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Superposition calculation for a circuit

  1. Nov 21, 2017 #1
    1. The problem statement, all variables and given/known data
    I have tried to answer the superposition question in the included file. I have gone through the workings as the same as in my john bird electrical book. I believe I have done nothing wrong in my approach and should be getting the same answer as my thevenins answer. the current through the load is completely different though

    2. Relevant equations
    total impedance when v2 removed = j4 in series with parallel j6 and 35 + j 35 .71
    v1 removed total impedance j6 in series with parallel j4 and 35 + j35.71
    then volts over resistances to find I then use the currents to find load current adding i2 and i5.
    i1 x j6/j6 + 35 + j 35.71 = I2
    i4 x j4/j4 + 35 + j35.71 = 1 5

    i5 + i2 = I load

    3. The attempt at a solution
    it is included in file
    I have added my attemp
     

    Attached Files:

    Last edited by a moderator: Nov 21, 2017
  2. jcsd
  3. Nov 21, 2017 #2

    gneill

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    Staff: Mentor

    Hi billyray,

    Welcome to Physics Forums!

    I've just taken a look at your work for the first part where you've suppressed the V2 source, and I find that I don't agree with your net impedance value. In particular, the imaginary component looks incorrect to me. I think things started to go off the rails when you were dealing with the numerator during the normalization of the complex fraction. This is the bit I'm referring to:

    upload_2017-11-21_9-42-0.png
     
  4. Nov 21, 2017 #3
    hi Gniell
    Thank you so much for listening. I am trying very hard and small mistakes and lack of confidence is frustrating me. I have studied a lot for this and thought I could do it. I have changed the numerator as I see I did not add j4 do you think it is more wrong than this?. I have attached my correction.
     

    Attached Files:

  5. Nov 21, 2017 #4
    I then multiplied by the conjugate of the denominator that was 35- j35.71
     
  6. Nov 21, 2017 #5

    gneill

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    Staff: Mentor

    You're good until the last two lines. You can't add the j4 to the numerator at that point: it's a separate term with a different denominator.
     
  7. Nov 21, 2017 #6
    Thanks gneill
    i will redo
     
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