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Electrical Engineering - Thevenin Equivalent

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data
    (a) For the circuit above find the physical parameters of the Thevenin equivalent source with respect to terminals “a” and “b”. Assume that the resistances and reactances of the circuit are given at signal frequency 1 MHz.
    (b) Find the impedance of the matched load for that source.
    (c) Assuming the signal frequency of 1 MHz, find the parameters of the series R- C or R- L physical equivalent circuit of the matched load.
    (d) Generate a plot showing frequency dependences of the active and reactive power delivered into the matched load in the frequency range 100 kHz – 10 MHz.


    2. Relevant equations



    3. The attempt at a solution
    I mostly just have a general question about the problem I am working on, not necessarily looking for a specific answer. For parts (a) and (b) of the problem I am not sure if I am supposed to do one of two things. For part (a) I think I just need to use the voltage divider equation to solve it. So:

    Eth = (L1*E)/(L1+R1). I think this is correct?

    For part (b) I am not sure if I am supposed to be finding the Zth by solving for the equivalence of just R1 and L1 or if I need to find the equivalence of R1, L1, C1, and R2. Any help would be much appreciated! Thanks.

    Note: R1 is 5ohm resistor and R2 is 2ohm resistor.
     

    Attached Files:

  2. jcsd
  3. Apr 17, 2012 #2

    gneill

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    Hi justbaum30, Welcome to Physics Forums.

    Not quite. You need to include the effects of the capacitor and R2 in the Thevenin equivalent, since they are part of the network "behind" terminals a and b.

    Note though that you can find the Thevenin equivalent in stages. It's okay to first find the Thevenin equivalent for the subcircuit comprising E, R1, and L. That should make it easier to then "swallow" the capacitor and R2 into a further Thevenin model.
    For part (b) you will make use of the Thevenin impedance for the network that you found in part (a), and then use the definition of 'matched load'.
     
  4. Apr 17, 2012 #3
    Hmm, I am somewhat confused then. First to find Zth I must:

    Zth = (R2+C1) || (L1) || (R1)

    I got this to equal, in phasor form, 18.42 < -172.38. Then isn't this the answer for (b) or am i misunderstanding the definition of "matched load"? As for (a) can I simply do:

    Eth = (((R2+C1)||(L1))*E)/Zth?

    I hope this all makes sense. And for a quick clarification, (c) is asking me to redraw the circuit diagram using a combination of a resistor and capacitor/inductor from the answer of (b) correct?
     
  5. Apr 17, 2012 #4

    gneill

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    Note that R2 is not in series with C. Looking from a-b into the circuit, R2 actually parallels what lies beyond it. Start with R1 and L, then work your way "forward" through the circuit, combining the impedances appropriately as you go.
    A matched load for impedance is similar to, but not exactly like that for pure resistance. Take a look at the wikipedia entry for impedance matching for a clue.
    Again, I think you're getting tripped up by how to combine the impedances. Note that the Thevenin voltage will be whatever voltage appears across R2.
    It's asking you to draw an equivalent circuit for whatever you find to be the matching load. First you need to sort out how to determine that matching load as explained above. As a hint, you might find it easier to determine that from the complex number form for that impedance.
     
  6. Apr 17, 2012 #5
    So are you saying to take E, R1, and L as a circuit, find the Eth and Zth of that, and then treat Eth, Zth, C, and R2 as a circuit and find the Eth2 and Zth2 of that? Then you would be able to use this second circuit to find the voltage across R2 using the voltage divider rule? Hopefully this makes sense.
     
  7. Apr 17, 2012 #6

    gneill

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    Yes, that would work admirably :smile:
     
  8. Apr 17, 2012 #7
    Haha, thank you very much. I will try this method and post back, hopefully understood!
     
  9. Apr 17, 2012 #8
    In the second circuit though wouldn't two of the three structures (Zth, C, and R2) have to be in series? I don't understand how all three could be in parallel?
     
  10. Apr 17, 2012 #9

    gneill

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    R2 cannot be in series with anything from the perspective of a-b; It is wired in parallel with (across) a-b.

    Sketch your circuit with the 'first' Zth in place of R1 and L. Now place your hand over R2. What do you see? Reduce it.

    Now sketch your circuit with this 'second' Zth in place of R1, L, and C. What is the relationship between R2 and this Zth?
     
  11. Apr 17, 2012 #10
    I am pretty sure it is supposed to be:

    Zth2 = (Zth1+C) || R2?
     
  12. Apr 17, 2012 #11

    gneill

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    Looks reasonable.
     
  13. Apr 17, 2012 #12
    Okay, I am hoping I did this correctly. I ended up getting Eth2 to equal 6.33<39.62 for part (a) and got Zth2 to equal 1.48<5.85 or 1.84 + 0.19j.

    Zth1 = (L)(R1)/(R1+L) = 5+6j
    Eth1 = (L)(E)/(R1+L) = 23.05<89.81
    Zth2 = (Zth1+C) || R2 = 1.84+0.19j
    Eth2 = (R2)(Eth1)/(Zth1+R2+C) = 6.33<39.62

    For part (b) my understanding was to take the conjugate complex so the answer would be 1.84-0.19j

    For part (c) then:
    R = 1.84ohm
    C = 1/(2*pi*10^6*0.19) = 8.37E-7 F

    I think this is right, at least I sure hope it is. Haha. Thanks again.
     
  14. Apr 17, 2012 #13

    gneill

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    The rest will be thrown off by the above Zth1 problem.
     
  15. Apr 17, 2012 #14
    Sigh, what a silly mistake. Here we go again:

    Zth1 = (L)(R1)/(R1+L) = 2.95+2.46j
    Eth1 = (L)(E)/(R1+L) = 23.05<89.81
    Zth2 = (Zth1+C) || R2 = 1.27-0.23j
    Eth2 = (R2)(Eth1)/(Zth1+R2+C) = 8.90<107.09

    Part (a) is Eth2

    For part (b) my understanding was to take the conjugate complex so the answer would be 1.27+0.23j

    For part (c) then:
    R = 1.27ohm
    L = 0.23/(2*pi*10^6) = 3.66E-8
     
  16. Apr 17, 2012 #15

    gneill

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    The method looks good, and the values are okay, although, you might want to carry a couple more digits of precision in intermediate values. By all means round results for presentation, but keep more digits in intermediate results to prevent rounding errors from creeping into the significant figures.
     
  17. Apr 18, 2012 #16
    Thank you very much for your help! It is very much appreciated. Have a good one
     
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