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i have studied different books and sites for the above question but did not get the right concept that why the complex power encounter current conjugate not the original Phasor.if someone explain it with good example in simple words i would be very thankful...

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  • \$\begingroup\$ You need to provide a link that explains the motive of your question I reckon. \$\endgroup\$
    – Andy aka
    Dec 2, 2016 at 19:03
  • \$\begingroup\$ @Andy aka quora.com/…* also i see this in alex zander and david irwin book of electric circuits... \$\endgroup\$
    – Wisal
    Dec 2, 2016 at 19:14
  • \$\begingroup\$ @Andy aka and this one also youtube.com/watch?v=Xo0-3LtVea0 \$\endgroup\$
    – Wisal
    Dec 2, 2016 at 19:16
  • \$\begingroup\$ It is explained in the second video. \$\endgroup\$
    – Tyler
    Dec 2, 2016 at 20:51

3 Answers 3

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The voltage and current signals have an angle associated to them, better known as \$\theta_v\$ and \$\theta_i\$, respectively.

In terms of power, you want the phase difference between those two parameters, an angle we can call '\$\theta\$'. That is, you are looking for:

$$ \theta=\theta_v-\theta_i$$

If you were to find \$P=\text{VI}\$, where \$\text{V}\$ and \$\text{I}\$ have the form \$\text{a}+\text{bi}\$, you are implicitly finding $$ \theta=\theta_v+\theta_i$$

instead of the difference. This can be easily seen if you look at this in terms of Euler's identity.

Let's say that \$\text{V}\$ and \$\text{I}\$ now have the form \$|{\text{V}}|\angle\theta_v\$ and \$|\text{I}|\angle\theta_i\$.

If you now try to find \$P\$ as \$P=\text{VI}\$, you get

$$ P=|{\text{V}}||{\text{I}}|\angle(\theta_v+\theta_i)$$

Instead of the correct way:

$$ P=|{\text{V}}||{\text{I}}|\angle(\theta_v-\theta_i)$$

What makes you have the phase difference instead of the sum, is the conjugate of \$\text{I}\$ or \$\text{I}^*\$

When you find the conjugate the magnitude stays the same but the angle has opposite sign. So when you multiply the complex voltage and current, you are also subtracting \$\theta_v\$ and \$\theta_i\$.

Hopefully that clears things up.

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  • \$\begingroup\$ How does this work when current has an angle of zero, but voltage is leading or lagging? Then I* has no effect. \$\endgroup\$
    – Damien
    Sep 4, 2023 at 11:23
  • \$\begingroup\$ In answer to my own question, I chose a value for the current phase angle so that the voltage angle is zero. \$\endgroup\$
    – Damien
    Sep 5, 2023 at 14:29
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I want to call out the assumption here that voltages and currents are sinusoidal for the purposes of this discussion. It doesn't matter whether they are sine or cosine (just a phase difference between them) so long as we are consistent.

So real voltages and real currents are generally described as either:

$$\begin{align*} V&=V_{0}\cdot \textrm{sin}\left(\omega t+\phi_{_V}\right) \\ I &= I_{0}\cdot \textrm{sin}\left(\omega t+\phi_{_I}\right) \end{align*}$$

or (as will turn out to be the convention):

$$\begin{align*} V &= V_{0}\cdot \textrm{cos}\left(\omega t+\phi_{_V}\right) \\ I &= I_{0}\cdot \textrm{cos}\left(\omega t+\phi_{_I}\right) \end{align*}$$

You get to call it out. But there is a reason for a specific convention to be used that prefers one of the above over the other.

Voltage and current can be represented as phasors based on Euler's:

$$\begin{align*} V&=V_{0}\cdot e^{j\phi_{_V}}\cdot e^{j\omega t} = V_{0}\cdot\left[\textrm{cos}\left(\omega t+\phi_{_V}\right)+i\cdot\textrm{sin}\left(\omega t+\phi_{_V}\right)\right] \\ I &= I_{0}\cdot e^{j\phi_{_I}}\cdot e^{j\omega t} =I_{0}\cdot\left[\textrm{cos}\left(\omega t+\phi_{_I}\right)+i\cdot\textrm{sin}\left(\omega t+\phi_{_I}\right)\right] \end{align*}$$

But where only the real part is taken and the \$e^{j\omega t}\$ is dropped:

$$\begin{align*} V&= V_{0}\cdot e^{j\phi_{_V}} \equiv V_{0}\cdot\textrm{cos}\left(\phi_{_V}\right) \\ I &= I_{0}\cdot e^{j\phi_{_I}} \equiv I_{0}\cdot\textrm{cos}\left(\phi_{_I}\right) \end{align*}$$

In short, to gain the power granted by Euler's, a decision is made where the cosine represents the observation reality we measure as voltage and current.

With all that said and done, now. The answer to your question is easier.

We will want to stay with the convention that real power is the real part of Euler's. Or, in short, the cosine part. So real power is:

$$V_{0}\cdot I_{0}\cdot\textrm{cos}\left(\phi_{_V}-\phi_{_I}\right)$$

or, if \$\theta=\phi_{_V}-\phi_{_I}\$ (and there is another convention here as to which is subtracted from what, but I'll avoid discussing that here), then:

$$V_{0}\cdot I_{0}\cdot\textrm{cos}\left(\theta\right)$$

Which means again we can easily represent this as:

$$\left(V_{0}\cdot I_{0}\right)\cdot e^{j \theta}$$

But return to our phasor representation of voltage and current. We want the convenience of just multiplying them. (No special rules here. We want the benefits of all of mathematics to apply.) Simply multiplying the two phasors of voltage and current, without first taking the complex conjugate of current, gives this:

$$V_{0}\cdot e^{j\phi_{_V}}\cdot I_{0}\cdot e^{j\phi_{_I}} = V_{0}\cdot I_{0}\cdot e^{j\left(\phi_{_V}+\phi_{_I}\right)}$$

If you look closely there, you will see \$\phi_{_V}+\phi_{_I}\$ and not the desired \$\phi_{_V}-\phi_{_I}\$. So the answer is just wrong and is therefore useless.

So we have to adjust the rules for the purposes of electronics.

The desired result is then achieved by nothing more than simply taking the complex conjugate of \$I\$, instead. Then the sign changes correctly and we have our result!!

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    \$\begingroup\$ The formula below "So real power is" is wrong. The wt shouldn't be there, otherwise the mean of real power delivering to load is zero. \$\endgroup\$
    – emnha
    Sep 27, 2021 at 21:41
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    \$\begingroup\$ @anhnha Thanks! I can see I was a little bit sloppy most everywhere above. I need to fix more than one thing. Appreciate the catch! :) \$\endgroup\$
    – jonk
    Sep 27, 2021 at 22:25
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The reason $${S=VI^*}$$ 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 $${\frac{VI}{2}e^{j(\theta_v-\theta_i)}}$$ $${\frac{V}{\sqrt{2}}\frac{I}{\sqrt{2}}e^{j(\theta_v-\theta_i)}}$$ $${V_{rms}I_{rms}e^{j(\theta_v-\theta_i)}}$$ $${V_{rms}I_{rms}e^{j(\theta_v)}e^{j(-\theta_i)}}$$ $${V_{rms}e^{j(\theta_v)}I_{rms}e^{j(-\theta_i)}}$$ $${V_{rms}I_{rms}^{*}}$$

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