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Tagged Originally Posted by GoldDigger Definitely with the angels. I'm just not getting the same solution for Power. So I don't know where I am going wrong.  Reply With Quote

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Tagged Originally Posted by BatmanisWatching1987 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.  Reply With Quote

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Tagged Originally Posted by BatmanisWatching1987 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.  Reply With Quote

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Tagged Originally Posted by Sahib 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).  Reply With Quote

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Tagged Originally Posted by BatmanisWatching1987 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!  Reply With Quote

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Tagged Originally Posted by BatmanisWatching1987 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.  Reply With Quote

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Tagged Originally Posted by BatmanisWatching1987 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 ).  Reply With Quote

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Tagged  Reply With Quote

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In this problem, they used peak value voltage and current phasors instead of RMS value phasors.  Reply With Quote

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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:     Reply With Quote

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