Talk:Volt-ampere

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Volt-ampere

Why does the article claim that the unit's use is restricted to AC circuits when it is perfectly meaningful for DC or any other current flow? 68.238.80.80 19:57, 26 July 2006 (UTC)[reply]

Because for Direct Current, we just use watts. A volt is one joule [of energy] per coulomb [of charge]. An ampere is one coulomb per second. Therefore, a volt-ampere (DC) is one joule per second, which is a watt.

The dynamics of Alternating Current, however (which I don't fully understand, and probably never will!), are far more complex. Apparently because current (in AC circuits) often doesn't track voltage directly (for capacitive loads, the current rises and falls ahead of the voltage; for inductive loads, the change in current lags behind it), you can't just multiply volts times amperes to get watts, because the volts and amps don't "line up." So volt-amperes are used instead of watts. See the article on AC power for more information. 24.6.66.193 21:42, 17 August 2006 (UTC)[reply]

DC, once it reaches its steady state, has no reactive component. You could say that Watts = VA in this case, but it is not particularly useful to do so. Note that unlike AC, RMS = peak for DC, so the numbers should match.

The distinction between real power and reactive VA does apply in non-sinusiodal power systems. Madkaugh (talk) 00:30, 16 October 2009 (UTC)[reply]

"The distinction between real power and reactive VA does apply in non-sinusiodal". VA, and reactive VA is not meaningful if not sinusoidal. For an arbitrary time-varying non-periodic waveform, what are the V and the A you are multiplying to give a value? You CAN compute real power by integration, but how do you define VA? I'm not sure of the situation for non-sinusoidal but periodic (eg square) waves, but note that a square wave is a Fourier series of sinewaves of odd harmonic frequencies (f, 3f, 5f, ... nf ... ); reactive circuit elements will affect the different frequencies differently. Conceivably the concept of VA could be meaningful, or made to be, but it would need to be documented, it's not obvious. Pol098 (talk) 15:12, 4 July 2010 (UTC)[reply]

Most often, RMS current and RMS voltage are used for VA. The more usual case is the rating of transformers, where the current and voltage are (usually) different between primary and secondary, but VA will be the same. Gah4 (talk) 08:20, 11 January 2017 (UTC)[reply]

Standard watthour meter

I just about understand the difference between a Watt and a Volt Amp. What I am curious about, is is whether a normal domestic electricity meter measures Watt hours or Volt Amp hours. I am finding, in some cases, a factor of six difference between the two statistics on some appliances.

The standard household electrical meter presently measures kWh (energy in kilowatthours) although some of the more advanced meters can also record other quantities such as kilovarhours, peak kVA demand, demand interval data, etc. Industrial and commercial facilities are often billed on VA demand in addtion to kWh. Nearly all households today are billed solely on kWh.

Rcdugan 17:42, 23 June 2007 (UTC)[reply]

By design, watthour meters integrate true power, that is, watts. (That is, the integral of power in watts with respect to time.) Connecting an ideal inductor or ideal capacitor should integrate zero watts. But it is possible that, due to nonlinearities, that the measurement isn't perfect. But note that reactive power has a cost to the power company, as it increases the current, and so ohmic loss in power lines. Also, it requires larger wire and larger transformers. Gah4 (talk) 08:26, 11 January 2017 (UTC)[reply]

Why 70%

The section about 70% and the example formulas appear to be a response to some specific question that has been left out. These formula would only apply to a device with a known power factor of .7. This section should be removed and probably just replaced with a reference to the article on Power Factor.

Nevermind, someone seems to already be fixing it.

D3z (talk) 04:50, 17 June 2008 (UTC)[reply]

The 70% is not due to power factor. It is because VA is based on peak values, and Watts is RMS. Madkaugh (talk) 00:30, 16 October 2009 (UTC)[reply]

VA is normally based on RMS current and RMS voltage. Other than as described by the US FTC, power (watts) are never RMS. Gah4 (talk) 11:45, 11 January 2017 (UTC)[reply]

Volt-ampere corrections

The vast majority of Wikipedia articles are very useful, and contain a minimum of errors.

Sadly, the above doesn't apply to the Volt-ampere article. It is wrong in most respects, and what remains is misleading. The article should be removed entirely.

