Talk:AC power

Real, reactive and apparent power

In the UK, the standard term, defined in British Standards, is "Active Power" rather than "Real Power". Most electrical engineering textbooks that I have used use this same term. Hence, I have edited this article to use "Active Power" as the preferred term, while retaining a mention of "Real Power". I hope this does not conflict with usage in other English-speaking countries. 86.142.3.205 (talk) 16:39, 27 June 2010 (UTC)

Real power, reactive power and apparent power should be combined into one article. This is a mess - repetitive and I'm not sure the three articles are even consistent with each other. --Wtshymanski 04:30, 3 Apr 2005 (UTC)

agreed, assuming appropriate redirects. fyi, you seem to imply that the apparent power is the product of the Vrms and Irms only for a sine wave. This is not true. The rms calculation is not waveform dependent, although the phase separation calculation is based on a sine wave. I also find your use of the term "linear" in this context to be somewhat quirky. Specifically, the juxtaposition with containing only... as you might recall, resistance, capacitance, and inductance is a basis set for a waveguide or conductor with linear response. Moreover, it is the equivalent transfer function that is linear, not the network itself. I admit this is a slight nitpicking of usage but linear is a dangerously overloaded term. (unsigned comment by Pearlg)

we probablly wan't to bring in complex power as well especially since apparent power is just the magnitude of complex power. Plugwash 00:39, 22 Apr 2005 (UTC)

ok well i've done it what do you think of the result?

I am very happy with all you've done with this article so far. I have one objection to "Apparent power is handy for rough sizing of generators or wiring, especially when the power factor is close to 1." As it happens apparent power is very important when the power factor is far from 1. Indeed, when the power factor is close to 1, an engineer need only consider the real power for sizing, but as the power factor declines, these calculations must be made on the basis of the apparent power. The turbines suppling energy to a generator operating at 100% efficiency MUST be conveying energy at a power equal to the real power + losses due to the reactive power. From the stand-point of a person or turbine turning the generator the apparent power affects how much power must be generated to deliver the real power. --Pearlg 03:10, 24 Jun 2005 (UTC), --Pearlg 23:07, 24 Jun 2005 (UTC) (note: this comment was edited by its authour AFTER it was replied to)

sorry but that doesn't make sense. reactive power is just a quirk of the way electrical cuircuits work it doesn't represent ANY net transmission of power. Indeed in a balaced 3-phase system it doesn't even represent any instantanious trasfer of power. so if the power flowing into the generator really is equal to the apparent power then the generator would make insane amounts of heat with low power factor loads. do you have any sources for the claim you just made?! Plugwash 03:40, 24 Jun 2005 (UTC)

No, it is more than a quirk of the way electrical circuits work. Though *ideally* it does not represent any net transmission of power, in practice it does. Unfortunately, if you look at the magnitude of Z of a material through which the power is conveyed or used, you will find that there is resistance (ac) at the oscillation frequency. Though experiment: take a ring of copper. Induce a current in it. The current wave will die out overtime. Repeat with a superconductor: you can store energy for a very long period of time. So it goes with reactive power. Pushing the reactive power back and forth along the conductor induces losses. Thus, as the power factor degrades, you must size up all your components... the amounts of your active materials and your dielectrics on the basis of S. The joule loss of an m-phase balanced system is given by dP = m*r*(S/V)^2 where V is the line-neutral voltage, S is the apparent power, and r is the magnitude of Z.

"Every elementary nonactive component of S has two basic attributes:

• It causes power loss in the system
• It is the amplitude of an oscillation of power which does not contribute to a continuous unidirectional transfer of energy" (Emanuel, A., "On the Definition of Power Factor and Apparent Power in Unbalanced Polyphase Circuits with Sinusoidal Voltage and Currents" IEEE Transactions on Power Delivery, Vol 8., No. 3, July 1993

"The hottest spot temperature in transforms, alternators and cables is a function of [the Apparent Power]" (ibid)

So, the load seen by the generator depends on not only the energy transferred but on the energy losses from both the energy transfer and the oscillation. --Pearlg 23:07, 24 Jun 2005 (UTC)

---

The concepts of reactive and apparent power may be extended to multi-phase systems by summing up the apparent powers of the individual phases. It should however be noted that in an unbalanced multi-phase system the overall apparent power may be higher than ${\sqrt {P^{2}+Q^{2}}}$ where P and Q are the total real and reactive powers. -- I'd like a cite for this.

The total VA is the sum of the VA figures for each phase, with the voltage taken between line and neutral. Historikeren (talk) 17:19, 14 November 2014 (UTC)

Apparent power is used to describe the power load as seen by a generating source. It is the vector sum of the real power, which represents energy transferred from the source to the load, and reactive power, which represents energy that circulates between the source and the inductive and capacitive energy storage elements of the load. It is typically of most interest in power transmission and distribution. $|S|^{2}=P^{2}+Q^{2}$

move later to better spot complex power is a complex quantity which captures information about both the magnitude and type of power consumed by a load. It can be defined as the product of the phasor representation of rms voltage and the complex conjugate of the phasor representation of current. Real power is the real part of complex power, reactive power is the imaginary part of complex power and Apparent power is the magnitude of complex power. $S=P+jQ$

All the math for calculating Watts, VARs, and VA, assumes that the voltage and current waveforms are purly sinusoidal. If any harmonic content is present, these calculation produce incorrect results. 173.165.52.29 (talk) 20:53, 17 January 2012 (UTC)

Who coined the misleading and incorrect term 'Apparent Power'? The power is the power, it needs no adjectives. If a term is needed for active power, name is what it is, the average power. Volts, amps and power are understood, VA is unambiguous; but power qualified by an obscure adjective confuses people who suspect quite correctly that adjectives usually mislead. Historikeren (talk) 17:19, 14 November 2014 (UTC)

---

"At every instant the product of voltage and current is positive or zero, the result being that the direction of energy flow does not reverse." What is this trying to say? It isn't a correct sentence; it's just two dependent clauses stuck together with a comma. It's missing the part, between the two clauses, that can stand alone as a sentence. For example: "At every instant...", [something happens], "the result being that...".Pbyhistorian (talk) 20:07, 28 April 2020 (UTC)

Reactive Power

Looking carefully you will see that the sign reversal of the product of V and I occurs every quarter cycle, not every half cycle. I had to draw it out because the 90 degrees does not allow for that to occur for a half cycle and it choked my senses, so I drew it out. Yes, I know that adds up to a half cycle per cycle, but when getting down to that level it is obvious that the sign of V and I are opposite on alternate quarter cycles.... Regards, -- Steve -- (talk) 01:23, 16 January 2017 (UTC)

Notation

"(In this section overline will be used to indicate phasor or complex quantities and letters with no annotation will be considered the magnitude of those quantities.)"

Is this a standard notation? - Omegatron 23:16, August 14, 2005 (UTC)

well its what my university lecturers use i dunno if its a standard beyond that though. Plugwash 23:19, 14 August 2005 (UTC)

It wasn't standard when I was studying or practising electrical engineering (although that may just indicate how old I am!). A one-side-head horizontal arrow over a symbol was sometimes used to indicate a vector quantity, but most texts used boldface symbols for that. I always used a simple overline in the statistics sense - a mean or average (rms of course). Magnitude was denoted with matrix notation e.g. |z|. Plain symbols were used only for DC. Of course we used e for electromotive force and i for current, which drove the mathematicians nuts because they had totally different meanings for them. To top it off, because i was already in use, we used j for what mathematicians call i. I'm certain we can never find a notation that satisfies everyone, so we have to clearly specify the one that we use, then keep it consistent across related articles. Boldface and || have the advantage that they can be expressed directly within WP - no CSS or PNG required. JohnSankey 17:54, 5 October 2005 (UTC)

That's standard enough. As long as we clarify in an article whether:

• V = phasor, |V| = magnitude ([1] [2] [3])
• V = magnitude, ${\overline {V}}$ = phasor

Actually, it seems the only uses of overline are WP mirrors... :-\ Including complex number, which uses overline to mean complex conjugate. - Omegatron 01:01, August 15, 2005 (UTC)

I suspect that the use of overline for phasors is borrowed from some kind of vector notation (after all you could consider a phasor to be a two dimensional vector rather than a complex number its just easier to do the required calculations with it treated as a complex number). As for overline not being used much on the web i suspect that is a case of simply noone knowing that its availible (its only availible through CSS not through the old style html tags). Plugwash 00:11, 25 August 2005 (UTC)

In the article are these three sentences:

If the load is purely reactive, then the voltage and current are 90 degrees out of phase and there is no net power flow. This energy flowing backwards and forwards is known as reactive power. A practical load will have resistive, inductive, and capacitive parts, and so both real and reactive power will flow to the load.

The first sentence states: If the load is purely reactive... there is no net power flow.

The third sentence seems to contradict: [... both real and] reactive power will flow to the load. As reactive power has no net flow, it is confusing to include reactive power when writing of a flow to the load. Alexander 420 (talk) 22:13, 23 March 2009 (UTC)

Math markup

I try and avoid png rendered math inside text because it looks so ugly inline. What do others think? Plugwash 23:22, 14 August 2005 (UTC)

Hehehehee. Are you sure you want to ask that? Wikipedia_talk:WikiProject_Mathematics/Archive4(TeX) I really don't think we should be using CSS for overlines, though, as it's not necessarily supported by all browsers.

Proposed mergers

Right now, the merges suggested look like this (i think):

Electric reactance -> AC power -> Alternating current

I'm okay with this, except may I suggest that we first merge Electric reactance -> Reactance, so that the mergers would look like this:

Electric reactance -> Reactance -> AC power -> Alternating current

Reactance appears to this layman as the better written article. The only reason for merging Reactance -> Electric reactance, is to distinguish this from the sense in which Reactance is used in chemistry. Just a thought. Vonkje 11:51, 3 October 2005 (UTC)

I consider the concepts of electrical current/voltage, resistance/reactance and power to be sufficiently different that they should not be merged, otherwise the combined article will become too long and very difficult to organize clearly. I note that there is inadequate discussion in these articles of non-sinusoidal waveforms, particularly of the substantial 3rd and 5th harmonics generated by the non-linearity of standard power transformers - that would make a combined article even longer. Then there are the square and 2-step waveforms of DC-AC power converters. I recommend keeping these three separate but consistent, with appropriate cross links. As a retired electrical engineer, I could try to start this process, but would do so only when agreement is reached on separation or merger. JohnSankey 15:31, 5 October 2005 (UTC)

I do not think it is a good idea to merge electric reactance with AC power. The concept of reactance is important in other areas of electrical engineering, such as RF circuit design. Merging would change its apparent position from a fundamental concept in EE to an apparent subtopic of power transmission. Future articles that refer to reactance would become extremely awkward. Mbset 07:23, 15 October 2005 (UTC)

Can we remove some of the merge templates then?

Watts, volt-amps, and power meters

This is a very nice article, but a bit more than I need to know. I'm trying to understand, something more basic, the difference between Watts and Volt-amps in the real world. I have a meter that measures both watts and VA when you plug in normal household appliances. If I plug in an incandescant bulb the watts and VA are virtually identical, if it is a compact flourescant bulb the VA are almost twice the watts. I understand why this is, but my question is more basic. Which is being measured by my electric meter? I'm under the impression that it counts VA hours but calls them watts. If this is so, why do all electric appliances come rated in watts and not VA? -- Samuel Wantman 06:35, 18 January 2006 (UTC)

I think at least here in the uk domestic meters do actually measure watts (meters for large commercial and industrial installs measure watts and VArs seperately). but i'm not entirely positive. Plugwash 14:16, 18 January 2006 (UTC)

VA hours would not be equivalent to watts. It's equivalent to joules. The difference between VA and W is real vs apparent power, as explained in the article. — Omegatron 17:20, 18 January 2006 (UTC)

Reactive Power

Reactive Power carries much more significance than what is highlighted in this article. I think that Reactive Power should not be combined with AC Power. A few major points about Reactive Power not discussed here, are:

1. Reactive Power - Power System Voltage regulation (Voltage Control)
2. Significance of Reactive Power in Power Industry today
3. Efforts on part of Regulatory/Governing bodies to ensure availability of reactive power for System Stability
4. Compensation of Electric Generators for providing Reactive Power in emergency situations

Other discussions will come up as soon as we open this topic. Therefore, I request you to create a separate topic for Reactive Power.

It certainly deserves mention in a general article on AC power but there is no reason there cannot be a seperate article on it if you belive you can produce enough content for one. Plugwash 21:25, 17 November 2006 (UTC)

I concur with the request to create a separate and detailed topic for "Reactive Power". The search on this phrase led me to this article, which was a good basic review (I am an MSEE and have recently returned to the Power field several decades after having studied it in college.) I specifically care about the four sub-topics related to Reactive Power as outlined above. RWVesel 16:31, 11 July 2007 (UTC)

Included a paragraph that explained how utilities work, and whether or not they charge for reactive power. Also introduced the idea of power factor. Nisha123456789 (talk) 19:14, 9 December 2015 (UTC)

Terminology

This entire article is inconsistant with standard terminology employed by electrical engineers. We speak of Volts, Amperes, and Watts, noting the differences between peak power (Watts) and root mean square power (Watts). This article doesn't mention any of these. Celestialmechanic 16:10, 11 August 2007 (UTC)

This is pretty normal terminology in the electric power industry. It knocked me for a spin when I started working in the industry 20 years ago, and took me a while to get up to speed. It seems to be more common these days: for example you'll find VA being used in preference to Watts (even though the units and magnitude are the same) on quite small UPSes these days. Resuna (talk) 23:34, 16 June 2008 (UTC)

Different power types are expressed in different units to not confuse them, although all of those units are dimensionally the same (W = J/s = VAr = VA.)

The standard is: Active power and instantaneous power are measured in watts, reactive power is measured in volt-amperes reactive, apparent power and complex power are measured in volt-amperes.

In sinusoidal steady state, instantaneous power can be decomposed into two components: instantanoeus active power and instantaneous reactive power. The the peak value of the former is 2·P (where P is active power or average power) and the peak of the latter is |Q| (the absolute value of reactive power.)

