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Solve for I and U using Thevenin's Theorem

  1. Oct 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Hello guys!

    I am getting really desperate here and I'd be very glad for any help.

    I have the following circuit with 5 resistors :(excuse the drawings, I am in a hurry, I hope they are sufficient)


    And I am supposed to solve for IR3 and UR3.

    2. Relevant equations

    3. The attempt at a solution

    So my guess was to first transform it into a linear circuit so I put R1 R2 and R4 and R5 together:

    R12 = R1*R2/R1 + R2
    R45 = R4*R5/R4 + R5

    So the circuit would be now simplified like this and I could apply the theorem:


    And now I removed R3, and solved for Rt = R12*R45/R12 + R45

    Then I figured out the current Ix*R12 + Ix*R45 = U
    Ix = U/R12 + R45

    And then the Ut = U - IxR12

    When I put it all together like this to figure out IR3:

    IR3 = Ut/Rt + R3

    I am not getting the right value so I must be doing something wrong.

    I solved the circuit using the Star Delta transformation and I got all the right values.

    I am not posting any numbers because I wanna solve this myself, I just need someone to point me in the right direction.

    I am a beginner and I would be extremely glad for any help I can get.

    Thank you and have a great day!
  2. jcsd
  3. Oct 6, 2015 #2


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    Well, that leaves 2R !

    You should make two Thevenin equivalents:

    A) For R1, R4
    B) For R2, R5

    Now setup:


    Calculate current through/voltage across R3.
  4. Oct 6, 2015 #3
    Hello there and thank you very much for your reply!

    I am not quite sure what you mean by "that leaves 2R".

    Was combining the parallel resistors not a right step?

    And I am not sure which equivalents do you mean right now. Could you please elaborate?

    Should I combine R1 and R4?

    Thank you again!
  5. Oct 6, 2015 #4


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    Perhaps this will hep you move forward. With R3 removed from the circuit it should be clear that you have two potential dividers providing potentials at the open terminals where R3 sat. Each potential divider is comprised of a pair of resistors and a driving voltage U. Redraw the circuit so that the two potential dividers are shown separately:


    You can provide each with its own voltage source U without changing how the circuit operates, since both potential dividers are driven by a potential U to begin with and they don't interact in any way with R3 out of the picture.

    Now you should be able to reduce each of the potential dividers to their Thevenin equivalents and then proceed to further simplification. In fact, you should be able to work out the overall Thevenin equivalent pretty much by inspection.
  6. Oct 7, 2015 #5
    Hello there, I am afraid I am not really getting it : (

    Could you please specify what was wrong with my method of simplifying the circuit?

    I was thinking, if I remove R3, I should then combine R1 and R2 and R4 and R5 like I have written above in order to get a linear circuit, I have no idea what you have done there where you put R4 and R1 and R5 and R2 together like that.

    Also, do I need to use another voltage source? Do I need it for figuring out Ut? How will I get its value? I have the value of the original source only.

    Thanks a lot!
  7. Oct 7, 2015 #6

    The Electrician

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    You will find gneill's suggestion explained in more detail here: http://www.hallikainen.com/rw/theory/theory6.html
  8. Oct 7, 2015 #7


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    Sorry, my fault. It should have been:

    R12 = R1*R2/R1 + R2 = 2*R2.

    Yes, as qneill has shown ( same with R2, R5 ).

    R1 and R2 will not become in parallel even if you remove R3. Both ends of R1 will not be connected to R2. Only if you short circuit R3, R1 and R2 are connected in parallel, but then you change the all over circuit.
    Last edited: Oct 7, 2015
  9. Oct 7, 2015 #8
    Okay guys, I think I might be on the right track!

    So, to figure out Rt, instead of treating R1,R2 and R4,R5 like parallel, I do this:

    R14 = R1 + R4
    R25 = R2 + R5

    And with R3 removed, R14*R25/R14 + R25 should give me Rt.

    Then to figure out Ut, I calculate the current Ix = U/R14 + R25 and then Ut = U - IxR14 while still treating the resistors as linear?

    Is that approach correct? I am not sure about the Ut because I am not sure whether the current should be splitting between UR14 and UR25 with the voltage source now present or not.

    I am getting an IR3 that matches the IR3 I calculated via the Start Delta method for 2 decimal figures so I am not sure whether I am still doing something wrong or not.

