These videos might help you:

https://youtu.be/Xo0-3LtVea0

https://youtu.be/XD36hh7mWLk

Dear Chintan,

S^ = VI* is not "apparent power", but "complex power". Apparent power can be gained as an absolute value of S^, thus |S^| = S.

Complex conjugate of current phasor is used because for S you need phase difference between the voltage phase and current phase. Thus the S^ angle (phi) is identical to the angle of the load impedance Z (obtained by Z = V/I) and under the asumption of zero harmonic distortion, cos(phi) is the power factor.

I recommend checking the IEEE 1459-2010 standard for further details about power definitions.

because the difference between the angles of voltage and current required .

I am not an expert of the field. But the following two sites may be useful to you:

https://www.quora.com/Why-do-we-take-a-conjugate-of-current-when-we-write-the-formula-of-complex-power-S-VI*

http://www.ee.ic.ac.uk/hp/staff/dmb/courses/ccts1/01400_AcPower.pdf

I think it is to justify the true meaning of transits of the current and power in the circuit.

I recommend you see the IEEE 1459-2010 where power definitions are well detailed.

For power calculation, we need phase difference between voltage & current, which will possible when we use conjugate of either current or voltage. Generally voltage is taken as reference phasor, so we use conjugate of current.

interested question, you can find the answer following website

https://www.quora.com/Why-do-we-take-a-conjugate-of-current-when-we-write-the-formula-of-complex-power-S-VI*

The conjugate of current is considerate so, because this current has an apparent value, with which it has module with corresponding angle.

Similarly, the voltage has a module and an angle and how result, the apparent power, S, whose unit is express in VA, is the same that voltage for ampere, understand?

Besides Chintan, I exposure other form, for example:

If the voltage is equal to 230 angle[0º] V and the current is 10 angle[30º] A

The virtual power will be 230 (10) angle[0º]+[30º]  = 2,3 angle[30º] KVA

In the other form:

230 angle[0º] = Ro cos(0º)+jRo sin(30º) =230 + j0 V

10 angle[30º] = Ro cos(0º)+jRo sin(30º) =8,678 + j4,968 A

Then, you can multiply with complex numbers thus:

(230 + j0) (8,678 + j4,968) = 1995,94 + j 1142,64 VA

In the  expression of the complex power, S=VI∗S=VI∗, Vand I are phasors and represent sinusoidal variables, respectively voltages and currents. That is, if V=Vefexp(jϕv)V=Vefexp⁡(jϕv) then it represents v(t)=Vefcos(ωt+ϕv)v(t)=Vefcos⁡(ωt+ϕv) and, along the same line, I=Iefexp(jϕi)I=Iefexp⁡(jϕi)represents i(t)=Iefcos(ωt+ϕi)i(t)=Iefcos⁡(ωt+ϕi). 

The 'ef' subscript indicates those are 'effective' or 'rms' (root-mean-square) values of the variables; for any sinusoidal with amplitude VMVM, its rms value is Vef=VM/2√Vef=VM/2.

The real power PP dissipated in a reactive port, where VV is applied at its terminals and IIflows into it, -- the port can be a simple resistor, capacitor or inductor, or it can be a circuit with a combination of these elements and, eventually, also with dependent sources ,-- is given by P=R{S}=VefIefcos(θ)P=ℜ{S}=VefIefcos⁡(θ), where θ=ϕv−ϕiθ=ϕv−ϕi is the difference between the phases of the voltage and the current. (θθ could be the symmetrical value, ϕi−ϕvϕi−ϕv, because the cosine is an even function, but the common standard is  θθ to be the voltage angle minus the current angle, as we defined before).   

Now notice that VI∗=VefIefexp(j(ϕv−ϕi))VI∗=VefIefexp⁡(j(ϕv−ϕi)) and its real part is precisely P=VefIefcos(θ)P=VefIefcos⁡(θ). However, the real part of the product VI=VefIefexp(j(ϕv+ϕi))VI=VefIefexp⁡(j(ϕv+ϕi))is VefIefcos(ϕv+ϕi)VefIefcos⁡(ϕv+ϕi) which is different from PP.

So, in order the real part of SS is equal to P,P, the real power dissipated in the port, the expression S=VI∗S=VI∗ must be used.  (Eventually one could use the conjugate S∗=V∗IS∗=V∗I, because R{S}=R{S∗}ℜ{S}=ℜ{S∗}, but conventions dictate to use the former).

However, it should never be used   S=VIS=VI, because in this case the real part of SS is not the real power dissipated in the port.

Conjugate of either voltage or current can be taken just to find phase angle difference between phase voltage and current in order to find power factor, because Power Factor is cosine of phase angle between voltage and current.

The conjugate of the current determined whether the load is capacitive, resistive or inductive. This will also help as to know if the reactive power is positive or not.

The conjugate of the current is used to calculate the correct phase difference between the voltage phasor and the current phasor.

Power factor is pf = cos (a-b) where a is the voltage angle, b is the current angle. Then we can say Real power P = VIxcos(a-b), and S= P+jQ=VIcos(a-b)+jVIsin(a-b).

If we don't use conjugate, we will have S= VIxcos(a+b) + jVIsin(a+b). This is not S = P+ jQ . where P, Q are defined as P = VIcos(a-b) , Q=VIsin(a-b)! Hence S = VI* where I* is the conjugate.

SO, that the real powers does not becomes negative, we take the conjugate of current which changes the positive or negative reactive power.

I was also searching for this answer, so far, I'm putting this with an example as follows:

We all know that we need to consider the phase difference between voltage and current.

Let's say, V=220

In simply way to find the logic is, a generator should deliver both active and reactive power to satisfy this (s=VI*) is always preferable otherwise -ve part of reactive power shows that absorbing reactive power.

It depends on the reference you take.

If you take Current as reference you don't need t use Conjugate.

If you take Voltage as reference - or an other reference - you don't need t use Conjugate.

So, it is a concept and convention.

S = (I^2).Z =(I.I*).(V/I) = V.I*