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Homework Help: Thevenin's theorem question - really need help

  1. Apr 7, 2010 #1

    I am wondering whether someone could help me with this question, attached below as an image. For some reason electronics has always been something I find difficult to comprehend and I have some deep misunderstandings, so I'm trying to work hard on the subject for my degree.

    My current approach is as follows. I know how to draw a Thevenin circuit, but I'm not sure about how to work out RTh and VTh in this circuit.

    The Thevenin voltage must be that across the 150 Ohm resistor, which is 15V (?).

    The Thevenin resistance is the parallel sum of the 300Ohm and 150Ohm resistor with the series sum of the 50Ohm resistor. This works out to be 150 Ohms.

    Sorry about the somewhat unclear wordy solution I have given.

    According to my (wrong?) method VTh = 15V and RTh = 150 Ohms.

    Am I right? Thevenin's theorem, applied to certain situations I'm confused about, is impeding me significantly in moving onto other topics so I'd really appreciate someone to help me with this question.

    Attached Files:

  2. jcsd
  3. Apr 7, 2010 #2


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    Welcome to PhysicsForums!

    You're correct that the Thevenin Voltage needs to be the voltage across the 150 ohm resistor, however, you're incorrect about its value. Remember that the 150 and and 300 ohm resistors are in series with one another, and form a voltage divider.

    As for the Thevenin resistance, you have arrived at the correct value using the same method that I would've used.

    EDIT: In general, this approach (for the Thevenin Resistance) works most of the time. However, you can't go wrong using Vth/Isc--Thevenin Voltage divided by the short-circuit current (i.e. the amount of current that would be flowing in a 0-ohm wire you connect between points A and B)
  4. Apr 7, 2010 #3
    I'm somewhat confused by the fact that you've said that my method for working out the Thevenin resistance - adding the 300 Ohm and 150 Ohm resistors in parallel and then adding the equivalent resistance in series with the 50 Ohm resistor - is correct, but have also said that the 300 Ohm and 150 Ohm resistors are in series.

    I can't quite see why the 300 Ohm and 150 Ohm resistors are in series. According to the textbook I am using (An Introduction to Modern Electronics, Faissler) "components are in series when there are no branching nodes between them." In this circuit there is a node between both resistors.

    I'm sorry that my questions are so elementary. For some reason I've become very confused by electronics during my studies. Hopefully I'll get onto some more interesting stuff soon.
  5. Apr 7, 2010 #4

    George Jones

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    But when there is an open circuit (infinite resistance) between A and B, no current enters/leaves the node through the 50 Ohm resistor, i.e., the node doesn't "act like" a node.
  6. Apr 7, 2010 #5


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    They are in series because the 300 ohm resistor is attached to the 150 ohm resistor. While there is a branch (the 50 ohm resistor going to terminal A), no current flows through it because it is open circuited, and there is no return path back to the other side of the voltage source. That's also why you can disregard any voltage drop across this resistor in the calculation of the Thevenin resistance--there is no voltage drop across it, because there's no current going through it!
  7. Apr 7, 2010 #6
    So, using the voltage divider equation, it is found that the Thevenin voltage is 5V. Is this correct?

    Also, for the calculation of the Thevenin resistance, using your method we must disregard the 50 Ohm resistor and add the 150 Ohm and 300 Ohm resistor in parallel? I did this and got 100 Ohms, but before, using the method I mentioned initially, I got 150 Ohms and you said it was right. I'm a bit confused now. I was thinking that when the load resistance is added, the 50 Ohm resistor should be included, resulting in an extra 50 Ohms of resistance, added in series with the other two resistors.
    Last edited: Apr 7, 2010
  8. Apr 7, 2010 #7


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    You have to be careful here, those are two SEPARATE operations, and you'll need to redraw the circuits twice. When determining the Thevenin Resistance (with INDEPENDENT voltage/current sources), you can short circuit voltage sources, and open circuit current sources, and then figure out the resistance of the purely resistive network.

    Or, the slightly more involved approach is to (WITHOUT removing the current and voltage sources) figure out what the short-circuit current would be (again, the current that would flow through a piece of wire connecting terminal A with terminal B), and then divide the Thevenin Voltage by the Short Circuit Current. This should give you the same value for R_th that you determined using the approach above.
  9. Apr 8, 2010 #8
    Wow, I think I understand this solution now. Thanks a lot for your help.

