Finding norton equivalent with impedances..

Discussion in 'Homework Help' started by nirvanaguy, Dec 1, 2009.

  1. nirvanaguy

    Thread Starter New Member

    Oct 3, 2009
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    I'm having trouble with this problem..

    So far I've tried combinging the 2 branches and get (4-j2)/5 for the combined impedance

    I found the open circuit voltage by taking all of the 20A flowing into its series circuit through the (4-j2)/5 and getting that voltage and subtracting 2Ib from it , therefore the open circuit voltage came out to be 25.3<-71.57 degrees

    then when solving for Isc, i just did a voltage division and treated the controlled source at the top as a 2 ohm resistor since its in a form of R*I and its a voltage source. so i used current division to find how much of the 20a goes to the short circuit and i got that to be 6.32<-18.44. after dividing Voc/Isc, i found Zth= 4 < -53.13 and then got the norton equivalent from just drawing a diagram of the Zth and Isc in parallel. can someone verify if my steps are valid? if not, how else do I approach finding the numbers?


    If someone could guide me on how to further approach this i'd REALLY appreciate it! thanks
    soo yea [​IMG]
     
    Last edited: Dec 2, 2009
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Everything was going well until this point. You can't make this simplification of converting the dependent source to a passive equivalent. You have to solve using the available techniques while treating the source as a dependent voltage source.
     
  3. tinach

    New Member

    Dec 2, 2009
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    I have problem in posting an article,could you tell me how to post one?
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    Perhaps there is a shortcut you can take with dependent sources, but I don't see one that's obvious to me. If you can't see one either, you can always do the problem by brute force.

    The Norton equivalent current (In) can be found by shorting the output and calculating the short circuit current using circuit equations.

    The Norton equivalent resistance (Rn) can be found by shorting the output through a load resistor equal to Rn (symbolically). Then calculated the current through the load Rn using circuit equations. Then set this current to In/2 and solve for Rn, which will be the equivalent Norton resistance.

    Basically, a Norton equivalent current source should output In when shorted and it should output In/2 when loaded with a resistance Rn.


    A similar idea can be used with Thevenin equivalent voltage sources. Calculate the open circuit voltage to find Thevenin equivalent voltage Vth. Then load the output through Rth and calculate the output voltage. Then set this output voltage to Vth/2 and solve for the Thevenin equivalent source resistance Rth.

    Basically a Thevenin equivalent voltage source should output Vth when open circuited and it should output Vt/2 when loaded with a resistance Rth.
     
  5. t_n_k

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    This turns out to be a rather interesting circuit.

    If terminals 'ab' are shorted the solution is indeterminate.

    It appears no solution exists for any terminating resistance at 'ab' less than 2Ω - which makes it an interesting class of problem with a limited range of solutions.
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    Can you double check your work please? If you still feel it's right, please post your work.

    My feeling is that when you short out the ab terminals. The shorted current (i.e. the Norton current) is 20 A. For this to happen, IB=0 and V1=0. Just visually looking at the circuit tells me this solution should be valid. I don't see how any other solutions are possible, but I guess I'd have to see your work to see if I'm overlooking something.

    Thank you in advance. :)
     
    Last edited: Dec 2, 2009
  7. steveb

    Senior Member

    Jul 3, 2008
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    Aside from your checking to see if this comment is right in this particular case, we can talk about what to do in any cases where the current is indeterminable when the output is shorted.

    In these cases, we can simply choose to load the supply with two different resistances and then solve for the source current value and the source impedance value. Since there are two unknowns, two independent equations will allow us to solve for them. There may be some isolated cases where shorting creates an indeterminate current. In these cases (which are probably not practical cases anyway), just choose another load resistance to get a usable equation.
     
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Using the admittance matrix method gives a ready solution for the open circuit voltage at a-b and the impedance there. V1 is the voltage at the top of the 1Ω resistor, and V2 is the voltage at a-b, with no load applied to a-b.

    \left[ \begin{array}{3}1+\frac{j}{2}&0 \\1-j&-1\end{array}\right]*\left[ \begin{array}{3}V1\\V2\end{array}\right]=\left[ \begin{array}{3}20\\ 0\end{array}\right]

    V1 = 16 - 8j
    V2 = 8 - 24j = Vth

    The impedance seen at a-b can be found by inverting the Y matrix; the result is Zth = Znorton = (2 - 6j)/5

    A load resistance RL can be added to the system like this:

    \left[ \begin{array}{3}1+\frac{j}{2}&\frac{1}{RL}\\1-j&-1\end{array}\right]*\left[ \begin{array}{3}V1\\V2\end{array}\right]=\left[ \begin{array}{3}20\\ 0\end{array}\right]

    Solving, we get for V2 = Vab with a load RL at terminals a-b:

    V2 = \frac{(40-40j)RL}{(2+j)RL+2-2j}

    We can divide by RL and get an expression for the current in the load resistor:

    I = \frac{(40-40j)}{(2+j)RL+2-2j}

    This doesn't appear to be indeterminate for any value of RL.

    If the output is shorted, we can take the limit as RL -> 0; the current into a short will be:

    Isc = \frac{40-40j}{2-2j} = 20

    which is what we would expect. We can see this by inspection.

    The Norton equivalent is a current source of 20 amps in parallel with an impedance of (2 - 6j)/5
     
  9. t_n_k

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    Mar 6, 2009
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    Yes thanks - steveb & Electrician I agree with your answers.

    My original "observation" came from a transient simulation - using PSIM. Not from a direct steady-state analysis.

    I suspect that the simulation didn't work with Rab less than 2Ω because the transient component takes the solution into numerical instability. Not sure why this would happen at such a "precise" non-zero condition. (I set the time step to a reasonable value.) If Rab=1.9999Ω or less the simulation fails. If I set it to 1.99999Ω or higher then no problems. The successful simulated values are consistent with a hand-worked solution.

    I guess it's possible that this circuit has a steady-state solution for all values of Rab but can also fail to give a transient solution for a range of positive values for Rab [including Rab=0].

    I'd be interested in any feedback on the simulation issue and whether it's possible to design a working simulation which allows an Rab value between 0 and 2Ω.

    Cheers.
    [​IMG]
     
  10. t_n_k

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    Image of my simulation circuit is attached if of interest .... set for case of Rab effectively open circuit (2MΩ).
     
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