nobody talking about switching PS wasting power on input filter caps? - Page 1
Forum section: electronics / beginners

author: stefon
author_rank: Newbie
timestamp: August 22, 2019, 12:09:08 pm
content: They say like PS are 80-90% efficient, but they all have filter capacitors on input. So if I calculate correctly 5W cellphone charger (230V/50Hz) with 220nf X2 filter cap, just plugged to wall will waste 3.5W of power...Some power supplies have like mains filter with dual 0.47uf cap... so just this fitler is wasting 16W???But also this cap is not getting hot at all... so I'm calculating something wrong? But I measured current and it matches calculations....

author: BravoV
author_rank: Super Contributor
timestamp: August 22, 2019, 12:10:47 pm
content: Quote from: stefon on August 22, 2019, 12:09:08 pmHow do we know if its wrong or not ? Show us the calculation.

author: capt bullshot
author_rank: Super Contributor
timestamp: August 22, 2019, 12:54:04 pm
content: This is called reactive power. The current is 90° phase shifted to the voltage, energy flows forward and back with each cycle of the mains frequency.You can measure this current, and it will match your calculation, but you won't get billed for itNo one talks about it in public, but yes, it's there and it loads the utilities and creates losses along the distribution network.

author: ArthurDent
author_rank: Super Contributor
timestamp: August 22, 2019, 01:28:55 pm
content: I believe you're wrong blaming the capacitors. If you check RC charging circuits and formula you will see that the current drops to zero when the capacitor is fully charged. For capacitors in a SMPS to dissipate 3.5 watts they would have to be defective. The voltage across those filter capacitors is still connected to the inverter circuit that could draw some current but with no output load, that is close to zero in small SMPSs. I Just checked the supply for my laptop and my wattmeter registered 0 watts or voltamps.I think the efficiency you mention is generally for a fully loaded supply and there are lots of components in a SMPS that contribute to losses under full load.

author: themadhippy
author_rank: Super Contributor
timestamp: August 22, 2019, 02:00:13 pm
content: Quoteyet,once there's a 100%enforcementtake up of smart meters the energy suppliers will have the ability to start charging for it ,as they already do with large power users.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 22, 2019, 02:12:19 pm
content: Quote from: ArthurDent on August 22, 2019, 01:28:55 pmThis is true after a rectifier bridge, but there's also mains filtering which puts a cap directly on L and N. But as others have stated: this is reactive power.

author: james_s
author_rank: Super Contributor
timestamp: August 22, 2019, 03:34:40 pm
content: Look up the difference between true power, apparent power and that will lead into power factor.While it does result in losses in the wiring, something that helps is the fact that inductive loads lag and capacitive loads lead so when you have a mix of these they tend to cancel out. With the demise of iron transformer power supplies there are a lot fewer inductive loads in the typical house though.

author: GerryR
author_rank: Frequent Contributor
timestamp: August 23, 2019, 12:36:06 am
content: In most industrial environments where there are many motors, they have huge capacitor banks, which switch automatically for pf correction, depending on the different loads throughout the day. Capacitors, for the most part, are considered "lossless."

author: Ysjoelfir
author_rank: Frequent Contributor
timestamp: August 23, 2019, 06:02:04 am
content: Quote from: capt bullshot on August 22, 2019, 12:54:04 pm... yet. But as themadhippi already wrote, smart meters can measure reactive power. I am working at a manufacturer for precision power analyzers, so we have rather good connections to many many energy providers. I assure you that they would absolutely LOVE to enforce the useage of smart meters in every house in germany, not for the added features like peak power reduction in combination with flexible plans or automatic transmission of readings, nooo, they would be happy to bill you all that reactive power as they can measure it then precisely.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 23, 2019, 07:56:11 am
content: Quote from: Ysjoelfir on August 23, 2019, 06:02:04 amYoko?

