Talk:Power factor/Archive 1

Archive 1

Archive 2

Archive 3

Archive 4

bill for kVAh

The article states "The significance of power factor lies in the fact that utility companies supply customers with volt-amperes, but bill them for watts." At least in Australia, this is not correct. The electricity meters measure volt-ampere hours. The bill incorrectly says kWh, but really you are charged for kVAh. This means that it is in the customer's interest to have unity power factor, since this reduces the amount they pay for electricity.

Illustration

I'm a rather bored power engineer, so i had a hack about, hopefully it makes a bit more sense now. let me know if it needs clearer explanation

Much better. It might be nice to illustrate the power triangle with something like the graphic at http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html.

Distortion power factor

This article only talks about "displacement" power factor. What about the effects of non-sinusoidal currents? --Wtshymanski 22:35, 16 Dec 2004 (UTC)

Non-sinusoidal (non linear)analysis is very difficult, and a bit too tricky to try and explain here. I have included some details about harmonics. --Euripides 15:53, 3rd Jan 2005

Explanation of Apparent and Reactive power

There does not appear to be any proper explanation of what apparent and reactive power are, or how to calculate them.

Mrwooster 02:42, 25 February 2007 (UTC)

Form Factor

In the case of non-sinusoidal currents and voltages, the term "form factor" is sometimes used to refer to the ratio between the mean value and the RMS (root mean square) value. The form factor can become large in the case of pulsed waveforms with a high mark to space ratio.

True, but I didn't think it clarified power factor at all. --Wtshymanski 03:04, 16 November 2005 (UTC)

Question re load sharing generators

dear sirs,

i've just ran into your discussion about <power factor>. would you participate in following:

two equal generators running in parallel load sharing mode, - both of them loaded equal due (as sample 2 x 150kW) but one has too less current than another one (difference is about 150 Amps).

if to rely on the fact that P = square root 3 x U x I x Cos Phi we face reducing power factor for one of generators ... it is clear.

while peaks of load, as a result, one with higher current is tripping due to overcurrent protection.

the practical question is where the the influence for <power factor> appeared from (?) if measurements taken around both generators' excitation systems shows no difference.

with other words: could it be that consumers' =low state of insulation= causes such a behaviour ?

thanks for yr participation brgds Alex

This looks like a question for http://eng-tips.com electric power engineering section --C J Cowie 15:24, 1 March 2006 (UTC)

Advertising links

The last four of the five external links seem to various extents to be advertising links. However each one seems to be a direct link to useful information related to the subject of this article. Should they all be removed? --C J Cowie 01:30, 23 August 2006 (UTC)

I was wrong about that. Learn about Power Factor provides forms to be filled out to request information by email. Very little information is directly available from the linked page. I am removing that one now. --C J Cowie 17:43, 23 August 2006 (UTC)

User:Wtshymanski acted while was still talking. --C J Cowie 17:46, 23 August 2006 (UTC)

Physical Explanation of Leading Power Factor

I was just wondering if anyone can explain how a leading power factor makes sense physically. Current is a result of voltage, isn't it? So how can an effect possibly precede its cause? Unless it is really lagging by >=3/4 of a period...? Anyone? GBMorris 22:00, 9 December 2006 (UTC)

I agree that the current through the power lines is caused by the voltage (in combination with other factors). The voltage is not "caused by" the current or any other factor in my house, since the voltage across the power lines to my house remains practically the same whether I turn everything on (lots of current) or turn everything off (zero current).

So it seems like a leading power factor would be impossible, since the effect (appears to) precede the cause.

Let me try to draw an analogy:

A mountain climber climbs a snow-covered slope for days, finally reaching the top at high noon on Wednesday. Journalists in the peaceful village nearby use binoculars to watch him plant a flag at the peak, rushing to print a story it ("Mountain Climber finally climbs Mount on Wednesday"). Alas, he dislodged a chunk of snow that triggers an avalanche on Tuesday. That avalanche roared for what seemed like hours, with the greatest volume around 11 am. How can the peak of an effect (the avalanche on Tuesday, peaking at 11 am) possibly precede the peak of the cause ("Mountain Climber finally climbs Mount on Wednesday" with the peak reached at noon)?

When we place a capacitor across the AC power lines, many things happen. I suspect the capacitor article has a better explanation. For simplicity, let's assume that at time T=0, we place a discharged ( V=0 ) capacitor across the power lines at the exact instant that the voltage across the power line is also zero (the "zero crossing"). Also, we watch "the voltage" v(t) (the voltage across the capacitor) and "the current" i(t). For the next few milliseconds, the voltage increases approximately linearly with a slope m = dv(t)/dt. However, during those milliseconds, the current stays approximately constant at i = C*m.

The *rising* voltage causes electrons to flow in one lead of the capacitor, and out the other lead.

Later, around the time the voltage reaches its peak, the voltage stays approximately constant for a few milliseconds.

The approximately *constant* voltage causes electrons in the capacitor to stay more-or-less in the same place. So during those few milliseconds, the current of electrons is approximately zero.

We can graph this voltage v(t) vs. current i(t) over several cycles.

