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Thread: Calculating Real Power

  1. #31
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    Quote Originally Posted by GoldDigger View Post
    Definitely with the angels.
    I'm just not getting the same solution for Power. So I don't know where I am going wrong.

  2. #32
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    Quote Originally Posted by BatmanisWatching1987 View Post
    Vdrop = IpZLoad = (8.567∠48.82)(7+j8) = 91.07∠2.06V

    S=3VpIp∠θ° = 3*(91.07)(8.567)∠2.06--48.82 = 1476.79+j1815.89

    So I get P equal to 1476.79 W.

    I'm not sure if I am computing the calculation correctly with the angels?
    You need to take only the impedance angle of (7+j8) and current magnitude of 8.567 ie Vp and Ip only magnitudes and power factor angle is the impedance angle of (7+j8).
    Last edited by Sahib; 02-28-19 at 03:39 AM.

  3. #33
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    Quote Originally Posted by BatmanisWatching1987 View Post
    Vdrop = IpZLoad = (8.567∠48.82)(7+j8) = 91.07∠2.06V

    S=3VpIp∠θ° = 3*(91.07)(8.567)∠2.06--48.82 = 1476.79+j1815.89

    So I get P equal to 1476.79 W.

    I'm not sure if I am computing the calculation correctly with the angels?
    The angle of current=<-46.75 and not <48.82. The angle of conjugate of Ip=<46.75. So the angle of S=<2.06+46.75=<48.81.

  4. #34
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    Quote Originally Posted by Sahib View Post
    The angle of current=<-46.75 and not <48.82. The angle of conjugate of Ip=<46.75. So the angle of S=<2.06+46.75=<48.81.
    I'm just a little confused by how to add the Voltage and Current for finding the Apparent Power Angle.

    I understand we need to take the conjugate of the Current Angle.

    So, without doing the conjugate of the Current Angle, would we add up the Voltage and Current Angle to get the Apparent Power to be <2.06+(-46.75), then we take the conjugate of the Current Angle to get <2.06+(+46.75).

  5. #35
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    Quote Originally Posted by BatmanisWatching1987 View Post
    I'm just a little confused by how to add the Voltage and Current for finding the Apparent Power Angle.

    I understand we need to take the conjugate of the Current Angle.

    So, without doing the conjugate of the Current Angle, would we add up the Voltage and Current Angle to get the Apparent Power to be <2.06+(-46.75), then we take the conjugate of the Current Angle to get <2.06+(+46.75).
    The easiest way to understand vector multiplication is to plot it on paper and deduce how much angle is there between the two vectors. Easy!

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    Quote Originally Posted by BatmanisWatching1987 View Post
    I'm just a little confused by how to add the Voltage and Current for finding the Apparent Power Angle.

    I understand we need to take the conjugate of the Current Angle.

    So, without doing the conjugate of the Current Angle, would we add up the Voltage and Current Angle to get the Apparent Power to be <2.06+(-46.75), then we take the conjugate of the Current Angle to get <2.06+(+46.75).
    Attachment 22390
    Look at your attachment above (from post#18). The angle of current phasor, as worked out, is <-46.75 degrees.

  7. #37
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    Quote Originally Posted by BatmanisWatching1987 View Post
    I'm just a little confused by how to add the Voltage and Current for finding the Apparent Power Angle.

    I understand we need to take the conjugate of the Current Angle.

    So, with conjugate of the Current Angle, would we add up the Voltage and Current Angle to get the Apparent Power angle to be <2.06+(+46.75)? .
    Yes. (I edited your quote as above in red to clarify ).

  8. #38
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    Click image for larger version. 

Name:	Chapter 12 Problem 38.JPG 
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Name:	Chapter 12 Solution 38.jpg 
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    I tried using the same method I just learned, but I am not getting the same solution at all with the following problem.

  9. #39
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    Quote Originally Posted by BatmanisWatching1987 View Post
    Click image for larger version. 

Name:	Chapter 12 Problem 38.JPG 
Views:	37 
Size:	139.2 KB 
ID:	22458Click image for larger version. 

Name:	Chapter 12 Solution 38.jpg 
Views:	25 
Size:	123.1 KB 
ID:	22459

    I tried using the same method I just learned, but I am not getting the same solution at all with the following problem.
    In this problem, they used peak value voltage and current phasors instead of RMS value phasors.

  10. #40
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    For the record only: in my post 26 from 02_26_2019 I did a mistake. Actually the attached note has to be:

    Click image for larger version. 

Name:	S calculation.jpg 
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ID:	22479

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