author: Ysjoelfir
author_rank: Frequent Contributor
timestamp: August 23, 2019, 08:40:55 am
content: Quote from: Ice-Tea on August 23, 2019, 07:56:11 amNo

author: radiolistener
author_rank: Super Contributor
timestamp: August 23, 2019, 10:24:02 am
content: it costs less, but it doesn't means that it is more efficient from point of view of power consumption

author: dmills
author_rank: Super Contributor
timestamp: August 23, 2019, 10:48:26 am
content: But it IS more efficient!The reactive power flow only causes losses in the distribution cabling (I^2 R), And the power company will correct large aggregate PFC issues) while real power flow is total loss.The better sort of small switcher these days has a PFC input stage anyway, but as ever that is not itself 100% efficient (NOTHING is), so from a power efficiency perspective PFC on a small DC supply may or may not be a win (From a power quality perspective it is very much a win, mainly for reduced flat topping).Regards, Dan.

author: radiolistener
author_rank: Super Contributor
timestamp: August 23, 2019, 02:28:22 pm
content: Quote from: dmills on August 23, 2019, 10:48:26 ambut you will pay real money for reactive power. So this is expenses for your wallet. See conversation above...

author: james_s
author_rank: Super Contributor
timestamp: August 23, 2019, 04:16:56 pm
content: In the USA residential consumers do not pay for reactive power, I can't speak for elsewhere.Also as I mentioned earlier, if you have any inductive loads they will cancel out the the capacitive loads.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 23, 2019, 04:47:59 pm
content: Haven't heard of any place that charges residential customers for reactive power, but as some others have pointed out the advent of 'smart' meters may lead to that.

author: soldar
author_rank: Super Contributor
timestamp: August 23, 2019, 04:50:57 pm
content: Let us not confuse the speed with the bacon.Reactive power is one thing. Power factor is a different thing. They are loosely related.A capacitor or an inductance across the line consume reactive power. This means they take power, store it and return it to the network. This means a power factor of less than unity.A rectifier bridge followed by a capacitor does not consume any reactive power. At all. None. It keeps everything it gets and returns nothing. And yet has a bad power factor.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 23, 2019, 06:30:10 pm
content: Quote from: soldar on August 23, 2019, 04:50:57 pmLoosely?

author: stefon
author_rank: Newbie
timestamp: August 23, 2019, 07:27:08 pm
content: > A capacitor or an inductance across the line consume reactive power. This means they take power, store it and return it to the network. This means a power factor of less than unity.I don't get it, it is AC, so current flows thru cap, so it is wasting power, because it is shorting L to N.> A rectifier bridge followed by a capacitor does not consume any reactive power. At all. None. It keeps everything it gets and returns nothing. And yet has a bad power factor.i don't care about caps after bridge.

author: soldar
author_rank: Super Contributor
timestamp: August 23, 2019, 07:41:05 pm
content: Quote from: Ice-Tea on August 23, 2019, 06:30:10 pmIn the sense that there is not a biunivocal relation. Reactive power lowers power factor. So do non linear loads. The problem is not reactive loads. The problem is low power factor, whatever the cause.

author: james_s
author_rank: Super Contributor
timestamp: August 24, 2019, 01:16:00 am
content: Quote from: stefon on August 23, 2019, 07:27:08 pmNo, you don't get it. It's not shorting anything and it's not wasting power.Current flows yes, but no power is consumed. Ok in reality nothing is ever 100% but for the sake of simplicity we can call it none. A reactive load draws energy during part of the cycle and dumps it back into the line on another part of the cycle. It is power that is just sort of bouncing around, it is passing through the wires but then getting returned to the source without being consumed.

author: soldar
author_rank: Super Contributor
timestamp: August 24, 2019, 07:03:07 am
content: Quote from: stefon on August 23, 2019, 07:27:08 pmCurrent flows through capacitor. Correct. Charging it and storing energy in the capacitor. This energy is returned back to the grid, it is not lost.Think of an elevator counterweight. As it is raised it consumes (and stores) energy. This energy is returned to the system the next time the elevator is raised and the weight falls.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 11:02:36 am
content: Quote from: soldar on August 24, 2019, 07:03:07 amyes, but this "returned back" energy creates impedance change for power source. So, a power source (power plant) needs to spend more energy in order to fight with this "returned back" energy. Because this "returned back" energy will works against power source. Isn't it?Also, there is additional power loss in the capacitor and power loss in the wire, due to higher current. All this energy is lost and cannot be consumed. Yes, not all this loss happens inside capacitor, but this energy is wasted due to capacitor

