# Power calculation

1. May 24, 2015

### 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?

2. May 24, 2015

### davenn

what is S ?

3. May 24, 2015

### 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.

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 $S=VI^\star$.

4. May 24, 2015

### davenn

Thanks Simon

was wondering what the heck he was talking about

5. May 24, 2015

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].

6. May 25, 2015

### Simon Bridge

... erk: here lets 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.