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Circuits - Variables dependent on R?

  1. Feb 5, 2013 #1
    The circuit of interest is attached.
    The question is to solve for I1, I2, I3, I4, and V_diode.

    My question is will the variables come out to be numbers or dependent on R?


    This is how I did it and it depends on R:
    I3=0 b/c of the open.
    The branch with the open is pretty much not there.
    Thus,
    KCL: I1+I2=I4
    KVL gives 2 equations:
    1-2R*I1+R*I2=0
    -R*I2-R*I4+1=0

    So I guess I1, I2, I4, and V_diode will not solve to be numbers and will depend on R? Or did I do this all wrong?
     

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    Last edited: Feb 5, 2013
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  3. Feb 5, 2013 #2

    rude man

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    How is the diode hooked up? Cathode to + or the other way?
     
  4. Feb 5, 2013 #3
    The diode is hooked up so with the way I_3 is drawn, it would cause the 'ideal' diode to become an open. I'm not sure how to describe that in terms of the cathode.
     
  5. Feb 5, 2013 #4

    rude man

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    So i3 can only be zero or negative?
     
  6. Feb 5, 2013 #5
    I_3 is 0 because that branch of the circuit technically doesn't exist.
     
  7. Feb 5, 2013 #6

    rude man

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    Then why does the question ask for I3? Are you sure you have stated the question accurately?
     
  8. Feb 5, 2013 #7
    I don't see why it can't ask for I3.
    I_3 is simply 0. That's the answer for I3, only 1 of the 5 variables to be solved.
    I still need to solve for I1, I2, I4, v_diode.
     
  9. Feb 5, 2013 #8

    rude man

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    OK, so the first thing is to simplify the circuit by removing irrelevant components.
     
  10. Feb 5, 2013 #9

    rude man

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    Do you know the superposition theorem?
     
  11. Feb 5, 2013 #10
    Yes, but everything in class and tests will be done from using KVL and KCL. So those are the things I would prefer to use.

    Wouldn't superposition give me the same answer as my original post?


    I attached a new drawing with the resistor removed. I kept the V_diode in there though.
     

    Attached Files:

  12. Feb 5, 2013 #11

    rude man

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    Good first step. Now, looking at the original schematic, can you identify one of your new nodes with Vdiode(-)? Let us call the node Vdiode(+) = zero volts, shall we.
     
  13. Feb 5, 2013 #12

    The node would be the wire between the 3 resistors.

    But I still think you answer will not solve to be a number and will still depend on R.
     
  14. Feb 5, 2013 #13

    rude man

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    You are correct on both counts. How could the answer not depend on R?
     
  15. Feb 5, 2013 #14
    What I mean by that is if I1, I2, I4, and Vdiode will come out as actual numbers (i.e R ends up canceling out somewhere in the equation and you're left with actual numbers) or if it will come out as involving the variable 'R.'

    i.e Vdiode maybe something like Vdiode=2R instead of something like 10V
     
    Last edited: Feb 5, 2013
  16. Feb 6, 2013 #15

    NascentOxygen

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    The equations for the currents will all involve variable R.
     
  17. Feb 6, 2013 #16

    rude man

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    Not possible. You can't get around I = V/R. V is a number so I must involve R.

    What level are you at in your studies?
     
  18. Feb 6, 2013 #17
    3rd year undergrad. Why?
     
  19. Feb 6, 2013 #18

    rude man

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    I ask that question a lot to determine how to respod to questions. Sometimes I have to revert to basics, other times I can assume more advanced knowledge.

    Are you in an engineering or science curriculum?
     
  20. Feb 6, 2013 #19
    Yes, my major is mechanical and aerospace engineering. I've taken quite a bit of physics and circuits courses when I was in high school and during my freshman year, but also forgot quite a bit. Currently, I'm taking a mechatronics course, which is where this problem came from.

    I just solved this system and it was much easier than I thought it would be.
     
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