On that graph, we see the power line voltage going up and down as always, completely oblivious to the capacitor. The current reaches its peak *before* the the power line voltage reaches its peak, so we say the current is leading the voltage. --68.0.124.33 (talk) 16:27, 15 August 2008 (UTC)

Amateur explanation of power lead/lag

Having never studied this stuff to Uni level but having been interested since childhood, here is my take on the logic of current lead/lag:

A capacitor is like a short-term rechargeable battery, only a rechargeable needs to be charged carefully. A capacitor will act effectively like a short circuit until it is storing the same voltage as is being applied. At the other end of the scale, a capacitor that is sitting across a 12v DC line and is charged to 12v presents an almost infinite (minus leakage) resistance across the 12v line. Should that 12v line waiver, the capacitor's charge will try to either boost the deficiency or grab the increase.

Now put this across rectified AC, empty capacitor, 0v line. Line voltage increases, capacitor absorbs. Load also starts to absorb. Although the load will not reach full current until the voltage peaks, the capacitor will grab every electron it can get its layers on and therefore the current through the capacitor will be hugely higher than that through the load. Therefore, the current the line supplies will be far more to do with what the capacitor wants than what the real load wants. By the time the AC peaks, the capacitor is fully charged and presents no load so the real load gets everything.

Now, the AC wave starts to diminish. The line starts to become less charged than the capacitor, so the capacitor starts to discharge into the line. The load is supplied by both the line and the capacitor. The current drawn by the line is therefore much less than would be expected, due to the capacitor's help.

This is why the current curve precedes the voltage curve by 0.25 of a full sinusoidal cycle with a capacitor across the line.

Coils are the opposite of capacitors in that they fight, or kick against, any change in voltage. Put a coil across the same rectified AC line and it will appear as a near infinite resistance whilst the voltage is rising. As soon as the voltage has peaked and started to decrease, however, the coil then becomes less and less resistant. The nearer the voltage gets to zero, the less resistant the coil is. Near zero it is almost a short circuit and so the line current is at its highest. Having said that, the strange part is that the coil is now fully charged. Once the voltage starts to increase again, the coil then starts to discharge. So although the load drawn by the load is xAmps, the large part of this will be supplied by the coil and only a small amount by the line. This is why the current curve lags the voltage curve.

Read up on loudspeaker crossovers and passive graphic equalizer units. Also look at power supplies which used to have a diode valve (American:Tube) followed by an inline choke (coil), which then became bridge rectifier diodes followed by capacitor across the line.

Then look at rechargeable batteries: put 4 in parallel and charge to 1.2v. Once charged, switch them to series and they will return 1.2v x 4 = 4.8v. Now apply this to capacitors and you have a switched mode power supply. —The preceding unsigned comment was added by APNorth (talk • contribs) 23:41, 21 December 2006 (UTC).

ELI the CIVIL ICE man

Having gone straight from ham radio in high school to physics in college I missed out on all the nice EE mnemonics like "Little star up in the sky, power equals R squared I" or whatever it is, so ELI the CIVIL ICE man is news to me. Mulling it over, I concluded there must be four types of EE.

Type 1. Depends on mnemonics to remember whether current leads or lags voltage in capacitors and inductors.

Type 2. Intuits that current lags voltage in an inductor by thinking of inductors as flywheels that respond sluggishly to an applied force. Doesn't intuit current lead for capacitors (how could a mere capacitor predict the future?) but does remember that capacitors are the opposite of inductors.

Type 3. Is equally at home with voltage causing current and current causing voltage. An AC current applied to a capacitor causes the voltage to rise from zero when the current is at its peak and causes the voltage to reach its peak just when the current has dropped to zero and is starting to go negative. Hence in a capacitor voltage lags current by 90 degrees.

Type 4. Understands flywheels by analogy with inductors. --Vaughan Pratt (talk) 08:21, 26 January 2008 (UTC)

I don't know if there are four types of EE; after a few years in the business you don't think about this any more than you think about how to tie a tie. The mnemonic is just there to backstop you on an exam. Analogies are tricky - best to learn the physics ( then you don't have to remember which property is which - is speed more like current or voltage? )--Wtshymanski (talk) 04:30, 27 January 2008 (UTC)

If you know that a capacitor tries to resist a change in charge (voltage), because it must receive or send electrons (current) in order to change it's internal voltage. Voltage (charge) is proportional to the amount of electrons stored within the capacitor. So, one can deduce that the current flow will be greatest when the voltage rate of change is greatest (dv/dt). This rate peaks out at the voltage zero crossing point in the AC sine wave, which is exactly 90° ahead of the the voltage peak (slowest rate of change).

The opposite is true for inductors. An inductor's current flow is lowest when the dv/dt is highest, because the expansion of it's magnetic field requires little current because the magnetic lines crossing the inductor during storage creates a back EMF that makes the inductor look high impedance. When dv/dt is at minimum at the top of the voltage wave, the magnetic field begins to collapse which induces current into the circuit that opposes the still positive voltage. Also, when there is little or no dv/dt the inductor looks like a piece of long wire across the voltage source (low impedance or short circuit) and thus highest current. —Preceding unsigned comment added by Moesdaddy (talk • contribs) 20:16, 24 December 2008 (UTC)

Distribution vs. Generation Losses

If a poor power factor is balanced by compensating complementary elements, the distribution losses only occur within that sub-system. Low power factors do not lead to generation losses unless they are not compensated somewhere, within the system as a whole. So, in general, small users with some equipment with poor power factor need not feel guilty. But if poorly designed computer power supplies are commonly used, then there could be significant system-wide issues.-69.87.193.242 12:49, 4 April 2007 (UTC)

