Please I need your help with this thevenin problem !!!!!????????

Discussion in 'Homework Help' started by ujnikm, Dec 28, 2012.

  1. ujnikm

    Thread Starter New Member

    Nov 15, 2012
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    Hello everybody,, I wanna your help aggressively:confused::confused:
    regarding this problem I managed to get the i from this equation(KVL):
    20 = 6*i - 2*i + 6*i
    i= 2 A.
    and Vth=i*6=12 V
    so far I don't have a problem ,
    but I wanna get Isc between A&B but I don't know how please help me :confused::confused::confused: !!!????



    [​IMG]
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Suppose you have a 6 ohm resistor in parallel with a 10 ohm resistor, and you know that the total current through the parallel combination is It, do you know how to use the current divider rule to find the current through the 6 ohm resistor as a fraction of the total current?
     
  3. ujnikm

    Thread Starter New Member

    Nov 15, 2012
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    do u mean the existing 6 ohm ???
     
  4. The Electrician

    AAC Fanatic!

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    It doesn't matter. If you want to consider the existing 6 ohm resistor, then do so.
     
  5. WBahn

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    Mar 31, 2012
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    Please provide a sketch of the circuit you are trying to analyze to get the Isc. It will really help if we know we are on the same page to start with.
     
  6. ujnikm

    Thread Starter New Member

    Nov 15, 2012
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    What I have understood is AFTER putting the sort circuit between A & B

    the 2A current would flow through the 2 parallel resistances 6 and 10

    using Current divider:
    the current through the 10 Ω and flow through the short circuit= ((10.6/16)/10).2=12/16 A ,,,,,,,,,,right ???????:confused::confused:
     
    Last edited: Dec 29, 2012
  7. ujnikm

    Thread Starter New Member

    Nov 15, 2012
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    it's the same on the post but only connect a short circuit between the 2 terminals and Isc flow through it
     
  8. ujnikm

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    Nov 15, 2012
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    I have a question when i connect a short circuit ,,,there will be an additional resistance (10Ω) how could We consider the same current would flow through the combination of (6 and 10 Ω) ...shouldn't the current to be decreased after connecting the short circuit ???!!!!!
     
  9. WBahn

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    Why do you think that there will still be a total of 2A flowing? You have changed the circuit, therefore your prior analysis, which only applied to the circuit that was analyzed, is no longer valid.

    That' why it's usually a good idea, particularly until you get good at it, to always draw the circuit you are actually analyzing and, any time you make a change, to start with a new sketch and a new analysis.

    I have no idea where this equation is coming from.

    PLEASE track your units!

    And always ask if the answer makes sense!

    Let's assume that there is a total of 2A flowing into the parallel combination of the 10Ω and 6Ω resistors. If the current were to split eveningly there would be 1A in each branch. But we know it won't split evenly -- one branch will have more than 1A and the other will have less than 1A. Your answer has more than 1A flowing in the 10Ω resistor, meaning that less than 1A is flowing in the 6Ω resistor. Does this make sense?
     
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  10. ujnikm

    Thread Starter New Member

    Nov 15, 2012
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    thanks for your effort,,,,but i understood u i have to sketch a new circuit and get the new current by getting the Rt and get It=Vt/Rt then use the Current divider to get the current through the short circuit,,,,does it make a sense ????!!!
     
  11. WBahn

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    If there were only constant voltage (and/or current) sources than this would be fine. But you have a dependent voltage source and so you can't just find some Rt that easily.

    We need to get a feel for what techniques you have been exposed to. Do you know how to analyze circuits using any of the following methods: branch current, loop voltage, Node Voltage Analysis, Mesh Current Analysis? The first two are just brute force application of KCL and KVL, respectively, and the latter two are just more systematic and elegant formulations of the first two.
     
  12. ujnikm

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    Nov 15, 2012
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    I know the methods you mentioned but don't know how to use any of them while having dependent source
     
  13. WBahn

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    Okay, so let's just pick one and see where it leads.

