# Confusion with derivation of average power formula [closed]

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?

• This would be more clear if you used Mathjax for the formulas. (EE uses \$ to start and end in-line math, instead of just $). – The Photon Oct 3 '18 at 4:36
• In particular, I'm not sure what you mean by "Vm" and "Im". Are these the magnitudes of V and I, or something else? – The Photon Oct 3 '18 at 4:41
• 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. – user6615434 Oct 3 '18 at 4:53
• Are you sure that the definitions of things are equivalent between courses? Prehaps you should check... – Solar Mike Oct 3 '18 at 5:18
• Be sure of pk,rms and avg units because 1/√2*1/√2=1/2 for V & I – Tony Stewart Sunnyskyguy EE75 Oct 3 '18 at 14:23

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 \$$.