0
\$\begingroup\$

I came up with the following example on a textbook I was given.

Let's suppose we have an R-L load. I want to calculate the active power P when the voltage and the current applied on the load are given with the following functions: $$u(t)=200 \sqrt2\sin(ωt)+ 200 \sqrt2\sin(5ωt+30^{\circ}) $$ $$i(t)=26.19\sin(ωt-68.2^{\circ})+ 6.22\sin(5ωt-85.4^{\circ}) $$

In the answer it is given that we get: $$ P = \tilde{V_1} \tilde{I_1} cosφ_1 + \tilde{V_5} \tilde{I_5} cosφ_5 \Rightarrow $$ $$ P = 200\frac{26.19}{\sqrt2}cos(68.2^{\circ}) + 200\frac{6.22}{\sqrt2}cos(85.4^{\circ}) $$

What I don't get is why the angle φ5 is not calculated as : $$ φ_5 = 30^{\circ} - (-85.4) = 115.4 ^{\circ} $$ since $$ φ_n = θ_v - θ_I $$ where θv is the angle of the voltage harmonic and θi is the angle of the current harmonic. Isn't this the right was to calculate it or am I missing something?

\$\endgroup\$
  • 1
    \$\begingroup\$ Yep you're definitely quite right! \$\endgroup\$ – carloc Sep 29 '19 at 12:48
  • \$\begingroup\$ @carloc Thanks for the response!So whenever I have higher order harmonics in the same frequency at both the voltage and the current waveform then for the angle φ I should always take the difference between the angle of the voltage harmonic and the angle of the current harmonic, right? \$\endgroup\$ – MJ13 Sep 29 '19 at 13:43
  • 1
    \$\begingroup\$ Just consider that different harmonics are orthogonal, i.e. they can be seen as separated processes going on in the same hardware, they don't exchange any average energy in a period. So, again yes each voltage harmonic versus its own current. \$\endgroup\$ – carloc Sep 29 '19 at 15:08
0
\$\begingroup\$

\$\small A\angle \theta \times B\angle \phi = AB\angle (\theta + \phi)\$, so phase angles should be \$\small -68.2^o\$ and \$\small -55.4^o\$. When taking the cosine, the phase angles can be regarded as positive if desired.

\$\endgroup\$
  • \$\begingroup\$ I don't think this is the correct way to find the angle φ.Because in the calculation of the reactive power Q the correct angle for the sinφ1 is 68.2 and not -68.2 which gives us a negative result. \$\endgroup\$ – MJ13 Sep 29 '19 at 14:30
  • \$\begingroup\$ \$\small 200\sqrt 2\:sin (\omega t)\$ looks like the \$\small 0^o\$reference phasor, hence \$\small 26.19\:sin (\omega t-68.2)\$ lags by 68.2 deg, hence negative. \$\endgroup\$ – Chu Sep 29 '19 at 14:43
  • \$\begingroup\$ I am pretty much sure that θ (Theta) is the phase angle between the voltage and the current (θv - θi). Check here: electronics-tutorials.ws/accircuits/power-in-ac-circuits.html \$\endgroup\$ – MJ13 Sep 29 '19 at 15:01

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.