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Thevinen equivalent

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    Find equivalent thevenin circuit

    3. The attempt at a solution

    I found the equivalent thevenin resistor already.
    The 1000 + 500 are in parallel. Come out to 333.333 equivalent.
    Equivalent is in series with 1000, then its 1333.333
    1333.33+1000 in parallel is 571.43 ohms

    Thevenin resistor equivalent = 571.43 ohms

    I am stuck on the thevenin voltage.

    From my understanding I can take the 5V and divide it by the 1k resistor to get the current at that point. Which is .005A

    Am I allowed to add the currents together since they are in parallel? Make it 2.005 A?

    Reduce the resistors down to one: 571.43 ohms

    V=IR ....571.43 ohms * 2.005 amp = 1000+ Volts...this seems horribly wrong.

    Help :(?
  2. jcsd
  3. Sep 5, 2011 #2


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    Staff: Mentor

    You have to make sure that currents really are in parallel before you can add them; No series resistances in the way!

    One way to approach this problem would be to do multiple transformations (Thevenin, Norton, as required), working your way from left to right through the circuit. You already began that process when you proposed converting the 5V supply and its series 1k resistance to a current supply with a parallel 1k resistance.
  4. Sep 5, 2011 #3
    5V / 1k = .005 A

    1k + 500 resistor in parallel = 333.33 ohms

    333.33 ohms * .005 A = 1.666V

    333.33ohms + 1k ohms in series = 1333.33 ohms

    1.666V/1333.33ohms = 1.25 mA

    Nothing in series between the currents so current is 1.25mA+2A = 2.00125A

    1.33k + 1k in parallel = 571.43 ohms

    571.43 ohms * 2.00125A = 1143.57428 V

    Should I be expecting something this massive in voltage? Feels like I am taking the wrong second step once I get the .005A
    Last edited: Sep 5, 2011
  5. Sep 5, 2011 #4


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    Staff: Mentor

    How did you get from 1.666V to 1666.66V ?
  6. Sep 5, 2011 #5
    Sorry misread my paper when I was working it out.

    Was suppose to still be 1.667 which reduces the the current I got to mA, not A. I updated above.
  7. Sep 5, 2011 #6


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    Staff: Mentor

    Okay, looks better now. The voltage is okay for that amount of current being pushed through several hundred Ohms :smile:
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