    I've got a few other Thevenin's theorem questions that I'm a bit confused about that I might post later on.

  10. Apr 8, 2010 #9
    Here's another question, attached below.

    Initially I was very confused by this. The only reason I have a clue about how to do this is because I've seen a similar example before.

    What I have so far is as follows.

    After the 50 Ohm resistor is removed, the infinite resistance between A and B means that no current will flow in the direction of the 50 Ohm resistor's place, therefore the two 150 Ohm resistors can be considered to be in series and summed to get 300 Ohms.

    At the node after the 300 Ohm resistor which is in series with the voltage source, the current splits equally because the resistance on either side of the node is equal. The voltage across the 150 Ohm resistor, and therefore the Thevenin voltage, can therefore be found by the following calculation.

    VTh = 1/2Itot*150 Ohms = 1/2*(9V/300 Ohms)*150 Ohms = 2.25V

    The Thevenin resistance can be found by adding the 300 Ohm and 150 Ohm resistors in parallel and then adding the equivalent resistance in series with the 150 Ohm and 300 Ohm resistors.

    RTh = 550 Ohms?

    I'm not sure about whether this is the right approach. I think I'm getting there...

    Attached Files:

  11. Apr 8, 2010 #10


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    It's even simpler than that. Note that the 300 ohm resistor on the left and the 300 ohm "simplified resistor" on the right are in parallel... This reduces down to a simple two-resistor voltage divider!

    When determining R_th, use the first technique I posted in post #7--you'll discover that things simplify down a fair bit.

    When I was first taking the circuits course, everything seemed so arbitrary--there always seemed like there was some 'trick' for each particular circuit that only the prof knew. And then, I realized the secret, there is a trick, but it's the same trick over and over again: identifying which elements are in parallel.

    Okay, that and redrawing the circuit after each and every simplification. That took me a while to get used to (since I started out by ruler-drawing wires, although a friend of mine actually used a resistor/capacitor/inductor template), but it helps a whole lot more than trying to simplify, visualize the resulting circuit in your head, and identify the next simplification.

    My big tip: don't skimp on paper, but also don't skimp on neatness of your drawing (I mean, in terms of making things recognizable, not necessarily drawings rivaling what your prof prints out--don't waste tons of time, just make it neat and recognizable). If your university is anything like mine, you can usually find tons of free scrap paper in the recycle bins near the photocopiers / printers (often only imprinted on one side, and sometimes neither!)
  12. Apr 8, 2010 #11
    Thank you for your invaluable advice.

    I've been doing the problem above, for which I have the answers RTh and VTh values given. Using the method you have outlined, I get the correct answer for RTh, but double the correct answer for VTh when I use the potential divider equation! I encounter the same problem for other questions and I'm not entirely sure why. Attached below is my attempt at the aforementioned question.

    Sorry this is taking so long. I almost understand! :D

    Attached Files:

  13. Apr 8, 2010 #12


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    Problem is in Step 2--Vth is now over the composite 300 ohm resistor, not the original 150 ohm resistor! Formally, you should redraw this portion with the V_th* as a voltage source, and then find out the actual V_th. The quick and dirty approach (with can sometimes fail you depending on where you're actually measuring relative to), is to note that since the two resistors are the same, it's going to be half the value for the voltage drop across just one of them.
  14. Apr 8, 2010 #13
    I completely understand what I did wrong now. Thanks a lot. I think I'll be okay with Thevenin's theorem problems now =]

    Just one last thing. You say that "you should redraw this portion with the V_th* as a voltage source." I'm a little confused about what you mean by this. Do you mean that V_th* should be the new voltage across the composite 300 Ohm resistor?
  15. Apr 8, 2010 #14


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    That's an affirmative--V_th* as composite resistor (sorry, I added that part then got interrupted before adding the explanation, and forgot)! Next step: practice, practice, practice! Even when you approach more complex networks (and, judging by the examples, more complicated analysis techniques like nodal and loop analysis) redrawing and "massaging" circuits until they're simpler is always step one.
  16. Apr 8, 2010 #15
    Thanks a lot for resolving this issue and helping me to understand this concept. It seems like the textbook I'm using skips over the finer points and implications of Thevenin's theorem somewhat. The one example given is quite a bit easier than some of the questions, which doesn't help much.
    Last edited: Apr 8, 2010
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