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Node voltage with dependent voltage source

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data
    http://i62.tinypic.com/245fwxu.jpg



    2. Relevant equations & Attempt to solve

    http://i62.tinypic.com/142v6mv.jpg

    Bottom: A
    Top left: B
    Top Right: C


    (1/100 + 1/5 + 1/25)VB - (1/25)VC + 6.25I = 450 mA

    (20.8)VB - 0.04 VC + 6.25I = .450 A

    VC = 45 + 6.25I

    I know I'm doing something wrong here. VC doesn't seem right to me.
     
  2. jcsd
  3. Jan 21, 2015 #2

    NascentOxygen

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    Staff: Mentor

    ➊ This line is full of mistakes. Start again, explaining each term. If you are trying to do a lot in one line to save writing, then that's why you are making mistakes.

    ➋ Is the 6.25I a dependent voltage source or current source? (It's difficult to deduce from your working.)
     
  4. Jan 21, 2015 #3
    The way we always did it in my circuits class was put the thing straight into a matrix from the circuit by inspection. But that doesn't work with dependent sources as easily so in this class we abandoned that.

    6.25I is a dependent voltage source.


    So I thought from node A at the bottom to the first Node B at the top:

    The resistors connected to it would be:

    (1/100 + 1/5 + 1/25) *VB

    And the common resistor is -(1/25) VC
     
  5. Jan 21, 2015 #4

    NascentOxygen

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    I don't understand this.

    Try applying KCL clearly, expressing current leaving the node = current entering the node
     
  6. Jan 21, 2015 #5
    We have to use the node voltages method for this.
     
  7. Jan 21, 2015 #6

    NascentOxygen

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    Assume a node voltage, then sum the currents into that node.
     
  8. Jan 21, 2015 #7
    So if I understand then I still get without adding the dependent source.
    VB/(100) + VB/(5) + (VB-VC)/25 = .45 A
     
  9. Jan 21, 2015 #8

    NascentOxygen

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    It will be something like that. But you haven't correctly expressed the current through the 5 ohm resistor.
     
  10. Jan 21, 2015 #9
    would the 5 also contain a term for the dependent source?
     
  11. Jan 21, 2015 #10

    NascentOxygen

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    Staff: Mentor

    Ohm's law always applies: current through a resistor = voltage difference across the resistor / R
    You need to involve the dependent source, yes, it controls the voltage at one end of the resistor.
     
  12. Jan 21, 2015 #11
    VB/(100) + (VB-6.25I)/(5) + (VB-VC)/25 = .45 A

    Is that what you mean?
     
  13. Jan 21, 2015 #12

    NascentOxygen

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    Yes. And you know I in terms of VB and VC. And you know the value of VC.
    So that seems like enough equations to be able to solve for all unknowns.
     
  14. Jan 21, 2015 #13
    VC = 45 V right?
     
  15. Jan 21, 2015 #14

    NascentOxygen

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    +45V, yes.
     
  16. Jan 21, 2015 #15
    So when I solve I get : VB = 5I + 9


    I'm not really sure what I had for I/what I is
     
  17. Jan 22, 2015 #16

    NascentOxygen

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    I is indicated on the diagram as the current through the 25Ω, with a direction going from R to L.
    So, by Ohms Law, I = (the voltage on the right - voltage on the left) / 25
    i.e., I = (45 - VB) / 25
     
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