1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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

    SGT

    User Avatar

    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

    SGT

    User Avatar

    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

    SGT

    User Avatar

    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:

    • 1.GIF
      1.GIF
      File size:
      2.4 KB
      Views:
      55
  17. Jun 26, 2005 #16

    SGT

    User Avatar

    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

    SGT

    User Avatar

    [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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?