Yes, I think this also applies to compact fluorescent lamps, which usually have a much lower power factor than incandescent lamps. Biscuittin (talk) 09:43, 2 November 2009 (UTC)

Power factor for AC

I know to calculate power in an AC circuit you have to multiply the current times the voltage times the power factor, but what if you are actually measuring current with a meter? the initial equation for power factor is based on ohm's law. If I stick a current meter in a hot AC circuit does it measure a current that can be multiplied by the actual voltage to get power? using cosine theta would seem to me to throw this off. Then again I'm not an expert by any means. Expert advise please, and also perhaps if I'm right and the power factor only accounts for inductance as a means of correcting ohm's law for an inductive load that it should stipulate that you can use ohm's law directly with measured variables on a physical circuit.

Can't give a definite answer, but I think it depends what sort of meter is used. A moving-iron meter would give a different answer from a hot-wire meter. As for electronic meters, I have no idea. Biscuittin (talk) 09:50, 2 November 2009 (UTC)

Not really, no. Average-responding meters are calibrated to display the RMS values of a sine wave (or close to it). The only time you'd see a difference in current measurement is if the waveform has a different form factor than assumed in the meter calibration. This has nothing to do with power factor - to measure power factor you must measure real and reactive power, or current, voltage and phase angle. --Wtshymanski (talk) 19:32, 2 November 2009 (UTC)

Graphs

The graphs are pretty nice, but I think they should show RMS (root mean square) power rather than average. RMS voltage, power and current is the quantity typically discussed with AC signals.

Showing that the average AC power is zero in the power factor 0 graph is true but misleading. The load does not see 0 watts, and this is what a casual reader may conclude. —Preceding unsigned comment added by 24.68.59.219 (talk) 21:22, 10 April 2008 (UTC)

I believe that average power is what the original illustrator intended to demonstrate; that is what I did when I reimplemented the graphs in Gnuplot/SVG. There is some justification in doing so; an ordinary integrating electric meter will also show zero energy used when presented with a zero-PF load, which is why industrial consumers are nearly always demand-metered. 121a0012 (talk) 06:24, 17 April 2008 (UTC)

merge

I suggest merging power-factor correction and power factor correction unit into power factor. That way we don't have to re-explain "power factor", its implications, and ways to handle it, in all 3 articles. --68.0.124.33 (talk) 17:53, 15 August 2008 (UTC)

Yep. Better one long (but well-structured!) article than a lot of repetition. --Wtshymanski (talk) 18:00, 15 August 2008 (UTC)

I see Wtshymanski has merged these 3 pages. Well done! --68.0.124.33 (talk) 16:53, 26 September 2008 (UTC)

Active PFC

Active PFC is becoming a more important technical characteristic when purchasing power supplies for personal home computers. Defining this term as it applies to PC's would be very beneficial to many people.

68.241.177.152 (talk) 05:43, 10 January 2009 (UTC)jonny

Umm...it's in the article, already - did you not get that far? --Wtshymanski (talk) 16:01, 10 January 2009 (UTC)

in the news

Have you seen what the people at Tom's hardware guide say about this "power factor" article: [1]? --68.0.124.33 (talk) 16:03, 4 November 2008 (UTC)

Hmm. What gets regulated. Powerfactor, of course. What's a "mobo" ? Is there a crazy gadget - No, of course not. What are the facts? See the article. Does it shut things off in low use? Does what shut what off?

Sure, the article needs work, but "Idununnastanit" is not a useful guide to the editor. --Wtshymanski (talk) 16:27, 5 November 2008 (UTC)

power factor correction for residential

some say it is a scam: http://nlcpr.com/Deceptions1.php

the places that sell these capacitors say they work for residential homes. Can someone help me out here? Thanks --OxAO (talk) 04:42, 11 July 2009 (UTC)

The world is a very big place, but in the few places I've seen residential electricity bills, the utility doesn't charge individual residences for reactive power. Perhaps an apartment complex with a large air conditioning plant would be charged for its reactive power, but otherwise if you're not paying for power factor, there's no benefit to wiring capacitors into your home. --Wtshymanski (talk) 14:26, 15 September 2009 (UTC)

At least in the U.S., it would be determined by tariff and state regulations. For example, my power company's general tariff (applicable to all customers) states:

Except as may otherwise be provided in a specific rate, a Customer taking service is expected to maintain a power factor of not less than 80% percent [sic]. The Company may require any Customer not satisfying this power factor requirement to furnish, install, and maintain, at no cost to the Company, such corrective equipment as the Company may deem necessary under the circumstances. Alternatively, the Company may elect to install such corrective equipment at the Customer's expense.

Boston Edison Company d/b/a NStar Electric (2006-01-20). "Terms and Conditions — Distribution Services (M.D.T.E. No. 100A)" (PDF). Retrieved 2009-09-15.

It would be fairly unusual for a residential customer to have a power factor below 0.8 given that most residential loads have historically been resistive.

