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Finding Thevenin Equivalents using Source Transformations

  1. Mar 2, 2013 #1
    FIRST OFF: Sorry for the multiple post on this subject, I couldn't figure out how to edit my first post on this topic (I promise I'm not spamming!)

    1. The problem statement, all variables and given/known data
    initial.png
    2. Relevant equations



    3. The attempt at a solution

    Step 1:
    Removed unnecessary resistors R1 and R3. R1 is removed because of the fact that it will not have any effect on the voltage because it is in parallel with the voltage source. R3 is the same in contrast, series with a current source has no effect on current.
    step1.png

    Step 2: Used several source transformations to convert the current source into a voltage source (combined the resistors R4 and R5 in series). Converted the voltage source back into a current source combining the R4\R5 series (now 1 resistor and in parallel to the current source) with R6 (2000 ohms || with 1000 ohms = 666.67 ohms)
    step3.png

    I'm not sure where to go from here. Am I some how able to rearrage the voltages and resistors and combine them that way? I feel like I'm really close (Or I've done this completely wrong).

    Any help would be appreciated! :D
     
  2. jcsd
  3. Mar 2, 2013 #2

    gneill

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    Staff: Mentor

    R4 and R5 are not in series; the junction between them is not shared by them alone.

    Consider instead converting V1 and R2 to a current source.
     
  4. Mar 3, 2013 #3
    Oh, I see what you're saying, I forgot to convert V4 (step 2) into a current source (that was how I combined the three resistors into 666.67ohms).

    So taking your advice and converting V3 (step 2), into a current source (10 mA parallel 1k), thus I get...

    step4-1.png

    So, from here, can I somehow combine my resistors, and use superposition to combine my current sources. If I could, then that would be a norton equivalent, which I could easily convert over to the desired thevenin equivalent?

    edit: so my text book has an example of this problem, element for element (but not value for value), and at this point in the circuit, they simply combined the resistors in parallel, and combined the current sources.

    So if I were to follow their lead, I would get:
    i = 10mA-50mA = -40mA
    R = (1\1000 + 1\666.67)^-1 = 400 ohms,

    convert it all into a thevenin equivalent, V = IR, V=-.04 * 400 = -16V

    so -16V in series with 400 ohms
     
    Last edited: Mar 3, 2013
  5. Mar 3, 2013 #4

    gneill

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    Staff: Mentor

    Your results are not correct since you've made several errors in combining and transforming things. R4 and R5 (in your original circuit diagram) are NOT in series. So you cannot treat them as such. As a result, the 666.7 Ohms you found and the 50 mA current source are not correct.

    If you're going to solve the problem by repeated source transformation and combination, then you should start from the "back end" of the circuit and work your way forward towards the output terminals. So begin by converting the voltage supply V1 to a current source and move on from there, left to right. Resistors get combined suitably as you go.
     
  6. Mar 3, 2013 #5
    Ok, so I reworked the problem using you're suggesting of working back-to-terminals, and hey! It worked! I started by converting my voltage source to a current source, combining 10mA - 100 mA = -90mA, and was able to easily work my way through the rest to get a voltage of -18V and 600 Rth, or -30mA, 600 Rn

    Thanks! :D
     
  7. Mar 3, 2013 #6

    gneill

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    Staff: Mentor

    Excellent. Well done.
     
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