What do you mean with peak watts and RMS watts? What is described in this other article? At least in power systems, I've never heard of peak and RMS watts. --Alej27 (talk) 01:48, 21 October 2020 (UTC)

ac power

dear sir how power(reactive/real )is scalar quantity while both are directional.thanks. —Preceding unsigned comment added by 119.73.111.178 (talk) 22:56, 10 February 2009 (UTC)

Reactive power and real power are always 90 degrees apart. In terms of linear algebra they form basis vectors of a two-dimensional complex space. The volt-amps are the vector sum of the Wh (real) and VARh (reactive) power. So, only volt-amps have a changing direction in the space. The power factor is actually the cosine of the "voltage to current angle." I.e., Wh/VAh Hope that helps. Ray Van De Walker 01:07, 7 March 2010 (UTC)

How is active power and reactive power scalars when they have direction? Good question. That's the same as asking why current is also a scalar when it has direction.

Well, in the case of current, I see it in the following way. Consider a circuit diagram consisting of a voltage source connected in series (or in parallel) with a resistor. The circuit diagram is drawn like this: a vertical voltage source to the left, a verticasl resistor to the right, a horizontal wire on the top connecting the voltage source and resistor, and a horizontal wire on the bottom. Now, in one side of the resistor, draw a tiny arrow that points down. This arrows tells which direction will be considered positive for the current, and is known as reference direction. After you apply Ohm's law you get a mathematical expression for the current. This value is the same for all of the currents in each cross-sectional area of the wires of this circuit, because it is a series connection and so the current through each device is the same.

This seems good, so it suggests to use scalars. But it doesn't answer why not use vectors instead? To answer that, first you could remember how working with vectors is more tedious than with scalars. Secondly and more important, there's no single vector that would describe the spatial direction of current in each segment of the circuit. For example in the resistor the vector would have to point down, but in the source it would have to point up; since a vector doesn't have two directions but only one, we would need to write the second vector with a different symbol. In other words, if we used vectors, the current in series connection would need to be written with different symbols depending on the spatial direction of current; this would be tedious.

A third and important way to see why treating current as a scalar is fine, is to think about what does its direction mean. When you see the waveform of a sinusoidal wave, you see negative and positive values. This means that on the positive values, positive electric charge is flowing in the reference direction; on the negative values, positive charge is flowing in the opposite direction. Et voilà! There's no need to use vectors, since the sign of current already tells us the direction of flow of charge!

By the way, in the case of LTI AC circuits we usually work with complex numbers, which are very similar to two-dimensional real vector (except the former has dvision as a defined mathematical operation.)

Okay, so now you should know why we use scalars and not vectors for current. What about active and reactive power? Well, let's go back once again to what their directions mean. In the case of active power, remember active power is a synonym of average power. Thus, the direction of average power indicates which is the average direction in which energy is flowing. For example, in a resistor, if you use the passive sign convention, a positive average power means energy that's flowing from source to resistor. In an inductor, its average power or active power is zero, meaning energy on average doesn't flow to either direction (which is true.) My point is, once again, the sign of active power already tells the direction of flow of energy, so there's no need to treat it as a vector. In the case of reactive power, a somewhat similar explanation applies.--Alej27 (talk) 02:26, 21 October 2020 (UTC)

Unbalanced Polyphase Systems

Could someone talk about the resolution that came about in the 1990s? That is an interesting story and should be further discussed in this article, or a link should be provided if it is discussed elsewhere. —Preceding unsigned comment added by 131.183.23.25 (talk) 15:39, 4 September 2009 (UTC)

Dubious

"Power is defined as the rate of flow of energy past a given point."? Goes against what I have beent taught/learned and read before. (Are they confusing energy and current where flow 'past a given point' is correct?)
However in Power (physics) it says "In physics, power is the rate at which work is performed or energy is converted" which seems much more correct. Especially 'work is performed' not just 'rate of flow of energy'. --220.101.28.25 (talk) 02:25, 26 November 2009 (UTC)

Did you read the article? The reason this is the case is because not all power in an eletrical system does work. Real Power does work, but Reactive Power does not. It is a transfer of energy to one place, and then back from whence it came, with no work done in between, besides heating up the physical elements (losses). Reactive power exists because all inductive and capacitive elements in a electrical system cause such "net zero" energy flows to occur. —Preceding unsigned comment added by 96.231.76.6 (talk) 05:41, 29 January 2010 (UTC)

Instananeous power can be defined either as the rate at which work is done (the second definition you used), or as the rate at which energy is traferred (basically the first defintion you used.) Both definitions are correct.

Some people might say this is wrong because, from the thermodynamics point of view, not all energy is work (for example, heat is also another form of transfer of energy.) But at least in electrical engineering we consider both definitions equivalent.

Regarding the reply the person above me said, just to make it clear, we shall distinguish between instantaneous work and net work. Both resistors and reactor (inductors and capacitors) perform instantaneous work, but only the former perform net work (i.e. the net work done by inductors and capacitors is zero.) --Alej27 (talk) 03:08, 21 October 2020 (UTC)

Clarification Needed

This seems like complete rubbish to me, "In alternating current circuits, energy storage elements such as inductance and capacitance may result in periodic reversals of the direction of energy flow.
If it's an AC circuit it's reversing anyway isnt it? Are they talking about inductance and capacitance in the power lines, causing unintended reversals that interfere with the normal AC, for example? --220.101.28.25 (talk) 02:50, 26 November 2009 (UTC)

Nope. Energy, not current. AC through a resistor has unidirectional power flow. AC to a reactive load has points where the power flow reverses on each cycle. --03:00, 26 November 2009 (UTC)

Agree w/ second poster. If you have a purely resistive load in an AC circuit, all of the energy is being dumped into the load (one direction of energy flow), regardless of the fact that the current is alternating directions. On the other hand, if you have a capacitive or inductive element, energy gets stored in the element, and then returns to the source, hence reactive power. —Preceding unsigned comment added by 96.231.76.6 (talk) 05:48, 29 January 2010 (UTC)

The two people above me are correct. In a resistor, energy flow is unidirectional. In an inductor or capacitor, energy flow is bidirectional.

It's important to distinguish between the two types of flow in circuits: flow of charge (i.e. current) and flow of energy (i.e. power). The question of the first person is the result of confusing these two flows (I'm not blaming them, though; electromagnetism is quite counterintuitive.)

I suggest watching this video (https:// youtu. be/agujzHdvtjc) from Eugene Khutoryansky on transformers to clarify the asked question. In that video, charges are represented as red and pink points (so, the flow of these points is current.) Energy is represented as boxes. Throughout the video, Eugene assumed a purely resistive load (although he didn't say it explicitly.) At the minute 1:30, you can see that, while charge flow is alternating, energy flow isn't; it's always going from source to (purely resistive) load.

As a side note, since we're talking about the direction of these two flows, it's also important to talk about their speed: charges move at drift speed which is very low, while energy moves at the speed of light. --Alej27 (talk) 03:16, 21 October 2020 (UTC)

Terminology

"The portion of power flow that, averaged over a complete cycle of the AC waveform, results in net transfer of energy in one direction is known as real power. The portion of power flow due to stored energy, which returns to the source in each cycle, is known as reactive power."

This terminology is wrong. Power cannot be power flow. Power is energy flow (Watts, energy / time). I don't even know what "power flow" would mean, perhaps the time rate of change of power. —Preceding unsigned comment added by 96.231.76.6 (talk) 05:43, 29 January 2010 (UTC)

I agree with the person above me.

Charge flow is current, yet people/electrical engineers use the phrase "current flow." This phrase is redundant because current is already a flow. Now, some people might say I'm acting dumb or I'm being too strict, but in science/engineering we shall be strict with the words! To support my point, below I quote a text from section 17.1 of the book College Physics, volume II by Raymond Serway et al. (p. 571):

"The phrases flow of current and current flow are commonly used, but here the word flow'- is redundant because current is already defined as a flow (of charge). Avoid this construnction!"

Similarly, energy flow is power, so saying "power flow" is redundant. I'm aware that in electric power engineering, there's a type of study known as power flow, but its name is redundant.

Maybe the reason why engineers use the phrase current flow and power flow is because current and power are terms more commonly used than charge and energy. Or maybe they don't even pay attention. --Alej27 (talk) 04:30, 21 October 2020 (UTC)

am I going to open a can of worms?

I've read the article and all the comments on this talk page and everything / everyone seems to be describing a full sinusoidal wave form but what about North America where 220-240vac sine wave gets split into two 110-120vac half sine waves?

for examle the picture of the blinking light trace describes 2 blinks per wave whereas in north america if the light was connected to 110-120 ac wouldn't there be just 1 blink per wave?

in general, I've been finding wiki articles regarding mains power lacking not adequately describing (or ignoring) North American household use of AC

I'm not an effective writer so I'm not qualified to offer anything more than a suggestion or a request that someone modify these articles to include single phase half-wave power usage :-) — Preceding unsigned comment added by 24.6.92.12 (talk) 05:49, 25 February 2012 (UTC)

<months later> Have a look at Split-phase electric power. It's not "two half sine waves". --Wtshymanski (talk) 13:29, 15 May 2012 (UTC)

Average power

«This method of calculating the average power gives the real power regardless of harmonic content of the waveform. In practical applications, this would be done in the digital domain, where the calculation becomes trivial when compared to the use of rms and phase to determine real power.

$P_{avg}={\frac {1}{T}}*\sum _{k=1}^{n}v[k]*i[k]$»

Shouldn't it be: $P_{avg}={\frac {1}{n}}*\sum _{k=1}^{n}v[k]*i[k]$ ?

89.181.225.20 (talk) 01:11, 15 May 2012 (UTC)

Complex power

The diagram is nice, but WHAT is the formula, or a precise definition of complex power? It is not stated anywhere in the article... --82.54.173.32 (talk) 20:53, 6 June 2012 (UTC)

Complex power is defined ("defined" means you can't prove it from any other equation or theorem) as one half the voltage phasor times the complex conjugate of the current phasor, or, as the RMS voltage phasor times the complex conjugate of RMS current phasor. Why? Because this leads to a very useful quantity: its real part is the average or active power, its imaginary part is the reactive power, its magnitude is the apparent power, and it's angle is the power factor angle.-- Alej27 (talk) 04:40, 21 October 2020 (UTC)

Wrong HTML Anchor

In the contents navigator, clicking on "Reactive power" leads to "1 Active, reactive, and apparent power". Shouldn't it lead to "3 Reactive power"? 129.247.247.238 (talk) 10:27, 18 February 2015 (UTC)

Reorganization of Reactive Power Section

I have edited the reactive power section to clarify the differences between inductive loads and capacitive loads and their effects on reactive power and power factor. Some statements were made in the form of "this causes the voltage to lag the current" and I changed them to reflect the convention that voltage is defined as 0 degrees. I also added more information about sourcing vs. sinking reactive power and added the fact that inductive loads are primarily made up of generators and motors. A more in depth discussion about why reactive power is useful or necessary for power system operation.

Regarding the reactive power control section, I looked around and couldn't find the formula that was referenced. Also the phrase "pre-fault reactive generator use" is confusing, and I'm not sure what it means.

A figure illustrating the relationship between complex power and voltage and current and their shared angle would also be illustrative and useful. — Preceding unsigned comment added by Tripleripple (talk • contribs) 07:30, 29 April 2015 (UTC)

Section: Capacitive vs. Reactive Loads

It says, "This voltage increases until some maximum dictated by the capacitor structure." There are other things that also affect the maximum voltage a capacitor will charge up to, such as other circuit elements, so the point of that sentence isn't clear, not in context or otherwise. I think the explanation of the current leading in a capacitor might be more concise if that sentence were simply eliminated. I might put more emphasis on the voltage of a capacitor being unable to change instantaneously (with an explanation of why can't it -- in the cases of increasing or decreasing applied voltage), thereby causing a phase lag in the voltage relative to current when the applied voltage changes, i.e. voltage-lagging in the capacitor (or current leading).

75.110.98.103 (talk) 15:52, 10 March 2016 (UTC)

I agree with you. Since the current through an inductor can't change abruptly but voltage can, current lags voltage here. Since the voltage across a capacitor can't change abruply but current can, voltage lags current (or also current leads voltage) here. This is the explanation of the phrase ELI the ICE man. --Alej27 (talk) 04:48, 21 October 2020 (UTC)

Comments - I am trying to understand this

Text says:

This method of calculating the average power gives the active power regardless of harmonic content of the waveform. In practical applications, this would be done in the digital domain, where the calculation becomes trivial when compared to the use of rms and phase to determine active power.

$P_{\text{avg}}={\frac {1}{n}}\sum _{k=1}^{n}V[k]I[k]$

The link between the equation and the preceding text isn't clear. Is it meant that P(avg) = P(rms)? What is the term k? Is it just an arbitary small increment or is it an increment of time (in reallity)? It isn't defined -just saying. Cinderella157 (talk) 10:20, 15 March 2017 (UTC)

more

: $pf={P_{a}+P_{b}+P_{c} \over |S_{a}|+|S_{b}|+|S_{c}|}$

that is, the quotient of the sums of the active powers for each phase over the sum of the apparent power for each phase.

$pf={P_{a}+P_{b}+P_{c} \over |P_{a}+P_{b}+P_{c}+j(Q_{a}+Q_{b}+Q_{c})|}$

Is that: "over the sum of the absolute values of the apparent power for each phase"? I had to think what those bars meant (it was a long time ago). Cinderella157 (talk) 10:26, 15 March 2017 (UTC)

more

product terms in more detail we come to a very interesting result.

${\begin{aligned}&A\cos(\omega _{1}t+k_{1})\cos(\omega _{2}t+k_{2})\\={}&{\frac {A}{2}}\cos \left[\left(\omega _{1}t+k_{1}\right)+\left(\omega _{2}t+k_{2}\right)\right]+{\frac {A}{2}}\cos \left[\left(\omega _{1}t+k_{1}\right)-\left(\omega _{2}t+k_{2}\right)\right]\\={}&{\frac {A}{2}}\cos \left[\left(\omega _{1}+\omega _{2}\right)t+k_{1}+k_{2}\right]+{\frac {A}{2}}\cos \left[\left(\omega _{1}-\omega _{2}\right)t+k_{1}-k_{2}\right]\end{aligned}}$

A, omega (and k) are not defined? Cinderella157 (talk) 10:31, 15 March 2017 (UTC)

Which omega is zero - w1 or w2? Cinderella157 (talk) 10:34, 15 March 2017 (UTC)

Cinders,

Some comments:

Be careful with your terminology. There is no such (useful) thing as "RMS Power". RMS only refers to voltage and current values. The RMS (Root Mean Square) values of Voltage and Current are the effective values. This means they are the values that have the same effect as a DC quantity of that value for calculating (average) power. A voltage of x volts RMS give the same power in a circuit as x volts of DC would. Therefore, P(avg) = V(RMS) * I(RMS)

Average power is pretty much what you think. It is the power over a relatively long term.

