Thevenin & Norton for AC cct dependent source

Discussion in 'Homework Help' started by Jess_88, May 2, 2011.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
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    Hey guys.

    I'm a little bit lost for this question.

    I need to fined the Thevenin and Norton equivalent cct's for the following cct. Given w = 10 rad/s
    [​IMG]

    As you can see, it has a dependent source and two A.C independent sources.

    From what I understand, I need to fined the open cct voltage (Vth) and short cct current (iN) across ab.

    When I have the Vth and iN, Rth = RN = Vth/iN
    right?

    I'm having a little difficulty analysing this cct.
    Should I be using nodal analysis to fined Vth?
    and Mesh analysis to fined iN?
    is this the best/easiest method?

    Sorry if my question sounds silly... I haven't covered this in my lectures yet

    this is what I have so far for iN
    I1 = 2<-90
    Vo = 10(I1 - I2)

    loops
    10(I1-I2) = 0
    -j2(I2) + 25j(I2 - I3) = 12<0
    25j(I3 - I2) - 2vo = 0
    2vo = 0

    10I1 - 10I2 = 0
    (10 + j23)I2 + (25j)I3 = 12<0
    -20I1 - (20 + j25)I2 + (j25)I3 = 0
    20I1 - 20I2 = 0

    thanks guys :)
     
    Last edited: May 2, 2011
  2. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    I'v just realised, I can probably do a source transformation on the current source.
    I can do that right?

    So this would be the cct for finding iN(norton current) at ab
    [​IMG]

    And the equations
    (10 + j23)I1 - (25j)I2 = 12<0 + 20<-90
    (-j25)I1 + (j25)I2 - 2vo = 0
    2vo

    dont realy know what I'm supposed to do with loop 3.
    do I fined it by assuming Vo = 10I1?
    so I3 = 20*I1

    I'm just a lil unsure about everything, because from what I'v been reading, you cant use super position or node for analysis (because its AC and there is a dependent source). I also read that you don't analyse it the usual way with dependent sources involved (applying 1V source ext).

    So from what I can gather this is the only way... right?
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Shorting a&b takes the inductor out of the analysis. Temporarily remove the dependent source.

    You then need to find the value of Vo and the current contributed to the short from the independent sources in series with the R & C.

    Having found Vo you then add the contribution to the short circuit current from the dependent source.

    The analysis for the open circuit voltage across ab is more complicated (since you can't ignore the inductance) but conceptually simpler when dealing with the dependent current source.
     
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  4. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
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    Oh goodness... I have just noticed the units of the current source....
    What the hell, why are they in Volts?
    Is that a mistake in the question?
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yeah - I saw that and assumed it was a typo.
     
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  6. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
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    ok so this is what I have.
    removing the inductor, transforming the current source
    there is a series cct consisting of
    two voltage sources (20<-90) and (12<0) and impedance = 12 - j2

    adding the two voltage sources
    (12 +j0) + (0 - j20) = 22.323<-59.03

    pretty unsure about this next part.
    total current is in the series cct is therefore
    (22.323<-59.03)/(10-j2) = current through Vo

    next part I'm realy unsure about
    so 2Vo = 2(22.323<-59.03)

    so current through the s/c = 3vo

    ... finding current through vo cant be right

    for finding the o/c voltage, im also a bit lost
    would I remove the 2vo again
    then fined the voltage across the vo resister
    then add 2vo??

    so Vab = 3vo???

    thanks guys
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Just on the matter of your calculated reactance values. Given ω=10rads/sec

    Xc=1/(jωC)=1/(j*10*10*10^-3)=1/(j*0.1)=10/j=-j10 Ω

    XL=jωL=j*10*(1/2)=j5 Ω

    With ab shorted the effective impedance would be (10-j10)Ω. You only have the resistor and the capacitor to deal with.

    In the case of the open circuit voltage across ab, you can't exclude the controlled current source and then add it later - since it is driving current back into the independent sources and not just into a short circuit. You can readily solve the circuit by using nodal analysis.

    The "complex" bit is forming an equation which relates the current flow in the capacitor branch to the voltage Vo - thereby allowing you to define the condition for the current flow from the controlled source. It's essentially an algebraic manipulation challenge.
     
  8. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
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    just one thing.
    In my one of my lecture notes there is given w = 10rads/sec. The impedance for a capacitor in his cct is was found by 1/50*C.
    I was really confused by this, and he didnt note why 10 became 50 so I assumed he converted rad/sec to just rads.

    The only was I could see the conversion was-
    [(rads/sec)^2]/2

    Dose that make any sense or is that an error in the notes?

    thanks a bunch :)
     
  9. jegues

    Well-Known Member

    Sep 13, 2010
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    The impeadance for a capacitor is simply,

    \frac{1}{j \omega C} = \frac{-j}{\omega C}
     
  10. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    yeah, its the steps in the notes I have that is confusing me.
    In my notes using 1/jwC with a w = 10rads/sec, he seems to use 1/j50C.
    is that a mistake in the notes? (is there a conversion from rad/s to just rads)
     
    Last edited: May 8, 2011
  11. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    ok, so I figure there must have been a mistake in the lecture notes so neglect my last post.

    For nodal analyses, I'm a little unsure about my equations.
    For my first
    Vo/10 + (Vo -12<0)/-j10 = 2<90

    Is that on the right track?
    Am I allowed to used a super-node for this kind of cct?

    thanks guys :)
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    What condition are you solving for here - was this the a-to-b short circuit case?
     
  13. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    ok, so my previous post is clearly wrong.
    this is what I have so far for finding voltage at o/c
    [​IMG]

    so the equations I have made are
    Vo/10 = (2<-90) + i1
    V2/j5 = 2Vo - i1

    and i1 = 2Vo - V2/j5
    so
    Vo/10 = (2<-90) + 2Vo - V2/j5

    dose that look ok?

    I i'm a little bit confused about the voltage dependent current source... would I be evaluating it as a current, as I have done in my equations?

    cheers guys
     
  14. t_n_k

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    Mar 6, 2009
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    So far so good. But what you've done isn't complete, as you need to set up another independent relationship between Vo & V2 to allow you to find a solution for V2. Note that V2 & V3 on your diagram are one and the same node voltage.
     
  15. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    can the next equation be found from ix?
    Vo = ix*10
    ix = (2<-90 + i1)
    and i1 = 2vo - V2/j5
    Vo = 10*(2<-90 +2Vo - V2/j5)

    giving
    10V2/j5 = (20<-90) + 19Vo

    Is that the equation for the relationship between Vo and V2 you were talking about?
    I cant see another way to develop another equation
     
  16. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    Because V2 = V3, could I say V2 = 10(2Vo)?
     
  17. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Consider formulating an equation which links nodes marked as Vo & V2 with current i1 flowing between the nodes.

    As follows ...

    V2=Vo-j10*i1+12<0°

    Does that help?
     
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