The individual service tariffs also allow the company to compute demand (for demand-metered, commercial customers) on the basis of either 90% of apparent power or 100% of actual power, at the company's option. Presumably they choose the higher-revenue option. 121a0012 (talk) 02:36, 16 September 2009 (UTC)

Right, the utility may require customers to stay above a certain limit, but residential meters only meter kwh, not kvah - so it would be difficult for the utility to detect an individual residential customer with low PF. I haven't seen anything on it but it would be interesting to find a paper describing what a modern home's power factor really is...with all the fluorescent lighting, computers, microwave ovens, etc., the load isn't purely linear resistive incandescent lighting any more. Anyway, unless your utility bill shows both wh and kvah or equivalent, no residential power factor improvement is going to help reduce your bill. --Wtshymanski (talk) 16:31, 16 September 2009 (UTC)

Metering

I wouldn't know where to put this in the article, so I'll say it here.

A Thompson watt-hour meter, still in widespread use, only measures the energy that is delivered to the load. Even though a purely reactive load does not use any energy and does not cause the meter to turn, it still is a drain on the utility because the current flowing in the distribution system causes resistive heating of the distribution wires. That's real power that's being lost outside the customer premesis, but it does not show up on the meter.

Newer electronic meters are able to directly measure the voltage and current waveforms, and the utility potentially can charge extra for a low power-factor load---if their billing system is set up that way. —Preceding unsigned comment added by 71.199.121.113 (talk) 14:59, 10 August 2009 (UTC)

why do active PFC SMPS need two stages?

Given that a SMPS already regulates variable input and output voltages and currents, why is it not feasible to add the PFC stage to the existing feedback circuit? —Preceding unsigned comment added by 69.236.89.96 (talk) 02:04, 6 September 2009 (UTC)

Yes, a single SMPS stage can be designed to regulate either input current, or input power, or output current, or output power, or output voltage. Unfortunately, a single SMPS stage cannot regulate all these things simultaneously -- in fact, it cannot even regulate 2 of them simultaneously.

For example, say we have a typical active PFC SMPS converting (more or less) 230 VAC to (regulated) 12 VDC. Also, say we connect it to a typical load that pulls roughly constant power from the SMPS.

The feedback circuit of the final stage chooses whatever duty cycle is needed to force the output voltage towards exactly 12 VDC -- a duty cycle that varies to pull more current when the input voltage is low, and less current when the input voltage is high, in order to transfer a constant output power.

If the feedback circuit chose any other duty cycle, then the output voltage would no longer be 12 VDC.

Unfortunately, most homes do not have 3 phase AC which can deliver constant power at good power factor. Instead, most homes have single-phase AC. To get a good power factor from single-phase AC requires pulling lots of power at the peaks and pulling zero power at the zero crossings.

The feedback circuit of the initial stage of a PFC SMPS chooses whatever duty cycle is needed to force the input current to be proportional to the input voltage -- a duty cycle that varies to pull zero current at the AC voltage zero crossings and lots of current at the AC voltage peaks.

It is impossible to do both in a single stage, because that one stage would need a duty cycle to be *both* "a duty cycle that varies to pull more current when the input voltage is low, and less current when the input voltage is high" *and* "a duty cycle that varies to pull zero current at the AC voltage zero crossings and lots of current at the AC voltage peaks."

We need (at least) 2 stages in order to get 2 independently-controllable duty cycles in order to independently control 2 things: output voltage and input current.

Is there some way we can clarify this in the article?

--68.0.124.33 (talk) 05:13, 9 September 2009 (UTC)

Article was/is a conflation of DC and AC theory

The idea of stored energy is appropriate for DC circuits but isn't useful for explaining reactance. I replaced the explanation that referred to the latter with traditional AC theory because reactance, leading/lagging current and a phase relationship are applicable to AC. A step function, as implied by use of the term "step" where "phase" was meant, is was used incorrectly. So-called stepper motors are driven by a square wave, which appears as the cross-section of a step. Theory about stored energy explains how a DC voltage doubler works but instantaneous V and I in AC explain reactance accurately. Kernel.package (talk) 12:00, 9 December 2009 (UTC)

(months later) Stored energy is precisely useful to explain reactance - that's why there is reactance. Fundamentally there is no "ac" or "dc" theory, it's all Maxwell....--Wtshymanski (talk) 14:48, 15 March 2010 (UTC)

Confusing

The article is confusing. What do the various values of the power factor mean? Are they the same (just magnitude) for both sources and sinks? A diagram with at least -1, 0, and +1 would be helpful. How does leading and trailing play into this? Do only sinks have leading or trailing characteristics or do sources too? Does the utilization depend on matching them, or is +1 always desirable for sinks no matter what the value of the source?

If this isn't the right place for comments about the entry I apologise in advance. Please direct me to the proper place.

