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Superposition current with 2 loops:

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data


    http://imageshack.us/a/img823/5397/homeworktest2prob4.jpg [Broken]



    Find the current i using superposition



    2. Relevant equations

    V= IR, KVL, KCL,

    Superposition procedures



    3. The attempt at a solution

    I' + I'' = i

    say I' then 7A source cut off

    say I'' then 24V suppressed / silenced

    For I': R tot = 5Ω

    24V - 5Ω*I' - 3Ω*I' = 0

    24V = 8Ω*I'

    I' = 3A


    but



    if I do the second part (where 24V is silenced / suppressed), am I still looking at the current through the 3Ω again as I''?

    If I try to do mesh equations with it, is the dependent source still called "3Ωi"?

    I also got 0 for it anyways. What can I do about this?

    http://imageshack.us/a/img22/3858/homeworktest2prob4edit.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 25, 2013 #2

    gneill

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    Staff: Mentor

    Yes.
    Yes. Although you might find it simpler to write a node equation and solve for the potential at the top of the current supply, then find the current through the 3Ω resistor using it.
     
  4. Feb 25, 2013 #3
    I ended up with

    Vtop/3Ω + 7A = (-3Ωi + Vb)/2Ω

    To end up with

    -Vtop = -42V - 9Ωi

    Then i plug in to i'' = Vtop/3Ω

    i'' = (42V + 9Ωi'')/3Ω

    and get i'' = -7A but I don't think this is right.

    and the current i would be -4A.


    Thanks though
     
    Last edited: Feb 25, 2013
  5. Feb 25, 2013 #4

    gneill

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    Can you explain your node equation? What's Vtop and Vb? Is there a difference? What are the current directions implied by your equations?
     
  6. Feb 25, 2013 #5
    I meant Vtop being the node above the 7A, and it was just supposed to be Vb, my bad
     
  7. Feb 27, 2013 #6

    NascentOxygen

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    You can check that you are doing it correctly by comparing your answer with that obtained by another method. I find the voltage at the junction of the two resistors to be 29¼ volts, and from this you can determine the currents. Superposition should give the same result. :smile:
     
  8. Feb 27, 2013 #7
    I get 20.25 volts at that junction.
     
  9. Feb 27, 2013 #8

    NascentOxygen

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    EDITED: see later posts

    [strike]
    ∑ currents to that node must = 0.[/strike]
     
    Last edited: Feb 27, 2013
  10. Feb 27, 2013 #9
    Are you getting 29.25 volts for the complete network, with all sources in place, or for the case where the 24 volt source is missing?
     
  11. Feb 27, 2013 #10

    gneill

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    I concur with The Electrician on the node voltage when both independent sources are active: Vb = 81/4 = 20.25 V.
     
  12. Feb 27, 2013 #11

    NascentOxygen

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    Very puzzling, as I already checked my working....

    Ah! I think you are going to tell me that the 3i dependent source is not the current source I took it to be! http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon9.gif [Broken]

    Changing it to a dependent voltage source, I calculate 20¼ volts also. :wink:
     
    Last edited by a moderator: May 6, 2017
  13. Feb 28, 2013 #12

    NascentOxygen

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    Right.
    It's a voltage of 3i”
    Try again.
    http://imageshack.us/a/img22/3858/homeworktest2prob4edit.jpg [Broken]
    i” + current source + current from dependent voltage source = 0

    where current from dependent source

    = (voltage across the 2Ω resistor)/2Ω

    = ............

    Can you fill in the dots here? We are aiming for only one unknown, viz., i”.
     
    Last edited by a moderator: May 6, 2017
  14. Mar 1, 2013 #13
    Still here.

    Can't seem to get the math behind this i''


    Doing what nascent said above (and with the node junction above the 7A as "Vb":

    Vb/3Ω + 7A + (3Ωi'' - Vb)/2Ω = 0

    i'' = Vb/3Ω

    Substituting the above in to i'' above that would give 3Ω*(Vb/3Ω) as the whole coefficient in the equation so it would just cancel out as Vb - Vb = 0

    Then i'' would end up as -3A and then i' + i'' would equal zero.
     
  15. Mar 1, 2013 #14

    gneill

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    Indicate on your circuit diagram the locations and directions of each of the current terms in the above equations. Do they agree with each other and the given information?
     
    Last edited: Mar 1, 2013
  16. Mar 1, 2013 #15

    NascentOxygen

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    You already have a direction for i” marked on your figure, so you are asking for trouble if you fail to adhere to that direction.

    Also, you are confusing me (and probably yourself) by including the Ω symbol here. It reads as though the term Vb/3 has units of Ohms. It has units of Amperes, so you should omit the Ω. (If it were essential, you could make use of parentheses, e.g., Vb/(3Ω), but I see no desirability for this.)

    Your diagram has directions already decided for the currents, so try redoing the above equations closely observing those directions.
     
  17. Mar 1, 2013 #16
    Can you say that the i'' and 7A are going in and then the dependent source current is going out? Or are they all "going" in somehow?

    For that one all 3 were going in.

    I would get 0 for the i'' current so not sure.

    I just can't seem to figure the right equations out for this part (which is probably because of that cursed dependent voltage source).
     
  18. Mar 1, 2013 #17

    gneill

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    I find it easiest to simply assume that all currents are entering or all currents are leaving a given node and write the equation accordingly; the math will sort itself out. That said, you must respect any fixed current sources and assign an appropriate sign. If the current source is flowing into the node and you choose to sum currents flowing in, the current will have a positive sign. If you choose to sum currents leaving, then it will have a negative sign.

    So what would you like to do, sum the currents as entering or leaving? Choose one or the other (and it doesn't make any difference in the long run).
     
  19. Mar 1, 2013 #18

    NascentOxygen

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    That's how they are shown, so you need to tenaciously stick with that.
    You get to choose. Mark in a current, and stay with that direction. Regardless, you can always write that the sum of the currents into the node = 0. (If any current has been marked as leaving the node, then it's equivalent to the negative of that current entering the node.)

    The only way to get current i” through the 3Ω in the direction shown is for the left hand side to be considered more positive than the right, i.e.,

    i” = (0 - Vb)/3

    So try those equations again. :wink:
     
  20. Mar 1, 2013 #19
    Ok, so how about:

    Vb/3Ω + 7A = (-3Ωi'' + Vb)/2Ω

    ?
     
  21. Mar 1, 2013 #20

    NascentOxygen

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    Are you representing KCL as―
    • the current into a node = the current leaving?

    But (Vb-0)/3 is the expression for current leaving the node via the 3 Ohm. Writing Vb-0 indicates that Vb is being considered the greater voltage (more positive) and it follows that the direction of the resultant current must be away from Vb towards 0. That direction corresponds to current leaving the node.
     
    Last edited: Mar 1, 2013
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