Circuit analysis with a dependent source (VCVS)

Discussion in 'Homework Help' started by dybben, Sep 23, 2016.

  1. dybben

    Thread Starter New Member

    Sep 23, 2016
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    Hi everyone. This is my first post, so bear with me.

    Im taking a course where I can take some quizzes for practice, and the current theme is operational amplifiers.

    On of the questions is this circuit where I shall determine the current through the 6 ohms resistor. (see picture).

    Our current knowledge is:
    KCL
    KVL
    Ohms law
    Voltage divider
    Current divider
    Nodal analysis
    Mesh analysis
    Superposistion

    I am a little lost, and have tried a lot of different approachs, but without any luck. I think it is the dependent source which gives me problems. I know for a fact that the answer is 895 mA, but since it is a quiz there is no explanation..

    If anyone have the time to help, it could be great! I would appreciate if you could explain the different steps, or maybe just give me some hints to how to determine the current through the 6 ohms resistor.

    Thanks.
    Unavngivet.png
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    You've said that you have made many attempts. So post the work for the one you think is the best. Then we can see how you are looking at the problem and where you are going wrong.

    There are several reasonable ways to go about it -- but one way that is ruled out right away is superposition. Can you see why?
     
  3. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi there,

    Well, that's funny because i was able to solve this one using superposition. Maybe my technique is superduperposition? :) The trick is to solve for V1 first.

    It works, but i am not sure if anyone would want to use it in this circuit.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    When doing superposition, you generally (and every text I have ever seen) superpose the solutions for the independent sources. This circuit only has a single independent source.

    If you try to treat the dependent sources as if they were independent, then you run into a problem because when you solve the circuit for a case when only an independent source is "on" then you don't get the contribution due to the dependent sources reacting to that independent source and when you solve the circuit for a case when only a dependent source is "on" then you don't get anything at all because there are no active independent sources for it to react to.

    There is a way to fake treating a dependent source as an independent source, but it is very error prone and, the few times that I've looked at it, significantly more complicated than superposing only solutions for the dependent sources.

    I'd be interested in seeing how you did your solution. I'm particularly curious about how you solved for V1 first (i.e., before solving the circuit using superposition). Did not solving for V1 involve analyzing the circuit by some means other than superposition?
     
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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  6. dybben

    Thread Starter New Member

    Sep 23, 2016
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    One of my attempts was using mesh analysis where I get three equations with three unknowns (see picture).

    According to the superpostion approach do I understand it that way, that you cant "kill" a dependent source - not at my level :)

    Udklip.png
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    What do you get if you solve your 3 equations?
     
  8. WBahn

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    Mar 31, 2012
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    Your second equation isn't consistent because you assigned a voltage polarity to the 6 Ω resistor that is inconsistent with the polarity of i2. You are summing up voltage DROPS around the mesh as you go clockwise around it -- but i2 results in a POSITIVE voltage drop as you go clockwise through it.
     
  9. WBahn

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    Mar 31, 2012
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    Yes, I've read that paper before, and I am still of the opinion that while it is of valid academic interest, it greatly complicates things unnecessarily with a corresponding likelihood of making errors, particularly at the introductory level.
     
  10. dybben

    Thread Starter New Member

    Sep 23, 2016
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    Sorry for not adding it!

    Actually when i write the equations into Mathcad I noticed that I was subtracting the voltage over 6 ohms resistor in equation 2, which i thought was incorrect, and therefore changed it to plus sign (see the red circles). - cant say why it might is incorrect.

    I can see that i_2 is the right result, but when i calculate with ohms law, it says different?
    Unavngivet1.png
     
  11. WBahn

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    Mar 31, 2012
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    Aren't the current in the 6 Ω resistor and the mesh current i2 exactly the same thing?

    If so, then why are they different in both of your solution?

    Get in the habit of asking yourself if the answer makes sense. You won't do it all the time, but it will let you catch lots of stupid mistakes. I just got burned (deservedly so) over in another thread because I failed to ask if my answers made sense before posting.

