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Homework Help: Norton Equivalent Circuit

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data

    See picture:

    http://img88.imageshack.us/img88/8319/87671967.jpg [Broken]

    The answer is given as phasors
    I = 0.3536[tex]\angle[/tex]45
    Z = 2.828[tex]\angle[/tex]-45

    2. Relevant equations

    3. The attempt at a solution

    ZL = jwL = 2j Ohms
    ZC = -j/(wC) = -4j Ohms

    Can't do it by zeroing the sources as there's a dependent source (if it's possible, we haven't been taught it).

    So then we'll have to find both the open source voltage VOC and the short circuit current ISC, correct?

    Starting with VOC with a node b/w the inductor and resistor (which is equal to VOC because there is no voltage drop through the resistor due to no current going through it):
    (groud at bottom node)
    VOC: [tex]\frac{Voc}{-4j}[/tex] + [tex]\frac{Voc - 2}{2}[/tex] = 1.5IL

    IL = [tex]\frac{Voc - 2}{2}[/tex]

    Subbing in and arranging gives:

    [tex]\frac{Voc}{-4j}[/tex] = [tex]\frac{Voc - 2}{4}[/tex]

    Solving gives:
    Voc = -1.414[tex]\angle[/tex]45,

    however I'm fairly sure it's wrong given the answers ( IZ != V).

    I'm not sure how to find ISC because of the dependent current source. Have tried and can't get the correct answer - not sure if I'm going about it the right way.
    i1 = top loop current
    i2 = left loop
    i3 = right loop

    Taking positive to be CW,
    i1 = 1.5iL
    iL = i1 - i2

    iL = 2i2

    Loop two (left):
    -2i2 - 4j (i2 - i3) = 2[tex]\angle[/tex]0

    Loop three (right):
    2i3 - 4j (i3 - i2) = 0

    Solving the two linear equations gives
    i3 = iN = 2j

    which isn't the correct answer.

    Any help would be greatly appreciated...
    Last edited by a moderator: May 4, 2017
  2. jcsd
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