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 11:30:24 am
content: Quote from: radiolistener on August 24, 2019, 11:02:36 amNo, there is no net effect. The generator needs to make an extra effort to charge the cap, that effort is returned a little bit after that.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 11:50:19 am
content: Quote from: Ice-Tea on August 24, 2019, 11:30:24 am"little bit after" amplitude in the mains grid will be changed, because this is AC. So, returned energy will work against power source. This energy will works to distort sine waveform. For power source it will looks the same as additional load. So, electricity generator will needs to spend more power in order to keep the same voltage on the grid.The difference with usual load is that this energy will be returned back and wasted on AC electricity generator side instead of capactior. But anyway it will be wasted.First generator needs to make an extra effort to charge the cap, and then generator needs to make an extra effort to discharge the cap. This is worse than just waste that energy on dummy load. If you spend that energy on dummy load, there is no need for generator to spend extra effort to discharge the cap.

author: soldar
author_rank: Super Contributor
timestamp: August 24, 2019, 11:54:37 am
content: Quote from: radiolistener on August 24, 2019, 11:02:36 amNo. You do not understand the process. Two voltage sources in parallel are not working "against each other". They are "helping" each other.Quote from: radiolistener on August 24, 2019, 11:02:36 amUnless the capacitor is placed to compensate inductive rectance. But yes, poor power factor uses energy in the transmission wires regardless of the cause.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 12:10:47 pm
content: Quote from: soldar on August 24, 2019, 11:54:37 amThere is no two voltage sources with the same voltage in parallel. On rise edge of sine wave, capacitor always have less voltage than mains and consume mains current for charging. On fall edge of sine wave capacitor always have higher voltage than mains and consume mains current for discharging. Mains voltage and capacitor voltage always shifted by some phase. So, they always working against each other.Capacitor doesn't consume that energy. It just returns it back to AC generator. And AC generator needs to make an extra effort to charge and discharge this capacitor. It just waste some part of energy in order to charge and discharge this capacitor.You can fix it by placing inductor, so it will make phase shift in the opposite direction. In such way AC generator will not needs to make an extra effort to charge and discharge capactior. But you will needs additional expenses on this inductor and heat loss in the wire between capacitor and inductor.Exactly the same thing will happens with inductor connected to the mains. AC generator will needs to make an extra effort to fight against inductor. Inductor don't consume energy it just will return it back to AC generator (we don't talk about heat loss in the wire). And AC generator will needs to fight agains "returned back" energy from inductor. In order to fix it you will need to place capacitor to make phase shift in the opposite direction.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 12:40:59 pm
content: Quote from: soldar on August 24, 2019, 11:54:37 amThe same you can place inductor to compensate capacitive resistance. But this compensating inductor or capacitor requires additional expenses and needs to be connected together with reactive load. If you don't place such capacitor/inductor, AC generator will needs to make additional effort to work against your reactive load. Who will pay for that?

author: JohnG
author_rank: Frequent Contributor
timestamp: August 24, 2019, 01:31:03 pm
content: Wikipedia is your friend:https://en.wikipedia.org/wiki/Power_factorJohn

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 02:00:19 pm
content: Quote from: radiolistener on August 24, 2019, 12:40:59 pmDepends on your definition of effort. A genset consists of some kind of motor and the actual generator. The motor doesn't care. Whatever extra effort it needs to do is returned. The generator *does* care, but only because of the current. It's the current that heats up the winding and it's the maximum temperature of that winding that will determine the maximum VAR of that generator. So in theory, you can load the generator up to its maximum while the attached motor is still idling (except for the I²R losses and provided the motor has sufficient inertia to load/unload).

