Power Calculation: S=VI & V*I Meaning

[smruti]

in the calculation of complex power S=VI we sometimes use S=VI* and sometimes as S=V*I.what is the physical significance of these two notations and in which case these two expression has different meaning?

 

[davenn]

what is S ?

 

[Simon Bridge]

The difference is in the phase of the reactive power.
To understand that - you need to understand the significance of the reactive power.
http://www.allaboutcircuits.com/vol_2/chpt_11/2.html

The two relations get you the same magnitudes for the components and you normally only need the power delivered to the load anyway.
If you derive the equation for S from the usual phasors, then you get $S=VI^\star$.

 

[davenn]

Thanks Simon

was wondering what the heck he was talking about

 

[Babadag]

I agree with Simon, of course.In my opinion we use the conjugate current in formula S=VxI* since in an inductive circuit the actual current lags the voltage but the reactive power is considered positive[conventional].So if S[apparent power]= VxI=(Vre)x(Ire-jIim)=(VrexIre)+(-VrexIim)j then Q=-Vre*Iim.
If S[apparent power]= VxI*=(Vre)x(Ire+jIim)=(VrexIre)+(VrexIim)j and Q=+Vre*Iim.
If the current leads the voltage [capacitive circuit] then the reactive power is negative[conventionally].

 

[Simon Bridge]

S[apparent power]= VxI*=(Vre)x(Ire+jIim)=(VrexIre)+(VrexIim)j and Q=+Vre*Iim.

... erk: here let's tidy that up a bit:

$S = VI^\star = V_{re}\big(I_{re}+jI_{im}\big) = V_{re}I_{re} + j V_{re}I_{im}$ and $Q=+V_{re}I_{im}$.

... better? But did you take the conjugate properly? - I decided not to wade through all those letters to check.

LaTeX: worth the learning curve.