1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thevenin equivalent with dependent current source

  1. May 10, 2014 #1

    Maylis

    User Avatar
    Gold Member

    ImageUploadedByPhysics Forums1399715323.025351.jpg


    For this problem, I added in an external source, and found an expression for the external current in terms of ix. Then I divided Vex/iex to get the Rth. Then I did superposition to find the Vth. I had some troubles getting rid of the extra currents, particularly i3 and i2. I was wondering if this is right or if I am incorrect. I noticed there is a supermesh.

    ImageUploadedByPhysics Forums1399715345.225654.jpg
    ImageUploadedByPhysics Forums1399715360.934328.jpg
    ImageUploadedByPhysics Forums1399715382.008644.jpg
     
  2. jcsd
  3. May 10, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    Just how "dependent" is the current source? Vs is a fixed source. What determines ix? Can it ever change?
     
  4. May 10, 2014 #3

    Maylis

    User Avatar
    Gold Member

    I don't understand your hint. It is a dependent current source. My interpretation is that for a circuit with a dependent source, one must add a external source to find the external current. Then divide by the external voltage to get the thevenin resistance
     
  5. May 10, 2014 #4

    gneill

    User Avatar

    Staff: Mentor

    Can ix EVER vary due to any external stimulus? In what way is ix dependent on anything external to the circuit as shown?
     
  6. May 10, 2014 #5

    Maylis

    User Avatar
    Gold Member

    I'm still not understanding the hint. Why wouldn't ix change if I added an external source? I don't see it.
     
  7. May 10, 2014 #6

    Maylis

    User Avatar
    Gold Member

    ImageUploadedByPhysics Forums1399773870.138757.jpg

    ImageUploadedByPhysics Forums1399773917.938791.jpg

    Okay, a few revelations here.

    First, I want to show with the second image describing the external source. The textbook says that when dependent sources are in the circuit, one must employ an external source to find the equivalent circuit. I just want to prove that I am not crazy for using an external source method.

    The first image is doing what I would do for a circuit normally, without a dependent source.

    Thirdly, I finally think I am understanding your hint. If I use the external source method, I will short out of where the voltage supply Vs previously existed, and ix = 0, no matter what the external source is.

    part (b) asks what value I would put a load resistor at terminals (a,b) to get maximum power transfer. That is just at Rth, R1 + R3, right?
     
    Last edited: May 10, 2014
  8. May 11, 2014 #7

    gneill

    User Avatar

    Staff: Mentor

    The external source method works anytime, there doesn't have to be a dependent source.

    On the other hand, the external source method is not required if the dependent source's value is set by fixed sources inside the circuit (as is the case in this problem). A "dependent" source that cannot vary is identical to a fixed source.

    Except that I don't see a valid calculation for Vth in the first image.

    Sure. Refer to the Maximum Power Transfer theorem.
     
  9. May 11, 2014 #8

    Maylis

    User Avatar
    Gold Member

    All I did for the thevenin voltage was turn off both sources and did superposition. So I did it incorrectly?
     
  10. May 11, 2014 #9

    gneill

    User Avatar

    Staff: Mentor

    The problem occurred when you suppressed the voltage source which supplies the stimulus to the dependent current source. Effectively you suppressed both sources at the same time.

    Two possible approaches come to mind. First, use superposition but replace the "3ix" for the current source with its value it normally has due to Vs driving ix. Second, leave both sources active and use nodal analysis.
     
  11. May 12, 2014 #10

    Maylis

    User Avatar
    Gold Member

    So I'm allowed to deactivate the voltage source (which inadvertently deactivates the current source) to find the thevenin resistance, but I am not allowed to do the same to find the thevenin voltage?
     
