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I was wondering Where I could find the derivation for the complex power formula, S=VI*/2, where S, V and I are complex phasors.

I've seen a whole bunch of verifications where people sub stuff into the equation to show that it happens to work.

Here's what I know so far, If \$ V=V_{M} ∠ \phi _{V} \$ and \$ I=I_{M} ∠ \phi _{I} \$ and \$ S = V_{RMS} \cdot I_{RMS} \$,
then \$ V_{RMS}= \dfrac{V_{M} ∠ \phi _{V}}{\sqrt{2}} \$ and \$ I_{RMS}= \dfrac{I_{M} ∠ \phi _{I}}{\sqrt{2}} \$ and S = Vm∠ø_v*Im∠ø_i/2 \$ S = \dfrac{V_{M} ∠ \phi _{V} \cdot I_{M} ∠ \phi _{I}}{2} \$

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    \$\begingroup\$ You'll have to define S, V, I, and whatever "*/" is supposed to mean. \$\endgroup\$ Apr 8, 2012 at 14:59
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    \$\begingroup\$ @OlinLathrop, it is I* for complex conjugate of I(current) and divided by two, since they are both sin waves(V and I*) so they both have their RMS conversion. \$\endgroup\$
    – Kortuk
    Apr 8, 2012 at 16:19

2 Answers 2

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Let V and I be the instantaneous voltage and current on a load. From the definition of power, voltage and current, we have the relation for instantaneous power:

\$ p(t) = v(t) \cdot i(t) \$

Which means that the power on a given instant \$ t \$ is equal to the product of the voltage and the current exactly on that instant.

I'll assume you're familiar with what the phasor representation actually means. Just to state that shortly: a phasor is a mathematical shorthand for representing a sinusoid at a given unknown frequency.

So, \$ V=V_{M} ∠ \phi _{V} \$ is a shorthand for \$ v(t) = V_{M} \cdot cos(\omega t+ \phi _{V})\$. Similarly: \$ I=I_{M} ∠ \phi _{I} \$ means \$ i(t) = I_{M} \cdot cos(\omega t+ \phi _{I})\$.

Multiplying \$ v(t) \cdot i(t) \$ for all \$ t \$, gives us the waveform of the instantaneous power for every \$t\$. Working on that multiplication:

\$s(t) = v(t) \cdot i(t) = V_{M} \cdot cos(\omega t+ \phi _{V}) \cdot I_{M} \cdot cos(\omega t+ \phi _{I})\$

As \$ cos(u) \cdot cos(v) = \cfrac{1}{2} \cdot [cos(u-v)+cos(u+v) ]\$, with \$ u = \omega t+ \phi _{V} \$ and \$ v = \omega t+ \phi _{I} \$, we can simplify the equation above to:

\$ s(t) = v(t) \cdot i(t) = \cfrac{V_{M}I_{M}}{2} \cdot [cos(\phi _{V} - \phi _{I}) + cos(2\omega t+ \phi _{V} + \phi _{I})] \$

This waveform is pretty interesting for itself: it is a constant value \$\cfrac{V_{M}I_{M}}{2} \cdot cos(\phi _{V} - \phi _{I}) \$ summed by a sinusoid \$ \cfrac{V_{M}I_{M}}{2} cos(2\omega t+ \phi _{V} + \phi _{I})]\$.

This clearly shows that the instantaneous power is not constant with time.

Based on that result, we can see that the mean power is equal to the non-varying component of \$ s(t) \$ (it's pretty straightforward to prove that mathematically, one just have to solve the integral \$ \cfrac{1}{T}\int_{t}^{t+T}{s(t)dt}\$ )

Motivated by this result, and by the pretty sweet geometrical interpretation of \$ VIcos(\phi _{V} - \phi _{I}) \$, that value has been defined as the real power, that is, the power that is actually delivered to the load. Now you know that this so called real power is nothing more than the mean power at the load.