In the US, the term "power" means "average power". This is the power averaged over (at least) one full cycle of the AC waveform. This term is used to distinguish it from instantaneous power or the power V*I at one single instant. The term "active power" is not used in any context I have ever been involved in.

The formula you question is a discreet step version of the integral formula just above it. The integral version uses the continuous integral of the voltage and current to calculate (average) power. The discreet formula is saying that you sum the instantaneous power at small time increments (k) over some total time (n). The term 'k' signifies the voltage and current values at the times 'k' as 'k' is stepped from time 1 through 'n'. I think it would have been better to use 't' instead of 'k' to indicate time steps here.

In the "Multiple frequency systems "cos" formulas, "A" is the amplitude (cos has a peak value of 1), omega is the frequency in radians per second, and 'k' is the phase offset of that component frequency. These are standard notations for sinusoidal signals.

I'd have to go back to school to address the Omega1, omega2 question. The first equation is expanded using a trig identiry that is lost to memory... I'm not sure just what that trig identity is doing, nor just what your question is asking. Sorry.

The vertical bars mean absolute value. I believe (but could be wrong) those are there because for reactive circuits there are some instantaneous power values that go negative due to the phase shift between voltage and current waveforms. That is, the voltage is positive when the current is still flowing in the opposite direction and visa-versa. In other words you have V*(-I) or (-V)*I. In a purely resistive circuit V and I are always in phase and therefore have the same polarity.

Regards -- Steve -- (talk) 05:42, 16 March 2017 (UTC)

@ -- Steve -- , Thanks for the reply. I realise once you said it that P(rms) is meaningless - I was meaning P calculated from the rms values of V and I ? I was using P(rms) to mean a value calculated this way rather than one derived by summation. I guess, with the cos function, it is using terms that aren't definded (as it does in the sum compared to the integral). In part, I was pointing out some things that weren't clear (to me), hoping that somebody could clarify the text. Cinderella157 (talk) 11:33, 16 March 2017 (UTC)

I don't think the vertical bars refer strictly to the absolute value of apparent power, but instead the magnitude of complex power. Apparent power is always positive, since Vrms and Irms are both positive (unless a convention is used, though I have never heard of one nor seen negative apparent power.) Alej27 (talk) 23:20, 22 October 2020 (UTC)

Asterisk

Please explain asterisk in formula. —DIV (120.17.54.174 (talk) 12:33, 5 August 2018 (UTC))

It stands for the complex conjugate of the complex number right before the asterisk. In the Complex power equation, it stands for the complex conjugate of the current phasor. --Alej27 (talk) 04:53, 21 October 2020 (UTC)

Is it okay for you if in the first paragraph I change "power" to "instantaneous power"?

The term "power" has a lot of meanings. If we're talking about transformers, it usually means the apparent power. If we're talking about motors, it usually means the output mechanical power. If we're talking about a load, it usually means the active power (also known as average power or real power.) If we're talking about a capacitor bank, it usually means the reactive power.

So, as you can see saying "power" alone might be misleading. Because of this, I suggest to change in the very first paragraph of this article, the word "power" for "instantaneous power", which is what whoever wrote it meant (the rate of transfer of energy or the rate of work being done.) --Alej27 (talk) 01:21, 21 October 2020 (UTC)

What's the point of the section "Unbalanced polyphase systems"?

In that section, it reads "While active power and reactive power are well defined in any system, the definition of apparent power for unbalanced polyphase systems is considered to be one of the most controversial topics in power engineering." Why is it controversial? In a two-terminal (i.e. single-phase) element/device/load/generator, apparent power is defined as the product of RMS voltage times RMS current. This definition applies whether in sinusoidal steady-state (no harmonics) or in non-sinusoidal steady-state (with harmonics), and whether in balanced or unbalanced systems.

In balanced three-phase systems, the previous definition is usually called single-phase apparent power or per-phase apparent power, while the three-phase apparent power or total apparent power is three times the previous quantity.

In unbalanced three-phase systems, the single-phase apparent power is still RMS voltage times RMS current.

Maybe the controversy the section talks about is regarding three-phase (and not single-phase) apparent power in unbalanced three-phase systems? Can someone clarify me this? If that's true, then my point of view is that people/engineers can't expect apparent power to be the sum of the single-phase apparent power of each leg of the device, not because the system is unbalanced, but because the apparent power is not a conserved quantity (unlike active/real power, reactive/quadrature power, and complex power). --Alej27 (talk) 01:03, 31 October 2020 (UTC)

On reactive power, instantaneous reactive power, instantaneous active current and instantaneous reactive current

(This section is not finished. It's under construction)

The concept of active and reactive components of current is important yet I don't see it being taught in popular Circuit Analysis and Power Systems Analysis textbooks. Moreover, reactive power is usually defined but not much discussion on its meaning is discussed. The purpose of this section is to understand the meaning of reactive power, but as well mathematically back-up that interpretation.

Previous concepts

In this subsection, relevant concepts that the reader should already know are summarized.

Recall the mathematical definition of real instantaneous electric current $i(t)$ (in coulombs per second or amperes, C/s = A), real instantaneous voltage $v(t)$ (in joules per coulomb or volts, J/C = V), and instantaneous power $p(t)$ (in joules per second or watts, J/s = W):

$i(t)={\dfrac {\mathrm {d} q(t)}{\mathrm {d} t}}$

$v(t)={\dfrac {\mathrm {d} e(t)}{\mathrm {d} q}}$

$p(t)={\dfrac {\mathrm {d} e(t)}{\mathrm {d} t}}$

where $q(t)$ is the instantaneous electric charge (in coulombs, C), $t$ is the independent time variable (in seconds, s), and $e(t)$ is the instantaneous energy (in joules, J). In the case of instantaneous electric power, from the first two equations above it is evident that it can also be computed as:

$p(t)=v(t)\,i(t)$

--Consider a lumped two-terminal (i.e. one-port or single-phase) linear time-invariant (LTI) device, where the reference direction of $i(t)$ and the reference polarity of $v(t)$ are chosen such that the passive sign convention is used. If the device is operating in sinusoidal steady-state, meaning that a sinusoidal voltage (constant amplitude, constant phase angle, constant frequency) with no DC offset is applied to the device, resulting in a sinusoidal current with same frequency but in general with different amplitude and phase angle than the applied voltage, then the real instantaneous voltage $v(t)$ across the device and the real instantaneous current $i(t)$ through the device are the following real-valued real functions:

$v(t)=V_{\text{max}}\cos {(\omega t+\theta _{v})}$

$i(t)=I_{\text{max}}\cos {(\omega t+\theta _{i})}$

where $V_{\text{max}}$ and $I_{\text{max}}$ are the amplitude or peak value or maximum value (in volts; and amperes), $\omega =2\pi f$ is the angular frequency (in radians per second, rad/s), $f$ is the cyclic frequency (in hertz, Hz), $\theta _{v}$ and $\theta _{i}$ are the phase angle or phase shift (in degrees, °), and $\theta =\theta _{v}-\theta _{i}$ is the power factor angle (in degrees). The corresponding phasor voltage ${\tilde {V}}$ (in volts) and phasor current ${\tilde {I}}$ (in amperes) of the sinusoidal signals $v(t)$ and $i(t)$ are defined respectively as the following complex constants:

${\tilde {V}}=V_{\text{max}}\,e^{j\theta _{v}}=V_{\text{max}}\cos {(\theta _{v})}+jV_{\text{max}}\sin {(\theta _{v})}$

${\tilde {I}}=I_{\text{max}}\,e^{j\theta _{i}}=I_{\text{max}}\cos {(\theta _{i})}+jI_{\text{max}}\sin {(\theta _{i})}$

Notice I’ll denote phasors with a tilde (~) on top, while amplitudes (either peak values or RMS values) without a tilde, for example $\|{\tilde {I}}\|=I_{\text{max}}$. The relationship between the real instantaneous values and the phasors are, considering that $v(t)$ and $i(t)$ have been defined using the cosine function and not the sine function:

$v(t)=\Re [{\tilde {V}}e^{j\omega t}]$

$i(t)=\Re [{\tilde {I}}e^{j\omega t}]$

where

$v_{\text{c}}(t)={\tilde {V}}e^{j\omega t}=V_{\text{max}}\cos {(\omega t+\theta _{v})}+jV_{\text{max}}\sin {(\omega t+\theta _{v})}$

and

$i_{\text{c}}(t)={\tilde {I}}e^{j\omega t}=I_{\text{max}}\cos {(\omega t+\theta _{i})}+jI_{\text{max}}\sin {(\omega t+\theta _{i})}$

are defined respectively as the complex instantaneous voltage and the complex instantaneous current, which are complex-valued real functions.

--In practice, it's common to work with RMS values instead of peak values; in such cases, since it can be proven that $X_{\text{rms}}=X_{\text{max}}/{\sqrt {2}}$, it makes sense to define the RMS phasor voltage ${\tilde {V}}_{\text{rms}}$ and RMS phasor current ${\tilde {I}}_{\text{rms}}$ as:

${\tilde {V}}_{\text{rms}}={\dfrac {\tilde {V}}{\sqrt {2}}}\implies \|{\tilde {V}}_{\text{rms}}\|={\dfrac {\|{\tilde {V}}\|}{\sqrt {2}}}\implies v(t)=\Re [{\sqrt {2}}\,{\tilde {V}}_{\text{rms}}\,e^{j\omega t}]=V_{\text{rms}}{\sqrt {2}}\cos {(\omega t+\theta _{v})}$

${\tilde {I}}_{\text{rms}}={\dfrac {\tilde {I}}{\sqrt {2}}}\implies \|{\tilde {I}}_{\text{rms}}\|={\dfrac {\|{\tilde {I}}\|}{\sqrt {2}}}\implies i(t)=\Re [{\sqrt {2}}{\tilde {I}}_{\text{rms}}\,e^{j\omega t}]=I_{\text{rms}}{\sqrt {2}}\cos {(\omega t+\theta _{i})}$

Phasor active current and phasor reactive current. Instantaneous active current and instantaneous reactive current

As explained in section 11.3 subsection Projections and vector components (pages 787-788) by Ron Larson and Bruce Edwards in their textbook Multivariable calculus (9th edition), a two-dimensional real vector $\mathbf {u} =\langle u_{1},u_{2}\rangle$ can be decomposed into two other vectors ${\mathbf {w} }_{1}$ and ${\mathbf {w} }_{2}$, such that ${\mathbf {u} }={\mathbf {w} }_{1}+{\mathbf {w} }_{2}$, where ${\mathbf {w} }_{1}$ is parallel to ${\mathbf {v} }=\langle v_{1},v_{2}\rangle$ and is known as the projection of $\mathbf {u}$ onto $\mathbf {v}$ or the vector component of $\mathbf {u}$ along $\mathbf {v}$, and ${\mathbf {w} }_{2}$ is orthogonal to ${\mathbf {v} }$ and is known as the vector component of $\mathbf {u}$ orthogonal to $\mathbf {v}$, where

${\mathbf {w} }_{1}={\text{proj}}_{\mathbf {v} }{\mathbf {u} }=\left({\dfrac {{\mathbf {u} }\cdot {\mathbf {v} }}{\|{\mathbf {v} }\|^{2}}}\right){\mathbf {v} }=\left({\dfrac {u_{1}v_{1}+u_{2}v_{2}}{{\sqrt {v_{1}^{2}+v_{2}^{2}}}^{2}}}\right){\mathbf {v} }=\left({\dfrac {u_{1}v_{1}+u_{2}v_{2}}{|v_{1}^{2}+v_{2}^{2}|}}\right){\mathbf {v} }=\left({\dfrac {u_{1}v_{1}+u_{2}v_{2}}{v_{1}^{2}+v_{2}^{2}}}\right){\mathbf {v} }$

${\mathbf {w} }_{2}={\mathbf {u} }-{\mathbf {w} }_{1}$

In our discussion, we're going to conceptually define the phasor active current ${\tilde {I}}_{\text{a}}$ as that component of ${\tilde {I}}$ that is in phase (parallel to) or 180° out of phase (antiparallel to) with ${\tilde {V}}$, and the phasor reactive current ${\tilde {I}}_{\text{r}}$ as that component of ${\tilde {I}}$ that is in quadrature with ${\tilde {V}}$ (i.e. it is 90° out of phase, or is perpendicular or orthogonal to it, either leading or lagging). Since complex numbers are similar to two-dimensional real vectors, we can use the same two previous equations but with phasors:

${\tilde {I}}_{\text{a}}=\left({\dfrac {\Re [{\tilde {I}}]\Re [{\tilde {V}}]+\Im [{\tilde {I}}]\Im [{\tilde {V}}]}{\Re [{\tilde {V}}]^{2}+\Im [{\tilde {V}}]^{2}}}\right){\tilde {V}}$

${\tilde {I}}={\tilde {I}}_{\text{a}}+{\tilde {I}}_{\text{r}}\implies {\tilde {I}}_{\text{r}}={\tilde {I}}-{\tilde {I}}_{\text{a}}=(\Re [{\tilde {I}}]+j\,\Im [{\tilde {I}}])-(\Re [{\tilde {I}}_{\text{a}}]+j\,\Im [{\tilde {I}}_{\text{a}}])=(\Re [{\tilde {I}}]-\Re [{\tilde {I}}_{\text{a}}])+j(\Im [{\tilde {I}}]-\Im [{\tilde {I}}_{\text{a}}])$

where, by definition,

$\theta _{i_{\text{a}}}={\begin{cases}\theta _{v}&{\text{if }}-90^{\circ }<\theta <90^{\circ }\\\theta _{v}+180^{\circ }&{\text{if }}-180^{\circ }<\theta <-90^{\circ }{\text{ or }}90^{\circ }<\theta <180^{\circ }\end{cases}}$

$\theta _{i_{\text{r}}}={\begin{cases}\theta _{v}+90^{\circ }&{\text{if }}-180^{\circ }<\theta <0^{\circ }\\\theta _{v}-90^{\circ }&{\text{if }}0^{\circ }<\theta <180^{\circ }\end{cases}}$

Since ${\tilde {I}}$ consists of a sum of the phasors ${\tilde {I}}_{\text{a}}$ and ${\tilde {I}}_{\text{r}}$, the three phasors must correspond to sinusoids with same frequency, namely $\omega$ (since only phasors with corresponding sinusoids of same frequency can be algebraically operated). So we can multiply both sides of equation ${\tilde {I}}={\tilde {I}}_{\text{a}}+{\tilde {I}}_{\text{r}}$ by $e^{j\omega t}$ and then take the real part on both sides, which yields:

$i(t)=i_{\text{a}}(t)+i_{\text{r}}(t)$

where $i_{\text{a}}(t)$ is the instantaneous active current and $i_{\text{r}}(t)$ is the instantaneous reactive current.