author: Circlotron
author_rank: Super Contributor
timestamp: August 24, 2019, 02:01:10 pm
content: Quote from: radiolistener on August 24, 2019, 12:10:47 pmNo. On the falling edge of the sine wave the capacitor supplies current back to the mains, it does not consume current. It only consumes current on the rising edge of the mains.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 02:15:30 pm
content: Quote from: Circlotron on August 24, 2019, 02:01:10 pmOn falling edge capacitor prevents AC generator to decrease voltage. Because when AC generator trying to decrease voltage, capacitor provides it's own voltage and prevents voltage drop in such way, so AC generator needs to apply more power in order to overcome capacitor resistance.Quote from: Circlotron on August 24, 2019, 02:01:10 pmIt consume current on both edges. You're forgot that we're talking about AC voltage. AC voltage means that current change it's direction in a loop.On rise edge AC generator provides +current, and capacitor consumes -current in order to charge.On falling edge AC generator provides -current, and capacitor consumes +current in order to discharge.In simple words, capacitor prevents AC generator to change voltage in both direction and AC generator needs to consume more mechanical power in order to make AC voltage.Capacitor don't consume energy (except heat loss in the wires), but it makes phase shift, so AC generator will needs to fight with it's own energy returned back (which creates additional load for AC generator). This returned voltage prevents AC generator to change voltage in order to make AC voltage.The same issue with inductor. Inductor don't consume energy (except heat loss in the wires), but it makes phase shift in opposite direction, and it leads to the same result as with capacitor - AC generator will needs to fight with it's own energy returned back (which creates additional load for AC generator).When you connect capacitor and inductors together, they will compensate phase shift of each other, and it will reduce power load on AC generator. But even in this case it still has higher heat loss on wires between capacitor and inductor.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 03:10:50 pm
content: I appologise for the crudity of the drawing (and thank you darling for lending me your crayons) but this may help you understand. Or not.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 03:33:14 pm
content: Ice-Tea, AC generator needs to change voltage from + to - and then again to + and then again to - in a loop.Capacitor prevents that voltage change, it trying to make constant 0 V from AC voltage. Try to understand that when AC generator trying to reduce voltage (on the falling edge of sine wave), the capacitor will prevent that by providing it's accumulated energy.Capacitor will use it's accumulated energy in order to prevent voltage change (basically to stop AC generator). This energy returned by capacitor will be consumed by AC generator to stop rotation of AC generator rotor. And you will need to apply more mechanical power to AC generator rotor. For example it will leads to higher gasoline consumption, if you're using gasoline AC generator.As you can see, capacitor doesn't consume this energy, but when you connect it to AC generator it lead to waste additional power in the AC generator, because capacitor prevents AC generator to make voltage change in any direction.In short, capacitor uses AC generator energy against AC generator. So, AC generator first needs to waste some power in order to charge capacitor and then it needs to waste some power again to discharge capacitor. And this happens 50 times per second.

author: soldar
author_rank: Super Contributor
timestamp: August 24, 2019, 03:34:57 pm
content: Quote from: radiolistener on August 24, 2019, 02:15:30 pmYou are totally wrong here but this thread has degenerated into such confusion that I am going to bow out and let others deal with the confusion because this is like shooting at clouds.I do not know what your qualifications are but I can tell you are not an Electrical Engineer. There are some of us on this board whoareengineers. You should listen to them.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 03:52:39 pm
content: Quote from: radiolistener on August 24, 2019, 03:33:14 pmI have no idea how to explain this to you. However, you will not find anyone on this board or on another tech-savvy forum that agrees with you. Unless it's called "Free Energy For All" or something. This should tell you something. And perhaps it should bring you to read up on the subject.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 04:02:01 pm
content: Quote from: soldar on August 24, 2019, 03:34:57 pmCan you please explain, what is wrong exactly?I'm talking that reactive load, such as capacitor or inductor will return some part of power back to the AC power source. This power will be used against AC power source.In simple words, capactior trying to filter AC voltage and make DC. On the contrary AC generator trying to make AC voltage. So they fight each other.What is wrong here?

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 04:09:49 pm
content: Quote from: radiolistener on August 24, 2019, 04:02:01 pmYes.QuoteNo. Look at the diagram I posted earlier. The power returned to the source takes load *off* it, it doesnt add to it.

