I am somewhat troubled by the 'basic' formula that describes the relationship between apparent power (S), active power (P) and reactive power (Q) in AC systems. This formula is S^2 = P^2 + Q^2 and makes sense to me in linear systems with or without phase shifts. But quite some modern loads contain large amounts of harmonic currents that in my view can not create any work (active power) if the voltage is purely sinusoidal. But if apparent power is defined as S = V_rms * I_rms, the harmonic content adds up in I_rms, thus making S > P. A simple way to think about this, is that the difference can be called reactive power Q by adhering to S^2 = P^2 + Q^2. But I wonder whether that would be correct. Since I always believed Q is defined as reactive power with the vector component of the current perpendicular to the voltage by the relationships P = V_rms * I_rms * cos(phi) and Q = V_rms * I_rms * sin(phi). I subsequently would believe higher harmonics in currents combined with the same harmonics in the voltage could have these relationships as well, such that higher harmonics could create both active and reactive power. But if the voltage is pure sinusoidal, it has no higher harmonics and the current harmonics will not create reactive power (nor active power) in my view. It would be great if this was generally seen as 'distortion power' D that adds to P and Q in a quadratic way (analogous to displacement and distortion power factors), but I have not seen this.

My question is whether there is a definition of power generated by harmonics that will fix the difference between active and apparent power. And I am wrong assuming that if that difference (in a quadratic way) is called Q, that the definition of reactive power should be different?

I have made a non-linear system that has a sinusoidal voltage and a square current that are in-phase and I did some calculations to get an idea of the 'missing' power. The active power is in-line with what I expect and I can calculate it in two ways. Firstly by integrating the voltage and current over a multiple of their period and dividing it by the elapsed time (simple average of the product in this case actually). Secondly by multiplying the RMS voltage by the RMS value of the fundamental harmonic of the current, knowing I do not have to multiply that by cos(phi) since these are in-phase. These values match and I have showed them in the output of my copied script below. There is no active power by higher harmonics, since the voltage does not have higher harmonics. And the same is true for reactive power of both fundamental and higher harmonics: this is zero for the fundamental harmonic since the voltage and current are in-line. And is zero for higher harmonics given that the harmonic values of the voltage are zero and thus the product will be zero.

import numpy as np
import scipy as sp
import matplotlib.pyplot as plt

def rms(x):
    return np.sqrt(np.mean(x**2))
def fundamental(x):
    N = len(x)//periods
    X = abs(2/N*sp.fft.fft(x, N))
    return X[1]*np.sin(w*t)

freq = 60
w = 2*np.pi*freq
periods = 4
t = np.linspace(0, periods/freq, 10000)
v = np.sin(w*t)

i = sp.signal.square(w*t)

plt.plot(t, v, label='v(t)')
plt.plot(t, i, label='i(t)')

print('RMS Voltage =', rms(v),
      '\nRMS Current =', rms(i))
print('Active Power:',
     '\n  integral of v*i of one period divided by that period = ', np.mean(v*i))
print('Active Power by the fundamental current (phase angle is zero):',
     '\n  V_1_rms * I_1_rms = ', rms(v) * rms(fundamental(i)))
print('Active Power by higher harmonic currents:',
     '\n  V_n_rms * I_n_rms for n>=2 = 0 * I_n_rms = 0')
print('Apparent Power:',
     '\n  V_rms * I_rms =', rms(v) * rms(i))

enter image description here

Based on this example I would come to the following result for 'missing' power: enter image description here

Is there a definition for this value that I calculated? It is quite substantial at almost 50% of the active power.

  • 1
    \$\begingroup\$ "But if the voltage is pure sinusoidal, it has no higher harmonics and the current harmonics will not create reactive power in my view." Did you mean active? You can indeed treat each harmonic independently: apply Parseval's theorem. Related, though not an exact answer: electronics.stackexchange.com/a/633117/311631 \$\endgroup\$ Commented Dec 20, 2022 at 6:19
  • 1
    \$\begingroup\$ S, P, and Q are only defined for fundamental... All other are used for "distortion". \$\endgroup\$
    – Antonio51
    Commented Dec 20, 2022 at 9:40
  • 2
    \$\begingroup\$ True power is what it is (harmonics or not). Apparent power is what it is (harmonics or not) and, if you want to call what makes up the difference reactive power, then so be it. Are you trying to fix something that isn't broken I wonder? Is there any use for reactive power that shouts at needing a strict definition? \$\endgroup\$
    – Andy aka
    Commented Dec 20, 2022 at 10:24
  • 1
    \$\begingroup\$ @JeromeBu1982 S^2=P^2+Q^2+D^2. I think this is correct (Parseval theorem or identity?). It is a long time ago, I used some power analyzer. Don't remember how all parameters were measured ... Will search. \$\endgroup\$
    – Antonio51
    Commented Dec 20, 2022 at 12:27
  • 1
    \$\begingroup\$ Here are, for reference, the measured parameters with Fluke 430 Power Analyser i.sstatic.net/ecbnI.png \$\endgroup\$
    – Antonio51
    Commented Dec 20, 2022 at 12:45

