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?

  • \$\begingroup\$ 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. \$\endgroup\$ – Andy aka Jan 14 at 11:36
  • \$\begingroup\$ @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. \$\endgroup\$ – 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.

  • \$\begingroup\$ Thanks! I did not know this. Also , are you aware of any other formula which uses conjugate and is simply not the product? \$\endgroup\$ – Ronald Becker Jan 14 at 3:37
  • \$\begingroup\$ I think you can find such formulas quite in any field which uses complex representation of physical quantities, e.g. signal theory. \$\endgroup\$ – DavideM Jan 14 at 11:06

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