Covert Voltage soure to current source

Discussion in 'Homework Help' started by celect, Jun 22, 2005.

  1. celect

    Thread Starter Member

    May 31, 2005
    18
    0
    I need to convert a voltage source to a current source.

    E = 20V <20 ˚
    R = 5.6 Ω
    L=8,2 Ω
     
  2. celect

    Thread Starter Member

    May 31, 2005
    18
    0

    I believe to solve this problem
    I would solve for I E/Rs = 20V/5.6ohms = 3.6A
    Then I redraw the circuit replacing the voltage source with a current soure of 3.6A

    Am I going in the right direction?


    I been informed that i need to solve for Z then redraw the circuit, question do i redraw the current source in series or parallel with the inpedeance?
     
  3. Semyazza

    Member

    Jun 25, 2005
    12
    0
    5.6 + j8.2 = 5.6 + j8.2 = 9.92@55.67˚ = Ztotal

    20@20˚/ 9.92@55.67˚ = 2A @ 35.67˚ = Itotal

    Draw the Current Source in Parallel with the impedance. I dont know if you have done thevenin or norton equivalent circuits but if you memorize their layouts thats the best way to remember what to do IMHO.....
     
  4. celect

    Thread Starter Member

    May 31, 2005
    18
    0

    How does the 5.6 become 9.92
    and how does j8.2 become 55.67
     
  5. Semyazza

    Member

    Jun 25, 2005
    12
    0
    http://www.allaboutcircuits.com/vol_2/chpt_2/5.html will give you the answer. Its just a conversion from rectangular form to polar form. I used my calculator to do the conversion but you can do it by hand.

    Rectangular form: x + ri --or-- x+jr ---or--- x+y

    Polar form:
    x = cos θ
    y = sin θ

    Im pretty bad at explaining that but I would look up information on vector math in a physics book (I have noticed its broken down well in the ones I have seen) or anywhere else you can find out about it.
     
  6. davidand

    Active Member

    Jun 2, 2005
    43
    0
    Thnaks this info has been very helpful.
    I'm glad you sent that link.
     
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