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For my exam I need to calculate the apparent power, active power and reactive power.

I know I get the active power from the real part and reactive power from imaginary part of the apparent power. However, I can't find any formulas for my specific problem.

I have

\$ U = 82.58 e^{j31.89°} \$ and \$ I = 1.65 e^{j31.89°} \$

The formula I found is

\$ S = \frac{1}{2} UI^* \$

But it starts with the problem that I don't know how to get from for example \$68e^{j30°}\$ to something like \$68.19 - j42.45\$

Used Euler. But know I don't get the correct solution.
I have \$S= 0.5* 82.58 e^{j31.89°} 1.65 e^{j31.89°}\$
That would be \$S=68.13cos(63.78)+j68.13sin(63.78)\$

Tried to conjugate "I" like that: \$I=1.65 e^{-j31.89°}\$
But then \$\Phi = 0\$

But the solution is \$S=68.19 - j42.45\$

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Use these identities :-

\$z = R.e^{j\theta}\$
\$Re(z) = R\cos(\theta) = a\$
\$Im(z) = R\sin(\theta) = b\$
\$z = a + jb\$
\$R = |z| = \sqrt{a^2 + b^2}\$
\$\theta = Arg(z) = \arctan(\frac{b}{a})\$

For example:
\$56.e^{j40} = 56\cos(40) + 56j\sin(40) = 42.9 + 36.0j\$
\$75 - j22 = \sqrt{75^2 + 22^2}.e^\arctan(\frac{-22}{75}) = 78.16.e^{-16.3j}\$

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    \$\begingroup\$ @MikeJ-UK, I think we need to avoid solving the full problem for someone when it is homework. How about you edit it out, or, even better, delete your answer and post a new one that does not give the solution? \$\endgroup\$
    – Kortuk
    May 5 '11 at 9:29
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    \$\begingroup\$ @Kortuk - You just suggested to JustJeff that he should have put some examples in his answer!. I have only addressed his last example using standard formulae found in any reference and I haven't answered his original question. I'll change some numbers anyway if it's not too late! \$\endgroup\$
    – MikeJ-UK
    May 5 '11 at 9:34
  • \$\begingroup\$ @MikeJ-UK, I misread your example and thought you did the exact problem he was assigned. My bad. This is why I should not review content when I have been awake for 13 hours during a night shift. Please accept my apology and up-vote. \$\endgroup\$
    – Kortuk
    May 5 '11 at 9:40
  • \$\begingroup\$ i was getting the sense Kortuk was hoping he could mash our answers together =P \$\endgroup\$
    – JustJeff
    May 5 '11 at 9:45
  • \$\begingroup\$ @Kortuk - No need to apologize. It's a fine line between helping & doing the work! \$\endgroup\$
    – MikeJ-UK
    May 5 '11 at 9:54
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Euler's Formula: \$ e^{j \theta} = cos(\theta) + j * sin(\theta) \$

edit
Euler helps you to separate the complex power of \$e\$ into its real and imaginary parts. These agree with your active and reactive power resp.

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    \$\begingroup\$ I think this answer is more of a comment. Yes, this formula is important, but more detail and explanation may be warranted. Clearly, we do not want to do someones homework for them, but an example or showing some example changes might be worth it. Just my opinion. \$\endgroup\$
    – Kortuk
    May 5 '11 at 8:59
  • \$\begingroup\$ @Kortuk - Euler is the answer to OP's question. I'm sure he learned it previously, but probably didn't think about it right away. I just wanted to put him back on the tracks without making the complete exercise for him. \$\endgroup\$
    – stevenvh
    May 5 '11 at 9:49
  • \$\begingroup\$ the thought of someone knowing euler's formula and not knowing to use it without a misunderstanding of the formula did not cross my mind. I was only saying it was more of a comment because a single equation with no explanation seems like more of a comment to me. Please do not take offense, I am just trying to help improve the site where I can while I have a moment to look at things. I rarely get time to wait around but I had a large amount of vacuum system work tonight and it never ever goes quickly. \$\endgroup\$
    – Kortuk
    May 5 '11 at 10:10
  • \$\begingroup\$ Thank you, i learned it previously but forgot it. \$\endgroup\$
    – madmax
    May 5 '11 at 10:37
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It's easier to multiply complex numbers in polar form, so find your UI* product that way. The conjugate in polar form is easy too, just negate the angle. Apply Euler's formula at the end to separate out real and imaginary components.

The relation between polar and rectangular forms can be illustrated like so:

relation of polar and rectangular coordinates

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  • \$\begingroup\$ @JustJeff, for a homework question I think that some examples of different steps or some guide on what it means to be in polar form compared to rectangular form might be helpful. I have met multiple people whom never had the triangle explanation given to them. \$\endgroup\$
    – Kortuk
    May 5 '11 at 9:21
  • \$\begingroup\$ @Kortuk - I was intentionally being sketchy specifically b/c I didn't want to just hand over the answer, due to the word 'exam' in the question. \$\endgroup\$
    – JustJeff
    May 5 '11 at 9:26
  • \$\begingroup\$ @JustJeff, Yes, I assumed. I was suggesting maybe putting in a picture that connects the polar and rectangular form. \$\endgroup\$
    – Kortuk
    May 5 '11 at 9:27
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    \$\begingroup\$ @Kortuk - if i figure out how to post pictures, i might do that. \$\endgroup\$
    – JustJeff
    May 5 '11 at 20:43
  • \$\begingroup\$ @Justjeff, there is a button for pictures near where the button for links is, just give it a link to your picture and it will handle it. \$\endgroup\$
    – Kortuk
    May 5 '11 at 22:18

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