I am trying to come up with a bit of a derivation for \$ P = \frac{1}{2}V_mI_m\cos(\theta) \$ and \$ Q = \frac{1}{2}V_mI_m\sin(\theta) \$, but using phasors (because I can visualise it better).

I know the following (mostly from another question that I asked): $$ \text{Let } v(t) = V_m\cos(\omega t) \text{ and } i(t) = I_m\cos(\omega t + \theta)\\ p(t) = v(t)i(t)\\ p(t) = V_m\cos(\omega t)I_m\cos(\omega t + \theta)\\ p(t) = \frac{V_mI_m}{2}(\cos(\theta) + \cos(2\omega t + \theta)) $$ Then from the above it is easy to see that the average value is: $$ \bar{p}(t) = \frac{V_mI_m}{2}\cos(\theta) $$

Now this is not at all with phasors, I tried to get a similar thing going with phasors...

Here is my (failed) attempt.

If I assume I have some vector I for the current and some vector V for the voltage such that $$ \textbf{I} = I_m\angle \theta\\ \textbf{V} = V_m \angle 0 = V_m $$

Now from my understanding of Real Power, the real power is the component of the current, I, that goes in the direction of the voltage, multiplied by the voltage, V.

So the component of I that goes in the direction of V is $$ \textbf{I}_v = |\textbf{I}|\cos(\theta) \cdot \frac{\textbf{V}}{|\textbf{V}|}\\ \textbf{I}_v = |\textbf{I}|\cos(\theta)\angle 0 $$

So now we have the component of I that goes in the direction of V.

Now I thought I'd just literally multiply that by V $$ \textbf{P} = \textbf{V}\textbf{I}_v\\ \textbf{P} = |V_m||I_m| \cos(\theta) \angle 0 $$

Now bringing that back into the time domain $$ p(t) = \left(V_mI_m\cos(\theta)\right)cos(\omega t) $$

I'm pretty sure that this is just wrong... However, I can't work out where it doesn't make sense!


2 Answers 2


The error is in that you're using the amplitude instead of the RMS value:

$$P = V_{RMS} \cdot I_{RMS} \cdot cos(\theta) $$

but for sinusoidal signals, \$V_{RMS}=\dfrac{V_{M}}{\sqrt{2}}\$ and \$I_{RMS}=\dfrac{I_{M}}{\sqrt{2}}\$ so multiplying:

$$P = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} V_{M} \cdot I_{M} \cdot cos(\theta) = \frac{1}{2} V_{M} \cdot I_{M} \cdot cos(\theta) =\\ = \sqrt{\frac{1}{T} \int_T [p(t)]^2 \:\mathrm{d}t} $$

  • \$\begingroup\$ Thanks for the answer. To get the RMS values, I believe that you're supposed to take the average value of the power. My issue is that above I have 'found' (from the phasor method) that the real power is \$p(t) = \left(V_mI_m\cos(\theta)\right)cos(\omega t)\$, so I thought that I could then use the average value integral to get \$P = V_{RMS} I_{RMS} \cos(\theta)\$; however, this does not seem to work. \$\endgroup\$
    – user968243
    May 9, 2012 at 0:49
  • \$\begingroup\$ @user968243 Because p(t) and P are two different animals: p(t) is again a sinusoidal signal, so you have to take the average of the square of p(t) and then apply the square root. \$\endgroup\$
    – clabacchio
    May 9, 2012 at 6:03

Point is that by multiplying phasors (input of two complex numbers) you will have product that is also a phasor (another complex number), and this one have same frequency as two input ones, so clearly you can't use phasors to calculate active and/or reactive power in that fashion. By multiplying phasors you can only easily calculate response of passive linear network on AC input signal. Read more:



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