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Finding the steady state expression of a circuit

  1. Mar 31, 2015 #1
    1. The problem statement, all variables and given/known data
    For the circuit in (Figure 1) , suppose
    va=10cos16,000tV
    vb=20cos4000tV.
    Suppose that R = 700Ω .

    Write the steady-state expression for io(t) as io=I′ocos(ω′t+ϕ′)+I′′ocos(ω′′t+ϕ′′), where −180∘<ϕ′≤180∘, −180∘<ϕ′′≤180∘, and ω′>ω′′. Find the numerical value of I′o.

    Find the numerical value of ϕ′.

    Find the numerical value of ω′.

    Find the numerical value of I′′o.

    Find the numerical value of ϕ′′.

    Find the numerical value of ω′′.

    2. Relevant equations
    Um . . .

    3. The attempt at a solution
    You lost me at Io . . . Seriously, I don't know what it's even asking . . .
     

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  3. Mar 31, 2015 #2
    For starters: what do you know about the steady state solution of a circuit?
     
  4. Mar 31, 2015 #3

    gneill

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    Since Io is not defined in the problem statement or the circuit diagram there is no solution to the problem as given.
     
  5. Mar 31, 2015 #4
    When you put it like that . . . sinusoidal? Capacitors are open and inductors shorts?

    This is all that they give me. The first question asks what analysis technique should be used (aka superposition). Then it asks about the I equation . . .
     
  6. Mar 31, 2015 #5
    I'm assuming Io is the current caused by the voltage vo.

    Bluestribute - capacitors acting as open circuits and inductors acting as short circuits is something that happens in DC circuits, but you have a sinusoidal input. Do you know anything about phasor analysis (i.e. Frequency domain analysis)?
     
  7. Mar 31, 2015 #6
    Yeah . . . I guess it just seems like a really broad question and I'm not understanding what you're looking for. Or what I'm supposed to do to find I'o
     
  8. Mar 31, 2015 #7
    It gives you a circuit with two sources and a few elements. It gives you equations to describe the sources, and it wants you to use all of that information to find an equation that describes the current in (I assume) the middle branch of the circuit (after the transient solution has died out).
     
  9. Mar 31, 2015 #8

    NascentOxygen

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    I think it's a case of using superposition to complete the expression for the current through the inductor.
     
  10. Mar 31, 2015 #9
    So what is I'o? What are these prime terms? I know the Cdv/dt formulas. But . . . do I use that to solve for i(t) and then . . . do . . . something?
     
  11. Apr 1, 2015 #10

    NascentOxygen

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    I' and I'' are an alternative notation to I1 and I2.

    At least, that's how I think they are using it here. Nothing to do with derivatives.

    And, despite my new April 1 avatar, the aforewritten lines are meant to be taken seriously. :smile:
     
  12. Apr 1, 2015 #11
    So is I1 and I'o the . . . left mesh?
     
  13. Apr 1, 2015 #12

    NascentOxygen

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    Two meshes. One carries I'o and the other I''o.
     
  14. Apr 1, 2015 #13
    So derive Va and multiply by 625E-9 and derive Vb, then add those together?
     
  15. Apr 1, 2015 #14
    I also tried doing V/Z (Z, for Va, being 700-100j). No luck.
     
  16. Apr 1, 2015 #15

    gneill

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    Why don't you show details of your work for one of the cases?

    You've recognized that you will need to use superposition here because the two voltage sources have different frequencies. So call "case a" the situation where you've suppressed vb and only source va is active. Show how you solve for the resulting current through the inductor.
     
  17. Apr 1, 2015 #16
    Inductor is shorted.

    1/(625nf*16000) = 100
    Z = 700 - 100j (resistor + conductor)
    Va = 10cos16000t

    Divide Va by Z with a calculator
     
  18. Apr 1, 2015 #17

    NascentOxygen

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    It's the inductor voltage which you are needing to find, so I can't see why you would want to short it.
     
  19. Apr 1, 2015 #18
    I want to short it because I have no idea what I'm doing and after 17 posts, I still have no idea what I'm doing. It'd be great if someone could just show me some work instead of having me run circles around my own head, because I CLEARLY am not getting ANYWHERE with anything that's been said. I CLEARLY don't know what's going on, what's being asked, or how to solve it.
     
    Last edited: Apr 1, 2015
  20. Apr 1, 2015 #19

    NascentOxygen

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    The circuit has two unrelated voltage sources, and you are required ultimately to find the current through the inductor. If it had only one voltage source then I think you would be able to solve that, correct?

    So, take the sources one at a time, and "pretend" that the other source is zero volts. (That doesn't mean it's not present, it means it is present and is supplying a voltage of 0.00V.)

    Determine the inductor current under each of those two conditions.
     
  21. Apr 1, 2015 #20
    So:

    (Va-Vo)/-j100 + -Vo/700 = Vo/j400
    Vo = 12.8665+2.4508j

    like that for Va? Is that even remotely correct? Then what do I do with it?
     
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