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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.

Deepak
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123
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  • The square of the magnitude of $a+bj$ is $a^2+b^2$ – saulspatz Oct 17 '20 at 23:30
  • Thanks! So what do you square to get -54.24 Vrfl? – 123 Oct 17 '20 at 23:51
  • 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. – Gerry Myerson Oct 18 '20 at 00:47
  • 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. – 123 Oct 18 '20 at 00:57
  • Also, I just uploaded the picture showing the whole part. – 123 Oct 18 '20 at 01:04
  • @GerryMyerson The OP's image shows the left hand side accurately. It is not just $441.7$ (real) but instead $441.7\angle\delta$, which denotes, in polar form, a complex quantity with magnitude $441.7$ and argument $\delta$. I'll edit this into the question. – Deepak Oct 18 '20 at 01:21
  • To the OP, I have edited your question for accuracy. Note that you are equating two (non-real) complex numbers and then equating their squared magnitudes. The term you were asking about simply arises from elementary algebraic expansion. – Deepak Oct 18 '20 at 01:28
  • You're not paying attention, 123. When you multiply out $(.9313V-30.04)^2$ the term in $V$ is $(2)(.9313)(-30.04)$. When you multiply out $(.0034V+251.97)^2$ the term in $V$ is $(2)(.0034)(251.97)$. Have you never seen the formula, $(a+b)^2=a^2+2ab+b^2$? And, who is "Gary"? – Gerry Myerson Oct 18 '20 at 06:31

1 Answers1

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$$|(.9313\text{Vrfl}-30.04)+j(.0034\text{Vrfl}+251.97)|^2=\\(.9313\text{Vrfl}-30.04)^2+(.0034\text{Vrfl}+251.97)^2=\\ (0.8673\text{Vrfl}^2-55.953\text{Vrfl}+902.4)+(0.00001\text{Vrfl}^2+1.713\text{Vrfl}+63488.88)=\\ 0.8673\text{Vrfl}^2-54.24\text{Vrfl}+64391.28 $$

Randy Marsh
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