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Consider an RC circuit which is supplied by a sine source

$$v(t)=V_m\cdot \sin(\omega t)$$

The current through the circuit will be

$$i(t)=I_m\cdot \sin(\omega t+\phi) \\ \phi=\arctan \left(\dfrac{1}{\omega RC} \right)$$

The instantaneous power equation is

$$p(t)=V_m\cdot I_m\cdot \sin(\omega t)\cdot\sin(\omega t+\phi)$$

enter image description here

Clearly from this image we can see that the power equation has some frequency component also.

The average of this instantaneous equation over one cycle will give us average power.

The real power is going to be the first term which, however, has a zero and a high frequency component in it and the reactive power however has a high frequency component only (by high frequency I mean the value of frequency is higher (i.e.) twice the supply frequency.)

Are real power, active power and average power the one and the same same?

If they are not, then we don't see text books mentioning this high frequency component in the real power/active power/reactive power.

They directly give the average value as zero frequency component. This reactive power that is mentioned here always has only a high frequency component here. In the case of power systems, if we are speaking about reactive power are we just speaking about the peak reactive load demands and quantifying it for load flows and in case of real power in power system will we be speaking only about average value of real power load demands and proceeding with the load flow?

I lose clarity here and I am sure that I am wrong at some basic point. I would be grateful if someone is able to clarify this.

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  • \$\begingroup\$ The first term is actually INSTANTANEOUS Real Power. if you take average of that continuous-time power function by integral from 0 to 2pi, you will end up with the Average Power = Vrms.Irms.cos(theta). This is what is called "Real Power" in ac circuits, which has physical significance. \$\endgroup\$
    – Mitu Raj
    Jun 3 '21 at 11:15
  • \$\begingroup\$ Thank you sir, But my question is not that sir. Why dont we use this same average approach to find the value of this reactive power. Clearly reactive power is the peak value of the second term. Why is this difference? \$\endgroup\$ Jun 4 '21 at 16:31
  • \$\begingroup\$ What's the need of taking it while its average value is zero. \$\endgroup\$
    – Mitu Raj
    Jun 4 '21 at 18:19
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What is the exact difference between active power, average power, and real power?

I think you're confusing reactive power with instantaneous reactive power, so let's clarofy the terminology.

  • Instantaneous active current (\$i_\text{a}(t)\$) of a two-terminal device or network: in sinusoidal steady-state, it is that component of the instantaneous current \$i(t)\$ that is in phase or 180° out of phase with the instantaneous voltage \$v(t)\$.
  • Instantaneous reactive current (\$i_\text{r}(t)\$) of a two-terminal device or network: in sinusoidal steady-state, it is that component of the instantaneous current that is in quadrature (i.e. 90° out of phase, either leading or lagging) with the instantaneous voltage.
  • Instantaneous power (\$p(t)\$) of a two-terminal device or network: it is the time rate at which energy is flowing or being transferred, or the time rate at which work is being done. That is, it's the rate at which energy flows into (or out of) a two-terminal lumped network or device. This is true whether in steady-state or transient state, without or with harmonics, in balanced or unbalanced conditions. In sinusoidal steady-state, just like instantaneous current, it can be decomposed into an active and reactive component.
  • Instantaneous active power (\$p_\text{a}(t)\$) of a two-terminal device or network: in sinusoidal steady-state, it is that component of the instantaneous power that corresponds to the instantaneous active current, and represents the time rate of the energy flow that is unidirectional (i.e. at all instants it flows into or out of the two-terminal device or network).
  • Instantaneous reactive power (\$p_\text{r}(t)\$) of a two-terminal device or network: in sinusoidal steady-state, it is that component of the instantaneous power that corresponds to the instantaneous reactive current, and represents the time rate of the energy flow that is bidirectional (i.e. for half a cycle it flows into the two-terminal device or network, then for the next half cycle it flows out of it).
  • Active power, average power or "real" power (\$P\$) of a two-terminal device or network: in steady-state, it is the average (mean) of instantaneous power. So active power is the average flow of energy. It is constant.
  • Reactive power, quadrature power or "imaginary" power (\$Q\$) of a two-terminal device or network: in sinusoidal steady-state, the absolute value of reactive power is the amplitude (maximum value) of the rate of flow of the energy oscillating between the two-terminal network or device under discussion and the external network. It is constant. I haven't seen this explanation of reactive power in any textbook or webpage or video, but is correct and I proved it here; I've only seen it in the IEEE standard 1459, version of the year 2010, as shown next:

IEEE std 1459-2010 defining reactive power in sinusoidal steady-state for a two-terminal device or network

Examples:

  • If a two-terminal device operating in sinusoidal steady-state has an instantaneous power of "+4.5 W" (where the voltage polarity and current direction satisfy the passive sign convention), it means that, at that instant in time, energy is flowing into the device at a rate of 4.5 joules each second.
  • If a two-terminal device operating in sinusoidal steady-state has an active power of "-15 W" (where the voltage polarity and current direction satisfy the passive sign convention), it means that, on average, energy flows out of the load at a rate of 15 joules each second.
  • If a two-terminal device operating in sinusoidal steady-state has a reactive power of "+-10 VAR", it means that the maximum rate of flow of the energy oscillating between that device and the external network is +10 joules per second.