--In practice it's common to use RMS values, so we define the RMS phasor active current ${\tilde {I}}_{\text{a,rms}}$ and the RMS phasor reactive current ${\tilde {I}}_{\text{r,rms}}$:

${\tilde {I}}_{\text{a,rms}}={\dfrac {{\tilde {I}}_{\text{a}}}{\sqrt {2}}}\implies \|{\tilde {I}}_{\text{a,rms}}\|={\dfrac {\|{\tilde {I}}_{\text{a}}\|}{\sqrt {2}}}\implies i_{\text{a}}(t)=\Re [{\sqrt {2}}\,{\tilde {I}}_{\text{a,rms}}\,e^{j\omega t}]$

${\tilde {I}}_{\text{r,rms}}={\dfrac {{\tilde {I}}_{\text{r}}}{\sqrt {2}}}\implies \|{\tilde {I}}_{\text{r,rms}}\|={\dfrac {\|{\tilde {I}}_{\text{r}}\|}{\sqrt {2}}}\implies i_{\text{r}}(t)=\Re [{\sqrt {2}}\,{\tilde {I}}_{\text{r,rms}}\,e^{j\omega t}]$

It's also common to choose the voltage as angular reference (this means that $\theta _{v}=0^{\circ }=\theta _{i_{\text{a}}}\implies \Im [{\tilde {V}}]=0\implies \theta =-\theta _{i}$.) Using RMS values and taking the voltage as angular reference, the expressions for the phasor active and reactive components of current reduce to:

${\tilde {I}}_{\text{a,rms}}=\Re [{\tilde {I}}_{\text{rms}}]=\|{\tilde {I}}_{\text{rms}}\|\cos {(\theta _{i})}=\|{\tilde {I}}_{\text{rms}}\|\cos {(\theta )}$

${\tilde {I}}_{\text{r,rms}}=j\,\Im [{\tilde {I}}_{\text{rms}}]=j\|{\tilde {I}}_{\text{rms}}\|\sin {(\theta _{i})}=-j\|{\tilde {I}}_{\text{rms}}\|\sin {(\theta )}$

Instantaneous active power and instantaneous reactive power

We can compute instantaneous power as:

$p(t)=v(t)\,(i_{\text{a}}(t)+i_{\text{r}}(t))=v(t)\,i_{\text{a}}(t)+v(t)\,i_{\text{r}}(t)=p_{\text{a}}(t)+p_{\text{r}}(t)$

where

$p_{\text{a}}(t)=v(t)\,i_{\text{a}}(t)$

and

$p_{\text{r}}(t)=v(t)\,i_{\text{r}}(t)$

are defined respectively as the instantaneous active power and instantaneous reactive power, which are real-valued real functions. As we see, instantaneous active power is due to the instantaneous active current, and instantaneous reactive power is due to the instantaneous reactive current. To distinguish $p$ from $p_{\text{a}}$ and $p_{\text{r}}$, we may call the former as total instantaneous power.

--Let us compute the net energy transferred in $N$ cycles of the instantaneous power, once again under the assumption of sinusoidal steady-state. Let $T'$ be the (fundamental) period of instantaneous power (which is in general different than the (fundamental) period $T$ of instantaneous voltage and instantaneous current). Since $p(t)=\mathrm {d} e(t)/\mathrm {d} t$ (i.e. instantaneous power is the derivative of energy), energy is the integral of instantaneous power:

${\begin{aligned}W&=\displaystyle \int _{0}^{NT'}p(t)\,\mathrm {d} t\\&=\underbrace {\displaystyle \int _{0}^{NT'}v(t)i_{\text{a}}(t)\,\mathrm {d} t} _{\text{net energy}}+\underbrace {\displaystyle \int _{0}^{NT'}v(t)i_{\text{r}}(t)\,\mathrm {d} t} _{\text{no net energy}}\\&=P\cdot N\cdot T'+0\end{aligned}}$

where

$\displaystyle \int _{0}^{NT'}v(t)i_{\text{a}}(t)\,\mathrm {d} t={\begin{cases}\displaystyle \int _{0}^{NT'}V_{\text{max}}\cos {(\omega t+\theta _{v})}\,I_{\text{a,max}}\cos {(\omega t+\theta _{v})}\,\mathrm {d} t=V_{\text{max}}I_{\text{a,max}}\displaystyle \int _{0}^{NT'}\cos ^{2}{(\omega t+\theta _{v})}\,\mathrm {d} t={\dfrac {V_{\text{max}}I_{\text{a,max}}}{2}}\cdot N\cdot T'&{\text{if }}-90^{\circ }<\theta <90^{\circ }\\\displaystyle \int _{0}^{NT'}V_{\text{max}}\cos {(\omega t+\theta _{v})}\,I_{\text{a,max}}\cos {(\omega t+\theta _{v}+180^{\circ })}\,\mathrm {d} t=-V_{\text{max}}I_{\text{a,max}}\displaystyle \int _{0}^{NT'}\cos ^{2}{(\omega t+\theta _{v})}\,\mathrm {d} t=-{\dfrac {V_{\text{max}}I_{\text{a,max}}}{2}}\cdot N\cdot T'&{\text{if }}-180^{\circ }<\theta <-90^{\circ }{\text{ or }}90^{\circ }<\theta <180^{\circ }\end{cases}}$

and

$\int _{0}^{NT'}v(t)i_{\text{r}}(t)\,\mathrm {d} t=\int _{0}^{NT'}V_{\text{max}}\cos {(\omega t+\theta _{v})}\,I_{\text{r,max}}\cos {(\omega t+\theta _{v}\pm 90^{\circ })}\,\mathrm {d} t=0$

As we can see, the reactive component of instantaneous current does not contribute to the net energy transferred; only the active component does. In fact, in sinusoidal steady-state and assuming the passive sign convention is satisfied, it can be shown average power can be computed as:

$P={\dfrac {V_{\text{max}}I_{\text{max}}}{2}}\cos {(\theta )}=V_{\text{rms}}I_{\text{rms}}\cos {(\theta )}={\begin{cases}{\dfrac {V_{\text{max}}I_{\text{a,max}}}{2}}=V_{\text{rms}}I_{\text{a,rms}}&{\text{if }}-90^{\circ }<\theta <90^{\circ }\\-{\dfrac {V_{\text{max}}I_{\text{a,max}}}{2}}=-V_{\text{rms}}I_{\text{a,rms}}&{\text{if }}-180^{\circ }<\theta <-90^{\circ }{\text{ or }}90^{\circ }<\theta <180^{\circ }\end{cases}}$

--Alej27 (talk) 06:16, 17 February 2021 (UTC)

Derivation of various formulas for electric energy in terms of electric power

The purpose of this section is to prove various formulas for electric energy in terms of electric power. Unfortunately, I haven't found some of these derivations in any classical textbook on circuit theory, so I can't provide references. Because of that, I decided to prove the formulas, so that the reader can be sure this method works and how to derive it. There’s a summary at subsection 4. There’re two examples at subsection 5. (You can check the table of content at the top of this talk page.) The most relevant equations are (3), (4), (5), (6), (7), (8), (9), (17), (22) and (25).

Clarification on instantaneous energy, instantaneous power, net energy, active/real power and average power

The instantaneous power of a two-terminal device or network is conceptually defined as the time rate of flow of energy into (or out of) the device or network, or the time rate at which energy is being transferred into (or out of) the device or network, or the time rate at which work is being done onto (or by) the device or network. It is denoted by $p(t)$, and it is mathematically defined as the time derivative of energy or work $w(t)$ ([1, 2]):

$p(t)={\dfrac {\mathrm {d} w(t)}{\mathrm {d} t}}$ (1)

where $t$ is the independent time variable, and the parenthesis indicate $p$ and $w$ are a function of $t$. As we can see, in general the instantaneous power varies with time. It can be positive or negative.

By applying the chain rule, the instantaneous power can be computed also as the product of the instantaneous voltage $v(t)$ across the two-terminal device or network and the instantaneous conventional current $i(t)$ through the device or network ([1, 2, 3]):

$p(t)=v(t)\,i(t)$ (2)

If in the circuit diagram the reference polarity (the plus and minus signs) of the voltage $v(t)$ and the reference direction (the sense in which the arrowhead points) of the current $i(t)$ are chosen such that the passive sign convention is satisfied (which means the arrowhead of the current points towards the device or network into the positive reference terminal), then $p=vi$ is the instantaneous power consumed by the two-terminal device or network, i.e. a positive value indicates energy flowing from the surroundings into the device or network under consideration.

—Rearranging equation 1, we can determine the associated instantaneous energy $w(t)$ as the definite integral of the instantaneous power from the past ($-\infty$) up to the present ($t$) ([1, 4]):

$w(t)=\displaystyle \int _{-\infty }^{t}p(\tau )\,\mathrm {d} \tau$ (3)

In general the instantaneous energy varies with time.

—We can determine the net energy $W$ transferred across the two-terminal device or network between the instants $t_{1}$ and $t_{2}$ (where $t_{2}>t_{1}$) as follows ([2]):

$W=\displaystyle \int _{t_{1}}^{t_{2}}p(t)\,\mathrm {d} t$ (4)

where:

$T=t_{2}-t_{1}$ (5)

is the time interval under consideration. The net energy is a constant.

Equations 1, 2 and 4 are extremely general: they are valid whether the two-terminal device or network is linear or non-linear, time-invariant or time-variant, bilateral or non-bilateral, passive or active, operating with constant or alternating voltage and current , in steady-state or transient state, without or with harmonics.

—When the instantaneous power is periodic with fundamental period $T'$ (for example in AC steady-state, i.e. when the voltage and current are both periodic, whether sinusoidal or not, and with or without the same fundamental frequency), the active power, also known as average power or real power, of a two-terminal device or network is conceptually defined as the time average (mean) of the instantaneous power over one cycle of it. It is denoted by $P$, and it is mathematically defined as the mean using integrals ([5]):

$P={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}p(t)\,\mathrm {d} t$ (6)

where $t_{0}$ is any instant of time (for example, $t_{0}=0{\text{ s}}$ to simplify the equation). The previous equation is also valid when the instantaneous power is constant, since the average of a constant is that constant (this also implies that when the instantaneous power is constant, we can write it as $p$ or $P$ because it is equal to the active power).

—On the other hand, the average of instantaneous power over the whole time interval $T$ is obviously ([6]):

$P_{\text{avg}}={\dfrac {1}{T}}\displaystyle \int _{t_{1}}^{t_{2}}p(t)\,\mathrm {d} t$ (7)

Substituting equation 4 into equation 7, and solving for $W$, we get:

$W=P_{\text{avg}}\cdot T$ (8)

which is always valid, even if $p(t)$ is not constant nor periodic, and even in transient state.

Particular cases for net energy

When instantaneous power is constant

Suppose the instantaneous power $p(t)$ of the two-terminal device or network is constant (which occurs, for example, when the instantaneous voltage $v(t)$ across the device or network and instantaneous current $i(t)$ through the device or network are both constant, which at the same time occurs e.g. in DC circuits operating in steady-state). Then, in equation 4 we can take $p$ out of the integral and get the following:

$W=P\cdot T$ (9)

where $T$ is given by equation 5, and where I’ve written the instantaneous power with capital letter (just like active power) because it is equal to active power in this case.

When instantaneous power is periodic

Suppose the instantaneous power $p(t)$ of the two-terminal device or network is periodic with fundamental period $T'$ (which occurs, for example, when the instantaneous voltage $v(t)$ across the device or network and instantaneous current $i(t)$ through the device or network are both periodic, which at the same time occurs e.g. in AC circuits operating in steady-state, whether sinusoidal [without harmonics] or non-sinusoidal [with harmonics]). Then the active power $P$ is the time average of $p(t)$, given by equation 6. In equation 4, we can separate the integral in two by considering the time interval $[t_{1},t_{3}]$ followed by $[t_{3},t_{2}]$, like this:

$W=\displaystyle \int _{t_{1}}^{t_{3}}p(t)\,\mathrm {d} t+\displaystyle \int _{t_{3}}^{t_{2}}p(t)\,\mathrm {d} t$ (10)

where $t_{1}<t_{3}\leq t_{2}$. Rearranging equation 5, we can write $t_{2}=t_{1}+T$. Substituting this into equation 10, we get:

$W=\displaystyle \int _{t_{1}}^{t_{3}}p(t)\,\mathrm {d} t+\displaystyle \int _{t_{3}}^{t_{1}+T}p(t)\,\mathrm {d} t$ (11)

So far, we haven’t assigned a specific value for the instant $t_{3}$, we’ve only required it to be greater than the instant $t_{1}$ (the beginning of the time interval $T$ under consideration) and less than or equal to the instant $t_{2}$ (the end of the time interval $T$). If we, however, choose $t_{3}$ in a smart way, we can simplify equation 11.

So how should we choose $t_{3}$ to simplify equation 11? Well, we can do it in four steps. As first step, I’ll recall the following equation, without proving it in this Talk page. If the instantaneous power $p(t)$ of a two-terminal device or network is periodic with fundamental period $T'$ (like in our case), then we can prove the net energy transferred across such device or network during $k$ cycles of the instantaneous power is:

$\displaystyle \int _{t'}^{t'+kT'}p(t)\,\mathrm {d} t=P\cdot k\cdot T'$ (12)

where $k$ is a positive integer number, and $t'$ is any instant of time (for example, $t'=0{\text{ s}}$ to simplify the equation). Notice how simple is the right-hand side of the previous equation; it is simply active power times the number of cycles under consideration times the fundamental period of the instantaneous power. Our goal will be to try to use that expression (equation 12) in equation 11.

