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Thevenin Theorem (where does Z Thevenin fit in?)

  1. Jul 25, 2016 #1
    1. The problem statement, all variables and given/known data

    (a) Calculate the load current using Thevenin's Theorem
    (b) Calculate the load current using Superposition


    2. Relevant equations

    3. The attempt at a solution

    There have already been a couple of historical posts of this question but those threads don't give me any clues to answer my question, and they're a bit disjointed.

    I have calculated the Thevenin equivalent voltage (with load disconnected), crude calculations:

    V1 + V2 = 415 - j 415 = 586.9 Volts
    Current though the inductors is 586.9 / j 10 = 58.6 amps
    VTh = V1 - Vj4 = 299

    Therefore, current through RL = 299 / (35 + j 35.707) = 5.98 A (or 4.18 + j 4.27)
    ZTh = (j 4 + j 6) / j10 = j 2.4

    So far so good? My problem is making the current derived from using Superposition match with that above. I shall wait to see if I have done this bit right first.

  2. jcsd
  3. Jul 25, 2016 #2


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    V1=√2*415*cos(ωt). This means peak value of V1 is √2*415 V. While taking the rectangular form of V1 and V2, you should consider the peak voltages.
    This is correct.
  4. Jul 25, 2016 #3


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    Staff: Mentor

    What happened to Zth? What role does it play?

    You should be indicating that you are showing a magnitude as a result. Technically, using an equals sign here is incorrect since a magnitude is not the same as a complex value.
  5. Jul 26, 2016 #4
    Understood, it was a conscious decision just for brevity at the time. Here is the recalculated current:

    Phase angle relative to V1

    Voltage across j4 and j6 = ##(415 - j415)##, then current is ##\frac {415 - j415}{j10}## which is ##41.5 - j41.5## Amps

    Then VTh is ## V1 - ((41.5 - j41.5) * j4)## which is ##249 - j166## (equals 299.26 Volts at -33.69 degrees)

    Then I RL = ##\frac {249 - j166} {(35 + j 35.707)} = 1.115 + j 5.88## which is 5.98 Amps at -79.26 degrees

    I understood that √2 though representing peak magnitudes, is simply a scaling factor, where I can do all of the calculations with RMS values and then add it as a prefix in the final answer. There isn't any requirement to calculate power values so I could have used this throughout, it is just to save the tedium of adding to each line.

    So the first part of the question is whether I correctly calculated the current through RL. I am to understand that the j 2.4 Thevenin equivalent is 'included' as part of the circuit 'inside the box'. Therefore, I do not need to include it in series with RL when computing RL current.

    If it is agreed that my RL current calculation is correct, then for the superposition part of the question:

    Suppress V2 with a short, which effectively puts j6 in parallel with RL. Calculating the parallel combination - using Wolfram Alpha and rectangular form, I get:

    For V1 only
    $$j6 + RL = \frac {(35 + j 35.707) * j 6} {(35 + j 35.707) + j 6} = 0.425 + j 5.494$$
    Then using a voltage divider to calculate potential across this parallel combination is:
    $$VRL = 415 * \frac {0.425 + j 5.494} {(0.425 + j 5.494) + j 4} = 240.502 + j 7.811$$
    Then $$(V1) IRL = \frac {240.502 + j 7.811} {35 + j 35.707} = 3.4786 + j 3.325$$
    Now we repeat for V2 only
    $$j4 + RL = \frac {(35 + j 35.707) * j 4} {(35 + j 35.707) + j 4} = 0.199 + j 3.773$$
    Then using a voltage divider to calculate potential across this parallel combination is:
    $$VRL = - j 415 * \frac {0.199 + j 3.77} {(0.199 + j 3.77) + j 6} = -5.1889 + j 160.244$$
    Then $$(V2) IRL = \frac {-5.1889 + j 160.244} {35 + j 35.707} = - 2.216 + j 2.318$$
    Finally, adding those currents together gives us 1.262 + j 5.643 = 5.78 Amps at 77.39 degrees

    Now, if I take this: I RL = ##\frac {249 - j166} {(35 + j 35.707)} = 1.115 + j 5.88## which is 5.98 Amps at -79.26 degrees[red] and add ZTh of ##j2.4## Ohms to the denominator, I get 5.78 Amps as the result!

    So I have 5.98 Amps and 5.78 Amps. Have I gone wrong or misunderstood something?
    Last edited: Jul 26, 2016
  6. Jul 26, 2016 #5


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    No, that's not correct. The Thevenin impedance is in series with the Thevenin voltage, hence in the path of the current supplying the load:

  7. Jul 26, 2016 #6
    So I do need to add ZTh to the calculation? i.e. include it in series with RL? That would mean the last bit in post #4 does then tie up?

  8. Jul 26, 2016 #7


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    Yes, The Thevenin impedance must be included.