(I'll assume sinusoidal steady-state [periodic waveforms with no harmonics].)

Is reactive power periodic in nature?

Reactive power? Not in time. Reactive power is a constant. If something is periodic in time, it must be variable in time. A constant is independent of time.

If instead you mean instantaneous reactive power, then the answer is yes, it is periodic in time. Instantaneous reactive power is sinusoidal with zero average value, and its amplitude is the absolute value of reactive power. Similarly, instantaneous active power is also periodic in time, but it is not sinusoidal (it's a sinusoid squared).


Edit as per comments from OP

What is the necessity to use Average value for the instantaneous real power and the peak value for the instantaneous reactive power. All could have been on the same scale that Real power could also be the average value, and Reactive power could also be the average value. Or both reactive as well as the real power could be peak value. Of course, you could instead use the amplitude

In a two-terminal device or network operating in sinusoidal steady-state, the following is true:

  • The time average of the instantaneous power and of the instantaneous active power is \$P\$.

  • The amplitude of the instantaneous power is \$|P|+S\$.

  • The amplitude of the instantaneous active power is \$2|P|\$.

  • The time average of the instantaneous reactive power is \$0\$.

  • The amplitude of the instantaneous reactive power is \$|Q|\$.

  • The instantaneous power, instantaneous active power and instantaneous reactive power are all periodic in time with fundamental cyclic frequency twice the fundamental cyclic frequency of the instantaneous voltage \$v(t)\$ and instantaneous current \$i(t)\$.

So, as you can, see with the current definition of active power, we can refer to both averages and amplitudes of instantaneous power and instantaneous active power in terms of the active power. There's no need to define active power as the amplitude of instantaneous active power or of instantaneous power.

And why don't we define (the absolute value of) reactive power as the time average of instantaneous reactive power? Because it'd be zero, which isn't useful; in fact, we also use amplitude (or RMS/effective value) for sinusoidal voltages and currents, since their average is zero (assuming the DC offset is also zero).

I made an online GeoGebra app here that shows the plot of the waveform of the instantaneous voltage (green curve to the left), instantaneous current (yellow curve to the left), instantaneous active current (blue curve to the left), instantaneous reactive current (red curve to the left), instantaneous power (purple curve to the right), instantaneous active power (gray curve to the right), and instantaneous reactive power (brown curve to the right), for a linear time-invariant RL network excited by a sinusoidal voltage and operating in (sinusoidal) steady-state (click image to amplify it):

Waveforms in RL networks

and for a linear time-invariant RC network also excited by a sinusoidal voltage and operating in (sinusoidal) steady-state:

Waveforms in RC networks

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  • \$\begingroup\$ Hello sir, Thank you so much for your reply. \$\endgroup\$ Jun 3 '21 at 3:25
  • \$\begingroup\$ What is the necessity to use Average value for the instantaneous real power and the peak value for the instantaneous reactive power. All could have been on the same scale that Real power could also be the average value, and Reactive power could also be the average value. Or both reactive as well as the real power could be peak value. Why is this discrimination seen between these two kinds of power for the AC supply? \$\endgroup\$ Jun 3 '21 at 3:33
  • \$\begingroup\$ I've updated my answer to try to answer your questions. \$\endgroup\$
    – alejnavab
    Jun 3 '21 at 9:17
  • \$\begingroup\$ Thank you so much sir. I have got an idea. So the only reason to use average value for Real Power is it will be a non-zero value, and for Reactive power, Voltage and Current profile the average value is zero and hence we always represent it by the peak amplitude. But when we represent Voltage and Current with RMS Value is there any specific for not representing the reactive power using the rms value. Also the real power could also have been represented uniformly with a peak value similar to current,voltage,reactive power. But it is not. Is there any specific reason for this sir? \$\endgroup\$ Jun 3 '21 at 16:02
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I will be talking about the ideal cases where there are no harmonics, as I presume you're talking about, given your derivation.

Are real power, active power and average power the one and the same same?

The active power is the same as the real power. And any power is instantaneous by nature, i.e. at any instance of time it has a value. Averaging the instantaneous power results in an average, and this average value, like the instantaneous value, can come from any power.

Therefore: active power == real power and they refer to a specific type of power, while the average is the mathematical average performed on any quantity. No averaging means instantaneous.

we don't see text books mentioning this high frequency component in the real power/active power.