As second step, I’ll recall the following equation, without proving it in this Talk page. To round down the number $a$ to the nearest multiple of the number $b$, whose result we’ll denote by $c$, we can use the following:

$c=b\cdot \left\lfloor {\dfrac {a}{b}}\right\rfloor$ (13)

where $\lfloor \cdot \rfloor$ is the floor function. Therefore, we can use equation 13 to round down $T$ to the nearest multiple of $T'$. As third step, we can use such result to define $t_{3}$, and adding $t_{1}$, that is:

$t_{3}=t_{1}+T'\cdot \left\lfloor {\dfrac {T}{T'}}\right\rfloor$ (14)

We can go back to our derivation. Substituting equation 14 into equation 11, we get:

$W=\displaystyle \int _{t_{1}}^{t_{1}+T'\cdot \left\lfloor {\frac {T}{T'}}\right\rfloor }p(t)\,\mathrm {d} t+\displaystyle \int _{t_{1}+T'\cdot \left\lfloor {\frac {T}{T'}}\right\rfloor }^{t_{1}+T}p(t)\,\mathrm {d} t$ (15)

As fourth and last step, now compare the left-hand side of equation 12 with the left integral of equation 15. Notice they have the same form, where $\lfloor T/T'\rfloor$ of equation 15 is a positive integer number and is equal to the $k$ of equation 12, and $t_{1}$ of equation 15 is equal to the $t'$ of equation 12. Therefore, the left integral of equation 15 is equal to the right-hand side of equation 12:

$\displaystyle \int _{t_{1}}^{t_{1}+T'\cdot \left\lfloor {\frac {T}{T'}}\right\rfloor }p(t)\,\mathrm {d} t=P\cdot T'\cdot \left\lfloor {\frac {T}{T'}}\right\rfloor$ (16)

Substituting equation 16 into equation 15, finally we get the net energy transferred across a two-terminal device or network of periodic instantaneous power between the time instants $t_{1}$ and $t_{2}$:

$W=P\cdot T'\cdot \left\lfloor {\frac {T}{T'}}\right\rfloor +\displaystyle \int _{t_{1}+T'\cdot \left\lfloor {\frac {T}{T'}}\right\rfloor }^{t_{1}+T}p(t)\,\mathrm {d} t$ (17)

By the way, at a first glance it may appear that equation 17 is valid only if $T\geq T'$, and not if $T<T'$. However, if $T<T'$, then $\lfloor T/T'\rfloor$ in equation 17 will be zero, and so the left term is also zero, and the lower bound of the right integral reduces to $t_{1}$, so equation 17 reduces to:

$W=0+\displaystyle \int _{t_{1}+0}^{t_{1}+T}p(\tau )\,\mathrm {d} \tau$

which is the same as equation 4 (remember that $t_{2}=t_{1}+T$). So equation 17 is valid even if $T<T'$.

When instantaneous power is periodic and the finite time interval is an integer multiple of the fundamental period of the instantaneous power

As a particular case of equation 17, if $T$ is an integer multiple of $T'$, then:

$T=l\cdot T'$ (18)

where $l$ is an integer, and so:

${\begin{aligned}T'\cdot \left\lfloor {\dfrac {T}{T'}}\right\rfloor &=T'\cdot \left\lfloor {\dfrac {(l\cdot T')}{T'}}\right\rfloor \\&=T'\cdot \left\lfloor l\right\rfloor \end{aligned}}$

$\implies T'\cdot \left\lfloor {\dfrac {T}{T'}}\right\rfloor =l\cdot T'$ (19)

Substituting equations 18 and 19 into the right-hand side of equation 17, we get:

$P\cdot (l\cdot T')+\displaystyle \int _{t_{1}+(l\cdot T')}^{t_{1}+(l\cdot T')}p(t)\,\mathrm {d} t=P\cdot l\cdot T'+0$ (20)

where the right integral is zero because the lower and upper bounds are the same instant. Substituting equation 20 into equation 17, we get the particular case:

$W=P\cdot l\cdot T'$ (21)

Rearranging equation 18, we get $l=T/T'$. Substituting this into equation 21, after simplifying we get:

$W=P\cdot T$ (22)

Note this is the same as equation 9.

When is the finite time interval an integer multiple of the fundamental period of the periodic instantaneous power?

The use of equation 22 instead of equation 17 whenever possible is of course desired, since the latter is very simple, while the latter requires to know the instantaneous power, which in practice isn’t usually known/measured. However, to use equation 22, the instantaneous power must be periodic, and the finite time interval $T$ under consideration must be an integer multiple of the fundamental period of the instantaneous power. So a question arises: when does that occur?

As we know, the period $T'$ is related to cyclic frequency $f'$ (the one measured in hertz) by the equation:

$T'={\dfrac {1}{f'}}$ (23)

In linear time-invariant circuits excited by sinusoidal voltage and current sources of one same frequency and operating in AC steady-state, we can prove the (fundamental) cyclic frequency $f'$ of the instantaneous power of any two-terminal device is twice the (fundamental) cyclic frequency $f''$ of the instantaneous voltage and current of such device:

$f'=2f''$ (24)

So if a linear time-invariant device or network is excited by a 60-Hz voltage source, then the (fundamental) period of the instantaneous power is 120 Hz in steady-state. Furthermore, in some (but not all) non-linear and/or time-variant two-terminal devices, such as LED bulbs, the instantaneous current has harmonics but has the same fundamental frequency as the instantaneous voltage, and the instantaneous power has also twice the frequency of the instantaneous voltage and current.

Substituting equation 24 into equation 23, we get the following:

$T'={\dfrac {1}{2f''}}$ (25)

If $T$ was an integer multiple of $T'$, then the division $T/T'$ should result in an integer (equation 18). So, if applicable, you can use equation 25 to compute $T'$ from the frequency $f''$ of the instantaneous voltage or current of the device; then, compute $T/T'$; if that division results in an integer, then you can use equation 22; if it doesn’t result in an integer, you must use equation 17 (or equation 4).

Summary

To compute the net energy transferred across a two-terminal device or network during a finite time interval:

• In general we can’t compute the net energy as the product of active power and the finite time interval.
• We can always use equation 4 (i.e. the net energy is the integral of instantaneous power over the time interval).
• We can always use equation 8 (i.e. the net energy is the product of the average of the instantaneous power over the time interval times the time interval).
• We can use equation 9 (i.e. the net energy is the product of the power [instantaneous or active or average] times the time interval) *if* the instantaneous power is constant (as in DC circuits operating in DC steady-state without ripple/harmonics).
• We can use equation 17 if the instantaneous power is periodic (as in AC circuits operating in AC steady-state without or with harmonics).
• We can use equation 22 (i.e. the net energy is the product of the active power times the time interval) if the instantaneous power is periodic (as in AC circuits operating in AC steady-state without or with harmonics) and the time interval under consideration is an integer multiple of the fundamental period of the instantaneous power.
• Equation 25 (i.e. the fundamental period of the instantaneous power is the inverse of twice the fundamental cyclic frequency of the instantaneous voltage and current) is always true for a two-terminal device or network whose instantaneous voltage and instantaneous current are sinusoidal of same frequency without DC offset, for example if the device or network is linear time-invariant and is excited by a sinusoidal voltage. (An example of a linear time-invariant network is an interconnection of resistors, capacitors and inductors of constant resistance, capacitance and inductance.) This equation is sometimes also true for non-linear and/or time-variant two-terminal devices excited by sinusoidal voltages, such as LED bulbs.

where:

• $W$ is the net energy transferred across the two-terminal device or network during the finite time interval under consideration.
• $p(t)$ is the instantaneous power of the two-terminal device or network.
• $P$ is the active power (i.e. the average or mean of the instantaneous power over one cycle of it) of the two-terminal device or network, if the instantaneous power $p(t)$ is constant or periodic.
• $P_{\text{avg}}$ is the average or mean of the instantaneous power over the time interval under consideration, of the two-terminal device or network, if the instantaneous power $p(t)$ is constant or periodic.
• $T$ is the finite time interval under consideration.
• $t_{1}$ is the instant of time at which the time interval $T$ begins.
• $T'$ is the fundamental period of the instantaneous power $p(t)$, if the latter is periodic.
• $\lfloor x\rfloor$ means “$x$ rounded down” and is the floor function.
• $f''$ is the fundamental cyclic frequency of the instantaneous voltage across the two-terminal device or network.

Now we can answer the question “Can we always compute the electric energy as the product of power and time?”. The answer is: it depends on what is specifically meant by power, energy and time:

• If time refers to time interval, power refers to active power, and energy refers to net energy, then the answer is “no, the equation (net energy) = (active power) ∙ (time interval) is not always true”. It is sometimes true, as in equations 9 and 22, but it’s not always true, as in equations 4 and 17.
• If time refers to time interval, power refers to average power over the whole time interval, and energy refers to net energy, then the answer is “yes, the equation (net energy) = (average power) ∙ (time interval) is always true”. This was shown in equation 8.
• If time refers to time interval, power refers either to active power or average power over the whole time interval, and energy refers to instantaneous energy, then the answer is “no, the equations (instantaneous energy) = (active power) ∙ (time interval) and (instantaneous energy) = (average power) ∙ (time interval) are not always true”. This was shown in equation 3.

Examples

Example #1

Statement: Consider a single-phase appliance in a house with a split-phase system, whose instantaneous voltage is $v(t)=120{\sqrt {2}}\cos {(120\pi t)}{\text{ V}}$, and instantaneous current is $i(t)=2{\sqrt {2}}\cos {(120\pi t-45^{\circ })}{\text{ A}}$. We’re asked to find the net energy transferred from $t_{1}=0{\text{ s}}$ to $t_{2}=2{\text{ h}}=7\,200{\text{ s}}$.

Solution: Since the instantaneous voltage and current are sinusoidal, the instantaneous power is periodic, so we can use equation 17. And since the voltage and current are sinusoidal of same frequency ($f''=\omega ''/(2\pi )=(120\pi {\text{rad/s}})/(2\pi )=60{\text{ Hz}}$), the instantaneous power has a period given by equation 25: $T'=1/(2[60{\text{Hz}}])=1/120{\text{ s}}$. The time interval is given by equation 5: $T=(7\,200{\text{ s}})-(0{\text{ s}})=7\,200{\text{ s}}$. Thus:

${\dfrac {T}{T'}}={\dfrac {(7\,200{\text{ s}})}{(1/120{\text{ s}})}}=864\,000$

The active power of the device is:

$P=(120{\text{ V}})(2{\text{ A}})\cos {([0^{\circ }]-[-45^{\circ }])}\approx 169.71{\text{ W}}$

Since $T/T'$ is an integer number, we can use equation 22, which is simpler than equations 17 and 4:

$W\approx (169.71{\text{ W}})\cdot (7\,200{\text{ s}})=1\,221\,912{\text{ J}}$

Below is a computation in Mathematica that shows the previous value is correct.

Example #2

Statement: Consider a resistor in an electronic circuit whose instantaneous voltage is $v(t)=5{\sqrt {2}}\cos {(2\,000\pi t)}{\text{ mV}}$, and instantaneous current is $i(t)=1{\sqrt {2}}\cos {(2\,000\pi t)}{\text{ μA}}$. We’re asked to find the net energy transferred from $t_{1}=0{\text{ ms}}$ to $t_{2}=5.9{\text{ ms}}$.

Solution: Since the instantaneous voltage and current are sinusoidal, the instantaneous power is periodic, so we can use equation 17. And since the voltage and current are sinusoidal of same frequency ($f''=\omega ''/(2\pi )=(2\,000\pi {\text{rad/s}})/(2\pi )=1{\text{ kHz}}$), the instantaneous power has a period given by equation 25: $T'=1/(2[1\,000{\text{Hz}}])=1/2\,000{\text{ s}}$. The time interval is given by equation 5: $T=(5.9\cdot 10^{-3}{\text{ s}})-(0{\text{ s}})=0.0059{\text{ s}}$. Thus:

${\dfrac {T}{T'}}={\dfrac {(0.0059{\text{ s}})}{(1/2\,000{\text{ s}})}}=11.8$

The active power of the device is:

$P=(5\cdot 10^{-3}{\text{ V}})(1\cdot 10^{-6}{\text{ A}})\cos {([0^{\circ }]-[0^{\circ }])}=5{\text{ nW}}$

Since $T/T'$ is not an integer number, we can’t use equation 22; we must use equation 17 or 4. Let’s use equation 17, recalling that $p(t)=v(t)\,i(t)$:

${\begin{aligned}W&=(5\cdot 10^{-9}{\text{ W}})\cdot (1/2\,000{\text{ s}})\cdot \left\lfloor {\frac {(0.0059{\text{ s}})}{(1/2\,000{\text{ s}})}}\right\rfloor +\displaystyle \int _{(0{\text{ s}})+(1/2\,000{\text{ s}})\cdot \left\lfloor {\frac {(0.0059{\text{ s}})}{(1/2\,000{\text{ s}})}}\right\rfloor }^{(0{\text{ s}})+(0.0059{\text{ s}})}[5{\sqrt {2}}\cos {(2\,000\pi t)}\cdot 10^{-3}{\text{ V}}][1{\sqrt {2}}\cos {(2\,000\pi t)}\cdot 10^{-6}{\text{ A}}]\,\mathrm {d} t\\&=2.5\cdot 10^{-12}{\text{ J}}\cdot 11+10\cdot 10^{-9}\displaystyle \int _{1/2\,000{\text{ s}}\cdot 11}^{0.0059{\text{ s}}}[\cos ^{2}{(2\,000\pi t)}{\text{ W}}]\,\mathrm {d} t\\&\approx 29.12{\text{ pJ}}\end{aligned}}$

If instead we had used equation 22, we would’ve incorrectly computed:

$W\approx (5\cdot 10^{-9}{\text{ W}})\cdot (5.9\cdot 10^{-3}{\text{ s}})=29.5{\text{ pJ}}$

Note the previous value (29.5 pJ) isn’t very far from the correct value (29.12 pJ); the true percentage relative error is 100 ∙ (29.12 – 29.5)/(29.12) ≈ −1.30%. Furthermore, even if we neglect the integral in equation 17, we’d get an approximated and close value:

${\begin{aligned}W&\approx P\cdot T'\cdot \left\lfloor {\frac {T}{T'}}\right\rfloor \\&\approx (5\cdot 10^{-9}{\text{ W}})\cdot (1/2\,000{\text{ s}})\cdot \left\lfloor {\frac {(0.0059{\text{ s}})}{(1/2\,000{\text{ s}})}}\right\rfloor \\&\approx 2.5\cdot 10^{-9}{\text{ J}}\cdot 11\\&\approx 27.5{\text{ pJ}}\end{aligned}}$

and if instead we used the ceiling function (to round up instead of rounding down):

${\begin{aligned}W&\approx P\cdot T'\cdot \left\lceil {\frac {T}{T'}}\right\rceil \\&\approx (5\cdot 10^{-9}{\text{ W}})\cdot (1/2\,000{\text{ s}})\cdot \left\lceil {\frac {(0.0059{\text{ s}})}{(1/2\,000{\text{ s}})}}\right\rceil \\&\approx 2.5\cdot 10^{-9}{\text{ J}}\cdot 12\\&\approx 30{\text{ pJ}}\end{aligned}}$

Note the two previous value (27.5 pJ and 30 pJ) aren’t very far from the actual value (29.12 pJ); the true percentage relative errors are, respectively, 100 ∙ (29.12 – 27.5)/(29.12) ≈ +5.56% and 100 ∙ (29.12 – 30)/(29.12) ≈ −3.02%. Below is a computation in Mathematica that shows 29.12 pJ is the correct value.