    For your superposition calculations you've got sign errors in some of your results. You'll want to check them. Also, you might want to keep a few extra digits in intermediate values during calculations, as some of the results are showing the effects of rounding/truncation "creep" into the significant digits.

    The result that I see for both methods is ##I_L = 5.784~A## with phase angle ##-81.124°~~## ( I've included three decimal places for each so that you can confirm your significant figures after calculation ).
  9. Jul 26, 2016 #8
    Thank you. Very grateful for your help. I estimate to have spent in excess of 20 hours on this question (really!) testing various methods of getting matching answers!
  10. Jul 26, 2016 #9

    The Electrician

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    I see gneill got in there just a few minutes before I did, but here's my response anyway.

  11. Jul 26, 2016 #10
    I actually used Wolfram and complex numbers so that the signs would be consistent and I wouldn't have to keep track of adding and subtracting angles. The mistake I made was that the negative sign for the imaginary part is appended to the real part, rather than being in front of the imaginary part. I assumed that the 'dash' sign at the end of the real part just means that the value can be expanded to expose more decimal places...

    I shall try it all again and hopefully adding ZTh to both calculations still provides the same answer. I shall be back if it doesn't!
  12. Jul 26, 2016 #11
    Can you guys tell me what you use for calculations? I'm using both Wolfram Alpha and my Texas Instruments TI-83.

    I'm using a fair few decimal places (say 4 or 5), but is there an easier way to do this in 'one chunk' ? I'm surprised that such large errors are creeping in due to the compounded rounding errors.

    I've got a trial of Mathematica but it looks waaaayy too complicated.
  13. Jul 26, 2016 #12


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    I use MathCAD. I assign given values to variables and then do the "calculations" symbolically. That way there's never a rounding/truncation issue, and it eliminates transcription errors that tend to pop up when copying long numbers.

    Your TI-83 should have several memory stores for holding intermediate values, no?
  14. Jul 26, 2016 #13
    It does, I should consider using it more. Maybe calculate on paper using a symbolic version and then store the intermediate values.

    Thanks for the tip.
  15. Jul 26, 2016 #14

    The Electrician

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    You should be able to do the calculations on a TI-83 without writing anything down. If you keep intermediate results in the calculator, they should be kept with 12 digit accuracy. Then all your calculations will be done with 12 digit accuracy and you round your final results to whatever you want, such as 4 digits.

    I did the calculations in my post with an HP-50G which also does 12 digit calculations.

    If the TI-83 can do matrix arithmetic you can set it up to do everything in "one chunk", but I'm not sure if the TI-83 is able to do that. I have a TI-86 and it can do matrix arithmetic, but I'm not sure about the TI-83.

    The big mathematics programs like Matlab, Mathcad, Mathematica, etc., can do it, but as you found out there is a learning curve to those.
  16. Jul 26, 2016 #15

    The Electrician

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    Your problem was to do it in several "chunks", but if you were allowed to do it however you wanted, here's a "one chunk" solution:

  17. Jul 26, 2016 #16
    Thanks for the help so far. Storing the intermediate values in the calculator, I get IRL = 5.783797359 but with an angle of -13.7436 degrees, so something is still wrong :frown:

    My complex value for this is ##5.618199032 - j1.374100259##
  18. Jul 26, 2016 #17

    The Electrician

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    You got a far better result than this in post #4, and with the corrections in post #9 you would be right on. Are you doing the same calculations from post #4, but keeping the intermediates in the calculator? You should be able to watch the intermediates and compare them to what you did in post #4.
  19. Jul 27, 2016 #18
    I finally got the same value for the angle as gneill in post #7.

    Though this appears illogical to me, if I enter ##415-(166-j166)## into the calculator, it gives me ##249+j166## as the result. If I enter ##415-166-j166## it gives me ##249-j166## as the result.

    I have a feeling that there's some multiplication rules going on when the brackets are there, such as minus * minus? I have to be really careful.
  20. Jul 27, 2016 #19
    I've been battling all morning with this again, and I've finally got the correct answer. Of course the problem I created this time was adding ZTh in series with the load when calculating using the superposition theorem!

    Don't know if someone can explain in simple terms why we add the 'internal resistance' in the Thevenin model but not when using superposition? I will assume that it's because the superposition method uses the real world measurements, whereas with Thevenin we are modelling the behaviour instead. It's not really that clear :confused:
  21. Jul 27, 2016 #20

    The Electrician

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    When you have an expression involving parentheses like this: a - (b + c) and you remove the parentheses you get a - b - c.

    When the starting expression has a minus sign inside the parentheses like this: a - (b - c) you get a - b + c. The minus sign in front of the left parenthesis applies individually to each term inside the parentheses. Think about how you would remove the parentheses from this: -(-j166).
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