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Thevenin's Theorem

  1. Oct 10, 2014 #1
    1. The problem statement, all variables and given/known data

    FIGURE 1 shows a 50 Ω load being fed from two voltage sources via
    their associated reactances. Determine the current i flowing in the load by:

    (a) applying Thévenin’stheorem
    (b) applying the superposition theorem
    (c) by transforming the two voltage sources and their associated
    reactances into current sources (and thus form a pair of Norton
    generators)
    upload_2014-10-10_17-15-52.png

    2. Relevant equations




    3. The attempt at a solution

    Could someone please help me get this started? The 'hand outs' I have been given are awful, and do not explain how to incorporate the reactance and p.f.

    If all of the loads were simply a resistance, I can see this being fairly easy. Although I could be wrong! Any help is appreciated.

    Many thanks
     
  2. jcsd
  3. Oct 10, 2014 #2

    gneill

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    Staff: Mentor

    Hint: Convert the load to a complex impedance. Then treat everything like resistors but use complex arithmetic.
     
  4. Oct 12, 2014 #3
    Thank you for your reply.

    Would this be using;

    ##Z = \sqrt{R^2+X^2}##

    When R = 50 Ohms
    And utilising the p.f. of 0.7 to get X?
     
    Last edited by a moderator: Dec 2, 2014
  5. Oct 12, 2014 #4

    gneill

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    Not quite. The 50 Ohms is the magnitude of the load impedance. The power factor tells you what the angle of the impedance is (for a complex number in magnitude / angle form). Remember that the power factor happens to be the cosine of the angle.
     
  6. Oct 12, 2014 #5
    Thanks again gneill,

    Cosine of 0.7 = 45.57

    arctan(51.01 / 50) = 45.57

    Therfore: Z = 50 + j51.01
     
  7. Oct 12, 2014 #6

    NascentOxygen

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    No. You are given Z=50 ohms, not R=50 ohms.
     
  8. Oct 12, 2014 #7

    gneill

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    With the angle and the magnitude you have the load impedance in polar form. Thus
    $$Z_L = 50 \Omega ~~\angle 45.57°$$
    You can convert this to rectangular form if you need it that way.

    By the way, before you dive into the calculations for the Thevenin equivalent, take a close look at the definitions of the voltage supplies. While they both have the same frequencies they won't have the same phase angle (why is that do you think?).
     
  9. Oct 12, 2014 #8
    In rectangular form, would it be written as;

    ##Z_L = 35 \Omega + j35.71##

    ?

    Then would I;

    Disregard the right portion of the circuit and and find the total impedance of the coil and load. Find the equivalent Voltage across the open circuit terminals (disregarding the j6 coil?)

    Really confusing! Apologies
     
    Last edited: Oct 12, 2014
  10. Oct 12, 2014 #9

    gneill

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    To apply Thevenin here you want to disregard the load for now and concern yourself with just the source network. That means finding the open circuit voltage and equivalent impedance of the source network.
     
  11. Oct 12, 2014 #10

    NascentOxygen

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    With the load removed, you are left with two equal-frequency voltage sources connected by a pair of inductances. Find the voltage at the junction of the inductors and that's going to be your Thévenin voltage.
     
  12. Oct 13, 2014 #11
    Hi guys - Sorry for the late reply, but I went back to the drawing board as I just wasn't understanding it. And to be honest - I'm still lost.

    From the notes I have:

    "First find the equivalent resistance. To do this we remove the load..." OK That bit makes sense.

    "Replace all sources with their internal resistance.." OK - Couldn't be simpler!

    "Next, find the Thevenin equivalent voltage, that is the voltage across the open-circuit terminals.." What open-circuit terminals?
     
  13. Oct 13, 2014 #12

    gneill

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    The open terminals that are where the load was connected before you removed it in a prior step.
     
  14. Oct 13, 2014 #13
    But, then the circuit would be in 2 parts, split down the middle.

    ##Edited##

    Oh dear. I believe I was taking it far too literally. Ignore the above statement!!

    Would the next step then be;

    To work out the current by adding the resistance in series. Then reconnecting the load, and work out the current flowing through the load, again in a series formation??
     
  15. Oct 13, 2014 #14

    NascentOxygen

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    The load stays right out of the picture while you are determining the Thévenin voltage.

    You are at the stage of trying to determine the voltage at the junction of the inductances.
     
  16. Oct 13, 2014 #15

    gneill

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    If you're concerned about how the circuit looks when the load is removed, redraw it so it looks more familiar:

    Fig1.gif

    Once again I encourage you to look at the specifications of V1 and V2. See post #7.
     
  17. Oct 14, 2014 #16
    Thanks again to you both for the above post's. It's been a great help so far.

    So, I believe I am at the point of calculating the equivalent impedance.

    Which I have calculated as;

    Z_t = (j4 * j6) / (j4 + j6)
    Z_t = -24 / j10
    Z_t = j2.4

    Then I need to find the equivalent voltage.

    So the current in the circuit is;

    I = (V1 - V2) / (j4 + j6)

    The terminal voltage is 'V2' plus the volt drop across the j6 coil.... Isn't it?

    ...Before I carry on, I'd like to make sure I'm on the right track!

    gneill - I have no idea why the phase angles would be different, and what the significance of that is. Completely dumbfounded thus far...*sigh*
     
  18. Oct 14, 2014 #17

    gneill

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    Take a close look at the definitions of those voltages. If you're going to use phasors to do the calculations you want to make sure that both of them are based on the same trig function, either sine or cosine. That or carry the full functions through the math and use trig identities to sort out adding or subtracting them. It's usually easier to just make them both either sines or cosines right from the start.
     
  19. Oct 15, 2014 #18
    OK.

    So cosine is phase shifted by 90°.

    V2 = √2 * 415 Cos(100π t - 90)
     
  20. Oct 15, 2014 #19

    NascentOxygen

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    I would have replaced sin by cos(wt+90°), since sine leads cosine by 90°

    EDIT: and I would have been wrong :oops:

    What you wrote is correct.
     
    Last edited: Oct 16, 2014
  21. Oct 19, 2014 #20
    Can anyone confirm how the current was achieved i.e.

    V1 (sqr2x415cos(100pi t) - V2 (sqr2x415cos(100pi t-90)/ j4 + j6

    as im not sure where to start concerning the voltages?
     
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