Bibliography

• Suresh Kumar, K. S. Electric Circuits and Networks: For GTU.
• Hayt Jr., William Hart; Kemmerly, Jack Ellsworth; Durbin, Steven M. Engineering Circuit Analysis (8th edition).
• Thomas, Roland E.; Rosa, Albert J.; Toussaint, Gregory J. The Analysis and Design of Linear Circuits (8th edition).

References

• [1] Suresh Kumar, p. 26.
• [2] Thomas et al., p. 6.
• [3] Hayt et al., p. 422.
• [4] Suresh Kumar, p. 33.
• [5] Hayt et al., p. 425.
• [6] Hayt et al., p. 424.

--Alej27 (talk) 17:59, 6 June 2021 (UTC)

Derivation of various formulas for (total) instantaneous power and (total) active power of a single-phase or three-phase network in terms of instantaneous voltages and currents

The purpose of this section is to prove various formulas for the (total) instantaneous power of a three-phase network in terms of line and phase voltages and currents. Three methods will be explained. The first method is applicable to any three-phase device. The second and third method are more general, because they can be applied to any three-terminal or four-terminal network, even if such network is not a three-phase network (e.g. it is a split-phase network), and even if the network is an interconnection of various three-phase devices. The most relevant equations are shown the summary subsubsection of each subsection, that is, equations (1), (2) [or (3)], (4) [or (5)], (6) [or (7)], (8), (17) [or (18)], (19), (24), and (25).

It is possible to obtain at least two more equations for the (total) instantaneous power of three-phase networks if we assume balanced sinusoidal steady-state conditions. I won't do that.

Recall that a device refers to a lumped element, while a network is more general and refers either to a single device or an interconnection of various devices. For example, a three-phase generator (induction or synchronous) is a three-phase device and a three-phase network; on the other hand, two three-phase generators connected in parallel are also a three-phase network, but not a three-phase device.

In this section, device is synonym of element, load, and component. Total power (whether it is instantaneous, active, apparent, etc.) is synonym of three-phase power. Line voltage is synonym of line-to-line voltage and of phase-to-phase voltage.

The concept of line voltages and line currents make sense in three-phase devices and networks, but the concept of phase voltages and phase currents only make sense in three-phase devices, not networks. Why? Consider a three-phase network (i.e. a three-wire or four-wire network) consisting of two three-phase devices connected in parallel. Now ask yourself what are the phase currents of such network. You'll realize it doesn't make sense. I mean, sure, we can talk about the phase currents of each three-phase device of such network. But phase currents of the whole network? Nope.

Just to make sure we're on the same page, below I illustrate the instantaneous line and phase voltages and currents of three-phase devices.

Figure 1. Three-phase three-wire wye-connected device showing instantaneous phase voltages and currents. Click on image to expand it

Figure 2. Three-phase four-wire wye-connected device showing instantaneous phase voltages and currents and the instantaneous neutral current. Click on image to expand it

Figure 3. Three-phase three-wire delta-connected device showing instantaneous phase voltages and currents. Click on image to expand it

Figure 4. Three-phase three-wire wye-connected device showing instantaneous line voltages and currents. Click on image to expand it

Figure 5. Three-phase four-wire wye-connected device showing instantaneous line voltages and currents and the instantaneous neutral current. Click on image to expand it

Figure 6. Three-phase three-wire delta-connected device showing instantaneous line voltages and currents. Click on image to expand it

First method: using phase voltages and phase currents, for a three-wire or four-wire device

Summary

This is perhaps the most obvious way. In this method, we obtain the total instantaneous power by simply summing the single-phase (or per-phase) instantaneous powers of the three phases of the device.

This method is very general: it is valid for linear or non-linear, time-invariant or time-variant, bilateral or non-bilateral, passive or active, three-wire or four wire, three-phase devices, in wye or in delta, whether in steady-state (with harmonics or without harmonics) or in transient state, in balanced or unbalanced conditions, with positive/abc or negative/acb phase sequence.

The only requirements are that (i.e. the assumptions made):

1. The three-phase device can be adequately modeled as a lumped-element (as opposed to a distributed-element, such as a transmission line).
2. The three-phase device doesn’t have magnetic (inductive) or electric (capacitive) coupling with other elements in the network, because in that case we’d have to also consider the corresponding power to such energy coupling.
3. The three-phase device is a device and not a network/circuit (consisting of the interconnection of various three-phase devices).

The total instantaneous power consumed by the three-phase device is:

$p_{\text{T}}(t)=v_{\phi {\text{,a}}}(t)\,i_{\phi {\text{,a}}}(t)+v_{\phi {\text{,b}}}(t)\,i_{\phi {\text{,b}}}(t)+v_{\phi {\text{,c}}}(t)\,i_{\phi {\text{,c}}}(t)\quad {\text{(1)}}$

where the reference polarity for the instantaneous phase voltages and the reference direction for the instantaneous phase currents are those shown in figures 1–3.

Proof for a three-phase three-wire or four-wire device

First, we find the instantaneous power consumed by each phase of the three-phase device. Whether the three-phase device is connected in wye (three- or four-wire) or in delta, such power is the product of the phase instantaneous voltage times the corresponding phase instantaneous current, where the reference polarity for the voltage and the reference direction for the current are chosen such that the passive sign convention is satisfied (that is, such that the arrowhead of the current points towards the phase of the three-phase device into the positive reference polarity), as shown in figures 1 to 3.

The instantaneous power consumed by each phase is (whether it is numerically positive or negative):

$p_{\text{a}}(t)=v_{\phi {\text{,a}}}(t)\,i_{\phi {\text{,a}}}(t)$

$p_{\text{b}}(t)=v_{\phi {\text{,b}}}(t)\,i_{\phi {\text{,b}}}(t)$

$p_{\text{c}}(t)=v_{\phi {\text{,c}}}(t)\,i_{\phi {\text{,c}}}(t)$

Then, the total instantaneous power consumed by the three-phase device is just the sum of the instantaneous power consumed by each phase, as stated on page 520 (section 12.7) of Alexander & Sadiku's Fundamentals of Electric Circuits (5th ed.):

${\begin{aligned}p_{\text{T}}(t)&=p_{\text{a}}(t)+p_{\text{b}}(t)+p_{\text{c}}(t)\\\implies p_{\text{T}}(t)&=v_{\phi {\text{,a}}}(t)\,i_{\phi {\text{,a}}}(t)+v_{\phi {\text{,b}}}(t)\,i_{\phi {\text{,b}}}(t)+v_{\phi {\text{,c}}}(t)\,i_{\phi {\text{,c}}}(t)\quad {\text{(1)}}\end{aligned}}$

Second method: using line voltages and line currents, for a three-wire device or network

Summary

This method states we can compute the total instantaneous power of the three-phase network by using two line currents and two line voltages, by choosing one “common” node, with respect to which we measure the two line voltages, and whose line current we don’t measure. The following proof of this method is quite interesting, which is explained on pages 19–21 (section 1.7) of Parodi & Storace's Linear and Nonlinear Circuits: Basic & Advanced Concepts, Volume 1.

This method is very general: it is valid for linear or non-linear, time-invariant or time-variant, bilateral or non-bilateral, passive or active, three-wire three-phase devices or networks, in wye or in delta, whether in steady-state (with harmonics or without harmonics) or in transient state, in balanced or unbalanced conditions, with positive/abc or negative/acb phase sequence. Furthermore, it is applicable to any three-terminal network (not just three-phase three-wire and split-phase networks), so it can even be used for BJTs, MOSFETs, SCRs. The only requirements (i.e. the assumptions made) are the first and second requirements of the first method, and additionally:

1. The three-phase device must be three-wire, or it can be four-wire as long as the neutral current is zero. Fortunately, we can extend this method to the case where the neutral current isn’t zero, which I won’t explain in this answer.

Unlike the first method, this method can be applied to three-phase networks (an interconnection of one or more three-phase devices), which is very helpful.

The total instantaneous power consumed by the three-phase device or network, taking phase b as common (as shown in the following figure), is:

${\begin{aligned}p_{\text{T}}(t)&=v_{\text{L,a}}(t)\,i_{\text{L,a}}(t)-v_{\text{L,b}}(t)\,i_{\text{L,c}}(t)\quad {\text{(2)}}\\&=v_{\text{ab}}(t)\,i_{\text{L,a}}(t)-v_{\text{bc}}(t)\,i_{\text{L,c}}(t)\quad {\text{(3)}}\end{aligned}}$

Figure 7. Three-phase three-wire network, whose total instantaneous power is measured using instantaneous line voltages and currents, taking the phase B as common. Click on image to expand it

Or taking phase a as common (as shown in the following figure):

${\begin{aligned}p_{\text{T}}(t)&=v_{\text{L,c}}(t)\,i_{\text{L,c}}(t)-v_{\text{L,a}}(t)\,i_{\text{L,b}}(t)\quad {\text{(4)}}\\&=v_{\text{ca}}(t)\,i_{\text{L,c}}(t)-v_{\text{ab}}(t)\,i_{\text{L,b}}(t)\quad {\text{(5)}}\end{aligned}}$

Figure 8. Three-phase three-wire network, whose total instantaneous power is measured using instantaneous line voltages and currents, taking the phase A as common. Click on image to expand it

Or taking phase c as common (as shown in the following figure):

${\begin{aligned}p_{\text{T}}(t)&=v_{\text{L,b}}(t)\,i_{\text{L,b}}(t)-v_{\text{L,c}}(t)\,i_{\text{L,a}}(t)\quad {\text{(6)}}\\&=v_{\text{bc}}(t)\,i_{\text{L,b}}(t)-v_{\text{ca}}(t)\,i_{\text{L,a}}(t)\quad {\text{(7)}}\end{aligned}}$

Figure 9. Three-phase three-wire network, whose total instantaneous power is measured using instantaneous line voltages and currents, taking the phase C as common. Click on image to expand it

The total instantaneous power consumed can also be computed taking an external node o as common (as shown in the following figure):

$p_{\text{T}}(t)=v_{\text{ao}}(t)\,i_{\text{L,a}}(t)+v_{\text{bo}}(t)\,i_{\text{L,b}}(t)+v_{\text{co}}(t)\,i_{\text{L,c}}(t)\quad {\text{(8)}}$

Figure 10. Three-phase three-wire network, whose total instantaneous power is measured using instantaneous line-to-common voltages and currents, taking an external node as common. Click on image to expand it

If we additionally assume the total instantaneous power $p_{\text{T}}(t)$ is periodic with fundamental period $T'$ (which occurs e.g. if the network is operating in AC steady-state, whether sinusoidal or not and balanced or not, but not in transient state), and if we also assume each power term of equations 2 to 8 also have that same fundamental period, then, by using integrals, the total active power $P_{\text{T}}$ consumed by the network is:

${\begin{aligned}P_{\text{T}}={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}p_{\text{T}}(t)\,\mathrm {d} t&={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{L,a}}(t)\,i_{\text{L,a}}(t)\,\mathrm {d} t-{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{L,b}}(t)\,i_{\text{L,c}}(t)\,\mathrm {d} t\quad {\text{(9)}}\\&={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{ab}}(t)\,i_{\text{L,a}}(t)\,\mathrm {d} t-{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{bc}}(t)\,i_{\text{L,c}}(t)\,\mathrm {d} t\quad {\text{(10)}}\end{aligned}}$

${\begin{aligned}P_{\text{T}}={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}p_{\text{T}}(t)\,\mathrm {d} t&={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{L,c}}(t)\,i_{\text{L,c}}(t)\,\mathrm {d} t-{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{L,a}}(t)\,i_{\text{L,b}}(t)\,\mathrm {d} t\quad {\text{(11)}}\\&={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{ca}}(t)\,i_{\text{L,c}}(t)\,\mathrm {d} t-{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{ab}}(t)\,i_{\text{L,b}}(t)\,\mathrm {d} t\quad {\text{(12)}}\end{aligned}}$

${\begin{aligned}P_{\text{T}}={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}p_{\text{T}}(t)\,\mathrm {d} t&={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{L,b}}(t)\,i_{\text{L,b}}(t)\,\mathrm {d} t-{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{L,c}}(t)\,i_{\text{L,a}}(t)\,\mathrm {d} t\quad {\text{(13)}}\\&={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{bc}}(t)\,i_{\text{L,b}}(t)\,\mathrm {d} t-{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{ca}}(t)\,i_{\text{L,a}}(t)\,\mathrm {d} t\quad {\text{(14)}}\end{aligned}}$

$P_{\text{T}}={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}p_{\text{T}}(t)\,\mathrm {d} t={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{ao}}(t)\,i_{\text{L,a}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{bo}}(t)\,i_{\text{L,b}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{co}}(t)\,i_{\text{L,c}}(t)\,\mathrm {d} t\quad {\text{(15)}}$

Proof for a three-phase three-wire device or network, taking node b as common

First, suppose we connect the three-terminal network of figure 7 to two two-terminal lumped devices that aren’t magnetically nor electrically coupled with the environment or any other external device, as shown in the following figure.

Figure 11. Three-phase three-wire network, whose total instantaneous power is measured using instantaneous line voltages and currents, taking the phase B as common, connected to two two-terminal elements. Click on image to expand it

Since we’ve assumed all two devices as well as the three-phase network are lumped and don’t exchange energy with the environment, the whole circuit (that is, the two two-terminal devices and the three-phase network) is an isolated system, and thus the law of conservation of instantaneous power is satisfied. According to this law (in one of its forms), the total instantaneous power of a circuit (considering the consumed as positive and the supplied as negative) is zero.

