Circuit with Simultaneous Independent and Dependent Current and Voltage Sources

Discussion in 'Homework Help' started by mjakov, Sep 16, 2014.

  1. mjakov

    Thread Starter New Member

    Feb 13, 2014
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    Hi all!

    I have been trying to solve this circuit with about every method I could think of: node voltage, mesh current, branch current, superposition, and source simplification. Although using one of these methods always worked for circuits with no dependent sources and all voltage or current sources, in this case I obtain a different solution with each method. So I assume, there must be something wrong with the assumptions I used when setting the circuit equations. The solution should be:  I = -12 (A)

    [​IMG]

    Question 1 - Node voltage method (NVL) :
    So what is correct, i.e. can we assume that the current in the left branch is constant 2A:
    a)  \frac{V_a}{3} = 2 , or
    b)  2 + \frac{V_a - 4}{2} + \frac{V_a - 3V_r}{5} = 0 , or
    c)  \frac{V_a}{3}+ \frac{V_a - 4}{2} + \frac{V_a - 3V_r}{5} = 0 , or
    d)  2 + \frac{V_a}{3} + \frac{V_a - 4}{2} + \frac{V_a - 3V_r}{5} = 0

    Question 2 - Superposition:
    At first the 2A source is suppressed resulting in the following circuit:

    [​IMG]

     -3V_r + 5I + 4 + 2I = 0
     V_r = 2I
     I = -4 (A)

    Then the 4V independent voltage source is suppressed resulting in the following circuit:

    [​IMG]

    By the branch current method:
     I = -2.4 (A)

    So, by superposition the total current should be:
     I = -6.4 (A) ,

    which differs from the solution in the book. What is the correct way to solve this?
     
  2. t_n_k

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    Mar 6, 2009
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    Question 1 NV method equation (a) is incorrect. It doesn't allow for the voltage across the 2A source.
    Use equation (b) and instead substitute either Vr=Va-4 or Va=Vr+4. Then solve for Va or Vr. And so on to the solution ...
     
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  3. t_n_k

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    In the superposition method your calculation for I with the 4V independent voltage source removed is incorrect. Since you haven't shown the method for this part we can't know where you've made the error.
     
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  4. MrAl

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    Jun 17, 2014
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    Hello,

    It looks like your (b) equation is correct as it sums the currents at the center node correctly. If you sum them and solve for Va then you can solve for I and the assumed result of -12 amps looks correct too. Of course Vr=Va-4 is your second equation (two unknowns requires two equations) or simply substitute.
    You may want to note that the 3 ohm resistor is superfluous in this problem because of the 2 amp current source. It can be almost anything (like 0 ohms, 5 ohms, 1000 ohms 2 zillion ohms) and it would not change Va or I.

    Using superposition on the two constant sources isnt too easy for this circuit. You have to be very careful with the polarities, and probably convert the 2amp source into a voltage source, and solve for the two currents through the 5 ohm resistor for example, then sum them to get the total current. I have to wonder if it even makes the task harder to do rather than easier mostly because shorting the 4v source doesnt help make the circuit too much simpler. Probably a better way to understand why this is so would be to look at the concept of a topological tree.
     
    Last edited: Sep 17, 2014
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  5. Vinetou

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    Sep 5, 2014
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    Here´s my contribution. I'm not 100% sure of these calculations.
     
  6. mjakov

    Thread Starter New Member

    Feb 13, 2014
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    This solution looks like a very elegant one. Thank you all for the suggestions!
     
  7. Vinetou

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    Sep 5, 2014
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    Forget about my post. I didn't notice the 2A source and did not take it into account.
     
  8. Jony130

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    Feb 17, 2009
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    But still your solution is wrong, because you are not consistent with your assumptions.
    First you assume that I1 current flows from VN node to VB. But then you write I1 = (Vb - Vn)/R3 ?? Can I ask why?
     
  9. Vinetou

    New Member

    Sep 5, 2014
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    You're absolutely right Jony.
    Making the corrections the current's would be

    VN = 2.52V
    I1 = -0.097A (which means that the flow adopted is not right. The correct flow is the opposite as the one adopted)
    I2 = 0.839A
    I3 = 0.742A

    This satisfies the equation I3 = I1 + I2 -> 0.742= -0.097+0.839.

    The voltages obtained are

    VR1 = 2.52V (the same as VN)
    VR2 = 1.48V
    VR3 = 0.48V

    The voltages for loop 1 are:

    VA-VR1-VR2 = 0
    4 - 2.52 - 1.48 = 0

    And for loop 2 are:

    VB - VR3 - VA - VR2 = 0
    -3 -(-0.48)-(-4)-1.48 = 0
    -3 + 0.48 + 4 - 1.48 = 0

    But again, that's not the solution for the circuit posted by mjakov as I did not take the 2A source into account.
     
  10. Jony130

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  11. MrAl

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    Jun 17, 2014
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    Hi,

    Did you also notice that the source on the far right is a dependent source, not a constant voltage source? Thus the voltage is 3*Vr not 3 volts.