VA is simply what it implies - RMS Voltage multiplied by RMS current. The waveform is immaterial, so any reference to 0.7 (which implies a sinewave) is irrelevant and grossly misleading. VA is commonly referred to as "apparent power", but this is not a physical property per se - Volts, Amps and Power are physical properties (in that they are defined quantities in physics textbooks).

"Volt-Amps" (VA) is simply a useful quantity in electrical engineering, and in most cases is far more useful than Watts because the power factor of the connected loads is often an unknown quantity. VA and Watts are only equivalent where the load has a unity power factor (i.e. is essentially resistive).

Rode sound (talk) 02:45, 21 October 2009 (UTC)[reply]

Major deletion

This major deletion made the article false. The first sentence says "A volt-ampere (VA) is the unit used for the apparent power in an electrical circuit, equal to the product of voltage and current." thus defining the apparent power as the product of voltage and current. This definition, in general, is false. The qualification that explains the limited circumstances under which this definition is true must be stated in the article, and not rely upon the reader to look at other articles to realize this article would be false on its face if the deletion is allowed to stand. Jc3s5h (talk) 15:06, 4 July 2010 (UTC)[reply]

Not false. And that deleted paragraph contains nothing that contradicts the definition of product of voltage and current. What makes you think the defintion of p=v * i is not correct? --Wtshymanski (talk) 15:12, 4 July 2010 (UTC)[reply]

Cut'n'paste from IEEE Standard 100: "apparent power (1) (rotating machinery) The product of the root-mean-square current and the root-mean-square voltage." If it's good enough for the IEEE, it's good enough for me. Of course one of the benefits of editing Wikipedia is that you get to associate with persons much more erudite than those people who merely write standards for trade associations. --Wtshymanski (talk) 15:18, 4 July 2010 (UTC)[reply]

First we have to suppose scalar numbers, because this is not described as any kind of vector product. That's easy for people who don't need to read this article to see, but might be a struggle for the intended audience. Then we have to ask, what kind of voltage: average, RMS, peak, instantaneous? The same questions apply for current. Real instantaneous power is the product of instantaneous current and instantaneous voltage. Apparent power is ordinarily the product of RMS current and RMS voltage. Although the math would work for peak current and peak voltage, I don't think anyone ever adopts that definition for apparent power, but the newbie would't know that.

It's possible that the deleted section could do a better job of describing the distinctions, but it's just wrong to flatly say that apparent power is the product of voltage and current. Jc3s5h (talk) 15:22, 4 July 2010 (UTC)[reply]

I see that whoever wrote that IEEE-100 definition was erudite enough to specify root-mean-square voltage and current. Do you suppose he or she just wanted to puff up the size of that already thick volume, or do you suppose the author thought the definition would have been wrong without those extra words? Jc3s5h (talk) 15:29, 4 July 2010 (UTC)[reply]

Conversion examples and article clarity

Please provide some details and examples of conversions and relation to better known units, esp. WATTS. I came here not recalling what a Volt-Amp or VA equals, though I hobby with electricals and electronics bits occasionally. I just don't see VA ratings that often to worry.

Clarity: I find some of the Wiki text confusing, more so now that I read the "talk" page. At this site: http://www.supercircuits.com/resources/tools/Volts-Watts-Amps-Converter they think that a VA directly equals a watt. Wiki should explain if this relationship is true or not, why and under what circumstances, and in language not requiring a EE (though it would be best to accommodate both levels of expertise). Here's a site which does a good job of that and I hope Wiki will model part of the article on it: http://www.power-solutions.com/watts-va.php . For the real techies (and I am at times), this is a great help: http://www.powerstream.com/VA-Watts.htm Jopower (talk) 23:15, 1 June 2012 (UTC)[reply]

I would like to add that I would have found some conversion forumlas useful on this page. I would like to see that added to this page by someone who knows what they are talking about :) Quinwound (talk) 21:04, 27 February 2013 (UTC)[reply]

As far as I know, VA is mostly used in rating transformers. Maybe also motors, but less often. For a motor, the current rating tells the size of wire and switches needed, and also for any transformers used. It is especially convenient for transformers, since it works in either direction. It also tells the size of wire and switches needed for the other side of a transformer. Gah4 (talk) 10:57, 8 December 2019 (UTC)[reply]

SI units

From the previously given reference: ^[1] section 2.2.2 A derived unit can often be expressed in different ways by combining base units with derived units having special names. Joule, for example, may formally be written newton metre, or kilogram metre squared per second squared. This, however, is an algebraic freedom to be governed by common sense physical considerations; in a given situation some forms may be more helpful than others. Seems to me that this does allow for volt ampere as an SI unit, in the same way that newton meter is allowed for torque. In this case, the volt ampere helps to describe the physical situation. Gah4 (talk) 08:37, 11 January 2017 (UTC)[reply]

By starting a new section, you have started a new topic. It is unclear what the motivation is for your comment, or whether it has anything to do with any other topic that appears on the talk page.

The volt-ampere is something of a special case; in power engineering (which usually deals with AC electricity at a frequency of 50 or 60 Hz), if the voltage is measured with with a meter that is only capable of measuring the RMS voltage, and the current is measured with a different meter that is only capable of measuring RMS current, and the two values are multiplied, the result is said to be in volt-amperes (VA). If a meter is used that can simultaneously measure the instantaneous voltage and instantaneous current, multiply them to find the instantaneous power, and compute the RMS power, that value is said to be measured in watts. Jc3s5h (talk) 10:11, 11 January 2017 (UTC)[reply]

The reason for this section is to explain the change to the article. Yes it is due to separate measurements of RMS current and voltage, but also that transformers have to be rated by the current supplied, independent of the phase of that current. So, transformers are properly rated as to VA that they can supply, and not Watts. Low power factor loads can be expensive to support. Gah4 (talk) 11:43, 11 January 2017 (UTC)[reply]

References

^ SI Brochure, 8th ed.

sinusoidal

The article seems to suggest that VA only applies when the current and voltage are sinusoidal. I believe that this isn't true. VA is the product of RMS voltage and RMS current, independent of the wave shape. Among others, it allows one to find the current in a wire, knowing the voltage and VA, which is necessary to determine the wire size. Non-linear loads, such as HID lamps, generate a large third harmonic in the current. This adds complication in three phase systems, more than just VA tells you. For even more non-linear loads, such as many switching power supplies, peak current could be high, while RMS isn't so high. Again, VA isn't all you need to know. Gah4 (talk) 07:34, 4 November 2017 (UTC)[reply]

I was about to write this, but then I find that I already did. The mention of sinusoids of the same frequency is still there. One common case is the third harmonic (maybe more) with non-linear loads, such as discharge lamps. So you can have a third harmonic current, and first harmonic voltage determining VA. Gah4 (talk) 10:59, 8 December 2019 (UTC)[reply]

How to?

Someone removed the statement: SI allows one to specify units to indicate commosn-sense physical considerations. The reason for that, was that previously someone claimed that SI didn't allow such units. I am not against the removal, but I don't understand the edit summary about a how to guide. Is it obviously an allowed SI unit now? Why does that statement make Wikipedia a how to guide? Gah4 (talk) 06:32, 3 December 2017 (UTC)[reply]

How did you not put this in?

VA = Watts divided by power factor. 2600:1700:4CA1:3C80:F037:D0E9:72E7:94FE (talk) 00:53, 16 April 2018 (UTC)[reply]

Volt-ampere reactive

Is the expression "reactive power" used in industry or measuring instruments? I can see what it's getting at, but it's oxymoronic. Doug butler (talk) 00:05, 23 September 2022 (UTC)[reply]

I suppose, but that is what it is called. If you write it as a complex number, it is the imaginary part, so it could be imaginary power. Gah4 (talk) 03:44, 23 September 2022 (UTC)[reply]

SI allows one to specify units to indicate common sense physical considerations

I believe I put: SI allows one to specify units to indicate "common sense physical considerations" in many years ago, along with an appropriate reference, replacing a statement that said the exact opposite. If there are no questions about it, I am fine with removing it. Gah4 (talk) 06:48, 27 August 2023 (UTC)[reply]

[1] SI Brochure, 8th ed.

[1]