Vo using Norton

Discussion in 'Homework Help' started by cheypr, Dec 15, 2009.

  1. cheypr

    Thread Starter Active Member

    Oct 13, 2009
    34
    0
    Hello!! I understand pretty much of this Problem but i dont understand something in step 12. From where we get the 3.578/-63.435? I was thinking that the problem was over when we found Vo= 2.4 - j0.8.
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    I can tell you what they did, but I can't tell you why they did it.

    In step 1, there is a dependent source connected to what is later labeled V3.

    In step 12, they have replaced that dependent source with an independent source whose voltage is the same as Vo determined in step 11, namely 2.4 - j0.8.

    Then the Norton impedance becomes 1/2 Ω, and the Isc becomes 1.6 - j3.2, or 3.578 < -63.435
     
  3. cheypr

    Thread Starter Active Member

    Oct 13, 2009
    34
    0
    So, are these steps OK or not?
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    They don't provide the answer for the circuit shown in step 1. That circuit is solved when you reach step 11, as you thought.

    Steps 12 and 13 provide the answer for a different circuit, one where the voltage source connected to V3 is no longer a dependent source, but is an independent source with the value 2.4 - j0.8 volts.

    Maybe they provided this as an example of how you must handle dependent sources differently than independent sources.
     
  5. cheypr

    Thread Starter Active Member

    Oct 13, 2009
    34
    0

    OK, thanks for your advise :)
     
Loading...