nodal KVL

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

  1. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    I1= V1/6,

    6-10I-6(I_{1}-I_{2}=0

    6(I_{2}-I_{1})+2=0

    did i miss anything?

    thanks
     
  2. ELECTRONERD

    Senior Member

    May 26, 2009
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    Using source transformations, I get an output voltage of 3.75V based on KVL. I'm not very experienced in this type of thing but I think that's the answer. Do you have an answer to verify my assumption?

    Austin
     
  3. ELECTRONERD

    Senior Member

    May 26, 2009
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    On second thought it could actually be 6.5V, I'm not sure.

    Austin
     
  4. thakid87

    Active Member

    May 23, 2009
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    What is that component in the second parallel branch supposed to be?

    It is labeled 1/4 V1.
     
  5. ELECTRONERD

    Senior Member

    May 26, 2009
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    I think it's 1.5V, or an actual quarter of 6V (V1). That's what I used at least.

    Austin
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    There are two things labeled V1; the 6 volt battery, and the voltage across the 6Ω resistor R2.

    I think it would make more sense for the 1/4 V1 controlled source to be referring to the voltage across R2.
     
  7. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    hi clarification.
    the electrician was right.

    (1/4)V1 is with respect to 6Ω not the voltage source 6v.

    is I2=2A?


     
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Denote the node at the top of R2 as node A.

    There are 4 branches connected to A, and we can sum the currents in those 4 branches to zero, using the convention that a current leaving a node is taken as positive. Each branch current can be given as a simple application of Ohm's law:

    (V1-6)/10 + V1/6 - V1/4 + 2 = 0

    Rearranging, we have:

    V1*(1/6 + 1/10 - 1/4) = 6/10 - 2

    The solution is V1 = -84 volts.

    From this, we can determine that I1 = 9 amps in the direction shown.

    I2 = 84/4 + 2 = 23 amps in the direction shown.

    The current from the V1/4 dependent source is in the direction opposite to that shown by the arrow in the source symbol, because V1 is negative.
     
  9. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    thank u, the electrician.
    is there any supermesh element in the circuit?

    if so, could V1/4 current source be the constrain?
    pls see attached.
    the formula by KVL,

    mesh 1 in clockwise direction
    6-10I1-6(I1+V1/4 - 2)=0-------eq1

    mesh 2 in clockwise direction
    6(2-V1/4 + 2 - I1 - V1/4 +2)=0---------eq2

    regards,
    stupid

     
    Last edited: Dec 4, 2009
  10. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I don't think there is a supermesh.

    I merged the 2 amp source into the dependent source, so that the dependent source becomes V1/4-2.

    Then the equation for the left hand loop is:

    -6 + I1*(10+6) - I2*(6) = 0

    The second equation is a constraint equation:

    I2 = -(V1/4 - 2)

    since V1 = 6*(I1 - I2), this becomes:

    I2 + 6*(I1 - I2)/4 - 2 = 0

    Solving these two equations gives:

    I1 = 9
    I2 = 23
     
  11. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    the electrician,
    i remember a thread posted here dealing supermesh.
    see attached.

    the working:
    combine meshes 2 & 3 by KVL,
    5(I2-I3) + 15I3=0

    it is quite similar to the one discussed here.

    could u kindly me tell what constitutes a supermesh?

    regards,
    stupid

     
  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Imagine that you have a 1 amp and a 2 amp current source in series, feeding a node of some network. What is the current being fed into that node? Is it 1 amp, or is it 2 amps? The current is indeterminate; it can't be solved if the current sources are ideal. But, if there is a resistance in parallel with both sources, or even one of them, then the current can be solved.

    On the other hand, if you have two current sources in parallel, then the current from the combination is just the sum of the two individual sources. No problem in this case.

    Similarly, if you have two ideal voltage sources in parallel feeding a network and one source is 10 volts and the other is 15 volts, what is the voltage feeding the network? It can't be determined if the sources are ideal. But, if there is a resistance in series with one or both, then the voltage can be determined. (In the real world, the resistance of the wires serves to make the voltage determinate. The currents may be very large, perhaps large enough to cause damage.)

    Looking at the network in post #1 of this thread, if you go around the right hand loop you have two current sources in series and the loop current is indeterminate. But, if you combine them as two current sources in parallel, the problem goes away. I wouldn't try to treat those two current sources as a supermesh.

    In the circuit of post #11, you don't have a situation where going around a loop puts two current sources in series; there's a resistor in there. Then the supermesh concept doesn't lead to an indeterminate current.
     
  13. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    the statement that " When a current source is contained in 2 meshes or is not connected in parallel with a resistance, a supermesh is created by excluding the current source & any element connected in series with it."

    compare cct in post 11 with respect to above dictate, arent those 2 resistors in parallel with the current source & thus not a supermesh?

    regards,
    stupid
     
  14. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I said "In the circuit of post #11, you don't have a situation where going around a loop puts two current sources in series; there's a resistor in there. Then the supermesh concept doesn't lead to an indeterminate current."

    I am saying, in effect, that the supermesh concept works in this case (the circuit of post #11 case).

    That is, the circuit of post #11 does contain a supermesh. Did you think I was saying that it does not contain a supermesh?
     
  15. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    hi the electrician,
    i have taken note of your post 12 & agree to that.

    however, i also look at a particular statement quoted from my course material on my last post.
    i wonder if that statement were true or i misinterpret?

    according to that statement the cct in thread 11 seems agree not with that.

    regards,
    stupid

     
  16. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I what way does it not agree? The statement says, in part, "...When a current source is contained in 2 meshes...a supermesh is created".

    In the circuit of #11, the current source is "contained in 2 meshes", isn't it?
     
  17. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    how about the part .."or is not connected in parallel with a resistance, a supermesh is created..."

    the dependent current source is parallel with R2 & R3, isnt it?



     
  18. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    It says "or is not connected in parallel with a resistance, a supermesh is created...", not "and is not connected in parallel with a resistance, a supermesh is created..."

    The first part of the description is satisfied, so we have a supermesh.
     
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