Power factor is confusing and is difficult to explain in simple terms. I don't fully understand it myself so the following should be treated with caution. When the PF is 1 it does not have a sign. If the PF is (say) 0.9, it can be "0.9 trailing" (-0.9) for a reactive load, or "0.9 leading" (+0.9) for a capacitative load. The question about sources is a good one but I don't know the answer. Biscuittin (talk) 09:35, 2 November 2009 (UTC)

Don't think so. Power factor meters in my experience say "lead" or "lag" and never "+" or "-" - this has been discussed quite a lot in connection with this page (see below under "Sign"). Where does the article get confusing to follow? --Wtshymanski (talk) 19:32, 2 November 2009 (UTC)

My mistake. I saw the + and - signs elsewhere on this page and assumed they referred to leading or trailing power factor. Biscuittin (talk) 19:55, 2 November 2009 (UTC)

Power Factor for the Modern World

Wtshymanski has removed the section I added with no adequate response to my explanation. I am new to contributing to Wikipedia pages and am trying to learn the ropes quickly. I appreciate the spirit and goals of Wikipedia and do not want to degrade their mission. I am a physicist who has probably spent more hours over the last 30 years designing and experimenting with active power factor compensation than anyone else. I have founded 3 companies that design, manufacture and market this equipment and am currently the CEO of one of these companies so conflict of interest is definitely a possibility here. I am a recognized expert and innovator on this subject and was recently asked to give a presentation at EPRI's annual conference which took place in Quebec last June. My work is mostly a labor of love as I am genuinely trying to contribute to solving the world's electrical power problems so that we evolve into something that is more sustainable. To do this, my companies have to become very successful financially and that has certainly been the case with Heart Interface, Trace and Xantrex (now Schneider) which all just reeks of conflict of interest. However, the world has changed and power factor, which used to be a more esoteric engineering subject, has become something that is effecting most people's lives in ways they don't understand. Understanding this, by people with no engineering background, is a legitimate function of Wikipedia. Although the grid is a complex mix of loads, each load, by itself, consumes some small amount of power and creates some small amount of transmission loss in the entire system, including the windings of the massive grid generators and transformers. The grid has to pay for all of this energy and the losses so you can see that low power factor loads cost the grid more per kWh to provide than do high power factor loads. With the massive migration to CFL lighting and the great abundance of computer power supplies the grid is having to absorb increased costs which they ultimately pass on to the customer. There are some articles about this that appear in EDN and EETIMES and other engineering publications but the general public is generally not aware that this is even an issue yet they are the ones who are buying and installing these low power factor devices. Since they are ignorant of the issue they do not create any pressure to address it. So, if they somehow hear that it might be an issue and don't even know what "power factor" means they might go to Wikipedia. If all they get is a very technical, engineering and very correct and precise article, they will most likely not even read it. So, what might seem redundant to someone educated in electronics, might be the only thing that a non-engineer even reads.

What are the guidelines to adding external links and references at the end of the article? Please respond. Heart141 (talk) 14:05, 27 July 2010 (UTC)

Wtshymanski's comment on his talk page is revealing: "Blanked the page again; tired of getting S*T upon by strangers." He makes quite a few deletions that make people angry, and has limited interest in engaging with them about the deletions. The thing is, most if not all of his deletions make Wikipedia better, and a lot of people really are unreasonable when they find their material deleted, so it is hard to find fault.

I reviewed the material that you added and he removed, and I would have removed it had I noticed it first. For example, "This power factor issue is so critical that the Transverter remote panel continuously displays power factor and the included software uses the internal data acquisition to display real time pictures of the wave shapes" sounds like an ad for the Transverter PS, not like an encyclopedia article about power factor. If you work for the company that makes the Transverter PS, you should avoid adding info about the Transverter PS to Wikipedia pages. Instead you should go to the talk page and say "I am considering adding the following..." and ask for discussion. Another clue is that your essay goes on for quite a while without any citations, and is written in a folksy manner. Example: "t doesn’t have to be this way..." That's an editorial opinion, not an encyclopedic fact supported by a reference to a reliable source.

Experts on a particular subject matter can be a huge help in improving Wikipedia articles, and I don't want to discourage you in any way, but you need to learn how to do things the Wikipedia way. Here is a great place to start:

http://en.wikipedia.org/wiki/Wikipedia:List_of_guidelines

If you are interested in discussing any of this with me, I welcome cold telephone calls from other folks who work in the area of product development. My phone number is on my webpage (www.guymacon.com). Guy Macon 10:25, 4 August 2010 (UTC)

I've made many edits, 95+% of which were without controversy. I'm interested in editing articles, not in carrying on debates. We don't debate on Wikipedia, we find references. I have finite time to spend, as do we all. Everyone who contributes to Wikipedia gets the notice that their contributions will be ruthlessly edited by strangers. "If you can't stand the heat...", etc. - I certainly have had edits I've made ruthlessly altered by others; and often for the better. --Wtshymanski (talk) 13:43, 4 August 2010 (UTC)

I agree with all of the above, which is why I wrote " most if not all of [Wtshymanski's] deletions make Wikipedia better, and a lot of people really are unreasonable when they find their material deleted, so it is hard to find fault [with his policy of editing articles rather than carrying on debates.]" What you do and how you do it is very valuable, and I wouldn't change you even if I could.