2 Answers 2


The missing power can simply be called 'Distortion Power' \$D\$, unit VA. Per the following equation: $$ V_{rms}^2I_{rms}^2 = S^2 = D^2 +\sum_h P_h^2+\sum_h Q_h^2 $$ , where h is the number of the harmonic (source: https://electronics.stackexchange.com/a/489895/199042)

In the somewhat extreme example in the question, with a sinusoidal voltage and an in-phase square wave current, the RMS values, active power (P) and the 'missing' power (D) were calculated by a 10,000 sample simulation. These values are: (rounded to 3 decimals): $$ \begin{matrix} V_{rms} & = & 0.707\\ I_{rms} & = & 1\\ S & = & 0.707\\ P & = & 0.637\\ Q & = & 0\\ D & = & 0.308 \end{matrix} $$

These values are correct and can also be derived mathematically: $$ \begin{matrix} \displaystyle V_{rms} &=& \frac{1}{2}\sqrt{2} \mbox{ (known quantity for sinusoid with amplitude 1)}\\ I_{rms} &=& 1 \mbox{ (sqrt of area under the squared values divided by period)}\\ S &=& \displaystyle V_{rms}*I_{rms}\\ &=&\displaystyle \frac{1}{2}\sqrt{2} \\ P^2 &=& \displaystyle\sum_hP_h^2 \\ &=& \displaystyle \sum_h V_h^2 I_h^2\cos(\theta_h) \\ &=& \displaystyle V_1^2I_1^2\\\ & & \mbox{since } V_h=0 \mbox{ for } h>=2 \mbox{ and }\theta_1=0^\circ \end{matrix} $$

That leaves the calculation of the fundamental current \$I_1\$. This can be done by the fourier series for a square wave f(t). By expressing a square wave f(t) as an infinite sum of sineways with harmonic frequencies per: $$ f(t)=\frac{4}{\pi}\sum_{h=1,3,5,...}^\infty\frac{1}{h}\sin{\frac{h\pi t}{T}} $$ Where \$I_1\$ is the RMS value of the fundamental current, we can call \$i_1(t)\$ the fundamental current in the time domain. This leads to:

$$ \begin{matrix} \displaystyle i_1(t) = \frac{4}{\pi}\frac{1}{h}\sin{\frac{h\pi t}{T}}\ \mbox{ with } h=1 \\ \end{matrix}$$ \$I_1\$ is then simply \$\frac{1}{2}\sqrt{2}\$ times the amplitude of \$i_1(t)\$: $$ I_1 = \frac{1}{2}\sqrt{2}\cdot \frac{4}{\pi} $$ Thus: $$ \begin{matrix} P &=& V_1\cdot I_1\\ & =& \frac{1}{2}\sqrt{2}\cdot \frac{1}{2}\sqrt{2}\cdot \frac{4}{\pi}\\ &=&\frac{2}{\pi}\\ &\approx & 0.637 \end{matrix} $$ \$D\$ is then simply: $$ \begin{matrix} D^2&=&S^2-P^2-Q^2\\ &=&{(\frac{1}{2}\sqrt{2})}^2 - (\frac{2}{\pi})^2\\ D&\approx &0.308 \end{matrix} $$ And for reactive power: $$ \begin{matrix} Q^2 &=& \displaystyle\sum_hQ_h^2 \\ &=& \displaystyle \sum_h V_h^2 I_h^2\sin(\theta_h) \\ &=& \displaystyle 0\\ & & \mbox{since } V_h=0 \mbox{ for } h>=2 \mbox{ and }\theta_1=0^\circ \end{matrix} $$

This confirms the notion that the equation \$S^2=P^2+Q^2\$ simply does not hold in non-linear AC systems. It is true that quite some power can be in the form of non-reactive, higher harmonic power while not creating any work. Still many will call the quadratic difference between \$S\$ and \$P\$ simply \$Q\$. This is incorrect for many modern non-linear AC systems. \$Q\$ is the reactive component, orthogonal to the active power.

  • 1
    \$\begingroup\$ P, Q, and D are all "orthogonal" in geometric representations (3 axes) ... \$\endgroup\$
    – Antonio51
    Commented Dec 22, 2022 at 14:12
  • \$\begingroup\$ Thank you for pointing me to orthogonality. I used it implicitly. Now with Parceval’s theorem it is explicit. It makes the formulas work. And yet, I think the classical thinking of Q is just related the displacement of the fundamental. With the risk of putting the reactive harmonics in the “D bucket”. That would break orthogonality again. I am interested still how modern text books separate the classical thinking of displacement by machinery with the (mostly?) non-displaced harmonics by modern switched power supplies \$\endgroup\$ Commented Dec 22, 2022 at 14:49

Ah, I happened across this reference which may be handy:

"Power measurement techniques for non-sinusoidal conditions", S. Svensson, doctoral thesis, Chalmers University of Technology, Electric Power Engineering, Göteborg, Sweden, 1999
A review of reactive power accounting methods is provided in the first section.

See also: How to calculate reactive power of a capacitor with a current waveform which consists of a sum of sinusoids?

  • \$\begingroup\$ 150 pages with vary variable input on this subject. Good link! \$\endgroup\$ Commented Mar 18, 2023 at 16:47

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