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I know that average power is equal to \$VmIm/2 * PF\$.

I can show that this is equivalent to \$(1/2)Re(VI^*)\$ by writing VI* in magnitude phase form.

However, I also know that complex power \$S = VI^* = P + jQ\$.

But then P = P/2!

What is going on here?

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  • \$\begingroup\$ This would be more clear if you used Mathjax for the formulas. (EE uses \$ to start and end in-line math, instead of just $). \$\endgroup\$ – The Photon Oct 3 '18 at 4:36
  • \$\begingroup\$ In particular, I'm not sure what you mean by "Vm" and "Im". Are these the magnitudes of V and I, or something else? \$\endgroup\$ – The Photon Oct 3 '18 at 4:41
  • \$\begingroup\$ Sorry about that, I am not too familiar with the formatting. Yes, Vm and Im are the maximum magnitudes of the voltage and current wave-forms ( v(t) = Vmax sin(wt + phi) ). Also in our power course we take V phasor to be Vrms in magnitude, or Vmax / sqrt(2) for pure sinusoid. The P = (1/2)Re(VI*) is an equation from my RF course. I think the problem may have something to do with the conversion between time and phasor domain, but I'm not sure how to prove it. \$\endgroup\$ – user6615434 Oct 3 '18 at 4:53
  • \$\begingroup\$ Are you sure that the definitions of things are equivalent between courses? Prehaps you should check... \$\endgroup\$ – Solar Mike Oct 3 '18 at 5:18
  • \$\begingroup\$ Be sure of pk,rms and avg units because 1/√2*1/√2=1/2 for V & I \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 3 '18 at 14:23
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I found the issue, thanks to your comments. In my power course, the voltage phasor is defined as having RMS magnitude \$ V_{max} / \sqrt2 \$, while in my RF course, the voltage phasor is defined as having magnitude \$ V_{max} \$, where \$ v(t) = V_{max}cos(wt + \theta_v) \$. So in my RF course, complex power would actually equal \$ (1/2)VI^* \$. Therefore, \$ (1/2)Re(2S) = P \$.

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    \$\begingroup\$ Thank you for taking the trouble to come back and answer your question. \$\endgroup\$ – K H Oct 4 '18 at 8:42

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