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Find the Thevenin Equivalent

  1. Oct 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the thevenin equivalent of the cicuit shown below with V=500 V and i=10 A
    upload_2016-10-3_14-50-56.png

    2. Relevant equations
    Voc=Vth
    Isc=VN
    3. The attempt at a solution

    So, to solve this, I used my instructors method of first finding the Rth. So, what I did was set the independent sources to 0(based on the fact that Rsource doesn't change based on Vth), and tried to solve for the resistance. I used the y-Δ transformation on the three at the bottom and got my Rsource. However, I am confused as to if this is right or if I am allowed to do that. I got Rsource to be 74.2 ohms, and that doesn't look correct, so I cant really move from there.
     
  2. jcsd
  3. Oct 3, 2016 #2

    gneill

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    Staff: Mentor

    Hi roinujo1,

    You're right, your result doesn't look correct. You'll have to show the details of your work if we're to spot what went wrong. I will mention, though, that if the sources are suppressed the 8 Ω resistor becomes parallel to the 12 Ω resistor, so the "Y" disappears. You might want to spend a bit more time redrawing and simplifying the circuit before bringing out the Y-Δ machinery :smile:.
     
  4. Oct 3, 2016 #3
    Thank you soo much for the reply.So, I did some reworking and got an answer to the problem. For the resistance, I got Rth=7.5 ohms.
    Before we move on, is this correct? And, is it in parallel because if we remove the independent sources the only "current that should be flowing is from the a to the b nodes? I really am having trouble using the open and short circuit stuff.

    So, after that, i used mesh current analysis to find what the Ishortcircuit is. To do this, I simplified the 3 resistors at the bottom to 10 ohms:
    upload_2016-10-3_17-3-47.png

    Sorr for the shitty drawing. So basically, I found i1 and i2 and said i1 is=ix and ix=Isc. So, I found the Vth by using Isc/Rth.
     
  5. Oct 3, 2016 #4

    gneill

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    Staff: Mentor

    Your result of 7.5 Ω for the Thevenin resistance looks good. But I'm having my doubts about your simplification in the latest diagram. In particular, the reduction of the three resistors to the 10 Ω value looks dubious, as the current source and 30 Ω resistor are connected to related nodes.

    Here's a trick you can try. Convert the current source and its 30 Ω resistor to their Thevenin equivalent. Then note that the "new" voltage source sits atop the 500 V voltage source which gives it a fixed potential of 500 V above the reference node b. It's perfectly legal to duplicate the 500 V source and move the connection of the new Thevenin model away from that node, provided that the new 500 V source is connected to node b as well. Like this:

    upload_2016-10-3_18-43-31.png Convert current source to voltage source, then:

    upload_2016-10-3_18-45-30.png

    Now you can proceed to simplify this new layout.
     
  6. Oct 3, 2016 #5
    Thanks for the response again! I really wish I could use this technique, I dont understand it enough and chicked out. What I ended up doing is mesh analysis, but this time using the original 3 resistors without combining them:
    upload_2016-10-3_18-22-50.png

    And I found i1=69.97 A,i2=56.67 A, and i3=26.67 A

    i2=ishort circuit

    I then used ohms law to get the Vth=Rth*ishort circuit which i got to =425.03 V approximately.

    How about that?
     
  7. Oct 3, 2016 #6

    gneill

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    Staff: Mentor

    That looks good.
     
  8. Oct 3, 2016 #7
    Thank you so much for the help! Could you by chance help me understnad why the 8 and 12 are in parallel?Are they in parallel because if we remove the independent sources the only "current" that should be flowing is from the a to the b nodes?
     
  9. Oct 3, 2016 #8

    gneill

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    Staff: Mentor

    The procedure for finding the Thevenin resistance involves suppressing the fixed sources. For voltage sources that means replacing them with a short circuit (basically a wire), and for current source, replacing them with an open circuit (removing them). In this problem, suppressing the voltage source places the 8 and 12 Ohm resistors in parallel.

    The circuit with the sources suppressed:
    upload_2016-10-3_21-20-53.png
     
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