    You can check your result by comparing the current you get for I with the current given of -12 amps because that is known to be correct.
     
  12. Jony130

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    thx MrAl , I did not notice that.
     
  13. t_n_k

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    Superposition using Marshall Leach Jr's approach is quite simple & quick.
     
  14. t_n_k

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    This is the method ....
     
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  15. MrAl

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    Jun 17, 2014
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    Hi,

    I meant the more common way when you kill each source one at a time and take the response, not another method.
     
  16. t_n_k

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    That's exactly what I did - including killing the dependent source. This killing of dependent sources is often stated as an invalid method, which turns out is not the case - as shown by Leach.
    There is only one proviso - the controlling parameter "source" cannot be removed in any of the individual cases where any one active source is under consideration.
     
    Last edited: Sep 18, 2014
  17. MrAl

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    Jun 17, 2014
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    Hi,

    I apologize but i have trouble reading your circuit equations. It looks like you did them all at once, when killing each source one at a time results in TWO different responses, then later they are added. Perhaps a typed equation would be better for reading here. Also, i wanted to emphasize the circuit more when the voltage source is killed...that's the more difficult situation. Also keep n mind i never meant to say that it would not work, although there COULD be times when it does not work with dependent sources in the circuit.
     
  18. t_n_k

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    No worries - please don't apologize.
    I'll attempt to elaborate what was a fairly brief summary of the method.
    Attached is an image summary of the various cases one would apply as part of a superposition solution.

    The equations for each case [a, b & c] are developed as follows:

    Case a : Current source acting alone
    By the current divider rule, the current flowing to create voltage Vr_a is
    I_{2 \Omega}=-2 \times \frac{5}{\( 5+2 \)}=-\frac{10}{7} \ \small{Amp}
    which will produce a voltage Vr_a across the 2 ohm resistor
    V_{r\_a}=-\frac{20}{7} \ Volt

    Case b: Independent voltage source acting alone
    Using the voltage divider rule with the 4V shared across the 5 ohm and 2 ohm resistors.
    V_{r\_b}=-4 \times \frac{2}{\(5+2\)}=-\frac{8}{7} \ \small{Volt}

    Case c: Dependent source of value 3*Vr acting alone
    Again by the voltage divider rule, the voltage 3*Vr is shared across the 5 ohm & 2 ohm resistors.
    V_{r\_c}=+3V_r \times \frac{2}{\(5+2\)} = +\frac{6}{7}V_r \ \small{Volt}

    The superposition of the 3 cases to give Vr is then:

    V_r=V_{r\_a}+V_{r\_b}+V_{r\_c}=-\frac{20}{7}-\frac{8}{7} +\frac{6}{7}V_r \ \small{Volt}

    Multiplying through by 7 and collecting terms in Vr gives

    7 \times V_r-6 \time V_r=-20-8 \ \small{Volt}

    or

    V_r=-28 \ \small{Volt}


    Superposition process.jpg
     
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  19. MrAl

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    Jun 17, 2014
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    Hello again,

    First, it was very nice of you to write that all out in a very readable form that makes it easy to see and follow. I thought i was the only one (he he) that would take the time to do that kind of thing, but it's nice to know there is someone else that appreciates this subject matter enough to do that too.

    Now it's plain to see that you treated the dependent source in an independent way and that's what makes all the difference, and i think it was a good idea.

    One thing i am not sure of though is if the dependent source can always be handled like that (not this circuit but perhaps another different circuit). So we could look into this to see if it is a good general purpose method.

    I've actually done that in op amp circuits, where the op amp is considered to be a dependent source. It usually works, but i seem to remember that there was one case where it didnt work. I cant remember the circuit now though unfortunately. I also remember another time we had a problem like that but it was so long ago i cant remember that circuit either :) Perhaps we can try to find a circuit where it doesnt work, if there really are any. I think you may have mentioned this by the fact that we are not able to remove "Vr" for example (the sense input).
     
  20. t_n_k

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    Thanks Mr Al

    I have included a link to Marshall Leach's unpublished paper. He submitted the paper to a couple of reputable journals but it was rejected for publication on the grounds the material lacked relevance or sufficient interest. This is curious since (as Leach points out), many (most?) teachers of the superposition method would not allow for the disabling of dependent sources from any intermediate stage of the superposition process.

    https://www.google.com.au/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CB0QFjAA&url=http://users.ece.gatech.edu/mleach/papers/superpos.pdf&ei=jRUcVI7ONM-B8gXbh4CwCQ&usg=AFQjCNHlzhvqsK138Nh9QAjpnu6C0AJEig&bvm=bv.75774317,d.dGc

    It is an interesting topic & Leach himself provides several worked examples.

    I have seen another paper by a different author who claims there is a slight error in Leach's theoretical basis which can be corrected without loss of the general validity of the method itself.

    Cheers!
     
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