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Thevenin equivalent in a simple but tricky circuit

  1. Feb 21, 2017 #1
    Hey everyone, this circuit in the attachment is making me think... I am asked to find the current I2 going through the branch of R2. ....Also to find the voltage in the current source using thevenin equivalent, I found the thevenin R but I'm not sure is right, and should I take the whole branch (R2 & U2 ) as the load? or just R2? if so how?
    I aprecciate your help!
     

    Attached Files:

  2. jcsd
  3. Mar 4, 2017 #2
    You have 2 way to calculate Rth. First way it is to follow this rule:
    We set all the independent sources to zero (voltage sources →short circuit, current sources→ open circuit). Then we find the equivalent resistance between the two terminals.
    Then the sketch will be:
    upload_2017-3-4_17-30-22.png
    upload_2017-3-4_17-29-11.png

    since V1=0 then point Z will be in point W and R3 will be
    parallel with R4. Rth=[(R1||R2)+R3||R4]
    Second way : you can calculate the sum of currents entering
    point X and point Y. Calculating X-Z and X-Y you'll get X-Y=V
    (X-Z)/R1+(X-Z-V1-V2)/R2-I=0 in point X
    (Y-Z)/R4+(Y-W)/R3+I=0 in point Y
    X-Y=V=I*Rth+Vth UI=V
    I2=Ixw=(X-W-V2)/R2 W=Z+V1
    I2=(X-Z)/R2-(V1+V2)/R2
     
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