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CIRCUIT ANALYSIS: 2 resistors, Indep. Current Source, V.C.C.S - find v0

  1. Jan 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Find [itex]v_0[/itex] in the circuit below and the power dissipated by the controlled source.

    [​IMG]


    2. Relevant equations

    KCL

    [tex]v\,=\,i\,R[/tex]


    3. The attempt at a solution

    KCL: [tex]i\,=\,2\,v_0\,+\,10\,A[/tex]

    [tex]v_o\,=\,i\,R[/tex]

    [tex]v_o\,=\,\left(2\,v_0\,+\,10\right)\,(4\Omega)[/tex]

    [tex]v_0\,=\,8\,v_0\,+\,40[/tex]

    [tex]v_0\,=\,-\frac{40}{7}\,V[/tex]

    Am I on the right track here?
     
    Last edited: Jan 12, 2007
  2. jcsd
  3. Jan 12, 2007 #2

    berkeman

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    Staff: Mentor

    Looks good so far. BTW, I don't like the wording of the other part of the question:

    "and the power dissipated by the controlled source."

    The power *dissipated* depends on the efficiency of the power supply. So it will be the power delivered to the circuit, divided by the efficiency of the power supply, which is not given. So if that really is the wording of the question, you might calculate the power delivered to the rest of the circuit, and point out that you need the efficiency of the power supply to know what its power "dissipation" is.
     
  4. Jan 12, 2007 #3
    If [itex]v_0[/itex] looks correct then I can calculate the rest.

    [tex]i\,=\,\left[2\,\left(-\frac{40}{7}\right)\,+\,10\right]\,A\,=\,-\frac{10}{7}\,A[/tex]

    [tex]p\,=\,v\,i\,=\,\left(-\frac{40}{7}\,V\right)\,\left(-\frac{10}{7}\,A\right)[/tex]

    [tex]p\,=\,\frac{400}{49}\,W\,\approx\,8.16\,W[/tex]

    Does this look right?
     
    Last edited: Jan 12, 2007
  5. Jan 12, 2007 #4

    berkeman

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    Staff: Mentor

    But only part of the total power delivered to the resistive load(s) comes from the controlled current source, right? I missed that my first time looking at the circuit as well.
     
  6. Jan 12, 2007 #5
    I guess, the resistors are the other part of the resistive load?

    How do I figure for that then?

    [tex]P\,=\,v\,i\,=\,\left(v_0\right)\,\left(2\,v_0\right)\,W[/tex]

    [tex]P\,=\,2\,v_0^2\,=\,2\,\left(-\frac{40}{7}\right)^2\,W[/tex]

    [tex]P\,=\,2\,\left(\frac{1600}{49}\right)\,W\,=\,\frac{3200}{49}\,W[/tex]

    [tex]P\,\approx\,32.7\,W[/tex]

    Correct?
     
    Last edited: Jan 12, 2007
  7. Jan 12, 2007 #6

    berkeman

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    Staff: Mentor

    I don't understand your reply. Calculate the total load resistor voltage and then the current contribution from the controlled source.
     
  8. Jan 12, 2007 #7
    [tex]2\,v_0[/tex] is the current contribution of the controlled source.

    [tex]2\,v_0\,=\,-\frac{80}{7}\,A\,\approx\,-11.43\,A[/tex]

    Do I then use the p = v i ?

    [tex]P\,=\,\left(-\frac{40}{7}\right)\,\left(-11.43\,A\right)\,=\,65.3\,W[/tex]
     
  9. Jan 13, 2007 #8

    berkeman

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    Staff: Mentor

    Nice job, Vinny. Keep it up and you will go far.
     
  10. Jan 13, 2007 #9
    Thank you!

    Just to clarify, [tex]v_0\,=\,-5.71\,V[/tex] and [tex]P_{V.C.C.S}\,=\,65.3\,W[/tex]. These are correct?
     
  11. Jan 13, 2007 #10
    I believe the question wants you to determine the P = VI for the dependent source. Since you have the I as 2v0, all you need next is to calculate the V across the dependent source and then multiplying the two terms for the power.
     
  12. Jan 13, 2007 #11
    Would that V be half of the [itex]v_0[/itex]?

    P is 32.7W and not 65.3W?
     
    Last edited: Jan 13, 2007
  13. Jan 13, 2007 #12
    No Vinny, that V would be the voltage across the VCCS. Equivalently, that is also the voltage across the two resistors, since the (4+6)ohm, 10A source, and VCCS are all in parallel.
     
  14. Jan 13, 2007 #13
    The voltage across the VCCS is what I need for the P = v i equation right? It is the v and the i is the current contribution of the VCCS? That would mean [tex]p\,=\,v\,i\,=\,\left(-\frac{20}{7}\right)\,\left[2\,\left(-\frac{40}{7}\right)\right]\,=\,32.7\,W[/tex]

    If that is not it, how is it calculated?
     
  15. Jan 13, 2007 #14
    As I have earlier said, it will be the V across the (6+4)ohm. Since you have already found the current across the resistors, this V shouldn't be too difficult to find I am sure.
     
  16. Jan 13, 2007 #15
    The current across the resistors is [tex]i\,=\,2\,v_0\,+\,10\,=\,-1.43\,A[/tex], right?

    Now to get the V across both resistors.

    [tex]V\,=\,i\,R\,=\,(-1.429\,A)\,(10\Omega)\,=\,-14.3\,V[/tex]

    I then use the P = vi.

    [tex]P\,=\,v\,i\,=\,(-14.3\,V)\,(-1.429\,A)\,=\,20.4\,W[/tex]

    I hope thats right:)
     
  17. Jan 13, 2007 #16
    yea the current right. I was looking at your other post and this one, and it seems you have the same question a current through a series resistor. Remember that the current is the same through each resistor when they are in series. THe method for getting P is right to me. Do you know why finding V through the resistors allowed you to get P for the controlled source?
     
  18. Jan 13, 2007 #17
    no! why? (sorry) I thought the power dissipated by the resistors were a result of BOTH sources, not just one.

    edit: Oh I meant using the same current through the resistors to find P of D.S
     
    Last edited: Jan 13, 2007
  19. Jan 13, 2007 #18
    Don't really understand your post...

    The question is asking for "power dissipated by the controlled source"
    NOT the resistors.

    But looking at the work again, he found the P for the series resistors and not the controlled source. Have to use the 2(V0)*V = P(controlled souce)
     
  20. Jan 13, 2007 #19
    Yes, What you said just now was what I was asking.
     
  21. Jan 13, 2007 #20
    Oh alright. So yea, he's just not finished with the problem yet. Hope he checks back. :uhh:
     
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