Recall the instantaneous power of a two-terminal device is consumed if the voltage reference polarity and current reference direction are chosen such that they satisfy the passive sign convention. Thus, with the help of the previous figure, the instantaneous power consumed by the device 4 is:

$p_{4}(t)=v_{12}(t)\,i_{4}(t)=[v_{\text{L,a}}(t)]\,[-i_{\text{L,a}}(t)]=-v_{\text{L,a}}(t)\,i_{\text{L,a}}(t)$

(whether it is numerically positive or negative), and the instantaneous power consumed by the device 5 is:

$p_{5}(t)=v_{32}(t)\,i_{5}(t)=[-v_{\text{L,b}}(t)]\,[-i_{\text{L,c}}(t)]=v_{\text{L,b}}(t)\,i_{\text{L,c}}(t)$

According to the law of conservation of instantaneous power, the total instantaneous power consumed by the circuit of figure 5 is zero. Thus, the sum of the instantaneous power consumed by the device 4, by the device 5, and by the three-phase network is zero:

${\begin{aligned}0&=p_{4}(t)+p_{5}(t)+p_{\text{T}}(t)\\\implies p_{\text{T}}(t)&=-p_{4}(t)-p_{5}(t)\end{aligned}}$

where $p_{\text{T}}(t)$ is the instantaneous power consumed by the three-phase network. Substituting:

${\begin{aligned}p_{\text{T}}(t)&=-[-v_{\text{L,a}}(t)\,i_{\text{L,a}}(t)]-[v_{\text{L,b}}(t)\,i_{\text{L,c}}(t)]\\\implies p_{\text{T}}(t)&=v_{\text{L,a}}(t)\,i_{\text{L,a}}(t)-v_{\text{L,b}}(t)\,i_{\text{L,c}}(t)\quad {\text{(2)}}\\&=v_{\text{ab}}(t)\,i_{\text{L,a}}(t)-v_{\text{bc}}(t)\,i_{\text{L,c}}(t)\quad {\text{(3)}}\end{aligned}}$

—We can derive another equation. Now consider the circuit diagram of figure 10. It's the same three-terminal network, except there's now an external node o. Applying KVL (Kirchhoff’s voltage law) around the loop aoba (starting at node a, then going through an open circut and ending at node o, then going through an open circuit and ending at node b, then going through an open circuit back to node a):

${\begin{aligned}0&=v_{\text{ao}}(t)-v_{\text{bo}}(t)-v_{\text{ab}}(t)\\\implies v_{\text{ab}}(t)&=v_{\text{ao}}(t)-v_{\text{bo}}(t)\quad {\text{(16)}}\end{aligned}}$

applying KVL around the loop bocb:

${\begin{aligned}0&=v_{\text{bo}}(t)-v_{\text{co}}(t)-v_{\text{bc}}(t)\\\implies v_{\text{bc}}(t)&=v_{\text{bo}}(t)-v_{\text{co}}(t)\quad {\text{(17)}}\end{aligned}}$

and applying KVL around the loop coac:

${\begin{aligned}0&=v_{\text{co}}(t)-v_{\text{ao}}(t)-v_{\text{ca}}(t)\\\implies v_{\text{ca}}(t)&=v_{\text{co}}(t)-v_{\text{ao}}(t)\quad {\text{(18)}}\end{aligned}}$

Applying KCL (Kirchhoff’s current law) at a closed boundary which encloses the three-terminal network including its terminals:

${\begin{aligned}0&=i_{\text{L,a}}(t)+i_{\text{L,b}}(t)+i_{\text{L,c}}(t)\\\implies i_{\text{L,b}}(t)&=-[i_{\text{L,a}}(t)+i_{\text{L,c}}(t)]\quad {\text{(19)}}\end{aligned}}$

Substituting equations 16 and 17 into equation 3:

${\begin{aligned}p_{\text{T}}(t)&=[v_{\text{ao}}(t)-v_{\text{bo}}(t)]\,i_{\text{L,a}}(t)-[v_{\text{bo}}(t)-v_{\text{co}}(t)]\,i_{\text{L,a}}(t)\\&=v_{\text{ao}}(t)\,i_{\text{L,a}}(t)-v_{\text{bo}}(t)\,i_{\text{L,a}}(t)-v_{\text{bo}}(t)\,i_{\text{L,c}}(t)+v_{\text{co}}(t)\,i_{\text{L,c}}(t)\\&=[v_{\text{ao}}(t)\,i_{\text{L,a}}(t)]-v_{\text{bo}}(t)\,[i_{\text{L,a}}(t)+i_{\text{L,c}}(t)]+[v_{\text{co}}(t)\,i_{\text{L,c}}(t)]\end{aligned}}$

Substituting equation 19 into the previous one:

$p_{\text{T}}(t)=v_{\text{ao}}(t)\,i_{\text{L,a}}(t)+v_{\text{bo}}(t)\,i_{\text{L,b}}(t)+v_{\text{co}}(t)\,i_{\text{L,c}}(t)\quad {\text{(8)}}$

Proof for a three-phase three-wire device or network, taking node a as common

Now suppose we connect the two two-terminal devices taking the phase a as common, to the circuit of figure 8, as shown next.

Figure 12. Three-phase three-wire network, whose total instantaneous power is measured using instantaneous line voltages and currents, taking the phase A as common, connected to two two-terminal elements. Click on image to expand it

With the help of the previous figure, the instantaneous power consumed by the device 4 is:

$p_{4}(t)=v_{21}(t)\,i_{4}(t)=[-v_{\text{L,a}}(t)]\,[-i_{\text{L,b}}(t)]=v_{\text{L,a}}(t)\,i_{\text{L,b}}(t)$

and the instantaneous power consumed by the device 5 is:

$p_{5}(t)=v_{31}(t)\,i_{5}(t)=[v_{\text{L,c}}(t)]\,[-i_{\text{L,c}}(t)]=-v_{\text{L,c}}(t)\,i_{\text{L,c}}(t)$

Thus, by the law of conservation of instantaneous power:

${\begin{aligned}0&=p_{4}(t)+p_{5}(t)+p_{\text{T}}(t)\\\implies p_{\text{T}}(t)&=-p_{4}(t)-p_{5}(t)\end{aligned}}$

and substituting:

${\begin{aligned}p_{\text{T}}(t)&=-[v_{\text{L,a}}(t)\,i_{\text{L,b}}(t)]-[-v_{\text{L,c}}(t)\,i_{\text{L,c}}(t)]\\\implies p_{\text{T}}(t)&=v_{\text{L,c}}(t)\,i_{\text{L,c}}(t)-v_{\text{L,a}}(t)\,i_{\text{L,b}}(t)\quad {\text{(4)}}\\&=v_{\text{ca}}(t)\,i_{\text{L,c}}(t)-v_{\text{ab}}(t)\,i_{\text{L,b}}(t)\quad {\text{(5)}}\end{aligned}}$

—We can derive another equation. Rearranging equation 19:

$i_{\text{L,a}}(t)=-[i_{\text{L,b}}(t)+i_{\text{L,c}}(t)]\quad {\text{(20)}}$

Substituting equations 16 and 18 into equation 5:

${\begin{aligned}p_{\text{T}}(t)&=[v_{\text{co}}(t)-v_{\text{ao}}(t)]\,i_{\text{L,c}}(t)-[v_{\text{ao}}(t)-v_{\text{bo}}(t)]\,i_{\text{L,b}}(t)\\&=v_{\text{co}}(t)\,i_{\text{L,c}}(t)-v_{\text{ao}}(t)\,i_{\text{L,c}}(t)-v_{\text{ao}}(t)\,i_{\text{L,b}}(t)+v_{\text{bo}}(t)\,i_{\text{L,b}}(t)\\&=[v_{\text{co}}(t)\,i_{\text{L,c}}(t)]-v_{\text{ao}}(t)\,[i_{\text{L,c}}(t)+i_{\text{L,b}}(t)]+[v_{\text{bo}}(t)\,i_{\text{L,b}}(t)]\end{aligned}}$

Substituting equation 20 into the previous one:

${\begin{aligned}p_{\text{T}}(t)&=[v_{\text{co}}(t)\,i_{\text{L,c}}(t)]+[v_{\text{ao}}(t)\,i_{\text{L,a}}(t)]+[v_{\text{bo}}(t)\,i_{\text{L,b}}(t)]\\&=v_{\text{ao}}(t)\,i_{\text{L,a}}(t)+v_{\text{bo}}(t)\,i_{\text{L,b}}(t)+v_{\text{co}}(t)\,i_{\text{L,c}}(t)\quad {\text{(21)}}\end{aligned}}$

Notice this is the same as equation 8.

Proof for a three-phase three-wire device or network, taking node c as common

Now suppose we connect the two two-terminal devices taking the phase c as common, to the circuit of figure 9, as shown next.

Figure 13. Three-phase three-wire network, whose total instantaneous power is measured using instantaneous line voltages and currents, taking the phase C as common, connected to two two-terminal elements. Click on image to expand it

With the help of the previous figure, the instantaneous power consumed by the device 4 is:

$p_{4}(t)=v_{13}(t)\,i_{4}(t)=[-v_{\text{L,c}}(t)]\,[-i_{\text{L,a}}(t)]=v_{\text{L,c}}(t)\,i_{\text{L,a}}(t)$

and the instantaneous power consumed by the device 5 is:

$p_{5}(t)=v_{23}(t)\,i_{5}(t)=[v_{\text{L,b}}(t)]\,[-i_{\text{L,b}}(t)]=-v_{\text{L,b}}(t)\,i_{\text{L,b}}(t)$

Thus, by the law of conservation of instantaneous power:

${\begin{aligned}0&=p_{4}(t)+p_{5}(t)+p_{\text{T}}(t)\\\implies p_{\text{T}}(t)&=-p_{4}(t)-p_{5}(t)\end{aligned}}$

and substituting:

${\begin{aligned}p_{\text{T}}(t)&=-[v_{\text{L,c}}(t)\,i_{\text{L,a}}(t)]-[-v_{\text{L,b}}(t)\,i_{\text{L,b}}(t)]\\\implies p_{\text{T}}(t)&=v_{\text{L,b}}(t)\,i_{\text{L,b}}(t)-v_{\text{L,c}}(t)\,i_{\text{L,a}}(t)\quad {\text{(6)}}\\&=v_{\text{bc}}(t)\,i_{\text{L,b}}(t)-v_{\text{ca}}(t)\,i_{\text{L,a}}(t)\quad {\text{(7)}}\end{aligned}}$

—We can derive another equation. Rearranging equation 19:

$i_{\text{L,c}}(t)=-[i_{\text{L,a}}(t)+i_{\text{L,b}}(t)]\quad {\text{(22)}}$

Substituting equations 17 and 18 into equation 7:

${\begin{aligned}p_{\text{T}}(t)&=[v_{\text{bo}}(t)-v_{\text{co}}(t)]\,i_{\text{L,b}}(t)-[v_{\text{co}}(t)-v_{\text{ao}}(t)]\,i_{\text{L,a}}(t)\\&=v_{\text{bo}}(t)\,i_{\text{L,b}}(t)-v_{\text{co}}(t)\,i_{\text{L,b}}(t)-v_{\text{co}}(t)\,i_{\text{L,a}}(t)+v_{\text{ao}}(t)\,i_{\text{L,a}}(t)\\&=[v_{\text{bo}}(t)\,i_{\text{L,b}}(t)]-v_{\text{co}}(t)\,[i_{\text{L,b}}(t)+i_{\text{L,a}}(t)]+[v_{\text{ao}}(t)\,i_{\text{L,a}}(t)]\end{aligned}}$

Substituting equation 22 into the previous one:

${\begin{aligned}p_{\text{T}}(t)&=[v_{\text{bo}}(t)\,i_{\text{L,b}}(t)]+[v_{\text{co}}(t)\,i_{\text{L,c}}(t)]+[v_{\text{ao}}(t)\,i_{\text{L,a}}(t)]\\&=v_{\text{ao}}(t)\,i_{\text{L,a}}(t)+v_{\text{bo}}(t)\,i_{\text{L,b}}(t)+v_{\text{co}}(t)\,i_{\text{L,c}}(t)\quad {\text{(23)}}\end{aligned}}$

Notice this is the same as equations 8 and 21.

Third method: using line voltages and line currents, for a four-wire device or network

Summary

This method is just the extension of the previous one to four-wire three-phase networks. It has the same requirements/assumptions (except that obviously this one can be applied to four-wire networks) and is valid for the same types of networks. Again, it is applicable to any four-terminal network (not just three-phase four-wire networks).

The total instantaneous power consumed by the three-phase device or network, taking the neutral n as common, is:

${\begin{aligned}p_{\text{T}}(t)&=v_{\text{LN,a}}(t)\,i_{\text{L,a}}(t)+v_{\text{LN,b}}(t)\,i_{\text{L,b}}(t)+v_{\text{LN,c}}(t)\,i_{\text{L,c}}(t)\quad {\text{(24)}}\\&=v_{\text{an}}(t)\,i_{\text{L,a}}(t)+v_{\text{bn}}(t)\,i_{\text{L,b}}(t)+v_{\text{cn}}(t)\,i_{\text{L,c}}(t)\quad {\text{(25)}}\end{aligned}}$

Figure 14. Three-phase four-wire network, whose total instantaneous power is measured using instantaneous line-to-neutral voltages and line currents, taking the neutral N as common. Click on image to expand it

The total instantaneous power consumed can also be computed taking an external node o as common (as shown in the following figure):

$p_{\text{T}}(t)=v_{\text{ao}}(t)\,i_{\text{L,a}}(t)+v_{\text{bo}}(t)\,i_{\text{L,b}}(t)+v_{\text{co}}(t)\,i_{\text{L,c}}(t)+v_{\text{no}}(t)\,i_{\text{n}}(t)\quad {\text{(26)}}$

Figure 15. Three-phase four-wire network, whose total instantaneous power is measured using instantaneous line-to-common voltages and line currents, taking an external node as common. Click on image to expand it

If we additionally assume the total instantaneous power $p_{\text{T}}(t)$ is periodic with fundamental period $T'$ (which occurs e.g. if the network is operating in AC steady-state, whether sinusoidal or not and balanced or not, but not in transient state), and if we also assume each power term of equations 24 to 26 also have that same fundamental period, then, by using integrals, the total active power $P_{\text{T}}$ consumed by the network is:

${\begin{aligned}P_{\text{T}}={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}p_{\text{T}}(t)\,\mathrm {d} t&={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{LN,a}}(t)\,i_{\text{L,a}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{LN,b}}(t)\,i_{\text{L,b}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{LN,c}}(t)\,i_{\text{L,c}}(t)\,\mathrm {d} t\quad {\text{(27)}}\\&={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{an}}(t)\,i_{\text{L,a}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{bn}}(t)\,i_{\text{L,b}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{cn}}(t)\,i_{\text{L,c}}(t)\,\mathrm {d} t\quad {\text{(28)}}\end{aligned}}$

$P_{\text{T}}={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}p_{\text{T}}(t)\,\mathrm {d} t={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{ao}}(t)\,i_{\text{L,a}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{bo}}(t)\,i_{\text{L,b}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{co}}(t)\,i_{\text{L,c}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{no}}(t)\,i_{\text{n}}(t)\,\mathrm {d} t\quad {\text{(29)}}$

Proof for a three-phase four-wire device or network, taking node n as common

First, suppose we connect the three-terminal network of figure 13 to three two-terminal lumped devices that aren’t magnetically nor electrically coupled with the environment or any other external device, as shown in the following figure.

