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Average power dissipated in resistor

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the average power dissipated in the 30 Ω resistor in the circuit seen in the figure if ig=7cos20,000tA.

    2. Relevant equations
    KVL/KCL equations
    P=VI* (possibly) or I^2(R)

    3. The attempt at a solution
    I tried to do a mesh, but got I(60-j130) = 0, which isn't right . . . I'm assuming I find the current through the resistor and just do I^2(R)?
     

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  3. Apr 8, 2015 #2

    gneill

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    You'll want to find the RMS current through the resistor if you're looking for average power. The source specification would appear to be a time-domain peak current...

    Show your work!
     
  4. Apr 8, 2015 #3
    I = I

    30I(-j40) + 30I(30) - I(j10) = 0

    Doesn't quite add up . . .
     
  5. Apr 8, 2015 #4

    gneill

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    I don't see the controlled voltage source in there, and I don't see a contribution from the independent current source ig from the first mesh. What are your mesh equations?
     
  6. Apr 8, 2015 #5
    That was my one mesh equation . . .
     
  7. Apr 8, 2015 #6

    gneill

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    But there are two loops...
     
  8. Apr 8, 2015 #7
    But wouldn't the other one just be Ig? Or is it 5mH(I-Ig) = 0? Converting mH to ohms of course
     
  9. Apr 8, 2015 #8

    gneill

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    Well, you can call it ig if you wish, but it's still a separate mesh current.

    Fig1.gif

    In the figure, i1 = ig and iΔ is the net current through the inductor.
     
  10. Apr 8, 2015 #9
    Right right. So across the inductor, it'd be (I1 - I)10j = 0? And then for the other mesh, 30I(-j40) + 30I(30) - I(j10) = 0?
     
  11. Apr 8, 2015 #10

    gneill

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    No, I is the net current through the inductor. The first mesh is already "solved" for its current since it's identical to that of the fixed source. Thus ##i_\Delta = i_g - i_2##.
    Remember that there are two mesh currents flowing through the inductor. And the source is dependent on ##I_\Delta##, not I.
     
  12. Apr 8, 2015 #11
    Wait, you are very confusing right now.

    So I = Ig - I2? Ok, I can dig that.

    So for my other mesh, do I replace I with I2? And solve for I2?
     
  13. Apr 8, 2015 #12

    gneill

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    You just wrote an equation for I above! Use that.
     
  14. Apr 8, 2015 #13
    Wait, you lost me again. What two equations am I using? Am I solving I∆? Do I solve I∆ then use I^2(R) to find power?
     
  15. Apr 8, 2015 #14

    gneill

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    What current is flowing through the resistor? It's ##i_2##, right? So that's what you need to find. That's the mesh current for the second loop. ##i_\Delta## is just the net current through the inductor, and happens to be the "sense current" controlling the dependent source. You don't necessarily have to solve for it explicitly, but you do need to replace it with its equivalent in terms of mesh currents otherwise you'll have another variable on your hands.


    The first mesh current is effectively already solved for since it's identical to ##i_g##. You don't need to write a mesh equation for it. But you do need to use the auxiliary equation for ##I_\Delta## since that's needed to write the equation for the second loop (for the controlled source).
     
  16. Apr 8, 2015 #15
    30(Ig - I2)(-j40) + 30(Ig-I2)(30) - (Ig-I2)(j10) = 0?

    I just replaced I∆ with the equivalent current of Ig - I2

    But I still have one less equation than variables . . . So this gives me one equation two variables. Adding in the Ig relationship gives two equations three variables.
     
  17. Apr 8, 2015 #16

    gneill

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    The only variable I see in your equation is ##I_2##.
     
  18. Apr 8, 2015 #17
    Oh yeah, it says what Ig is.

    So when I do my calculations, I can use 7 for Ig, right, since its angle is 0?

    I just need to make sure my first attempt is correct. Since, you know, Mastering Engineering. I already failed one for putting the magnitude of j instead of the complex form of the impedance (it didn't ask for the complex form, just magnitude).
     
  19. Apr 8, 2015 #18

    gneill

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    You can. Just keep in mind that 7 is the peak value of the source current, not the RMS value. So you'll end up calculating the peak value of the current through the resistor.
    Yeah, I'm not a big fan of computer based quizzes myself.
     
  20. Apr 8, 2015 #19
    So use 7/root2 as I go through just to be safe . . . got it.

    Yep, this is pretty much the only online place where I actually learn something. Which is a shame 'cause that's what Mastering is SUPPOSED to do.
     
  21. Apr 8, 2015 #20

    gneill

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    If you want you can use 7 and find the peak current, then convert peak to RMS at the end. Saves carrying the √2 through all the maths. It's just a scaling factor for a sinusoidal waveform.
     
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