Question

Explain powar arpresston in phasor domain VA = V z real componant +rectve componant

Answer 1

Given the question is to explain the power expression, I assume the main doubt would be the explanation behind using a conjugate for calculating power.

Consider a case where we are not using the conjugate and power = V*I

elaborating it, Power = V$\angle \alpha$ I $\angle \beta$ and this product turns out to be VI (cos($\alpha$ + $\beta$) + j cos( $\alpha$ - $\beta$)). For suppose if $\alpha$ + $\beta$ > 90^o then power comes out to be of the form -P + jQ (P and Q are just representations, they hold no significant values)which is unacceptable for the real part to be a negative quantity

On the other hand, if we use the conjugate of current which is I^*, then Power = V $\angle \alpha$ I
- B
and this turns out to be VI (cos ($\alpha$ -  $\beta$) - j cos ($\alpha$ + $\beta$)). Here, we have to analyse the problem by considering to situations,

Case 1: if $\alpha$ > $\beta$(Voltage leads current), then the difference is positive and less than 90⁰ hence the Power comes out to be of the for P - jQ which by the fundamentals we know is an inductive one.

Case 2: if $\alpha$ < $\beta$(Voltage lags current). Could you guess what would happen? Yes, it becomes a capacitative one.

Hence this summarizes the use of phasor in the power expression and precisely why we use a conjugate for our calculation of power.