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Nodal Analysis Problem

  1. Jun 2, 2012 #1
    I need to find V0 in the given circuit. I've uploaded both the problem, and an annotated one with the variables I chose.

    These are the equations I got for the system:
    [tex]V_{0}=30-V_{2}-V_{3}[/tex]
    [tex]V_{2}=30-i_{1}(10\Omega )[/tex]
    [tex]V_{2}=I_{X}(2\Omega )[/tex]
    [tex]V_{3}=30-V_{2}-(-i_{1})(1\Omega )[/tex]
    [tex]V_{0}=i_{4}(5\Omega )[/tex]
    [tex]I_{X}=i_{1}+i_{2}[/tex]
    [tex]4I_{X}=i_{2}+i_{4}[/tex]

    Does this look right? I'm fairly certain I'm doing this wrong, as I got -15/7A for the current i1.
     

    Attached Files:

  2. jcsd
  3. Jun 2, 2012 #2

    gneill

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    Staff: Mentor

    You V2 and V3 represent node potentials, not potential differences across the 10 and 1 Ohms resistances. So your first equation is not correct; The same problem exists in several of your other equations, and you've introduced current variables I1 and I2 which are not necessary for writing the node equations.

    Clearly V3 and Vo must be the the potentials at the same node (they are connected by a wire), so V3 is in fact equal to Vo and there's no need to have duplicate variables.

    How many independent nodes do you count for your circuit? You only need that many node equations, plus one equation for each controlled power supply.
     
  4. Jun 3, 2012 #3
    Okay. Well then I can count 2 nodes. One after the 10Ohm resistor, and one after the 1Ohm resistor. I also want to count the ground rail.
     
  5. Jun 3, 2012 #4

    gneill

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    Staff: Mentor

    Okay, so then you should write two equations for the two independent nodes that you've identified (the ground rail is the reference node and doesn't require a separate equation) and one equation that gives the controlled current source value in terms of node potentials.
     
  6. Jun 3, 2012 #5
    Thus far I have:

    Node1
    [tex]I_{x}=i_{1}+i_{2}[/tex]
    [tex]\frac{V_{1}}{2}=\frac{30-V_{1}}{10}+\frac{V_{2}-V_{1}}{1}[/tex]

    Node 2
    [tex]4I_{x}=i_{2}+i_{3}[/tex]
    [tex]4I_{x}=\frac{V_{2}-V_{1}}{1}+\frac{V_{0}}{5}[/tex]

    I should be able to eliminate a variable since V0 will be equal to V2.
     
  7. Jun 3, 2012 #6

    gneill

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    Staff: Mentor

    Yes. You also don't need variables i1 or i2, since Ix can be easily obtained from the node voltage V1. That's the "other" equation I was referring to with regards to the controlled source.
     
  8. Jun 3, 2012 #7
    Idea! I pulled a third equation:
    [tex]30=V_{1}+V_{2}[/tex]

    and therefore:
    [tex]30=V_{1}+V_{0}[/tex]

    I got some numbers by using this form.

    V0=222/5
    V1=237/8

    I posted this just as you replied. I'll review what you said and take a look at it.
     
  9. Jun 3, 2012 #8

    gneill

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    Staff: Mentor

    This is not true! V1 and V2 are NODE POTENTIALS, not resistor voltage drops.
    You should be able to find Ix from a node potential though... (in fact, it's included as a term in your Node1 equation :wink:)
     
    Last edited: Jun 3, 2012
  10. Jun 3, 2012 #9
    Why not though? At V2 the only two branches are the resistor and the current source. Therefore V2 is the only potential left to drop before it hits the ground rail.

    I do have Ix there, but there are only two equations. I have three variables...
     
  11. Jun 3, 2012 #10

    gneill

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    Staff: Mentor

    Node 2 has three connections: the path to node 1, the path to the controlled current source, and the path to the load resistor.

    NOTE: Node voltages ARE NOT RESISTOR DROPS. They are the potentials, w.r.t. ground, of those nodes. If a resistor happens to be connected between a given node and ground, then it will THEN also be that resistor's potential drop. But you should know this already by HOW you write the node equations, right? You take the DIFFERENCE between node potentials for the resistance that lies between the two nodes.
    Yes, so you want to write an equation for Ix. Hint: You've already incorporated an expression for Ix in your equation for node 1.
     
  12. Jun 3, 2012 #11
    I see what you mean there. I was looking at it the wrong way. I think this is the equation you're referring to:

    [tex]V_{1}=I_{x}*2\Omega[/tex]
    Since the drop across that will be equivalent to V1.

    Doing that substitution I found V0 to be 15/19
     
    Last edited: Jun 3, 2012
  13. Jun 3, 2012 #12

    gneill

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    Your expression for Ix looks fine, but I think your reduction for Vo from the node equations does not look correct. Can you detail your work?
     
  14. Jun 3, 2012 #13
    On review I botched something along the way. Revised work:

    Rearranged node 1:
    [tex]\frac{8}{5}V_{1}=3+V_{2}[/tex]
    [tex]V_{1}=\frac{5}{8}(3+V_{2})[/tex]
    [tex]V_{1}=\frac{15}{8}+V_{2}[/tex]

    Node 2 rearranged to:

    [tex]3V_{1}=\frac{6}{5}V_{0}[/tex]

    after first using this equation:
    [tex]3V_{1}=2I_{x}[/tex]
    to sub out the Ix term from it.

    Combining them together, and replacing any instances of V2 with V0:
    [tex]3(\frac{15}{8}+V_{0})=\frac{6}{5}V_{0}[/tex]
    I then got V0 to be -25/3
     
  15. Jun 4, 2012 #14

    gneill

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    That looks better :smile:
    Be sure to add the appropriate units to the final value.
     
  16. Jun 4, 2012 #15
    Yes, V0 = -8.33V, so that's correct.

    With questions involving nodal analysis, I find it easiest to write out the set of node equations really carefully as you're going to need them to solve for the other variables and the slightest error will obviously follow through and then it's a mess having to troubleshoot 3 or 4 equations. Along the way you will see relevant aspects like that Ix = V1/2 which you're going to need to sub in later for your second node equation.
     
    Last edited: Jun 4, 2012
  17. Jun 4, 2012 #16
    Well thanks for the help! This has me off to a good start. I'll just practice the method now.
     
    Last edited: Jun 4, 2012
  18. Jun 5, 2012 #17
    Lancelot,

    Two equations with two unknowns. No need to figure out sub currents.

    Ratch
     

    Attached Files:

    Last edited: Jun 5, 2012
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