    Hint -- Ohm's Law relates the current through THAT resistor to the voltage across THAT resistor. Is 3·V1 the voltage ACROSS the 6 Ω resistor?
     
    dybben likes this.
  12. dybben

    Thread Starter New Member

    Sep 23, 2016
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    No, it is 4 ⋅ V1, of course!
    Udklip.png
    Much better! Thanks for helping me!

    Just for curious, how would you analyze this circuit?
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    In general I would analyze a circuit like this using mesh analysis, although nodal analysis isn't a bad choice, either.

    For this specific circuit a slightly ad hoc approach jumps out.

    If we call the voltage at the top center node V1, then the voltage increase over to the right node is 3·V1 making it 4·V1.

    Thus we know the current in the 6 Ω resistor is

    i6 = (4·V1) / 6 Ω = V1 / 1.5 Ω

    Applying KCL at the top center node we then have

    [(6 V - V1) / 4 Ω] = (V1 / 5 Ω) + (V1 / 1.5 Ω)

    6 V / 4 Ω = (V1 / 5 Ω) + (V1 / 1.5 Ω) + (V1 / 4 Ω)

    So now I multiply both sides by 60 Ω (since 60 = lcm(3, 4, 5)) to get

    15 · 6 V = V1 · (12 + 40 + 15) = 67 · V1

    V1 = (90/67) V

    i6 = V1 / 1.5 Ω = (60/67) A = 896 mA
     
  14. The Electrician

    AAC Fanatic!

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    I also read the paper a while ago, and re-reading it now, I don't agree that the method greatly complicates things. In every example Leach gives, using superposition with dependent sources is no more complicated than a mesh or nodal analysis, and is usually less complicated. He often only needs one equation.

    For the problem in this thread, here's a solution for V1 using superposition with the dependent source:

    Superpos.png
     
  15. WBahn

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    Perhaps I need to go back and revisit it. Right now I just don't have the time.
     
  16. RBR1317

    Active Member

    Nov 13, 2010
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    It is much simpler to solve with a single node equation.....
    CirAnlys-VCVS.png
     
  17. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Lots of replies since i was here last, and i didnt read them all yet but i can reply to this one first anyway. I guess this is an interesting topic for people.

    The trick is that the dependent source is treated as an algebraic source, that's the only difference, In other words, it is a variable that has not yet been solved for, yet fits into the equations as a real source would anyway.
    The solution doing this algebraic superposition is then the solution to V1, and from there we can get the numerical value of the dependent source.
    So we proceed in the same way as any other superposition problem but carry the unknown through the equations and at the very end we get the value for the source numerically. So for an independent source V1=5v the contribution is 5v, and the contribution from an independent source V2=Vx*7 is just that same thing, Vx*7, which of course holds the variable as an unknown. I am sure if you tried this you'd get the same result too. Of course we might need Ohm's Law as well.

    LATER:
    I went back and looked at Electrician's solution and that looks the same. It's basically superposition for V1 though, not the output voltage, and that is what i did too, and so maybe an argument can be made that the OUTPUT was not calculated directly using superposition therefore it is not true superposition. For me sometimes i just jump on a circuit and dont think too much about how i am solving it as to the exact technique unless it's a more complicated circuit.
    If we compare it to a solution using two solid independent and constant voltage sources, we see that the only difference is that we can calculate the output immediately without the need to carry a variable through. However, we are shorting out each source in turn so there are two views about this i guess. It's just that one of the 'contributions' has to be made a variable to start with.
    I do this all the time with op amp circuits though, i guess that it why it did not seem strange to me to short out the dependent source, because with most op amp circuits i can short the output to ground to help solve it, if i do decide to do it that way. I have read that you cant do that, but i've done it so many times and got the right result that i have to ignore that reading :)

    LATER THAN THAT:
    I found a small reference in an old EE book that shows two independent sources and one dependent source, and they basically do the same thing we did here.
    Of course we could also look at more examples too.
    I guess it doesnt matter if we show the answer now :)
     
    Last edited: Sep 24, 2016
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