ac - Why we use conjugate of current than the original phasor in the calculation of Complex Power i-e S=VI*

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...

Answer

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.