1
\$\begingroup\$

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?

\$\endgroup\$
2
  • \$\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
0
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.