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Confused about 2 Equivalent circuits

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    lel.png


    2. Relevant equations
    There's not much to write here except parallel resistors equation. This is supposed to be a superposition problem. How did he use a source transformation because I have no idea why and where he did the equivalent resistor and why did a current source turn into a voltage source...? Especially the placement of that voltage source. What happened?

    3. The attempt at a solution

    Don't say stuff like "figure it out yourself" or give me hints because I have NO clue what to do. I don't care about the answer since it's given , I want to UNDERSTAND what is happening.

    Thanks.
     
  2. jcsd
  3. Oct 5, 2014 #2

    phinds

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    He replaced the current source and 4 ohm resistors with their Thevinin equivalent. Do you understand how to do that?
     
  4. Oct 5, 2014 #3
    No I do not. I tried to read my book and online tutorials but it does not make sense.. Especially in this case. Do you mind offering an explanation please?
     
  5. Oct 5, 2014 #4

    phinds

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    If you do not understand Thevenin equivalent circuits then this problem is too advanced for you. First learn the basics of Thevenin and Norton equivalent circuits before you tackle something that uses them. It's like you are trying to solve a math problem with algebra when it requires calculus and you want me to teach you calculus.
     
  6. Oct 5, 2014 #5
    I know what Thevenin's and Norton's theorem says but I don't really know it can be applied. For example , in this case , where is R_load and where are we short-circuiting to find Vth and Rth? From there , I know Thevenin's circuit can be built be a more precise question would be which elements are used in the transformation. Thanks.
     
  7. Oct 5, 2014 #6

    phinds

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    Did the current source have a value in the original problem statement? Without it, there is no numeric solution to the problem as stated.

    You say you understand equivalent circuits, so apply that knowledge as I suggested in post #2
     
  8. Oct 5, 2014 #7
    Current source is 1A .I don't know what to do in this case. To me , it looks like a source transformation but since the 4 ohms is not in series with the CS then , it is not possible.

    Edit: Nevermind , it does seem like the resistor is in parallel with the CS , then why is Thevenin's theorem relevant in this problem?
     
    Last edited: Oct 5, 2014
  9. Oct 5, 2014 #8

    phinds

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    I say again, go study equivalent circuits until you understand them before you try to take on a problem that uses them.
     
  10. Oct 5, 2014 #9
    With all due respect, studying is one way to learn but it doesn't hurt to simply explain instead of pointing me to go back to the books. I understand the concept and you saying to not do this problem is basically saying go fail your exam that's next week because you are not smart enough to read from a book. This is a superposition problem. Try solving that without using Thevenin's theorem because we didn't learn that yet.
     
  11. Oct 5, 2014 #10

    phinds

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    If the problem is to be solved without equivalent circuits, why is the equivalent circuit even shown in the first place?

    In any case, if you want to solve the problem without equivalent circuits, it is a trivial current loop problem. Just write the equations for the current in the three loops of the original circuit and solve for the current through the 3 ohm resistor and you'll get Vo2
     
  12. Oct 5, 2014 #11
    I just tried doing mesh analysis and Vo2 gives me 2V? The current in the lower right loop is (2/3)A.
     
  13. Oct 5, 2014 #12

    phinds

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    show your work if you want help.
     
  14. Oct 6, 2014 #13

    gneill

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    The problem statement was an image only (strictly, not something that's generally allowed); no text of the actual problem was included so it was not clear what was to be solved or if there were restrictions on the methods to be used in reaching an unspecified goal. As Phinds points out, the image provided showed a source transformation (turning a Norton source into its Thevenin source equivalent). So it seems to be a permitted technique for addressing whatever the actual problem is.

    We cannot read minds here. If the problem description that you provided is lacking information on what is to be done or found or solved, that is not the fault of the volunteers who offer to help you. It shouldn't be nine posts into a thread where it is discovered that it is a superposition exercise or that there are restrictions on techniques that can be used.

    Regarding the source transformation that was performed: The basic form of the Norton Equivalent is a current source in parallel with a resistance. The basic form of the Thevenin Equivalent is a voltage source in series with a resistance. These simple forms are easily changed from one to the other and perform identically as far as the greater circuitry around them is concerned. Simply draw an example of one or the other by itself and apply the usual methods to find the desired 'other' model. You'll find that it is a very simple operation. The source transformation technique can be quite useful in circuit analyses and simplifications.
     
  15. Oct 6, 2014 #14
    lol23.png

    First part:

    When the current sources are shut OFF , I find that Vo is 10V (correct).

    Second Part: (Shutting off 2A CS and 20V VS)
    This is the part that we were discussing. From what gneill told me ,
    Since the current source is 1A and the resistor connected in parallel with it is 4 ohms , It is turned into its Thevenin equivalent circuit resulting in Vth= I_No x R , Where I_No is 1A and R is 4 ohms. Hence, Vth is 4V and is connected in series with a 4 ohms resistor and a 2 ohms resistor ( middle branch transformed circuit in post #1). I do not know how to simplify the circuit from there on... My guess is to add the resistors in series and do a source transformation from VS to CS.

    Part 3: (Shutting off VS 20V and CS 1A)
    This results in 6 ohms (top branch) being in parallel with resistors 4 ohms and 2 ohms which are themselves in series with each other so adding the resistors in parallel simplifies to [6*(4+2) ]/ [12] = 3 ohms. I guess the other part is just a matter of mesh or nodal analysis ? (Correct me if I'm wrong).
     
  16. Oct 6, 2014 #15

    gneill

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    Yes, that is certainly one way to go. You'll then find another opportunity to combine resistors to simplify things. Alternatively, apply nodal analysis at this point; there's only the one essential node and it conveniently has vo as its potential. Presumably since the objective is to use superposition, once you select a single source and suppress the others you are free to apply any techniques that you know in order to find that source's contribution.
    Fig1.gif
    Part 3: (Shutting off VS 20V and CS 1A)
    This results in 6 ohms (top branch) being in parallel with resistors 4 ohms and 2 ohms which are themselves in series with each other so adding the resistors in parallel simplifies to [6*(4+2) ]/ [12] = 3 ohms. I guess the other part is just a matter of mesh or nodal analysis ? (Correct me if I'm wrong).[/QUOTE]
    Fig2.gif
    Plenty of choices there. Certainly nodal analysis will work. Since everything is in parallel you could do current division, or convert the whole thing to its Thevenin equivalent.
     
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