I, on the other hand, am willing to take the time to try to educate well-meaning newbies who don't understand why Wikipedia is the way it is in the hope of turning them into valuable contributors. To do this I had to address Heart141's stated objection that "Wtshymanski has removed the section I added with no adequate response to my explanation." Heart141 needs to learn that you have no responsibility to respond to him, but rather it is his responsibility to learn and follow the Wikipedia guideline that he violated, leading to you removing his edit. I am sure he means well, and could become a valuable contributor if he is willing to learn what is and is not acceptable here. Guy Macon 22:12, 14 August 2010 (UTC)

Cosinus phi mentioned in image caption

Cosinus phi mentioned in image caption, but nowhere in the article (unless I missed it). Can someone clean this up? Thanks. --Xerces8 (talk) 09:15, 8 September 2010 (UTC)

It's "cosine" in English, and it's described under the heading "Definition and calculation". --Wtshymanski (talk) 13:43, 8 September 2010 (UTC)

Archived threads

Anything that hasn't had any commentary for more than a few months, I've moved to an archive page. Re-reading old comments sometimes is useful as a check on the "progress" of the article's contents. --Wtshymanski (talk) 15:45, 8 October 2010 (UTC)

The section on Negative Power Factor is probably ripe for archiving. No consensus was reached, and nobody new has chimed in with any input, so the article remains as it is. (That's fine by me, BTW; it's how Wikipedia works.) Anyone disagree? Guy Macon 01:59, 11 October 2010 (UTC)

Just in case any more contributors had something to add, I left it here. I was more concerned about stuff that in some cases was years old. --Wtshymanski (talk) 16:20, 11 October 2010 (UTC)

energy returns to the source

I think that this sentence could do with a bit more explanation: "Since this stored energy returns to the source and is not available to do work at the load" - why is this exactly? --TimSmall 13:16, 1 September 2006 (UTC)

The energy does not go into the load; it is reflected back (down the power lines) to the source, and so is wasted. — Omegatron 13:53, 1 September 2006 (UTC)

Current that is not in step with the voltage does not transfer energy from the source to the load but continually circulates energy back and forth between the source and the load. This energy circulation is not 100% efficient. During each "trip" from source to load or load to source, some energy is lost as heat in the wires and other parts of the power generation, transmission and distribution equipment. Does that help? --C J Cowie 14:24, 1 September 2006 (UTC)

Sorry, but this is still not clear. Why would any current not be in step with the voltage? Utilities control the voltage, don't they? I understand some energy is lost as heat, etc., but if we are talking about energy that circulates back and forth, then clearly what is being talked about is something different from that. Or is it? Is reactive power merely the additional amount needed to allow for power lost to heat on the wires etc.? And, why would a device send ANY energy back? I doubt most devices are even designed to do that. And why would wave forms have anything to do with this? 216.239.88.89 (talk) 04:36, 8 October 2010 (UTC)

Could you look at the part that goes

In a purely resistive AC circuit, voltage and current waveforms are in step (or in phase), changing polarity at the same instant in each cycle. All the power entering the loads is consumed. Where reactive loads are present, such as with capacitors or inductors, energy storage in the loads result in a time difference between the current and voltage waveforms. During each cycle of the AC voltage, extra energy, in addition to any energy consumed in the load, is temporarily stored in the load in electric or magnetic fields, and then returned to the power grid a fraction of a second later in the cycle.

and tell us what is unclear in it? --Wtshymanski (talk) 15:07, 8 October 2010 (UTC)

For what it's worth, I found the paragraph above to be exceptionally clear.

The problem User 216.239.88.89 is having is, in my opinion, not caused by the article being unclear but rather from some misconceptions he has picked up along the way. The question "Why would any current not be in step with the voltage? Utilities control the voltage, don't they?" shows this. If a utility outputs a sinewave at a fixed voltage, that means it has it has zero control over the current. Alas, this sort of misconception is not something that can be addressed in a Wikipedia article about power factor. He needs to go back and get a good understanding of Ohm's law, first in a DC circuit with a resistive load, then in an AC circuit with a resistive load, then in an AC circuit with a capacitive load. Ohm's law covers all of this quite nicely. Guy Macon 02:50, 9 October 2010 (UTC)

Perhaps ignorance rather than misconceptions. See, for example, I don't understand about a utility outputting a sine wave. I think they output a fairly controlled range of voltages and currents, matching output with load fairly closely. Anyway, I don't have all that education and I suppose it is unrealistic to expect the explanation to make sense when I don't have that. 216.239.78.204 (talk) 22:19, 10 October 2010 (UTC)

What I would hope for is the article having links that lead you to the science that the article is based upon. For an example, look at the article on string theory; it of course cannot explain string theory to someone who has no knowledge of physics or math, but in theory you could drill down through the many links and gain enough knowledge to understand it. I think that the same is true here; power factor links to real power, which in on the page explaining AC power, which links to Mains electricity, which links to Voltage, which links to Ohm's law, which will explain why it is that your belief that utilities "output a fairly controlled range of voltages and currents" is completely and utterly wrong. You will never understand power factor unless you abandon your basic misconception that it is possible to control both the voltage and the current into a varying load. That is simply not possible.