We certainly do, since it's part of the very nature of the multiplication: there are two sines, multiplied, which gives the trigonometric equivalent of \$\cos(2\omega)\$. But the average of it is a fixed, non-oscillating value.

They directly say that the [???] and they directly give only the average value as real/active/average power.

It seems you missed a few words ther, but even so, the part where "they" give only the average values is the part where only those matter for the counter, or for load-flow analysis. Remember that the counter performs an averaging in time. The result at the end comes out a fixed number.

This reactive power that is mentioned here always has only a high frequency component here

You are misleading yourself by not continuing the derivation:

$$\begin{align} p(t)&=\dfrac{V_pI_p}{2}\{[1-\cos(2\omega t)]\cos(\theta)+\sin(2\omega t)\sin(\theta)\} \tag{1} \\ &=\dfrac{V_pI_p}{2}[\cos(\theta)-\cos(2\omega t)\cos(\theta)+\sin(2\omega t)\sin(\theta)] \\ &=\dfrac{V_pI_p}{2}[\cos(\theta)-\cos(2\omega t+\theta)] \tag{2} \\ &=\qquad{\bar p(t)}\qquad +\qquad{\tilde p(t)} \tag{3} \end{align}$$

Now you can see that there is a fixed value, \$\cos(\theta)\$, and an oscillating value at twice the frequency, which is naturally occuring when two sines are multiplied. The fixed value is nothing but the average. Since the cosine is an even function, the average is never negative, while the oscillating part never goes beyond twice the amplitude.

In the case of power systems, if we are speaking about reactive power are we just speaking about the peak reactive load demands and quantifying it for load flows

Remember that the total power, S, is made of both active, P, and reactive, Q, powers, and their relation is orthogonal: \$S=\sqrt{P^2+Q^2}\$. And S is calculated based on the RMS values of the voltage and current. This means that no matter what displacement exists, the RMS values will be divided by \$\sqrt2\$, and their multiplication will always be one half of the peak values. The instantaneous values will have a frequency twice the fundamental, and its peaks will never be more than twice the \$\bar S\$. For example, if V=3 and I=2, S=3 and the peak will never be above or below ±6. Here shown for an angle varying from 0 (blue) to π/[2,3,4,6 (red)]):

S

and in case of real power in power system will we be speaking only about average value of real power load demands and proceeding with the load flow?

The load-flow assumes a behaviour in time, so the instantaneous values make little sense here. Therefore the only quantities of interest are magnitude and phase, which give the relevant average values.

As a final note, when talking about systems with harmoncs, the same reasoning as above can be used. For instantaneous values the displacement is now relevant only to one harmonic, and the total harmonic distortion (THD) takes place for the overall effect, while for a load-flow analysis, the same averages in time are done with the help of square-root sums of powers.

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  • \$\begingroup\$ I've updated my answer, hopefully it addresses all your questions now. \$\endgroup\$ Apr 30 '21 at 8:48
  • \$\begingroup\$ Thank you sir. My actual question is, the expression that you have written can be expanded in the form given in the question. And while writing in this form we see that active and reactive power are periodic. In books they say average/real power to be the first part(dc value) of the active power. And reactive power to be the peak value. Why is it so different that one is average and the other is peak. Why cant both be same? \$\endgroup\$ May 6 '21 at 12:02
  • \$\begingroup\$ @SangeerthPrabakar Those "books" are mixing instantaneous values with averages and RMS values. What you're showing is the result of p(t), or the instantaneous value of the active power. It has a DC term, \$\cos\theta\$, and an oscillating term at twice the frequency. The average of that whole expression is \$\cos\theta\$, or P. As shown in the answer \$Q=\sqrt{S^2-P^2}\$. If you consider the instantaneous formula, you need to consider all of p(t). If you're interested in P, only, then \$Q=\sqrt{1-\cos^2\theta}=|\sin\theta|\$. \$\endgroup\$ May 6 '21 at 14:35
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Both AC current and voltage are complex numbers. If you say the voltage is now 230 volts, it's actually 220∠0 (in degrees). You may write it as 220 + 0j. If you apply this voltage across an autotransformer, you may see the current is .3 0r 0.4 amp. But it's actually 0.3∠40 (as an example). You may write it as 0.223 + 0.1928j.

Now if you want to find the power consumed by the autotransformer you need to multiply the voltage with the complex conjugate of current. i.e. (220 + 0j)(0.223 - 0.1928j). The product will be 50.55 + 42.42j.
In simple form, Power $$ s = vi^* = |vi|∠\theta =re^{j\theta} = r(cos \theta + jsin\theta) = a+jb $$
The real part of s is the real power and the imaginary part is the reactive power. They are not the same thing. And from Euler's equation, you can see that they are periodic.

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  • \$\begingroup\$ It's periodic with θ (the phase difference between voltage and current) but is not periodic with time, am I right? \$\endgroup\$ Apr 29 '21 at 2:01

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