    Let's use Mesh Current Analysis. For starters, forget that the dependent voltage source is a dependent voltage source. Instead, assume that it is just a voltage source having a voltage of Vd. You have two meshes. What are the mesh equations? Get that far and we will make sure there aren't any mistakes to that point and then take it from there.
     
    Last edited: Dec 30, 2012
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  14. ujnikm

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    Nov 15, 2012
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    for the right mesh >>20V+Vd=I1(6Ω+6Ω)

    for the left one there is no any voltage source(v=0)
    0=Isc(10Ω)+6Ω(Isc+I1)

    does that make sense ????
     
  15. WBahn

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    Well, this isn't quite Mesh Current Analysis, it is basically branch current analysis, but that's fine. Let's go with that. You need to define the direction of Isc since it isn't given on the diagram. From your second equation, it wouild appeat that the I1 is the loop current going clockwise in the left loop and Isc is the loop current going counterclockwise in the right loop. That's fine, but be sure that you are careful when it comes time to finding Rth since the passive sign convention is being violated -- which is fine, as long as you are careful.

    Now, in your first equation, you are assuming that whatever current flows in the left 6Ω resistor (which you have called I1) is the same current that is flowing down through the right 6Ω resistor. Is that valid? What about Isc (the loop current in the right loop)?

    So patch that up (overall, you are close) and post updated equations. Also, if I'm made bad assumptions about what you means I1 and Isc to be, please clarify.
     
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  16. ujnikm

    Thread Starter New Member

    Nov 15, 2012
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    Actually,,,,you are very respectful man.,,,respect that I have never seen before....And I wonder How you endure my silly question.....Great Thanks
    -Your assumption is totaly correct.
    My updates are :
    -For mesh#1 20V+Vd=I1(6Ω)+(I1-Isc)(6Ω)
    where I1 is the current of Left mesh

    -For mesh#2 0=Isc(10Ω)+6Ω(Isc-I1)
     
  17. sleet1986

    New Member

    Dec 29, 2012
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    you are right now
     
  18. WBahn

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    I can endure them because they aren't silly and you are making an obvious effort to learn. I'll bend over backwards to help a willing student or colleague learn whatever I can pass along to them. I have several colleagues that are willing to put up with my "silly" questions, too -- and some of them are the same ones that I teach other things to.

    Okay, you appear to have reversed the direction of Isc. If you aren't going to provide a labeled diagram and, instead, rely on a verbal description, it is important that you clearly define your variables.

    I1: CW current in the left mesh
    Isc: CW current in the right mesh

    Left mesh: 20V+Vd=I1(6Ω)+(I1-Isc)(6Ω)
    Right mesh: 0=Isc(10Ω)+6Ω(Isc-I1)

    Notice that you have two equations and three unknowns, namely Vd, I1, and Isc. So you need a third equation (that doesn't introduce any additional unknowns).

    Can you see how you can write Vd in terms of the other unknowns? Hint: Technically, this requires two steps in which you write Vd in terms of an additional unknown (giving you three equations and four unknowns) but then writing the new unknown in terms of the existing unknowns, giving you four equations and four unknowns. But, these additional steps can be done in your head pretty easily -- but if you don't see them, do it explicitly.
     
  19. ujnikm

    Thread Starter New Member

    Nov 15, 2012
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    isn't Vd=2i =4 V
    so we have 2 unknown only with 2 equations
    Does that make sense ??
     
  20. WBahn

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    Only if i=2A, but there is no basis for making such a claim. Remember, that result is for a different circuit!

    If i=2A, then the voltage across that 6Ω resistor (which is also the voltage across the 10Ω resistor) would be 12V, making the current in the 10Ω resistor 1.2A. Thus the current in the other 6Ω resistor would be 3.2A which would result in it dropping 19.2V. So you would have 24V of gain due to the supplies and 31.2V of drop due to the resistors, violating KVL.
     
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