Monday, 28 November 2016

How to calculate apparent power?


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\$



Answer




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}\$


No comments:

Post a Comment

arduino - Can I use TI's cc2541 BLE as micro controller to perform operations/ processing instead of ATmega328P AU to save cost?

I am using arduino pro mini (which contains Atmega328p AU ) along with cc2541(HM-10) to process and transfer data over BLE to smartphone. I...