Diving into this concept a little bit (it's a pitty I can't draw here, but I'll try):

Let v be a vector with magnitude ||v|| and phase \$ \phi_v \$, and i be a vector with magnitude ||i|| and phase \$ \phi_i \$ If you multiply ||i|| by \$cos(\phi_v-\phi_i)\$ you have the projection of i over v. On the other hand, \$||i||sin(\phi_v-\phi_i)\$ is said to be the component of i in quadrature with v.

Now you can understand why the mean power has a cool geometric interpretation: the mean power is the voltage multiplied by the projection of the current over the voltage, on the phasor space.

This motivated the creation of the complex power S as:

S = P + jQ

With this definition, the real part of the vector is exactly the mean power delivered to the load, and the complex part is the power said to be in quadrature, called reactive power (google for Power Triangle to see the geometrical interpretation of this result).

Ok, now going back to the \$ s(t) \$ definition, we see that \$ P = \cfrac{V_M I_M}{2} \cdot cos(\phi_v - \phi_i) \$ and \$ Q \$, by definition, and to comply with the definition of S, is equal to \$ \cfrac{V_M I_M}{2} \cdot sin(\phi_v - \phi_i) \$

So, as we wanted to prove at the begining:

\$ S = P + jQ = \cfrac{V_M I_M}{2} \cdot cos(\phi_v - \phi_i) + j\cfrac{V_M I_M}{2} \cdot sin(\phi_v - \phi_i)\$

\$ S = \cfrac{V_M I_M}{2} \cdot [cos(\phi_v - \phi_i) + jsin(\phi_v - \phi_i) ]\$

\$ S = \dfrac{V_{M} ∠ \phi _{V} \cdot I_{M} ∠ -\phi _{I}}{2} \$

\$ S = \cfrac{V \cdot I*}{2} \$

So, there you go, what you wanted to see ;)

edit: What's the physical interpretation of Q?

I've shown above what's the physical interpretation of the real part of the complex power, P, that is, the mean power delivered to the load. But what's exactly Q, how can one visualize it? It's based on the fact that cos and sin are orthogonal, and the principle of superposition can be applied to power if the two waveforms involved in the calculation are orthogonal. Let's go into the math, because that's really what matters.

Using the result obtained above: \$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [cos(\phi _{V} - \phi _{I}) + cos(2\omega t+ \phi _{V} + \phi _{I})] \$

  • First case: purely resistive load, so that \$ \phi _{V} - \phi _{I} = 0\$

    \$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [1 + cos(2(\omega t+ \phi _{V}))] \$

    That is a sinusoid centered on \$ \cfrac{V_{M}I_{M}}{2} \$ with that same amplitude (its minimum value is 0 and its maximum value is \$ V_{M}I_{M} \$ ). Let's call it P

  • Second case: purely inductive load, so that \$ \phi _{V} - \phi _{I} = \cfrac{\pi}{2} \$

    \$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [0 - cos(2(\omega t+ \phi _{V}) - \cfrac{\pi}{2} )]\$

    \$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [sin(2(\omega t+ \phi _{V}))] \$

    That is a purely oscillatory waveform with mean value equal to 0. Let's call this result Q.

  • Third case: the generic case \$ \phi _{V} - \phi _{I} = \theta \$

    In this case, s(t) is exactly the general equation we found on the discussion above. But we can rewrite that to make use of the result of the two previous cases, like this:

    First, we rewrite the equation in terms of \$ \theta \$ (notice that \$ \phi_V + \phi_I = \phi_V -\phi_V + \phi_V + \phi_I = 2\phi_V - \theta \$): \$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [cos(\theta) + cos(2(\omega t+ \phi _{V}) - \theta)] \$ Knowing that: \$ cos(x-y) = cos(x)cos(y) + sin(x)sin(y)\$, letting \$ x = 2(\omega t+ \phi _{V}) \$ and \$ y = \theta \$

