Thevenin Circuit Question

Discussion in 'Homework Help' started by stefan6582, May 18, 2016.

  1. stefan6582

    Thread Starter New Member

    Mar 10, 2016
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    Hi guys, i'm basically stuck in this question for a while now. I have to calculate the Thevenin equivalent voltage and resistance, while considering R in the diagram as the load we have to apply this to. Could anyone give me a hint on solving this. Also sorry if this is in the wrong section, it was hard to decide if it's a circuits question or a math question.

    Thanks,
    Stefan. Screen Shot 2016-05-18 at 19.19.16.png
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Sounds like this should be in the Homework section.

    It's a matter of divide and conquer.

    First, the 8Ω and 4Ω are in parallel so you can solve for their equivalent resistance.

    Then the 2Ω and 4Ω are also in parallel (since a voltage source has zero resistance) so solving for that will give you one equivalent resistance in series with the source.

    The Thevenin equivalent resistance is the sum of those two equivalent resistances since they are in series.

    The Thevenin open circuit voltage is the voltage generated at the junction of the 2Ω and 4Ω resistor voltage divider with no output load.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    It definitely belongs in the Homework Help forum, which is where it appears it has been moved to.

    How are you circuit analysis skill leading up to this point?

    If you were given this same circuit, except with the load resistor removed entirely, and were asked to find the voltage across the resulting gap, could you do it? If so, that is the "open-circuit voltage" which is also the Thevenin equivalent voltage.

    If you were given this same circuit, except with the load resistor replaced by a short (a wire), and were asked to find the current through the wire, could you do it? If so, that is the "short-circuit current", which is also the Norton equivalent current.

    The equivalent resistance in either case (Thevenin equivalent circuit or Norton equivalent circuit -- which you may or may not have been introduced to yet) is the ratio of the Thevenin voltage to the Norton current.

    This approach, which is not always the quickest or easiest, has the advantage that it will always work (for any circuit for which a Thevenin equivalent circuit exists, anyway).
     
  4. omar-rodriguez

    Member

    Jun 24, 2015
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    I know the best way to find any thevenin exercise.. This never fails

    this is the concept:

    Let's Imagine any circurcuit of two terminals represented by a box:

    Imagen2.png

    As you should know, because of the Thevenin's theorem you can represent the box as a voltage source Vth in series with a resistor Rth


    Imagen3.png

    Now the problem is how can I find Vth and Rth ... easy just put a test load voltage Vab and choose a current test sense just like that:

    Imagen4.png

    Now find an equation for Vab in terms of Iab

    Imagen5.png

    In summary to solve any thevenin problem you have to do this:

    1. Replace the component by a Vab test load voltage and draw the sense of the current Iab just like the image
    2. Solve the circuit and find Vab in terms of Iab, the coefficient of Iab is the thevenin's resistor and the constant term is the thevenin voltage
     
  5. omar-rodriguez

    Member

    Jun 24, 2015
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  6. WBahn

    Moderator

    Mar 31, 2012
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    The approach you describe is incomplete.

    For instance, let's say that use Vab = 10V and find that Iab = 2 A. What are Rth and Vth? There are an infinite number of combinations that will satisfy those two values for Vab and Iab.

    You need to use TWO different values of Vab so that you get TWO equations of that form, since you are solving for TWO unknowns: Rth and Vth.

    Choosing two non-zero values for Vab (and, as a consequence, Iab) is perfectly legitimate and will work, but the analysis is often noticeably more complicated than doing one open-circuit analysis (which is the same as setting determining the value of Vab that forces Iab to be equal to zero) and one short-circuit analysis (which is the same as setting Vab equal to zero), which will also always work (as long as the circuit HAS a Thevenin equivalent).
     
  7. omar-rodriguez

    Member

    Jun 24, 2015
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    No you don't have to replace Vab to a numeric value, you just have to analyze the circuit knowing that Vab is something known in the circuit it is not an unknown variable, it is a source!!..... the answer that I wrote is correct, just do it yourself and prove that there are something wrong
     
  8. omar-rodriguez

    Member

    Jun 24, 2015
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  9. WBahn

    Moderator

    Mar 31, 2012
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    The way you are doing it is perfectly fine (I missed part of what you were describing), but it is quite a bit more complicated than doing simple open-circuit and short-circuit analyses. But it is valid.

    However, what part of the rules for posting in the Homework Help forum were unclear? Perhaps the part about NOT just solving people's homework for them?

    Since you've already done so, let's look at what is involved in doing open-circuit and short-circuit.
    By inspection you can see that the open-circuit voltage is 8V.

    The short-circuit current is a little more involved, but it too can be solved by inspection. A 4 Ω resistor is the same as two 8 Ω resistors in parallel. So when the output is shorted there are effectively five 8 Ω resistors in parallel, three of which go through the short (so 3/5 of the total current will go through the short). The total resistance is 2 Ω + (8/5) Ω or (18/5) Ω. The total current is therefore (12 V)/(18/5 Ω) which is (60/18) A = (10/3) A and 3/5 of that is 2 A. So Rth is 8 V / 2 A = 4 Ω.
     
  10. omar-rodriguez

    Member

    Jun 24, 2015
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    I'm sorry for break the rule, but I think that with one solved exercise he is going to be able to solve any problem later
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    Really?

    Consider that he has almost certainly seen several similar exercises solved in his text and in class. Clearly seeing someone else work exercises has left him unable to solve similar problems (and this is not at all uncommon). Why is seeing yet one more problem worked for him by someone else going to make such a huge difference? For most people struggling in this situation the best (and sometimes pretty much only) way to get over that hurdle is to struggle and fight with the problem themselves. We can help point out where he is going wrong (which is why we require that people show their best attempt) and we can nudge in the right direction, but just working for them usually accomplishes nothing more than possibly making them think that they've learned how to do the problem only to discover on the exam that the didn't learn anything more than they did from all the prior worked exercises they've seen.
     
  12. stefan6582

    Thread Starter New Member

    Mar 10, 2016
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    Thank you for the help guys :)
     
  13. crutschow

    Expert

    Mar 14, 2008
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    Sorry, but I think your unnecessarily long and complicated solution likely confused him more than helped. :rolleyes:
    Please stick to the rules.
     
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