and the bottom of the current source is on the other side.
    From there you can calculate the equivalent resistance, after which you can calculate the Vo voltage drop.

    For nodal analysis you should label your top 3 (not 4!) arrows as, say V1, V2 and V0.
    The bottom arrow would be 0 V.
    From there you can set up a set of 3 equations applying Kirchhoff's current law, (combined with Ohm's law).
    Solve the set of equations and you have Vo.
     
  6. Sep 25, 2011 #5
    Here's some work in trying to set up the simultaneous equations. Please show me where I messed up?

    g1Hzs.jpg
     
  7. Sep 26, 2011 #6
    Can someone please look at this for me? Thanks
     
  8. Sep 26, 2011 #7

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    Just got up. :zzz:

    You're on the right track! :smile:

    But what did you do with the current source?
    The 5 mA is supposed to be in there somewhere.
    You should add it in your 3rd equation.

    And in your 2nd equation the sign of (V2-V1)/4k appears to be wrong.
     
  9. Sep 26, 2011 #8
    Is it supposed to be (V1 - V2)/4k.

    For the third equation, how am I supposed to add the current source?
    Does the third equation look like this?

    For V': (V1-V')/6k + 5mA = 0

    I really appreciate your help! Thanks :D
     
  10. Sep 26, 2011 #9

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    Yep! :smile:


    If you look at the diagram in your problem statement at the point you have effectively labeled V', you need to apply KCL there.

    Current in is 5 mA, current out is V'/4k + (V'-V1)/6k.
     
  11. Sep 26, 2011 #10
    So then according to KCL, it should be something like V'/4k + (V'-V1)/6k = 5mA ?
     
  12. Sep 26, 2011 #11

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  13. Sep 26, 2011 #12
    Wait, was there a negative sign missed there?

    is it supposed to be (-)V'/4k + (V'-V1)/6k = 5mA ?? If so, would you mind checking the equations that I have so far to make sure it's right?

    (V1-V2)/4k - V2/6k = 0 or (V1-V2)/4k + Vo/6k = 0

    (V'-V1)/6k -V1/10k + (V1-V2)/4k = 0

    (V'-V1)/6k - V'/4k = 5mA

    Thank you!
     
  14. Sep 26, 2011 #13

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    No.
    Following what you did earlier, it would be:
    -V'/4k + (V1-V')/6k + 5mA = 0


    eq1: Since you chose V2 equal to Vo, Vo should have the same sign.

    eq2: Your sign of (V1-V2)/4k is still wrong.

    eq3: The sign of V'/4k is wrong.


    Aaaaaaand... I'm off to work. ;)
     
  15. Sep 26, 2011 #14
    Does that mean that Vo=4?
     
  16. Sep 26, 2011 #15

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    Yup.
    That's what I got too! :)
     
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