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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?

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    \$\begingroup\$ Yep you're definitely quite right! \$\endgroup\$
    – carloc
    Commented Sep 29, 2019 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
    Commented Sep 29, 2019 at 13:43
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    \$\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
    Commented Sep 29, 2019 at 15:08

2 Answers 2

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\$\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.

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  • \$\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
    Commented Sep 29, 2019 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
    Commented Sep 29, 2019 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
    Commented Sep 29, 2019 at 15:01
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You are right ...

Here is a Maple sheet that shows ... see results: pow vs p1__p2.
pow is calculated as usual, averaging over 1 s.
p1__p2 is calculated with the formulas in the Q, angles corrected.

enter image description here

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