# What am I doing wrong in finding the phase angle of power?

Consider a voltage supply $$\V\angle \phi_1\$$ and a circut with impedance $$\Z \angle \phi_2\$$.

Now the Apparent power($$\S\$$) is given by: $$\bar S =\frac{(\bar V)^2} {\bar Z}$$ Thus phase angle of $$\S\$$ is $$\(2\times\phi_1)-(\phi_2)\$$ but shouldn't it be just $$\(\phi_2)\$$? As $$\\cos(\phi_2)=\frac R Z=\frac P S\$$ .

Can someone tell me where I made a mistake?

• The frequency of S is twice that of V so, I'm unsure what phase really means to you because it is changing twice as fast as V phase changes. – Andy aka Jan 14 at 11:36
• @Andy now that you've said that I see that S and V canot have a constant phase difference. What I meant as phase was probably the power factor angle from the power triangle. And using phasor analysis I got wrong value. I see the error now though :D. Ps I am just introduced to phasors so I didn't know much about terminology. Sorry for the confusion. – Ronald Becker Jan 15 at 12:54

The mistake is in the definition of apparent power, which should be: $$\overline{S} \triangleq \overline{V}\overline{I}^*$$ In this way, by substituting for the current you get $$\overline{S}= \overline{V}\frac{\overline{V}^*}{\overline{Z}^*}=\frac{|V|^2}{\overline{Z}^*}$$ so that the phase of the complex power is simply the phase component of the impedance.