author: james_s
author_rank: Super Contributor
timestamp: August 24, 2019, 04:44:37 pm
content: Eventually perhaps it will sink in and you will understand what we are trying to say. Until then just accept that a reactive load does not cause the generator to draw more mechanical energy from the engine, turbine or whatever is driving it, beyond the small percentage that becomes heat in the windings and wiring. The fact that you do not grasp how this can happen does not make it any less true. It is easily demonstrated though if you can find a portable AC genset and some capacitors or inductors.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 04:45:52 pm
content: Quote from: Ice-Tea on August 24, 2019, 03:52:39 pmThere is no "Free Energy For All" at all, there is only wasted energy.Quote from: Ice-Tea on August 24, 2019, 04:09:49 pmJust imagine DC battery with + and - terminals. And capactior with A and B terminals.Now connect DC battery to capacitor: + to A and - to B. The capacitor will be charged.Now connect DC battery to capacitor with reversed polarity (this is how AC current works): - to A and + to B. The capacitor first will be discharged through DC battery and then charged to reversed polarity.Let's assume that capacitor charge takes energy E from battery. So, during experiment you spent 3*E energy (charge+discharge+charge) from DC battery. But capacitor stores just 1*E energy. So, you're lose 2*E energy.And note, this energy is NOT consumed by capacitor. It is consumed by DC battery. This is wasted energy of DC battery.The same things happens with AC generator with connected capacitor on the output.On rise edge of sine wave capacitor will be charged from AC generator.On falling edge of sine wave capacitor will be discharged to AC generatorand then charged again in reverse polarity. It happens in a loop.As you can see, on falling edge capacitor will be discharged to AC generator.It means that current from capacitor (which flows through AC generator) will make resistance to mechanical rotation. And you will spend more gasoline for AC generator rotation. You're just wasted that energy.This is something like use gasoline to move car from point A to point B and then move it back to point A. Gasoline wasted and the car is staying at the point where it was before

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 05:07:42 pm
content: Quote from: radiolistener on August 24, 2019, 04:45:52 pmOne more try. Look at the diagram. What you are describing is 2 phases out of 4. There where you have a positive voltage and you're charging the cap (first 90° section, postive voltage positive current) and there where you have a negative voltage and charging the cap (to a negative voltage) (third 90° section, negative current and negative voltage). This does not take into account to the two other sections that have postive voltage and negative current and the part where the voltage is negative and the current positive.

author: james_s
author_rank: Super Contributor
timestamp: August 24, 2019, 05:26:59 pm
content: There is no wasted energy. Rather than making an ass of yourself by insisting there is against several people who know what they're talking about, go try it yourself. Find a small portable backup generator and connect some large capacitors across it. You can monitor the engine load by listening to the sound it makes.Another method to prove there is no wasted energy is take a bunch of capacitors, wire them in parallel and place them inside an insulated styrofoam cooler with a temperature sensor and connect it to the mains. Measure the current draw and calculate the apparent power, then record the temperature rise over a period of say 30 minutes.Now find an incandescent lamp that will draw roughly the same current as the capacitor bank and place it in the same box, repeating the experiment in place of the capacitor bank. You will find that the lamp results in a dramatic temperature rise while the capacitors do not and this can only mean one thing, the lamp is consuming energy and the capacitors are not. Energy cannot be created or destroyed, for energy to be wasted something has to heat up.

author: Zero999
author_rank: Super Contributor
timestamp: August 24, 2019, 05:40:19 pm
content: Quote from: radiolistener on August 24, 2019, 04:45:52 pmYour battery analogy is flawed because it's a square wave, which will theoretically cause infinite current to flow when the polarity is reversed. In reality the power will be dissipated in the internal resistance of the battery and capacitor. The mains is sinusoidal, so you're not comparing like with like.Look at what happens when the power drawn from V1 is plotted in LTSpice. It starts at zero, gradually goes more negative (in this case a negative number is power being taken from V1)), reaches a peak at 45º, then falls to zero again and climbs to positive, as the energy stored in the capacitor is returned to V1. Note, that average power taken from V1 is zero, over the entire waveform. No one is claiming perpetual motion.In real life there will be some resistance which will dissipate power and cause the generator to work harder, but it won't be as extreme as with a square wave. The mains also typically has many inductive loads, such as motors and lightly loaded transformers, so a little bit of capacitance will help to bring the power factor closer to unity.

author: soldar
author_rank: Super Contributor
timestamp: August 24, 2019, 05:43:13 pm
content: Quote from: james_s on August 24, 2019, 04:44:37 pmYou know, it just occurred to me that what he is saying is, in a way, the same as the free energy crowd. They say they can make energy appear out of nowhere and radiolistener is saying energy disappears which is just as impossible. Energy can be stored but it cannot be made to disappear. Energy can be converted to heat and that is precisely the question of the OP: why do capacitors that are apparently consuming power do not get hot? And the answer is because they do not consume power, they store power and then return it. If they "wasted' it then it would be converted to heat.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 05:44:36 pm
content: Quote from: Ice-Tea on August 24, 2019, 05:07:42 pmOk, let's look at diagram.We have 311 Vpk 50 Hz AC source loaded with 31.83 uF capacitor (100 Ohm at 50 Hz).So, let's look at diagram of power consumption for AC source:green diagram = voltage on AC sourcered diagram = power consumption on AC sourceAs you can see, AC source first feeds capacitor with generated power (negative power consumption), and then AC source consume power from capacitor (positive power consumption).What happens with power consumed by AC source?I provided you with answer - this power will be used against AC generator. In simple words this power will be used to stop rotation of AC generator rotor. In order to continue AC generating you're needs to apply more mechanical power to AC generator.This is my explanation. So, what is your explanation, what is happens with power consumed by AC source?

