Thevenin Equivalent

Discussion in 'Homework Help' started by hugoleon, Jun 16, 2011.

  1. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    derive the thevenin equivalent of the circuit, I understand the equation but i don't know were to start on this equation.

    ......................r1 2.2kΩ r2 3kΩ
    -----------------------------------------1kΩ
    ..................... r3 1.2kΩ r4 300ohm
    .........9v.........................................................RL

    how do i find the rth for this equation?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Are you able to post a schematic drawing? I don't understand what you have entered.
     
  3. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    thanks i hope this helps a little better
     
  4. jegues

    Well-Known Member

    Sep 13, 2010
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    Start by determining the open circuit voltage.

    To do this, remove the load and calculate the voltage across the terminals of the load.

    Note that when you remove the load, the other end of the circuit is no longer connected to ground. What effect will this have on the circuit? (Think current)

    After determining the open circuit voltage, you need to find the equivalent Thevenin resistance (Rth) as seen by the load.

    To do so, short out the voltage source and determine the equivalent resistance as seen by the load terminals.

    If you are still confused, post your work and an attempt at the solution and if need be we can further assist you.
     
    Last edited: Jun 16, 2011
  5. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    From your notes, it looks like your having difficulties determining series and parallel circuits.
     
  6. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    my question is how do i start? i know that i have to make r1 to r4 into 1 resistor but do i divide by one or add?? what do i do to the r5?
     
  7. Adjuster

    Well-Known Member

    Dec 26, 2010
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    You would start by reducing R1 and R2 to a single resistance: R1 is in series with R2, although you seem to have written R1 // R2.

    Similarly, get the total value for R3 and R4. Now work out the parallel combination of the two branch totals.

    You now have a single resistance equivalent to R1...R4, in series with R5. Work out the total.

    You must get clear in your mind the difference between series and parallel connections, and how to calculate their equivalent values. http://en.wikipedia.org/wiki/Series_and_parallel_circuits
     
  8. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    I combined r1//r2 r3//r4// and i get 201kohm and now and 1kohm?
     
  9. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    i get total of 201.83ohms from r1 to r4
     
  10. Adjuster

    Well-Known Member

    Dec 26, 2010
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    I cannot see how you got the values in your last two posts. 201KΩ is very much too big, while 201.83Ω is too small.

    Try again: you are looking for an answer in the region of a few KΩ.
     
  11. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    now i got 1.164kOhm from r1 to r4
     
  12. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    now i got 1.164kOhm from r1 to r4
     
  13. jegues

    Well-Known Member

    Sep 13, 2010
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    That is correct.
     
  14. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    is this right (9*1kohm)/(1.164kohm+1kOhm)=4.1v?
     
  15. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    What exactly are you trying to calculate?
     
  16. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    DERIVE THE THENVENIN EQUIVALENT OF THE CIRCUIT SHOWN IN FIGURE 7.47a THIS IS THE QUESTION
     
  17. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    Yes I'm aware of that.

    I'm asking you, what were you trying to calculate?

    Was that an attempt at calculating the Equivalent Thevenin voltage (Vth)? (i.e. the open circuit voltage)

    As I mentioned in my previous post,

     
    Last edited: Jun 16, 2011
  18. hugoleon

    Thread Starter New Member

    Jun 15, 2011
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    Yes that was for (vth)
     
  19. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    It is not correct.

    When you remove the load to find the open circuit voltage, the circuit is no longer complete.

    What does this mean in terms of the current flowing throughout the circuit?

    What can you conclude about Vth from this? (Hint: Write a KVL)
     
  20. hugoleon

    Thread Starter New Member

    Jun 15, 2011
    13
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    Ok so its 9v
     
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