# Calculation of active power using two waveforms with harmonics

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?

• Yep you're definitely quite right! – carloc Sep 29 '19 at 12:48
• @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? – MJ13 Sep 29 '19 at 13:43
• 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. – carloc Sep 29 '19 at 15:08

$$\\small A\angle \theta \times B\angle \phi = AB\angle (\theta + \phi)\$$, so phase angles should be $$\\small -68.2^o\$$ and $$\\small -55.4^o\$$. When taking the cosine, the phase angles can be regarded as positive if desired.
• $\small 200\sqrt 2\:sin (\omega t)$ looks like the $\small 0^o$reference phasor, hence $\small 26.19\:sin (\omega t-68.2)$ lags by 68.2 deg, hence negative. – Chu Sep 29 '19 at 14:43