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 05:58:16 pm
content: Quote from: Zero999 on August 24, 2019, 05:40:19 pmYes. And my explanation is that this energy is consumed by AC source and it makes rotation of AC generator more hard. You're needs to apply more mechanical energy to keep AC voltage.ButIce-Tea,soldarandjames_sclaiming that this energy is not consumed by AC source and it doesn't change load resistance for AC source. So technically they claimed that there is perpetuum mobile or some kind of free energySo who is right?

author: Zero999
author_rank: Super Contributor
timestamp: August 24, 2019, 06:12:54 pm
content: Quote from: radiolistener on August 24, 2019, 05:58:16 pmIf the power is zero, then why would more mechanical energy be required?QuoteYou're wrong.No, they're not claiming free energy.QuoteThere's a clear gap in your understanding. You haven't considered the fact that the alternator can also work as a motor. Take energy out of it and put it into the capacitor and it will increase the load on the engine. Put energy back into it, from the capacitor and it will reduce the load on the engine. The mechanical inertia will smooth this out so it will behave as if there's no mechanical load.

author: soldar
author_rank: Super Contributor
timestamp: August 24, 2019, 06:20:44 pm
content: Quote from: radiolistener on August 24, 2019, 05:58:16 pmThis is wrong.Photo of capacitors in substation:https://en.wikipedia.org/wiki/File:Condensor_bank_150kV_-_75MVAR.jpgThey don't put them there so they can waste energy.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 06:38:09 pm
content: Quote from: Zero999 on August 24, 2019, 06:12:54 pmBecause energy generated with AC generator rotation was accumulated by capactior and then returned back to AC generator. AC generator consumed returned energy to make rotor rotation more hard.So, if you don't apply more energy, rotor will slow down and stop and AC voltage will drops down to zero. In order to keep AC voltage at 311 Vpk 50 Hz, you're needs to apply more power.Quote from: Zero999 on August 24, 2019, 06:12:54 pmI considered that. But it can works as a motor if you feed it with in-phase AC. Which is not the case for capacitor. Capacitor feeds AC generator with phase shifted AC.If AC generator rotating clock wise, the capacitor will feed AC generator in such way so it will rotate counter clock wise.Just look, if it works like you say, it will turns into perpetuum mobile, which is not possible. It doesn’t bother you?

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 06:42:50 pm
content: Quote from: soldar on August 24, 2019, 06:20:44 pmThey needs to remove reactive component from the load impedance. If load already don't have reactive component, they don't needs and don't used.Do you know that load may have reactive component in it's impedance?

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 07:07:08 pm
content: Quote from: soldar on August 24, 2019, 06:20:44 pmWhat is exactly wrong? Can you explain please?You don't agree that AC source consumes power from capacitor?Or you don't agree that power consumed by AC generator is used to make rotation more hard?

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 07:07:41 pm
content: Quote from: radiolistener on August 24, 2019, 06:38:09 pmThere isn't a textbook or forum member that will support your theory. It doesn't bother you?If it were a lossless system, it *would* go on forever. Just like this:It's easy enough to think up a system that would stay in motion forever if there were no losses. The impossibility of perpetuum mobile machines lies in the fact that they can not keep going if you extract energy from it.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 07:12:43 pm
content: Quote from: radiolistener on August 24, 2019, 07:07:08 pmA generator that consumes power is called a motor. Again: there are two options: either every forum member, text book writer, grid operator,... is plain wrong and you're on to something or your time is put to better use trying to understand what we are trying to tell you or to do some actual research rather than... whatever it is you're doing in this thread.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 07:18:05 pm
content: Quote from: Ice-Tea on August 24, 2019, 07:07:41 pmYou're wrong, I already say thatthere is a loss. Energy from capacitor will be consumed by AC source. You cannot make lossless circuit in such way and perpetuum mobile is not possible at all.Free energy and lossless - this is your words and your ideas. Believe me, power which is accumulated in capacitor will be returned back to AC source and consumed by AC source. It will be consumed and used to stop rotation of your AC generator. So, you won't be able to get infinite rotation by connecting capacitor to AC generator.So, believe me, your perpetuum mobile with infinite rotation won't works

