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Thread: AC formulae and power analysis etc

  1. #1
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    AC formulae and power analysis etc

    Hi

    Is my categorization of complex power S into phasor form, polar form, and rectangular form correct? Did I miss any category? Thank you for your help.

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  2. #2
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    The first and the second line formulae are correct.
    The third line result has to be: P=Vrms^2*R/(R^2+X^2) and Q=Vrms^2*X/(R^2+X^2) since: (R-jX)*(R+jX)=R^2+X^2

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    He did not need to use the conjugate for S = v^2/Z



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    Quote Originally Posted by Ingenieur View Post
    He did not need to use the conjugate for S = v^2/Z
    Quick question- what is "conjugate"?
    What is esoteric knowledge today will be common knowledge tomorrow.

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    Quote Originally Posted by mbrooke View Post
    Quick question- what is "conjugate"?
    If I = 100/60 deg = 50 + 86.6 j
    I* = 100/-60 deg = 50 - 86.6 j
    changing sign on complex portion

    S = V I* iirc





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    Quote Originally Posted by Ingenieur View Post


    If I = 100/60 deg = 50 + 86.6 j
    I* = 100/-60 deg = 50 - 86.6 j
    changing sign on complex portion

    S = V I* iirc


    (*) as in degrees?
    What is esoteric knowledge today will be common knowledge tomorrow.

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    Quote Originally Posted by mbrooke View Post
    (*) as in degrees?
    No it's the symbol for the conjugate of the qty
    like '!' is the symbol for factorial

    if you had phasor analysis you used conjugate

    S = V (conjugate of I) = V I*



  8. #8
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    [QUOTE=Ingenieur;1858884]He did not need to use the conjugate for S = v^2/Z[/QUOT
    IEEE Std 1459-2010, IEEE Standard Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions. 3.1.1.3 Reactive power (var)
    “NOTE 1— If the load is inductive, then Q > 0. If the load is capacitive, then Q < 0. This means that when the current lags the voltage θ > 0 and vice versa.”
    Using Z=R+jX instead of Z*=R-jX the resulted Q<0 [capacitive].Usually X is inductive.

  9. #9
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    [QUOTE=Julius Right;1858996]
    Quote Originally Posted by Ingenieur View Post
    He did not need to use the conjugate for S = v^2/Z[/QUOT
    IEEE Std 1459-2010, IEEE Standard Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions. 3.1.1.3 Reactive power (var)
    “NOTE 1— If the load is inductive, then Q > 0. If the load is capacitive, then Q < 0. This means that when the current lags the voltage θ > 0 and vice versa.”
    Using Z=R+jX instead of Z*=R-jX the resulted Q<0 [capacitive].Usually X is inductive.
    Yes he did!
    Actually, the resulting Q2 is always positive, allowing Q to be either positive or negative.

    If you do simple complex multiplication without the conjugate you get a negative value for
    q2, which is impossible, and you get cross terms which still contain j, also unphysical for power.

  10. #10
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    Quote Originally Posted by pg1995 View Post
    Hi

    Is my categorization of complex power S into phasor form, polar form, and rectangular form correct? Did I miss any category? Thank you for your help.

    Nice cut and paste job (twice) but what point were you, the OP, endeavouring to make?
    Si hoc legere scis nimium eruditionis habes.

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