Figure 16. Three-phase four-wire network, whose total instantaneous power is measured using instantaneous line-to-neutral voltages and line currents, taking the neutral N as common, connected to three two-terminal elements. Click on image to expand it

With the help of the previous figure, the instantaneous power consumed by the device 4 is:

$p_{4}(t)=v_{14}(t)\,i_{4}(t)=[v_{\text{LN,a}}(t)]\,[-i_{\text{L,a}}(t)]=-v_{\text{LN,a}}(t)\,i_{\text{L,a}}(t)$

the instantaneous power consumed by the device 5 is:

$p_{5}(t)=v_{24}(t)\,i_{5}(t)=[v_{\text{LN,b}}(t)]\,[-i_{\text{L,b}}(t)]=-v_{\text{LN,b}}(t)\,i_{\text{L,b}}(t)$

and the instantaneous power consumed by the device 6 is:

$p_{6}(t)=v_{34}(t)\,i_{6}(t)=[v_{\text{LN,c}}(t)]\,[-i_{\text{L,c}}(t)]=-v_{\text{LN,c}}(t)\,i_{\text{L,c}}(t)$

Thus, by the law of conservation of instantaneous power:

${\begin{aligned}0&=p_{4}(t)+p_{5}(t)+p_{6}(t)+p_{\text{T}}(t)\\\implies p_{\text{T}}(t)&=-p_{4}(t)-p_{5}(t)-p_{6}(t)\end{aligned}}$

and substituting:

${\begin{aligned}p_{\text{T}}(t)&=-[-v_{\text{LN,a}}(t)\,i_{\text{L,a}}(t)]-[-v_{\text{LN,b}}(t)\,i_{\text{L,b}}(t)]-[-v_{\text{LN,c}}(t)\,i_{\text{L,c}}(t)]\\\implies p_{\text{T}}(t)&=v_{\text{LN,a}}(t)\,i_{\text{L,a}}(t)+v_{\text{LN,b}}(t)\,i_{\text{L,b}}(t)+v_{\text{LN,c}}(t)\,i_{\text{L,c}}(t)\quad {\text{(24)}}\\&=v_{\text{an}}(t)\,i_{\text{L,a}}(t)+v_{\text{bn}}(t)\,i_{\text{L,b}}(t)+v_{\text{cn}}(t)\,i_{\text{L,c}}(t)\quad {\text{(25)}}\end{aligned}}$

—By the way, as shown in Parodi & Storace's textbook mentioned earlier, we can choose any terminal as common (so instead of the neutral, we can choose phase a, or b, or c).

—We can derive another equation. Now consider the circuit diagram of figure 15. It's the same four-terminal network, except there's now an external node o. Applying KVL around the loop aona (starting at node a, then going through an open circut and ending at node o, then going through an open circuit and ending at node n, then going through an open circuit back to node a):

${\begin{aligned}0&=v_{\text{ao}}(t)-v_{\text{no}}(t)-v_{\text{an}}(t)\\\implies v_{\text{an}}(t)&=v_{\text{ao}}(t)-v_{\text{no}}(t)\quad {\text{(30)}}\end{aligned}}$

applying KVL around the loop bonb:

${\begin{aligned}0&=v_{\text{bo}}(t)-v_{\text{no}}(t)-v_{\text{bn}}(t)\\\implies v_{\text{bn}}(t)&=v_{\text{bo}}(t)-v_{\text{no}}(t)\quad {\text{(31)}}\end{aligned}}$

and applying KVL around the loop conc:

${\begin{aligned}0&=v_{\text{co}}(t)-v_{\text{no}}(t)-v_{\text{cn}}(t)\\\implies v_{\text{cn}}(t)&=v_{\text{co}}(t)-v_{\text{no}}(t)\quad {\text{(32)}}\end{aligned}}$

Applying KCL (Kirchhoff’s current law) at a closed boundary which encloses the four-terminal network including its terminals:

${\begin{aligned}0&=i_{\text{L,a}}(t)+i_{\text{L,b}}(t)+i_{\text{L,c}}(t)+i_{\text{n}}(t)\\\implies i_{\text{n}}(t)&=-[i_{\text{L,a}}(t)+i_{\text{L,b}}(t)+i_{\text{L,c}}(t)]\quad {\text{(33)}}\end{aligned}}$

Substituting equations 30, 31 and 32 into equation 25:

${\begin{aligned}p_{\text{T}}(t)&=[v_{\text{ao}}(t)-v_{\text{no}}(t)]\,i_{\text{L,a}}(t)+[v_{\text{bo}}(t)-v_{\text{no}}(t)]\,i_{\text{L,b}}(t)+[v_{\text{co}}(t)-v_{\text{no}}(t)]\,i_{\text{L,c}}(t)\\&=v_{\text{ao}}(t)\,i_{\text{L,a}}(t)-v_{\text{no}}(t)\,i_{\text{L,a}}(t)+v_{\text{bo}}(t)\,i_{\text{L,b}}(t)-v_{\text{no}}(t)\,i_{\text{L,b}}(t)+v_{\text{co}}(t)\,i_{\text{L,c}}(t)-v_{\text{no}}(t)\,i_{\text{L,c}}(t)\\&=[v_{\text{ao}}(t)\,i_{\text{L,a}}(t)]+[v_{\text{bo}}(t)\,i_{\text{L,b}}(t)]+[v_{\text{co}}(t)\,i_{\text{L,c}}(t)]-v_{\text{no}}(t)\,[i_{\text{L,a}}(t)+i_{\text{L,b}}(t)+i_{\text{L,c}}(t)]\end{aligned}}$

Substituting equation 33 into the previous one:

$p_{\text{T}}(t)=v_{\text{ao}}(t)\,i_{\text{L,a}}(t)+v_{\text{bo}}(t)\,i_{\text{L,b}}(t)+v_{\text{co}}(t)\,i_{\text{L,c}}(t)+v_{\text{no}}(t)\,i_{\text{n}}(t)\quad {\text{(26)}}$

Fourth method: using voltages and currents, for a two-wire device or network

Summary

This should've been the first method I explained, not the fourth, sorry for this.

This method states we can compute the total instantaneous power of the single-phase network by using one current and one voltage, namely the current through the network and the voltage across the network.

This method is very general: it is valid for linear or non-linear, time-invariant or time-variant, bilateral or non-bilateral, passive or active, two-wire single-phase devices or networks, whether in steady-state (with harmonics or without harmonics) or in transient state. Furthermore, it is applicable to any two-terminal network (not just single-phase two-wire networks), so it can even be used for two-terminal resistors, two-terminal capacitors, two-terminal inductors, any two-terminal diode, independent or dependent ideal voltage or current sources. The only requirements (i.e. the assumptions made) are the first and second requirements of the first method, substituting the word "three-phase" for "single-phase".

The total instantaneous power consumed by the single-phase device or network, taking the neutral N as common, is:

$p_{\text{T}}(t)=v_{\text{LN}}(t)\,i_{\text{L}}(t)\quad {\text{(34)}}$

Figure 17. Single-phase two-wire network, whose total instantaneous power is measured using instantaneous line-to-neutral voltages and line currents, taking the neutral N as common. Click on image to expand it

The total instantaneous power consumed can also be computed taking an external node o as common (as shown in the following figure):

$p_{\text{T}}(t)=v_{\text{LO}}(t)\,i_{\text{L}}(t)+v_{\text{NO}}(t)\,i_{\text{N}}(t)\quad {\text{(35)}}$

Figure 18. Single-phase two-wire network, whose total instantaneous power is measured using instantaneous line-to-common voltages and currents, taking an external node as common. Click on image to expand it

If we additionally assume the total instantaneous power $p_{\text{T}}(t)$ is periodic with fundamental period $T'$ (which occurs e.g. if the network is operating in AC steady-state, whether sinusoidal or not and balanced or not, but not in transient state), and if we also assume each power term of equations 34 to 35 also have that same fundamental period, then, by using integrals, the total active power $P_{\text{T}}$ consumed by the network is:

$P_{\text{T}}={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}p_{\text{T}}(t)\,\mathrm {d} t={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{LN}}(t)\,i_{\text{L}}(t)\,\mathrm {d} t\quad {\text{(36)}}$

$P_{\text{T}}={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}p_{\text{T}}(t)\,\mathrm {d} t={\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{LO}}(t)\,i_{\text{L}}(t)\,\mathrm {d} t+{\dfrac {1}{T'}}\displaystyle \int _{t_{0}}^{t_{0}+T'}v_{\text{NO}}(t)\,i_{\text{N}}(t)\,\mathrm {d} t\quad {\text{(37)}}$

Proof for a single-phase two-wire device or network, taking node N as common

Instantaneous power is defined as the rate of flow or transfer of energy, or the rate of doing work:

$p(t)={\dfrac {\mathrm {d} w(t)}{\mathrm {d} t}}\quad {\text{(38)}}$

On the other hand, instantaneous current is defined as the rate of flow of charge:

$i(t)={\dfrac {\mathrm {d} q(t)}{\mathrm {d} t}}\quad {\text{(39)}}$

And in electric circuit theory/analysis, instantaneous voltage is defined as follows:

$v(t)={\dfrac {\mathrm {d} w(t)}{\mathrm {d} q}}\quad {\text{(40)}}$

Now, from calculus, specifically the chain rule for single-variable scalar functions, we know that if $y=f(u)$ ($y$ is a function of $u$) and at the same time if $u=g(x)$ ($u$ is a function of $x$), then:

${\dfrac {\mathrm {d} y}{\mathrm {d} x}}={\dfrac {\mathrm {d} y}{\mathrm {d} u}}\,{\dfrac {\mathrm {d} u}{\mathrm {d} x}}$

In our case, the energy being transferred ($y=w$) depends on how many charges are being transferred ($u=q$), which at the same time depends on the instant of time ($x=t$), so equation 38 can be expanded as:

$p(t)={\dfrac {\mathrm {d} w}{\mathrm {d} t}}=\left({\dfrac {\mathrm {d} w}{\mathrm {d} q}}\right)\,\left({\dfrac {\mathrm {d} q}{\mathrm {d} t}}\right)$

Substituting equations 39 and 40 respectively in the first and second derivative (read from left to right) of the previous equation:

$p(t)=v(t)\,i(t)$

Whether the previous equation tells the instantaneous power consumed or supplied depends on the reference polarity of the voltage and reference direction of the current, which is discussed in the article of passive sign convention. In our case, considering the reference polarity of voltage and reference direction of current in the previous figure, the power consumed by the two-terminal network is:

$p_{\text{T}}(t)=v_{\text{LN}}(t)\,i_{\text{L}}(t)\quad {\text{(34)}}$

—We can derive another equation. Now consider the circuit diagram of figure 18. It's the same three-terminal network, except there's now an external node o. Applying KVL around the loop LONL (starting at node L, then going through an open circut and ending at node O, then going through an open circuit and ending at node N, then going through an open circuit back to node L):

${\begin{aligned}0&=v_{\text{LO}}(t)-v_{\text{NO}}(t)-v_{\text{LN}}(t)\\\implies v_{\text{LN}}(t)&=v_{\text{LO}}(t)-v_{\text{NO}}(t)\quad {\text{(41)}}\end{aligned}}$

Applying KCL (Kirchhoff’s current law) at a closed boundary which encloses the two-terminal network including its terminals:

${\begin{aligned}0&=i_{\text{L}}(t)+i_{\text{N}}(t)\\\implies i_{\text{L}}(t)&=-i_{\text{N}}(t)\quad {\text{(42)}}\end{aligned}}$

Substituting equation 41 into equation 34:

${\begin{aligned}p_{\text{T}}(t)&=[v_{\text{LO}}(t)-v_{\text{NO}}(t)]\,i_{\text{L}}(t)\\&=v_{\text{LO}}(t)\,i_{\text{L}}(t)-v_{\text{NO}}(t)\,i_{\text{L}}(t)\\&=[v_{\text{LO}}(t)\,i_{\text{L}}(t)]-v_{\text{NO}}(t)\,i_{\text{L}}(t)\end{aligned}}$

Substituting equation 42 into the previous one:

$p_{\text{T}}(t)=v_{\text{LO}}(t)\,i_{\text{L}}(t)+v_{\text{NO}}(t)\,i_{\text{N}}(t)\quad {\text{(35)}}$

Example

Below I show the simulation of a non-linear time-variant unbalanced ungrounded three-phase three-wire device connected in wye (so the fourth and fifth methods are not applicable). The total instantaneous power of the device was calculated using the formula of the first method (turquoise curve) as well as the three formulas of the second method (pink curve, gray curve, yellow curve).

Figure 19. Simulation in LTspice of a three-phase three-wire wye-connected unbalanced ungrounded non-linear time-variant device supplied by a three-phase wye-connected balanced voltage source with positive phase sequence. Click on image to expand it

The fact that in the previous figure you can only see the yellow curve but not the other three curves (turquoise, pink, gray) shows they are actually equal to the yellow curve for each instant in time, and are “beneath” it.

--Alej27 (talk) 20:28, 9 June 2021 (UTC)