Thevenin equivalent schematic with 2 sources and junction in between

Discussion in 'Homework Help' started by punctualmanx375, Jun 3, 2018.

  1. punctualmanx375

    Thread Starter New Member

    Jun 3, 2018
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    I have to determine a Thevenin equivalent of this schematic (see attachment). (not as homework but preparation for an exam)
    I've been trying for over 3 hours but have no idea how to do this.
    I've tried to redraw the circuit but doesn't help much...
    Any help would be appreciated.
    (answers are also in attachment).

    schematic.png

    solution.png

    Moderators note: shown images full size
     
  2. Jony130

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    But did you manage to solve for Vth?
     
  3. punctualmanx375

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    Jun 3, 2018
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    Eth = Vth in the solution. I know how to draw a thevenin equivalent schematic. but what I'm asking is how did he(the teacher) get to Eth. Rth is no problem, I know how to do that but I'm stuck trying to find Vth/Eth
     
  4. Jony130

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    What type of circuit analysis technique do you know?

    Can you solve for Vx in this circuit ?

    1.png
     
  5. punctualmanx375

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    Jun 3, 2018
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    not sure what you mean by the first question, i've seen kirchoff in the past but want to solve this only using thevenin.
    and for the second question:
    -first calculate voltage over 2ohm resitor =(50*(2/(4+2)))=16.666667V
    -second calculate Vx(other 2ohm resitor) =(16.666667*(2/(5+2)))=4.7619V
    so Vx=4.7619V
    next?
     
  6. Jony130

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    Do you know nodal or mesh analysis? Or maybe Kirchhoff's law only?

    No, wrong.
    From what I see you trying to apply a voltage divider rule. But it won't work because yor voltage divider is loaded by (5Ω + 2Ω).
    And your equation apply only to unloaded voltage divider ( circuit without 5Ω + 2Ω resistor).

    Try agin.
     
    Last edited: Jun 3, 2018
  7. punctualmanx375

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    Jun 3, 2018
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    owww yeah, totally missed that.
    so then:
    calc replacement resistor for 2ohm//7ohm = 14/9 = 1.55555556 ohm
    calc voltage over replacement resitor = 50*((14/9)/((14/9)+4)) = 14V
    calc voltage over 2ohm resistor = 14*(2/(2+5)) = 4V
    so Vx = 4V
    (i use "//" to indicate parrallel resitors)

    and to answer you're first question, I think only kirchoff's law in other words current matrix and then use the current to calc all voltages.
     
  8. Jony130

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    Very good. So now find Vy in this circuit

    s23.png
     
  9. punctualmanx375

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    Jun 3, 2018
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    I thought you were going that route... :D
    then:
    calc 4//2 = 1.333333 (4/3) ohm
    calc voltage over 2 ohm resistor = -20*(2/(1.333333+5+2) = -4.8 ?
    Vy=-4.8
     
  10. Jony130

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    Wrong. Why 2Ω in the numerator?
     
  11. punctualmanx375

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    Jun 3, 2018
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    I think I see what you mean,
    calc 4//2 = 4/3 ohm
    calc voltage over 4/3 and 5 ohm = -20*(((4/3)+5)/((4/3)+5+2)) = -15.2?
    Vy = -15.2?
    and if you combine 1 and 2 you get -15.2+4=-11.2?

    (sorry for the long response time, only have 5 posts/hour apparently. So I had to wait...)
     
  12. Jony130

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    Yep, and this the Vth voltage equal to Vth = -11.2V
    And this technique of solving the circuit is called a superposition.
     
  13. punctualmanx375

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    Jun 3, 2018
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    oooh that rings a bell.
    thanks so much!
     
  14. Jony130

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    But are you familiar with nodal or mesh analysis?
     
  15. punctualmanx375

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    Jun 3, 2018
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    no idea, if i look it up on google mesh analysis rings a bell,
    determine the currents (by formula)
    then put them in a matrix and calculate?

    nodal analysis I haven't heard of though

    btw, any way to get rid of the 5 posts per hour?
     
  16. bertus

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  17. Jony130

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    For such a simple circuit matrix is not needed

    mesh.png

    For blue mesh (I1) we have

    (I1 + I2)*4Ω + I1*5Ω + I1*2Ω - 20V - 50V = 0


    And for red mesh (I2)

    (I2 + I1)*4Ω + I2*2Ω - 50V = 0

    And the soution is

    I1 = 4.4A
    I2 = 5.4A


    So Vth = I1*2Ω+(-20V) = 4.4A*2Ω-20V = -11.2V (KVL)
     
    Last edited: Jun 3, 2018
  18. WBahn

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    Mar 31, 2012
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    There are too many techniques that just "ring a bell" for you to succeed. You need to stop where you are and go back and review basic circuit analysis until you are proficient using

    Kirchhoff's Voltage Law
    Kirchhoff's Current Law
    Mesh Current Analysis
    Node Voltage Analysis
    Superposition

    to analyze circuits containing multiple sources, including all four of the basic dependent sources.

    If you don't, you will just be digging yourself a deeper and deeper hole that will be progressively more difficult to get out of.
     
    Jony130 likes this.
  19. MrAl

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    Jun 17, 2014
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    Hi,

    A nice way to do this is to first collapse the left side of the network (left of a,b)
    into it's Thevenin equivalents and then go from there. That leaves you with two
    voltage sources and just two resistors and the advantage is then we can easily
    find both the Thevenin voltage and resistance just by considering those two
    sources and those two resistances.

    To start on the left, the Norton current is i=50/4 and that is in parallel with the
    Norton resistance of 4 in parallel with 2, and 2 is like two 4 ohm resistors in parallel
    so the resistance is r=4/3. i*r=50/3 so the first Thevenin voltage is v=50/3 in series
    with resistance r=5+4/3=19/3.
    Now to find the superposition contribution from the 50v source we just use the voltage
    divider formula with upper R1=19/3 and lower R2=2 and voltage source V=50/3, and that gives
    us 4 volts. To find the total Thevenin voltage we just have to consider the -20v
    source next, along with the first Thevenin voltage and resistance.
    Since we also have the left hand side resistance too, we can then easily find the total
    Thevenin resistance by placing that in parallel with the 2 ohm resistor and that gives
    us the 1.52 ohms.
     
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