    \$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [cos(\theta) + cos(\theta)cos(2(\omega t + \phi_V)) + sin(\theta)sin(2(\omega t + \phi_V))]\$

    Rearranging the terms:

    \$ s(t) = cos(\theta) \cdot \cfrac{V_{M}I_{M}}{2} \cdot [1 + cos(2(\omega t + \phi_V))] + sin(\theta) \cdot \cfrac{V_{M}I_{M}}{2} sin(2(\omega t + \phi_V))\$

    Using the result of the two first cases above:

    \$ s(t) = cos(\theta)P + sin(\theta)Q \$

    An amazing result, right? What does that mean?

    Let's go back to what we are doing: calculating the power for the generic case where \$ \phi _{V} - \phi _{I} = \theta \$, that is, solvig the equation:

    \$ s(t) = V_{M}cos(\omega t + \phi_V) \cdot I_{M}cos(\omega t + \phi_I) \$

    Can we rewrite \$ i(t) = I_{M}cos(\omega t + \phi_I) \$ in the form of \$ i(t) = K_1 cos(\omega t + \phi_V) + K_2 sin(\omega t + \phi_V) \$?

    Let's try:

    \$ \phi_I = \phi_V - \theta \$ \$ i(t) = I_{M}cos(\omega t + \phi_V - \theta \$) \$

    Letting \$ \omega t + \phi_V = u \$ and \$ \theta = v \$

    With the relation:

    \$ cos(u-v) = cos(u)cos(v) + sin(u)sin(v) \$

    We have:

    \$ i(t) = I_{M}cos(\theta)cos(\omega t + \phi_V) + I_{M}sin(\theta)sin(\omega t + \phi_V) \$

    Just what we wanted, to rewrite i(t) as a sum of two components: one in phase with v(t), and one in quadrature with v(t)!

    Now the result of the case 3 can be explained: i(t) can be decomposed in two components, as shown above, and the power generated by i(t) is equal to the power generated by each one of these components individually. Whoa, just like superposition but for power! (Remember that this is only true, and it was proven above, because cos and sin are orthogonal)

    So Q is the amount of power generated by the component of i(t) that's in quadrature with v(t). It is purely oscillatory and has no mean value.

    P is the amount of power generated by the component of i(t) that's in phase with v(t). It is oscillatory but has a mean value that's equal the mean power delivered to the load.

    And the complex power S, the total power, is exactly the sum of these two components