BTW, saying that you don't understand what utilities output would be ignorance. Saying that utilities control voltage and current is not ignorance; it is a misconception. It is the difference between not knowing something and knowing something that isn't true. No shame in either, of course; only a fool thinks that having technical knowledge of a particular topic or never being in error makes them somehow superior. Guy Macon 01:51, 11 October 2010 (UTC)

The energy doesn't necessarily get bounced back and forth, either. It can disappear into the source. — Omegatron 14:42, 1 September 2006 (UTC)

Most of the reactive power flows back and forth between the source and load such that "On one half-cycle, the source supplies energy to the energy-storage element, and on the next half-cycle the energy-storage element returns energy to the source....currents required to supply the stored energy produce losses in the generating and transmission system..." Scott, Ronald E. (1960). Linear Circuits. Reading, MA: Addison-Wesley. {{cite book}}: Cite has empty unknown parameter: |coauthors= (help) “Low power factor means more current and greater $I^{2}R$ losses in the generating and transmitting equipment.” Fitzgerald, A. E. (1983). Electric Machinery (4th ed. ed.). Mc-Graw-Hill, Inc. ISBN 0-07-021145-0. {{cite book}}: |edition= has extra text (help); Unknown parameter |coauthors= ignored (|author= suggested) (help)

Again, I would have to say that it must be ignorance on my part, rather than any misconception. I do believe that utilities try to control the voltage, current, and frequency of the electricity. They deliberately change the voltage at various points, to achieve efficiencies in transmission. They must keep the supply and load reasonably closely matched, so that generators aren't over or under loaded. I know they do try to keep frequency aligned in the grid. So I'm not at all clear why that's a misconception. As for 'apparent power' and 'reactive power', I think it is something you have all learned about so long ago, that it isn't at all obvious to you that the explanation isn't clear to anyone without all that education. Why would a device, say a radio or toaster, reflect any power back? I'm sure it wasn't designed to do that. It might be something inherent in any electrical device, but what? Or, for that matter, why would a device temporarily store the apparent power when no doubt it wasn't designed to do that. They draw a certain amount of real power to do the work they need to do. Where does the other power come in? Obviously there is something called apparent power and reactive power, but it seems it needs a lot of education to be able to understand it. I'm not sure why that is the case. I would have liked a clear explanation. But if it isn't possible, it isn't possible. What, I wonder causes the reactive power to be at various levels? 216.239.77.231 (talk) 04:36, 6 November 2010 (UTC)

216.239.77.231, This is a bit over simplified, but perhaps it will help you. Loads other than 100% resistive loads such as electric heaters or incandescent lights require reactance. A standard motor or capacitor need an injection of energy before they will perform the intended function. For a capacitor it needs to be charged up. For a motor the stator and rotor coils will need to absorb an electrical field. A motor will not turn until all the coils internally are charged up. When you shut off a motor all the store energy in the coils discharge back into the system. This energy is not useful power, but rather a temporary electric field. —Preceding unsigned comment added by 158.106.48.10 (talk) 13:54, 21 November 2010 (UTC)

Merge from Distortion power factor

Distortion power factor should be merged here, it's small and it needs to be given a contenxt which this article already supplies. --Wtshymanski (talk) 15:45, 16 March 2010 (UTC)

Agree --Chetvorno^TALK 22:20, 17 March 2010 (UTC)

Agree, and note that the merge has been done. Guy Macon 01:40, 22 November 2010 (UTC)

Splatco

So what does Splatco tell us that's not already in the article and that can't be written into the article? External links are not a substitute for encyclopedia articles - everybody knows how to use Google already. --Wtshymanski (talk) 14:50, 21 February 2011 (UTC)

First, the edit comments are upside down on the burden of proof. The person who wants to add material to an article needs to justify it -- not the person who challenges material.

Second, the SPLat EL does not add to a technical understanding of the article. Most topics are already addressed, so the link is largely redundant. The "It ain't necessarily so!" nonlinear load section is problematic. I would exclude the link on technical grounds.

Third, the SPLat EL has a consumer-related discussion at the bottom (Power factor, consumer electricity costs, and scams) that the current article does not address. That is the EL's advantage (an EL should have material that is not yet included in an article). The section is short and has some issues. The EL page's first link is circular: the SPLat page refers to the open4energy page and open4energy refers to the SPLat page. They even have some identical text and figures. Consequently, I'd be leery of including either - but I'd prefer the open4energy page because it doesn't have the nonlinear load problems. The consumer issue, however, would be adequately served by including the NIST page[2] as an EL, and I've added the NIST link to the article.

Consequently, there is no reason to include the SPLat EL.

Glrx (talk) 17:03, 21 February 2011 (UTC)

Load

The article would benefit by adding a definition of or explanation of 'LOAD'.-- 11:59, 7 March 2011 Twinkletoos

Misleading Statement

"Correction equipment may be installed by individual electrical customers to reduce the costs charged to them by their electricity supplier" I believe this statement is very misleading, Implying that if you increase your pf to unity you will reduce your current consumption. Whole current meters, which are generally used in residential installations measure only active power. Any change in power factor does not change the active power measured by the meter, therefore there is no increased cost in having a low pf. Some utility might charge you for having a low pf, but generally only applies to industrial installations. What are other peoples thoughts? --202.168.24.162 (talk) 02:15, 5 April 2011 (UTC)

Costs, not current consumption. For the sort of customer that gets charged for reactive power, power factor improvement also reduces costs. If you're being billed only for kwh, your consumption is so small that the utility doesn't care (within broad limits) what your power factor is. If you're an industrial customer whose bill includes kva as well as kwh, then you have incentive to improve power factor. --Wtshymanski (talk) 03:04, 5 April 2011 (UTC)

as it reads, one might think "me being a individual electrical customer, i therefore can reduce my electrical costs by installing correction equipment" which is incorrect. I believe the statement needs to be reworked to incorperate that as an electrical customer being billed for kVA, generally industrial, it is beneficial to correct your pf above the minimum outline by the utility to avoid penalties (increased costs). Just recently in Australia power factor equipment was being advertised to the residential consumer as a means to reduce your power bill. It came out in the wash after an investigation from TV show, Today Tonight that active power consumption, which is what was being billed, did not change and that power factor correction did not reduce power consumption. Misleading, i think so --Mpleets (talk) 04:05, 5 April 2011 (UTC)

