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I am taking a power engineering class and I am not sure about taking the squared magnitude of the complex numbers. I figured out most of numbers, except -54.24Vrfl. Please click on the picture: enter image description here. Can someone tell me how did I get it? Thanks a bunch!

$441.7\angle\delta=(.9313\text{Vrfl}-30.04)+j(.0034\text{Vrfl}+251.97).$.

Taking the squared magnitudes of both sides,

$441.7^2=.8673\text{Vrfl}^2-54.24\text{Vrfl}+64,391.$

Btw, Vrfl is voltage.

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  • $\begingroup$ The square of the magnitude of $a+bj$ is $a^2+b^2$ $\endgroup$ – saulspatz Oct 17 at 23:30
  • $\begingroup$ Thanks! So what do you square to get -54.24 Vrfl? $\endgroup$ – 123 Oct 17 at 23:51
  • $\begingroup$ If voltage is a real number, then the equation, $441.7=(.9313V-30.04)+j(.0034V+251.97)$ makes no sense, as it equates a real number on the left to a nonreal number on the right. Do you mean to say that the magnitude of the complex number on the right equals the real number on the left? In any event, have you multiplied out $(.9313V-30.04)^2+(.0034V+251.97)^2$? I think if you do that you'll see where the $-54.24V$ comes from. $\endgroup$ – Gerry Myerson Oct 18 at 0:47
  • $\begingroup$ Thanks Gary! I squared .9313 and got .8673. I added (-30.04)^2 to (251.97)^2 and got 64,391 but still missing this -54.24. Btw, it is an example from the book. $\endgroup$ – 123 Oct 18 at 0:57
  • $\begingroup$ Also, I just uploaded the picture showing the whole part. $\endgroup$ – 123 Oct 18 at 1:04
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$$|(.9313\text{Vrfl}-30.04)+j(.0034\text{Vrfl}+251.97)|^2=\\(.9313\text{Vrfl}-30.04)^2+(.0034\text{Vrfl}+251.97)^2=\\ (0.8673\text{Vrfl}^2-55.953\text{Vrfl}+902.4)+(0.00001\text{Vrfl}^2+1.713\text{Vrfl}+63488.88)=\\ 0.8673\text{Vrfl}^2-54.24\text{Vrfl}+64391.28 $$

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  • $\begingroup$ Thank you! I really appreciate it. $\endgroup$ – 123 Oct 18 at 1:19

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