    Thanks for all your help guys!
  10. Oct 7, 2015 #9


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    Staff: Mentor

    Nope. Consider the diagram I gave in post #4. You are looking for the resistance between terminals a and b in that diagram, with the sources suppressed of course. Can you sketch the diagram with the sources suppressed?
  11. Oct 7, 2015 #10
    Like this?


    I do appreciate your picture you posted earlier but it rather confused me, especially since you added a second source which was not in the original circuit.

    I am trying to follow the instructions the professor showed us in the class but it was just one, very simple example with not much explanation how exactly it works, thus I struggle with applying it to this more complex circuit.

    What did I do wrong when calculating the resistance?
  12. Oct 7, 2015 #11


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    A Thevenin equivalent looks like this (dashed box).
    It's an emf ( Voc ) in series with the inner resistance, Rth ( like a battery ).
    So you must calculate Voc and Rth as for A and B ( separately ). Now attach R3 in between:

  13. Oct 7, 2015 #12

    The Electrician

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    Suppressing the sources in the diagram of post #4 means to replace the sources with a short circuit (a simple wire, in other words). If you replace the left source U with a wire, what is the resistance from point A to ground? Do the same for the resistance from point B to ground; what is that resistance?
  14. Oct 7, 2015 #13
    Okay guys, I think I better show you the example I have been given, so you know what knowledge I am supposed to be using while solving this problem.




    I' d say this one example is not sufficient in the least but thats what I have been given to work with.

    So I am not quite sure what you mean by saying I need to figure the resistance of A while in this simple example, I could just figure both at once which is what I thought I should do in the more complex one as well once I combine together R1,R4 and R2,R5.
  15. Oct 7, 2015 #14


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    Splitting a node by duplicating the source that drives it is okay so long as it doesn't change any of the potentials in the circuit. I did it to try to make clear that there are essentially two separate voltage dividers:
    The potentials driving the voltage dividers are the same after the source is duplicated. So you have one voltage divider providing the potential at terminal a, and another providing the potential at terminal b. The figure in post #4 was just a rearrangement of the the dividers on the page to highlight their independence and make it more obvious how to get at the Thevenin equivalent. And in fact you've successfully determined what the circuit would look like from the point of view of the open terminals with the sources suppressed.

    Now, how might you determine the open potential across the terminals a-b?[/quote][/quote]
  16. Oct 7, 2015 #15
    Hey there qneill, please check out what I have posted just above your post.

    I really do not get the "You must calculate the Voc and Rth as for A and B ( separately )" part since I was presented with an equation that did both at once.

    Did I at least get any of the steps right?
  17. Oct 7, 2015 #16


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    In the example you presented above, one of the target terminals was also the reference node, so you only had to determine the potential at one location (with respect to that reference node). The method and results were fine.

    In the present case you have two terminals, neither of which is the reference node. Yet you want to find the potential difference between them. The simplest way to go about that is to determine their potentials separately with respect to a common reference, then take the difference.

    You can do this in several different ways. One is to recognize that the terminals in question are each part separate voltage dividers and use the voltage divider formula. Another is to render them both into their own Thevenin equivalents, then combine those into a single Thevenin equivalent. That would get you to your overall result of both Vth and Rth in one set of steps, although you've already determined Rth so it might be extra work for little gain (except perhaps practice).

    I'd suggest going the voltage divider route at this point. Work with the bottom figure in post #14. Find the potentials at a and b.
  18. Oct 7, 2015 #17
    I am afraid I have no idea what a reference node means or what it is supposed to do : /

    My intuition tells me that its the "ground" towards which the arrow was pointing in the pictures I posted and since the b terminal was there, it was easier to determine the potential.

    And what I understood from what you wrote is that I need to figure out the current for each circuit and then use it to calculate the Ua and Ub ?

    But maybe I got it all wrong.

    The drawings above were all I was given with almost zero explanation so if you could please explain which one was it and why, and why here it is not the case, I would be extremely grateful.

    The class appears to be very poorly organized and as much as I would love the professor introduced us to more complex scenarios and explained this stuff thoroughly, I doubt this will be the case.

    And I would really prefer a human being to tell me why something is the way it is rather than searching on google.