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      \$\begingroup\$ Thank you for your good explantation! I have a few questions though: 1. I don't follow what happened to \$ -\cfrac{V_{M}I_{M}}{2} cos(2\omega t+ \phi _{V} + \phi _{I}) \$. I thought this term would be the reactive power, Q; however, \$ Q = ||i||sin(\phi_v-\phi_i) \$. 2. I don't understand how you went from \$ S = \cfrac{V_M I_M}{2} \cdot [cos(\phi_v - \phi_i) + jsin(\phi_v - \phi_i) ] \$ tp \$ S = \dfrac{V_{M} ∠ \phi _{V} \cdot I_{M} ∠ -\phi _{I}}{2} \$. It's as though \$ \cos(\phi_v - \phi_i) \$ is a phasor, but it's just a constant. Thanks again for your answer! \$\endgroup\$
      – user968243
      Apr 9, 2012 at 1:49
    • \$\begingroup\$ Yep. you're right, that's NOT Q. The reactive power is defined only in terms of the phase difference between voltage and tension, and it's a value that's directly related to the definition of S as a phasor. It's the power that would be delivered by the current in quadrature with the voltage. The time varying component is not taken into account, because in this sense what really matters is the mean power at the load. The varying part EXISTS, is really there (watch a incandescent light bulb, for example), but, over time, the power is related only to the static part of s(t). ;) \$\endgroup\$
      – Castilho
      Apr 9, 2012 at 2:10
    • \$\begingroup\$ Okay, so does this varying part have a special name? Anyway, so if I understand it correctly, the amount of I in the direction of V is the real power, and the amount of I, perpendicular to V is the complex power. \$\endgroup\$
      – user968243
      Apr 9, 2012 at 4:48
    • \$\begingroup\$ almost that, the amount of I in the direction of V multiplied by V is the real power P, the amount of I perpendicular to V multiplied by V is the REACTIVE power Q, P+jQ is the complex power, or apparent power ;) \$\endgroup\$
      – Castilho
      Apr 9, 2012 at 9:37
    • \$\begingroup\$ Okay, that makes sense! Actually in my previous comment, I was asking what the name for this is: −VMIM2cos(2ωt+ϕV+ϕI) I really thought that it was the reactive power... Thanks for your reples by the way, I'm grateful! \$\endgroup\$
      – user968243
      Apr 9, 2012 at 12:11
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    The way to derive $${S=VI^*}$$ is as follows.
    Let there be a voltage $${V\angle\theta_v}$$ and a current $${I\angle\theta_i}$$ in a phasor equivalent circuit in the steady state. In the time domain, the voltage and current phasors mean $${V\cos(\omega t+\theta_v)}$$ and $${I\cos(\omega t+\theta_i)}$$ Multiplying these two quantities gives $${V\cos(\omega t+\theta_v)*I\cos(\omega t+\theta_i)}$$ Letting the reference angle be taken with respect to the current phasor, we get $${VI\cos(\omega t+\theta_v)\cos(\omega t+\theta_i)}$$ $${VI\cos(\omega t+\theta_v-\theta_i+\theta_i)\cos(\omega t+\theta_i)}$$ Using the identity $${\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)}$$ We get: $${VI(\cos(\omega t+\theta_i)\cos(\theta_v-\theta_i)-\sin(\omega t+\theta_i)\sin(\theta_v-\theta_i))\cos(\omega t+\theta_i)}$$ $${VI(\cos(\omega t+\theta_i)\cos(\theta_v-\theta_i)\cos(\omega t+\theta_i)-\sin(\omega t+\theta_i)\sin(\theta_v-\theta_i)\cos(\omega t+\theta_i))}$$ Using the identities $${\cos^2(x)=\frac{1+\cos(2x)}{2}}$$ and $${\sin(2x)=2\sin(x)\cos(x)}$$ we get $${\frac{VI}{2}((1+\cos(2(\omega t+\theta_i)))\cos(\theta_v-\theta_i)-\sin(2(\omega t+\theta_i))\sin(\theta_v-\theta_i))}$$ We like to use phasors, so we like to express the waveform in terms of cosines, so another identity to use is $${\cos(x+\frac{\pi}{2})=-\sin(x)}$$ $${\frac{VI}{2}((1+\cos(2(\omega t+\theta_i)))\cos(\theta_v-\theta_i)+\cos(2(\omega t+\theta_i)+\frac{\pi}{2})\sin(\theta_v-\theta_i))}$$ Observe that the second waveform leads the first by a quarter of a period; this is equivalent to multiplying with $${j}$$ in the phasor domain. So, we can write $${\frac{VI}{2}(\cos(\theta_v-\theta_i)+j\sin(\theta_v-\theta_i))}$$ in the phasor domain. This is $${\frac{VI}{2}e^{j(\theta_v-\theta_i)}}$$ $${\frac{I}{\sqrt{2}}\frac{V}{\sqrt{2}}e^{j(\theta_v-\theta_i)}}$$ $${V_{rms}I_{rms}e^{j(\theta_v-\theta_i)}}$$ $${V_{rms}I_{rms}e^{j(-\theta_i)}e^{j(\theta_v)}}$$ $${V_{rms}e^{j(\theta_v)}I_{rms}e^{j(-\theta_i)}}$$ $${V_{rms}I_{rms}^{*}}$$

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