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Nodal voltages

  1. Jun 23, 2005 #1
    Can someone direct me to a tutorial for solving network
    nodal voltages.
    Also need to learn how to convert voltage source to a current source.
     
  2. jcsd
  3. Jun 23, 2005 #2
    a text book should have everything you need..your in philly they must have a kick a*s Library there.. :smile:
     
  4. Jun 23, 2005 #3
    Yea I was there today, will be back tomorrow.
     
  5. Jun 24, 2005 #4

    SGT

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    You cannot convert a voltage source into a current source. What you can do is to convert a voltage source in series with a resistor into a current source in parallel with the same resistor. This is Norton's theorem.
     
  6. Jun 24, 2005 #5
    Ok I'm reading that chapter now, I attached a sample I'm working on now.
     

    Attached Files:

  7. Jun 24, 2005 #6

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    The solution you posted is wrong. The value of the current source is V/Z and what must be in parallel with the source is the series impedance, not the parallel you draw.
     
  8. Jun 24, 2005 #7
    Regarding the right most circuit component, the symbol represents an inductor; yet the units is given in ohms.
     
  9. Jun 24, 2005 #8
    inductor are coils , therefore i believe they can have resistance,
    anyway,Checkout OpenCourseWare at MIT
     
  10. Jun 24, 2005 #9
    So if I need to find E/Z = 20/9.9 = 2A

    Now all I do is redraw circuit with 2A current source with
    impedance parallel in circuit?
     
  11. Jun 25, 2005 #10
    If you want to do a source transformation on this circuit, you should probably look up phasors.
     
  12. Jun 25, 2005 #11

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    No, you must make
    [tex]I=\frac{20[cos(20)+jsin(20)]}{5.6+j8.2}[/tex]
     
    Last edited by a moderator: Jun 25, 2005
  13. Jun 25, 2005 #12
    I'm confused on how to get proper result

    I have now I = 20V <20 degree/ 5.6+J8.2
    and I found in book the way to draw circuit.
    but I don't understand how to use the equation with cos and j 20 sin.
     
  14. Jun 25, 2005 #13
    That's just Euler's formula and phasor notation:
    "A-angle-theta" is phasor notation for A*ej*theta, which by Euler's formula is A[cos(theta) + j*sin(theta)], where A is a scalar value.
     
  15. Jun 26, 2005 #14

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    You can also write 5.6 + j8.2 as 9.93 <55.67
    So, [tex]I=\frac{20 <20}{9.93 <55.67}[/tex]
    That is an easiest form to evaluate the quocient.
     
  16. Jun 26, 2005 #15
    Here is a problem I got wrong, with the solution.
    I don't understand how they got answer.


    I don understand how you get
    5.6 + j8.2 as 9.93 <55.67
    how does 5.6 become 9.33
    and j.2 = 55.67
     

    Attached Files:

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  17. Jun 26, 2005 #16

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    I have transformed the complex number from its cartesian form to its polar form.
    Z = x + jy = R<φ
    Where R is the modulus of the complex:
    [tex]R = \sqrt{x^2 + y^2}[/tex]
    and
    [tex]\phi = tan^{-1}\frac{y}{x}[/tex]
     
  18. Jun 26, 2005 #17
    Thanks its coming together now.

    so my current = 20/9.93 = 2A

    I don't understand how to get degrees yet
    [tex]\phi = tan^{-1}\frac{y}{x}[/tex]

    do I take 8.3/5,5 tan and then use above formula?

    what is the -1 how do i use on calculator?
     
  19. Jun 26, 2005 #18

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    [tex]tan^{-1}[/tex] means arctan. It is the arc whose tangent is y/x.
    And your current is I = 2 <-35.67.
    When dividing complex numbers in the polar form you divide the moduli and subtract the phases.
     
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