  12. May 12, 2014 #11

    gneill

    User Avatar

    Staff: Mentor

    The correct concept is that to apply superposition you need to suppress all but one source at a time. In this circuit if you suppress Vs you also suppress the current source by proxy. That's not good. What you need to do is have the current source behave as if Vs were still there, which you can do because for this circuit Vs is fixed and causes a fixed "sampling" current through R4, thus making the current source "fixed" also. Note that this will not be the case in general! It's a quirk of this particular circuit.

    To find the Thevenin resistance you want to suppress all sources at once, so it's not a problem here.
     
  13. May 12, 2014 #12

    Maylis

    User Avatar
    Gold Member

    Here is my next attempt. This time, I employed the short circuit technique, creating a short between terminals (a,b) in order to calculate the short circuit current. I noticed that when I shorted the terminals, I had a supermesh and I wrote my KVL equations and did some substitutions to find Vth. Hopefully this time it is right !!

    ImageUploadedByPhysics Forums1399924417.047809.jpg

    ImageUploadedByPhysics Forums1399924431.781266.jpg
     
  14. May 12, 2014 #13

    gneill

    User Avatar

    Staff: Mentor

    Your supermesh equation doesn't look right. Surely i2 is flowing in the same loop direction through R4 as it is through R3. Why would their terms have different signs?

    But with a little thought you should be able to simplify your analysis greatly. You know that Vs fixes the current through R4 and thus fixes ix. You can do away with "ix" as a variable immediately by using this relationship.

    You can also forget about the third loop.... You know that you can switch the order of parallel components (it doesn't matter how they're depicted graphically as long as their connections remain the same). So swap the positions of Vs and R4 on your diagram. Now your "i3" becomes identical to ix and it's already solved for. Helpful hint: a voltage source that borders two loops effectively isolates them (makes the effects of their currents independent of each other). No matter what happens in either loop, the shared KVL term is fixed for all time by that voltage source.

    Another hint: Find the open circuit Thevenin voltage using nodal analysis. There's only one node since the voltage source makes the junction of R3, R4, and Vs into a supernode with the bottom rail.
     
  15. May 12, 2014 #14

    Maylis

    User Avatar
    Gold Member

    Is this what you mean? I don't understand the term bottom rail
    ImageUploadedByPhysics Forums1399926704.728315.jpg
     
  16. May 12, 2014 #15

    gneill

    User Avatar

    Staff: Mentor

    Getting there. But you haven't written the node equation properly. Look here:

    attachment.php?attachmentid=69726&stc=1&d=1399929181.gif

    There are two nodes (labeled V1 and Vs in red). But Vs is already solved (it's fixed by the voltage source). So only V1 is left to solve for. Note that ix has been done away with as a variable. It's not required if it's a fixed value depending on the fixed voltage source. Now, write the node equation for node V1 and solve for V1.
     

    Attached Files:

  17. May 12, 2014 #16

    Maylis

    User Avatar
    Gold Member

    ImageUploadedByPhysics Forums1399930544.582610.jpg

    I am hesitant to write a nodal equation at the node with the current source. I thought you weren't allowed to do that. There is a voltage drop across the current source that cannot be calculated??
     
  18. May 12, 2014 #17

    Maylis

    User Avatar
    Gold Member

    This is another way I may be interpreting what you are saying
    ImageUploadedByPhysics Forums1399930765.464696.jpg
     
  19. May 12, 2014 #18

    Maylis

    User Avatar
    Gold Member

    Sorry I am having a brain meltdown the final exam is tomorrow. Is this what you meant?

    ImageUploadedByPhysics Forums1399930948.588261.jpg
     
  20. May 12, 2014 #19

    gneill

    User Avatar

    Staff: Mentor

    No, that's incorrect. A node equation is just KCL, a sum of currents at a node. A current source makes for a trivial term in the equation! Current sources go very well indeed with nodal analysis.

    Also note that R2 plays no role whatsoever in the nodal equation: the current source sets the current in the branch no matter what value R2 has. Forget R2.
     
  21. May 12, 2014 #20

    gneill

    User Avatar

    Staff: Mentor

    Yes! Much better :approve:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted