# Volt Amps Reactive? **URL:** \ **Category:** Factual Questions\ **Created:** [September 26, 2003, 2:15am UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883 "2003-09-26T02:15:55Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![barking\_frog](https://avatars.discourse-cdn.com/v4/letter/b/96bed5/32.png) [@barking\_frog](https://boards.straightdope.com/u/barking_frog)\ **Post date:** [September 27, 2003, 11:27pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/21 "2003-09-27T23:27:34Z")
> [@](#): > > \*Originally posted by Q.E.D. \* > \*_A minor nitpick: there will always_ capacitance present, both between lines and from line-to-Earth. At low frequencies, the capacitive reactance is high and therefore has little effect on the circuit. \*\* Yeah, my wording was not that good. What I meant to say is that at low frequences the capacitive impedence can be neglected for engineering purposes. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [September 28, 2003, 6:33am UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/22 "2003-09-28T06:33:49Z")
A few other reasons for using HVDC transmission: a) In some situations, a long cable is required, e.g. because it has to be underground or underwater. In these situations, the capacitive charging current is very high, causing huge thermal losses and reducing the useful current carrying capacity of the cable. If the cable is operated using DC rather than AC, the charging current problem disappears. Examples of this include the cable connecting the north and south islands of New Zealand, and the cable connecting England and France. b) You can’t use AC transmission at all in some cases. This applies when connecting systems that operate at different frequencies, e.g. in South America where some countries use 50 Hz and others use 60 Hz. It also applies when you want to create a relatively low capacity link between two otherwise separate systems - if you used AC, you’d have to make the link big enough to force the two systems into permanent synchronism. This situation crops up between different countries in Europe, and also (for a short time) between New South Wales and Queensland in Australia (there’s now a high capacity AC link between the two states, but the low capacity DC link still operates in parallel). ---
**Author:** ![casdave](https://avatars.discourse-cdn.com/v4/letter/c/c6cbf5/32.png) [@casdave](https://boards.straightdope.com/u/casdave)\ **Post date:** [September 28, 2003, 1:12pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/23 "2003-09-28T13:12:24Z")
I always wondered how we in the UK imported power from France and vice versa, frequencies being differant. I assume there are some honkin big thyrister rectifiers at either end. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [September 28, 2003, 1:22pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/24 "2003-09-28T13:22:01Z")
> [@](#): > > \*Originally posted by casdave \* > \*\*I always wondered how we in the UK imported power from France and vice versa, frequencies being differant. > > I assume there are some honkin big thyrister rectifiers at either end. \*\* As far as I’m aware, both the UK and France operate at 50 Hz. I could be wrong. You might want to check here: [International Mains Voltages](http://www.user.fh-stralsund.de/~emasch/800x600/Dokumentenframe/Kompendium/Internationale%20Spannungen/0921271x.pdf) (warning, .PDF) to be sure. 😉 And yes, there would be some honkin great thyristors. It’s a 2000 MW link. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [September 28, 2003, 1:28pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/25 "2003-09-28T13:28:53Z")
An interesting side-note: the current trend is away from thyristors (SCRs and triacs) and toward MOSFETs. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [September 28, 2003, 1:57pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/26 "2003-09-28T13:57:01Z")
And here’s a photo of one of Toshiba’s [honkin great thyristor](http://www.atals.com/tic/hvdc_home.htm) (arrays). ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [September 28, 2003, 3:01pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/27 "2003-09-28T15:01:00Z")
> [@](#): > > \*Originally posted by Desmostylus \* > \*\*And here’s a photo of one of Toshiba’s [honkin great thyristor](http://www.atals.com/tic/hvdc_home.htm) (arrays). \*\* That was fascinating! It reminds me of the 1 farad capacitor I once saw. It was part of the Univac I and actually consisted of a bank of 2000 electrolytic capacitors, each 500 micro-farads. I assume it was in the power supply filter and often wondered why they didn’t just use a large battery. This would have been around 1956 or 57. Rem-Rand had given an obsolete computer to Penn. It must have been one of the largest vacuum tube computers ever built and had a MTBF of about 5 minutes. This whole thread has been extremely interesting. There is a lot about high power transmission that I don’t understand and you don’t get to understand just by knowing about low voltage things. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [September 28, 2003, 3:06pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/28 "2003-09-28T15:06:44Z")
The[Pacific Intertie](http://www.usbr.gov/dataweb/html/pninter.html) which connects Bonneville power station to power grids in the southwest uses a couple of high voltage dc tranmission lines. One of them runs down the eastern side of the Sierra Nevada range and has been in use since 1970. The other will run to the Phoenix, AZ vicinity. I believe the line mentioned as running to the Sylmar, CA converter station now operates at 1 million volts. The system still uses [Siemens tyristers](http://tdworld.com/ar/power_new_thyristor_valve/) at the latest report to which I have access. ---
**Author:** ![bbeaty](https://avatars.discourse-cdn.com/v4/letter/b/c0e974/32.png) [@bbeaty](https://boards.straightdope.com/u/bbeaty)\ **Post date:** [September 30, 2003, 6:36am UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/29 "2003-09-30T06:36:50Z")
**Attn Hari Seldon:** see the last bit at the end. > [@](#): > > \*Originally posted by Desmostylus \* > **More than one, I’d say. One in particular I find noteworthy, i.e., the utter confusion regarding terminology** “Utter confusion”, hardly. I was using the terms “real” and “unreal” in their everyday sense, but I did let inconsistent usage slip in when I mentioned “non-reactive or ‘Real’ Power.” OK, let’s delete that one. Read it again, and DON’T assume that I’m talking about Real/Imaginary numbers. I wasn’t. \*\*Reactive power is real. However, the reactive part of electric power reverses its direction of flow. It reverses twice per cycle. Normal (non-reactive) power does not reverse: it flows smoothly from the generator to the load. In other words, when the power is 100% reactive, they send you some energy, but you send it right back again 120th of a second later. Reactive power is “unreal” because it averages out to zero in the long run. That’s why electric meters don’t detect it and why the utility companies don’t charge you for it. On the other hand, reactive power is still real because it involves voltage and current just like any normal watt. Therefore the generators, transmission lines, transformers, etc., still get hot. If a utility company sends out a KWH of reactive energy, and a few percent is lost because of inefficiency… and then the customer’s load sends all the received energy back again 120th of a second later (and again a few percent is lost,) then the customer isn’t charged, since the customer didn’t consume any energy at all. Yet the utility companies have to shovel fuel into the boilers, since some electrical energy was consumed on the fly by the miles of power lines. Reactive power represents a financial loss to the utility companies.\*\* Is any part not clear? If so, please point it out. > [@](#): > > \*\* > It’s only comprehensible to people who know more about the subject than **bbeaty** himself. > \*\* Hmm. Is that supposed to be an insult? > [@](#): > > \*\*It’d be better to shut up altogether rather than juxtapose two entirely different senses of the word “real” in trying to explain what’s real and what isn’t. \*\* I disagree. As long as Hari Seldon gained anything significant from it, then the explanation is succesful. On the other hand, if someone with conventional power-engineering training has no idea what I’m talking about, then too bad. My explanation wasn’t aimed at utility network engineers, it was aimed at a knowledgable non-expert. Also if a person with no math training read my original explanation, they’d be forced to assume that I was using the word “real” in its everyday sense, and in that case my silp with “Real Power” might be mildly confusing but would hardly destroy the explanation. After all, the main concept is very simple. With purely reactive power, the energy vibrates back and forth between the generator and the distant load. The energy flowing from generator to the load will average out to zero. (It’s sloshing back and forth, it makes zero headway, and the load consumes zero energy.) On the other hand, with normal or purely non-reactive power, there is no back-and-forth sloshing of electrical energy at all, and the energy flows smoothly from generator to distant load. In real-world networks the loads are motors which contain both coils and resistances, so we’d see a combination of the undesirable “sloshing” energy flow and the desired “constant forward” energy flow. I’ve repeatedly found that engineering textbooks neglect to mention these concepts. They teach the math, but don’t say anything about parcels of energy moving uselessly back and forth between generator and load. Without that key concept, the math behind VArs and Power Factor seems far more obscure than it actually is. Without that key concept, you can’t explain what’s happening in plain english, and you end up spewing a bunch of equations while insisting that the audience take some engineering courses so that they can grasp the math. PS, here’s a little numerical toy I wrote in Excel. [http://amasci.com/amateur/pulsreflA.xls](http://amasci.com/amateur/pulsreflA.xls) It’s essentially a power line with a resistor at the far end. You send a pulse down it and see if the resistor absorbs the pulse. **Attn Hari Seldon:** You can fill the cable with higher capacitance or higher inductance then watch what happens. ---
**Author:** ![casdave](https://avatars.discourse-cdn.com/v4/letter/c/c6cbf5/32.png) [@casdave](https://boards.straightdope.com/u/casdave)\ **Post date:** [September 30, 2003, 7:53pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/30 "2003-09-30T19:53:35Z")
**bbeaty** Lets clear up something that you may not realise you appear to be saying. It appears you are saying that reactive power bounces around a system at twice mains frequency, and that it almost seems you mean that true power does not. Trueish but not all that clear. Try this, plot out the curves for current and voltage and assume they are in phase. Next plot the instantaneous product of those two curves. Looking at the curves you might imagine you just get another sinewave whose amplitude is probably very much larger than either of the other two, and crossing the zero axis at the same instant. That would be to forget the physical reality though, because all true power has a positive value, in other words - true power is not sent to the load and then returned, which is the only way you could get negative power, it is always a supply of energy, therefore the power curve must always be +ve or above the zero axis. Since the power is delivered on both + and - cycles, it has to be supplied in pulses running at twice mains frequency. This is the long way of saying that **all power** is suplied at twice mains frequency. If you now plot the power curve with a time differance between V and I, and what happens is, (if you are using trigonometrical functions) that the power curve, instead of being wholly above the zero axis, sinks lower and lower across the zero axis. The power curve below the zero axis is the reactive power, and if you were just to examine this on its own you would get a pulse of reactive power twice as often as mains frequency, but it would definately not be a continous wave. The fact that it is in pulses that can be of short duration can itself be a source of other problems. Using the trigonometrical double angle formulae is an unwieldy way of looking at this, but plotting it out illustrates things more readily than trying to go through phasors or j notation. This does not contradict what you are saying, and my explanation might still not be too clear to interested but non EE types but I tried anyway. ---
**Author:** ![bbeaty](https://avatars.discourse-cdn.com/v4/letter/b/c0e974/32.png) [@bbeaty](https://boards.straightdope.com/u/bbeaty)\ **Post date:** [October 1, 2003, 2:03am UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/31 "2003-10-01T02:03:24Z")
> [@](#): > > \*Originally posted by casdave \* > \*\ ***bbeaty** > > Lets clear up something that you may not realise you appear to be saying. > > It appears you are saying that reactive power bounces around a system at twice mains frequency, and that it almost seems you mean that true power does not. Trueish but not all that clear. > \*\* Why “truish?” Either it’s true or it’s not. And the things I appear to be saying… _you’re right, that’s what I’m saying_. 🙂 And I do realize it. 🙂 Also, if I’m saying something that’s NOT true, I certainly want to know about it. > [@](#): > > **Try this, plot out the curves for current and voltage and assume they are in phase.** That’s how I originally created my VArs explanation years ago. I had to explain VArs to the general public back in the late 1980s. No math allowed. Explain it so your grandmother can understand. Fortunately, once you can “see” what the energy is doing, you can dispense with the math, and simply describe it in plain english. > [@](#): > > \*\* > It appears you are saying that reactive power bounces around a system at twice mains frequency\*\* Why “appears?” I said it. And… reactive energy DOES bounce around the system at twice mains frequency. For each cycle of 60Hz, the energy makes two trips, going from generator to load and back again. > [@](#): > > **and that it almost seems you mean that true power does not.** Right… since it doesn’t bounce. Well, to be perfectly clear, no “power” flows at all. Electric power isn’t like a substance that moves from place to place. The stuff that moves through the network is called “energy.” Electrical energy flows from place to place, and power is the flow rate of energy. Joules flow, but watts do not. To avoid confusion, we shouldn’t mention power, but instead say that _non-reactive energy_ doesn’t bounce around inside the system. It’s not too difficult to visualize a flow of electric energy, or to describe it in everyday language. On the other hand, it’s impossible to visualize a flow of electric power, and indeed, power never flows. > [@](#): > > **…This is the long way of saying that all power is suplied at twice mains frequency.** Good point. The energy isn’t smoothly distributed in the power lines. If we could see it, electrical energy would look like a long chain of fast-moving puffs of fuzz which follow the wires. > [@](#): > > **This does not contradict what you are saying.** Right. My description was based on the math, but my goal was to remove the math afterwards, leaving behind a plain english description. ---
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [October 1, 2003, 3:22am UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/32 "2003-10-01T03:22:56Z")
> [@](#): > > \*Originally posted by bbeaty \* > > \*\*My description was based on the math, but my goal was to remove the math afterwards, leaving behind a plain english description. \*\* Right. And you very admirably succeeded in doing just that. I don’t know what **Desmostylus’** problem is, but if he wasn’t able to understand what you were saying then maybe he should refrain from posting in this type of thread. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [October 1, 2003, 8:07am UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/33 "2003-10-01T08:07:04Z")
Reactive power _does_ reverse direction four times per cycle. Take a capacitor and connect it to an AC source. For the first 1/4 cycle the capacitor is being charged and getting energy from the source. For the second 1/4 cycle the capacitor returns the stored energy to the source. For the third 1/4 cycle the capcaitor is again being charged and for the fourth quarter cycle it discharges. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [October 1, 2003, 10:16am UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/34 "2003-10-01T10:16:55Z")
> [@](#): > > \*Originally posted by bbeaty \* > **Is any part not clear? If so, please point it out.** Let’s just deal with _wrong_ first, and worry about _unclear_ later. > [@](#): > > \*Originally posted by bbeaty \* > **Normal (non-reactive) power does not reverse: it flows smoothly from the generator to the load.** This is what **casdave** was trying to point out to you. It doesn’t reverse, but it’s not _smooth_. It’s a sinusoid of frequency 2f. It’s a waveform exactly as “smooth” as the reactive power waveform. “Smoothness” isn’t something that differentiates between active and reactive power. > [@](#): > > \*Originally posted by bbeaty \* > **Reactive power is “unreal” because it averages out to zero in the long run. That’s why electric meters don’t detect it and why the utility companies don’t charge you for it.** Electric meters (and I assume here that you mean Watt-hour meters) don’t detect it _because they’re intentionally designed not to_. There are simpler ways to set up “electric meters” that _will_ detect reactive power. There’s nothing magical about reactive power that makes it undetectable to meters. Also, whilst residential customers don’t pay a _metered_ charge for reactive power, it’s false to suggest that they aren’t charged for it at all. The cost of it is averaged out over the residential customer base, and built into the rates for active power and into the service charge. Larger industrial customers _do_ pay a metered charge, and may also have to pay amortised capital costs as well. This was also pointed out by **bobk2** in the very first response to the OP. > [@](#): > > \*Originally posted by bbeaty \* > **On the other hand, reactive power is still real because it involves voltage and current just like any normal watt.** Involvement of voltage and current is not a defining factor of “reality”. E.g., “Superman is real, because you can see him on television, which involves voltage and current.”, or “Things that I imagine are real, because they involve voltage and current in my neurons.” > [@](#): > > \*Originally posted by bbeaty \* > **If a utility company sends out a KWH of reactive energy,** kWh is a “made up” unit which specifically refers to the integral of active power. Reactive energy has its own corresponding “made up” unit, the VAr-hour or VArh. It’s not correct to integrate instantaneous power over a time period of a quarter cycle, and then use the unit “kWh” to refer to the energy so derived. Look up the definition of kWh in a dictionary. “One hour” is the integration period. Sure, you can get away with shorter time periods than an hour, but _certainly not_ time periods shorter than one cycle. > [@](#): > > \*Originally posted by bbeaty \* > **…then the customer isn’t charged, since the customer didn’t consume any energy at all. Yet the utility companies have to shovel fuel into the boilers, since some electrical energy was consumed on the fly by the miles of power lines. Reactive power represents a financial loss to the utility companies.** Refer to the remarks above regarding metering and charging. It’s also incorrect to equate “cost” with “loss”, because that equation incorrectly assumes that there’s no method of cost recovery. Happy now? **Ring** (or **Sacroiliac** , or whoever happens to be logged in as **Ring** today): :rolleyes: ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [October 1, 2003, 12:08pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/35 "2003-10-01T12:08:45Z")
Desmostylus, the rest of us do not seem to have any problem interpreting bbeaty’s posts. I think you are just nitpicking the way he expresses himself and I do not think that is really helping clarify anything. I think he understands the concepts pretty well and his posts are helpful. I do not think it is helpful to pick them apart and misinterpret them. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [October 1, 2003, 12:58pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/36 "2003-10-01T12:58:53Z")
**sailor** , you are quite welcome to your opinion. I don’t think any of the EEs in this thread have any problem interpreting **bbeaty** ’s stuff; why would they? And I believe I already made that point a couple of days ago anyway. And I’m quite sure he means well, but that doesn’t automatically translate into being _correct_ now, does it? And if you want to refute any of my actual “picking apart” or “misinterpretation”, please do so. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [October 1, 2003, 5:04pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/37 "2003-10-01T17:04:21Z")
**Desmostylus** , his posts are correct if you interpret what he means and not looking for ways to misinterpret it. You are just playing games with words. I understand what he means by “smooth” but you prefer to play with the word and define a sinusoid as “non-smooth”. Your whole post is like that. When interpreted as he means them his posts are correct so let’s us not play a silly game of finding incorrect meanings in them because you can do that with any post. i do not think my refuting your posts adds anything to the thread just like I do not think your picking on his posts like that adds anything of value to this thread either. Better post _your_ explanations and leave his posts alone. ---
**Author:** ![casdave](https://avatars.discourse-cdn.com/v4/letter/c/c6cbf5/32.png) [@casdave](https://boards.straightdope.com/u/casdave)\ **Post date:** [October 1, 2003, 5:15pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/38 "2003-10-01T17:15:07Z")
> [@](#): > > Why “appears?” I said it. And… reactive energy DOES bounce around the system at twice mains frequency. For each cycle of 60Hz, the energy makes two trips, going from generator to load and back again. Your reply above was in response to this quote you made from me, however you quoted out of context, because you left out the final part, and this is important to the whole sentence, which, to quote myself was > [@](#): > > … and that it almost seems you mean that true power does not. Take that in its whole, I was saying that your statement was sightly misleading as it might imply that only reactive power is distributed at twice mains frequency, simply by the fact you did not mention all power, why state VAr runs at twice mains without mentioning this - unless it was a peculiarity of VAr and not of true power also ? That’s also why I said your statement was ‘truish’ because what you left out could infer to non EE’s VAr only has this property, but those of us who also know, will realise it whilst does have that property , it is not the entire picture. **Sailor** I believe you are highly aware that engineering language tries to rid itself of implications, and inferances that can be open to misinterpretation. You have explained on this board matters where you want to get a specific point across and you had to choose your words carefully, knowing that non-experts in the field could get a misleading impression. Certain terms have been mixed and matched here and EE’s like other engineer types prefer to keep such things precise, it might be nitpicking to some, but those who are perhaps entering the field with some learning may find problems later on with certain concepts if they are starting not with a blank slate, but one upon which slightly distorted ideas are already writ large. I still come across EE’s who do not understand power factor, or even basic maintenance costs who believe it is more economic to keep flourescant lights burning all the time, all because of a misunderstanding of terms, add to this that this sometimes gets passed on to middle managers who have far less technical knowledge and thereby trust the word of their experts and you have the blind leading the blind. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [October 1, 2003, 8:15pm UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/39 "2003-10-01T20:15:34Z")
casdave, I agree that in engineering using a precise vocabulary is essential but here we are explaining things to lay people and a more descriptive terminology is to be expected. If someone posts something which may be unclear or may be construed mistakenly I think the best thing is to just clarify the concepts without picking apart the post. If I say pi = 3 I have clearly stated something which is false and needs to be called out as false. But is I state that pi is the ratio of the periphery to the cross-section of a circle I have just expressed myself poorly and a helpful clarification may be in order. ---
**Author:** ![bbeaty](https://avatars.discourse-cdn.com/v4/letter/b/c0e974/32.png) [@bbeaty](https://boards.straightdope.com/u/bbeaty)\ **Post date:** [October 2, 2003, 8:11am UTC](https://boards.straightdope.com/t/volt-amps-reactive/203883/40 "2003-10-02T08:11:06Z")
> [@](#): > > \*Originally posted by casdave \* > **Take that in its whole, I was saying that your statement was sightly misleading as it might imply that only reactive power is distributed at twice mains frequency, simply by the fact you did not mention all power, why state VAr runs at twice mains without mentioning this - unless it was a peculiarity of VAr and not of true power also ?** Ah, I see the problem. I was pointing out that any given hunk of reactive energy BOUNCES at twice the mains frequency, it makes two trips per cycle. The “normal” or non-reactive energy doesn’t do this: **it doesn’t bounce.** I wasn’t talking about pulses of energy, I was talking about round trips. Yes, you’re right, electrical energy is distributed as pulses at twice the AC frequency. Perhaps this added detail might make it easier to visualize the situation? Or perhaps it makes the explanation slightly more complicated? Hard to say. [Previous page](https://boards.straightdope.com/t/volt-amps-reactive/203883.md?page=1) [Next page](https://boards.straightdope.com/t/volt-amps-reactive/203883.md?page=3)