You should only watch television for the jokes. We should probably rely on electrical customers being able to read their bills - if they aren't paying for low power factor, then they don't get any cost saving from improving it. People who don't edit encyclopedia articles are amazingly pragmatic about not doing things that don't matter. --Wtshymanski (talk) 14:21, 5 April 2011 (UTC)

about Synchronous condensors

In the paragraph "Power factor correction of linear loads", the Synchronous condensors are mentioned "Synchronous condensors are often used in connection with high-voltage direct-current transmission projects......". Is the power factor (and power factor correction) valid for the DC power network ?--Wolfch (talk) 15:08, 5 June 2012 (UTC)

You cannot correct power factor in a DC circuit, because there is no power factor to correct (which I suspect you had already figured out). However, where the DC is produced from AC using a rectifier station, it is often necessary to correct the power factor for the load provided by the rectifier to the AC supply. 86.159.159.194 (talk) 13:03, 14 October 2012 (UTC)

An illustration of distortion power factor

Current probe facing the light socket. The load draws 29 watts at 0.61 pf

Current probe facing away from the socket. Shock, horror, the load is drawing minus 29 watts, but still 0.61 power factor

Evidently whoever did the firmware for the Fluke 192C thinks power factor only goes from 0 to 1 inclusive. I had a chance to try one out and set up this demonstration mostly because I was curious about CFL lamps and also curious as to what happens when the current is reversed. I was delighted at the results, of course. (My wall plug power shows some flat-topping - about 2.5% 3rd harmonic according to the Fluke.) --Wtshymanski (talk) 02:11, 17 October 2012 (UTC)

I suspect that you have hit the nail on the head. The most obvious and simplest way of designing the firmware is to multiply the instantaneous voltage by the instantaneous current at each sampling point and average over the cycle (this gives the actual power). Then calculate the RMS voltage and RMS current over the cycle. Multiplying these together gives the VA. Dividing the former by the later gives the power factor. Now the only problem with this arrangement is that power will become negative if it flows 'the wrong way', but VA will remain positive. Thus the power factor will also be negative.

But if instead you substitute an algorithm that will report a negative VA if the power flows the wrong way (the one above will never do so), then the power factor will always come out positive. Having the sign of the VA matching that of the power makes more sense than having a mismatch and the power factor reporting as negative.

Trying a similar experiment to you, I tried it with one of those plug in power and energy monitors. At first, I thought it was not going to tell me anything because: why would it have a negative sign in the display? But it does. Lashing it up the wrong way round, it displays negative power, but positive VA. It displays negative power factor and thus (perhaps not so unsurprisingly) negative energy. At least it didn't display negative time! The specification sheet does not mention its ability to indicate negative quantities. But: this is all original research. 86.159.159.194 (talk) 11:41, 17 October 2012 (UTC)

Two excellent examples above! In the upper example, we have a firmware engineer who mistakenly designed an instrument that only reports PF in the range 0-1, despite reporting correct negative power and correct positive volt-amps. Clearly, that firmware engineer didn't use the correct formula PF=W/VA; perhaps he was mislead by the opening line in this Wikipedia article that says PF is in the range 0-1? Myself,m I have had to correct young instrumentation firmware engineers who have cited this article... In the second example, the instrument is behaving exactly as it should. The discussion, though, is incorrect. There's no such thing as negative volt-amps. Volt-amps is defined as RMS volts times RMS amps, and RMS values are always positive. The concept of volt-amps is, roughly speaking, the maximum possible rate of energy transfer in either direction. A not-very-precise analogy: you can have water flowing into a bucket -- let's call that positive flow -- and you can have water flowing out of that bucket -- let's call that negative flow. But the capacity of the bucket can only be positive. Not a perfect analogy, but maybe helpful in explaining why VA is always positive, but W can be either positive or negative. AMcEachern (talk) 14:42, 18 October 2012 (UTC)

And just a few screens above this one, some annonymous Fluke programmer was being cited as the authority for the existence of negative power factor. Which Fluke instruments were programmed by the guys with the right definition? --Wtshymanski (talk) 22:02, 14 November 2012 (UTC)

Well, we don't seem to have got the part the problem of exactly which is the right definition nailed down. All that has been established is that opinion is divided on the subject. I managed to get my hands on a Tektronix 2014B which has the ability to multiply the A and B channels together. I was initially disappointed to discover that it only gave the result from my illicit 100 Watt light bulb as 98.5 VA. However, turning the current clamp the other way around gave an indication of -98.9 VA. Hmm! The reality was revealed when a compact fluorescent lamp was used instead. What the instrument labels as VA is in reality the average of the instantaneous values of current * voltage over the samples in one cycle (or in other words - Power in Watts). This was confirmed when a capacitive load drew as near 0 VA as makes no difference. The user manual adds nothing over that the result is VA. Not helpful at all. 86.159.159.194 (talk) 16:52, 15 November 2012 (UTC)