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Circuit network with multiple ports

  1. Oct 21, 2011 #1
    Find V2/V1 for the network below:
    2prwbki.png

    What I did was try to solve by reducing to thevenin equivalents for V1 and V2 separately
    For V1:
    I got Vth = V1
    Zth1 = 41/15
    V1 = I1*(41/15)

    For V2:
    I got Vth = V2
    ...but I am having trouble trying to find the Zeq2:
    2dikap3.png

    I know once I find that V2 = I2*Zeq2
    and i also need help with relating I1 to I2..

    thanks in advance
     
    Last edited: Oct 22, 2011
  2. jcsd
  3. Oct 21, 2011 #2

    gneill

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    I think you'll find it better to just find the Thevenin equivalent for the circuit looking into the V2 port. That is, find the open circuit voltage there due to V1 and the Thevenin resistance looking into that port. That will reduce the circuit to a single voltage source (which will be V1 scaled by some factor) and a single series resistance. Then you can use these to determine the output voltage V2 across the load.

    You can perform a series of Thevenin transformations, working your way across the circuit until you've "consumed" all of the circuit between V1 and the load.

    An alternative is to use mesh analysis to solve for the current in the loop containing the load.
     
  4. Oct 22, 2011 #3
    I am attempting to do the problem in the first way you described by looking into the V2 port. I know if I do that I'll get something like V2 = (some resistance)*V1...but my problem is how to get that resistance. I'm not sure what's in series/parallel with each other

    i know V2/V1 should *supposedly* be 1/41 but i am not getting that answer

    when i try to solve for equivalent resistance looking into V2 i get 15/11
     
    Last edited: Oct 22, 2011
  5. Oct 22, 2011 #4

    gneill

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    You can proceed stepwise across the circuit from V1, creating a series of Thevenin equivalents as you go. As you go you'll take up the resistances in bite-size portions that are easy to digest. Here's one such suggested set of 'slices', the first one broken out for inspection. Hint: Be careful what you do with that first 1 Ohm resistor in parallel with the voltage source when you are calculating Rth1.

    attachment.php?attachmentid=40220&stc=1&d=1319285547.gif
     

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  6. Oct 22, 2011 #5
    in the last step, i get an equivalent resistance of 11/4 in parallel with the 1ohm resistor.
     
    Last edited: Oct 22, 2011
  7. Oct 22, 2011 #6

    gneill

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    Can you show your work step by step? Let's begin with the very first "slice". What do you get for the Thevenin voltage and resistance?
     
  8. Oct 22, 2011 #7
    Starting from the left...

    For my first slice, I get 3 (1 in series with 2)
    Combining this with slice 2, i get 11/4 ((3 || 1) + 2)
    Combining this with the last slice, I am getting 41/15 ((11/4 || 1) + 2)
     
  9. Oct 22, 2011 #8

    gneill

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    For the first slice, the 1Ohm resistor is in parallel with the voltage source. When you short out the voltage source to find the Thevenin resistance, that resistor is also shorted out (remember I said to be careful with that one!). So for the first slice the Thevenin resistance is just 2 Ohms, and the Thevenin voltage is V1.
     
  10. Oct 22, 2011 #9
    ok. now i got 10/3 as the equivalent resistance
     
  11. Oct 22, 2011 #10

    gneill

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    Is that your Thevenin resistance for the whole network (up to but not including the load resistor)? If so, that's not what I calculate it to be. I think you'll have to show your work.
     
  12. Oct 22, 2011 #11
    1) i have the 2 ohm resistor in series with 1 || 2 = 8/3
    2) i have the 8/3 in series in 1 || 2 = 10/3
     
  13. Oct 22, 2011 #12

    gneill

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    I think you're mixing up what's in series and what's in parallel.

    Presuming that you're calculating the Thevenin resistance for the second "slice", then the circuit you're looking at should look something like this:

    attachment.php?attachmentid=40250&stc=1&d=1319343156.gif

    When you short out Vth1, then Rth1 will be in parallel with the 1Ω resistor, and then the 2Ω resistor will be in series with them. A similar situation will hold for the next "slice".
     

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  14. Oct 22, 2011 #13
    series/parallel is my enemy.

    so..
    1) (2||1) + 2 = 8/3
    2) (8/3 || 1) + 2 = 30/11
     
  15. Oct 22, 2011 #14

    gneill

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    :smile:
    That looks better!
     
  16. Oct 22, 2011 #15
    ok..so after that i get
    V2 = 1/(1+(30/11))V1
    which gives V2/V1 = 11/41.. this is what i had originally :\
     
  17. Oct 22, 2011 #16

    gneill

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    What did you get for your final Thevenin voltage? (it's not V1!). You should calculate a new Thevenin voltage for every "slice" as you move across the circuit. So you have a new Thevenin resistance value and a new Thevenin voltage at the completion of each "slice".
     
  18. Oct 22, 2011 #17
    *le sigh* i did not do that
     
    Last edited: Oct 22, 2011
  19. Oct 22, 2011 #18

    gneill

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    "Alas" or "at last"? The result looks okay to me :smile:
     
  20. Oct 23, 2011 #19
    is this correct for the ratio of the current (I2/I1)
    V2 = -I2
    I1 = 56V2
    I2/I1 = -V2/(56/V2) = -1/56 ??
     
  21. Oct 23, 2011 #20

    gneill

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    Yes, that looks okay for the current ratio.
     
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