    Thank you very much for your time and patience! As I have said, I am a beginner and this is all very new stuff for me and can be quite hard to wrap my head around at times.
  19. Oct 7, 2015 #18


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    Yes, the reference node is one that you choose for convenience and is usually designated by the little triangle symbol (there are other symbols that are used occasionally, but the triangle is quite prevalent). Think of it as the spot in the circuit where you might attach the negative lead of a voltmeter and then use the other lead to measure various potentials elsewhere in the circuit: it is thus the reference point to which you refer other potentials. You are free to choose any node as a reference node, but most often a particular choice will be more convenient than others. It's largely a matter of experience to develop the insight and intuition to choose a node that will make your analysis easier.
    No, that's a reasonable way to proceed. Find the current then use it to determine the potential drops across the resistors in order to establish the potential at the node in question (do a "KVL walk" from the reference node to the node in question, summing the potential changes along the way). The voltage (or potential) divider rule that I've mentioned several times really just combines these two steps into one operation. The situation comes up so often in circuit analysis that the resulting reduced formula has been named. There's an analogous rule for current division.
    In the example you gave the reference node of choice happened to coincide with the lower connection point for the load for convenience. The reference node is always assumed to be the zero potential reference point in a circuit (unless otherwise specifically stated), so that (conveniently) gives you potential at that node without having to do any work to find it. That left only the potential at the top connection to find (where the top of R3 connects when it's in the circuit). A KVL walk from the reference node to that terminal might pass through V2 and R2 (or the other way around the loop, through V1 and R1).

    Sorry to hear about your class "style". Fortunately we are here to help where we can :smile:
  20. Oct 8, 2015 #19
    Okay I am just about to give up. Today's class was even more disappointing and we didn' t do any of these circuits so I would like to ask you if you could provide me with a step by step solution (with pictures that have arrows symbolizing the Ut and Rt I am looking for if possible, I would be very grateful for that) which I will then try to understand.

    Anyways, my last attempt was this:

    Calculating the current I:

    After removing the R3 resistor, I am left with 2 parallel resistors R1 + R4 and R2 + R5, so I calculated the current that goes through the lower half of the circuit like this:

    Ix = U/R2 + R5

    Then the upper current:

    Iy = U/R1 + R4

    And because the I coming from U is split between the upper and lower part of the circuit, the final I would be Ix + Iy.

    Calculating the Ut:

    Then using the current, I would determine UR1, UR2, UR4 and UR5.

    Then UA = U - UR1 and UB = U - UR2

    My final Ut would then be the difference: UA - UB

    Calculating the Rt:

    If I supress the source, I am left with a parallel circuit and I either put together R1 + R2 and R4 + R5 or R1 + R4 and R2 + R5.

    My intuition tells me the first one but I have drawn the picture earlier with the second option and you said that was correct.

    So I then do Rt = R14*R25/R14 + R25 to figure out Rt.

    I have tried both ways but neither gave me an answer that would match the star delta method one.

    I am however very close (I am supposed to be getting something like 0.22...... and I am getting 0.20.....).

    I' d love to hear your thoughts!
  21. Oct 8, 2015 #20


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    Unfortunately we can't solve problems for you in that manner. It's against the forum rules to do the work for you or provide complete solutions. But we can give guidance, make suggestions, point out errors, and so forth.
    Okay. You should insert parentheses to group terms in order to fix the order of operations so that there can't be misinterpretations:

    Ix = U/(R2 + R5)
    Iy = U/(R1 + R4)

    and that would be correct.
    Do you need that total I for anything? Certainly not for the potential calculations...
    Overall you've got the right idea. But I'm concerned about some details:

    You didn't explicitly specify a reference node, so It's hard to confirm your formulas. But something looks off to me since both R1 and R2 are connected to the "-" terminal of the voltage supply U and if you took that as your starting point a KVL walk from there through U to the terminals wouldn't pass through those two resistors.

    Perhaps you should take your initial circuit diagram and indicate the currents and potential drops that you're using (their polarities in particular).
    For two components to be in parallel both of their leads must be directly connected, nothing in between. For each of the components you need to be able to trace a path via wires only from one component to the other for both of their leads. Note that drawn placement or orientation on the page does not determine series or parallel... only the topology of the circuit counts. Components can be turned, moved, or otherwise mucked about with on the diagram so long as their node connections are not altered. When in doubt, number each node in the circuit and then list each component along with their node associations. Parallel components will have the same associations.
    That doesn't look right (even taking into account the lack of parentheses). Go back to the figure in post #4. Once you've reduced R14 = R1||R4 and R25 = R2||R5, how are R14 and R15 connected? Going from terminal a to terminal b through R14 and R25 do you need to split the path or just carry on through them in a single path?
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