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 07:24:44 pm
content: Quote from: Ice-Tea on August 24, 2019, 07:12:43 pmBut this motor will works to stop your generator, it won't works to keep rotation.So, you're think that AC source will not consume energy from capacitor?

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 07:33:35 pm
content: Quote from: radiolistener on August 24, 2019, 07:24:44 pmIT DOESNT!!This *electric motor* GIVES back the energy to the *mechanical motor* just as before the "electric generator* TOOK energy from *mechanical motor*

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 07:36:15 pm
content: Quote from: Ice-Tea on August 24, 2019, 07:12:43 pmAgain, there are two options:1) you're right, LTSpice works wrong and there is no power consumption by AC source from capacitor2) you are wrong and AC source consumes power from the capactor as it shown in LTSPiceI'm completely agree with LTspice and I think that AC source consumes power from capacitor. What is your choice?Quote from: Ice-Tea on August 24, 2019, 07:33:35 pmPreviously you're said that there is no power loss due to capacitor. Let's clarify things step-by-step.First answer - do you agree that AC source consumes power from the capacitor? Yes or no?

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 07:49:50 pm
content: Quote from: radiolistener on August 24, 2019, 07:36:15 pmFor the love of...Now integrate the power over time.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 08:03:39 pm
content: Quote from: Ice-Tea on August 24, 2019, 07:49:50 pmit will be zero.But you didn't answered on the question. Do you agree that AC source consumes power from the capacitor? Yes or not?

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 24, 2019, 08:14:16 pm
content: Quote from: radiolistener on August 24, 2019, 08:03:39 pmYou understand this demonstrates no real effort from the generator side is required, right? You know, the point you've been denying for two pages?QuoteI'm not sure what you mean by 'consume', but if you mean that the generator gets the power back/consumes the power it delivered before then yes, that's what I've been saying all along.

author: soldar
author_rank: Super Contributor
timestamp: August 24, 2019, 08:44:49 pm
content: radiolistener, what happens in real life is not under question. It is well understood by engineers and engineers all agree.Either you are totally wrong about it OR you do understand it and you are using vocabulary, language and explanations which are totally confusing and understood to be wrong. In either case the problem is with you, not with the rest of the world.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 08:51:53 pm
content: Quote from: Ice-Tea on August 24, 2019, 08:14:16 pmYes, I mean the power (which previously is delivered to the capacitor) is consumed back by AC source.So, we applied some amount of mechanical power to AC generator, it produce some amount of electrical power which is used to charge capacitor. Then capacitor applied this power back to AC generator and it was consumed by AC generator.So what happens to this power next? Does it affects mechanical rotation of AC generator? Or what happens with this power?Does this power just disappears? Does it slow down the rotor of AC generator? Does it speed up t he rotor of AC generator?

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 09:05:58 pm
content: Quote from: soldar on August 24, 2019, 08:44:49 pmYou're said that there is no energy loss due to capacitor. But if there is no energy loss, then you're invented perpetuum mobile. I don't believe in perpetuum mobile. And this is why I don't agree with you.And the rest of the world also don't believe in perpetuum mobile. Therefore do not attribute your opinion to the whole world. That's just your opinion.

author: Zero999
author_rank: Super Contributor
timestamp: August 24, 2019, 09:09:52 pm
content: Quote from: radiolistener on August 24, 2019, 06:38:09 pmNo it won't, the current is leading the voltage, which will be driving the alternator in the same direction.QuoteIn a real machine there are losses, so of course that won't happen. Adding a capacitor will increase the mechanical load on the generator, but not to the same degree as the equivalent resistive load will, which is what you think will happen. The increase in load will be minimal, inline with I2R.If the alternator, capacitor and wiring had no losses, then it would run forever, but it wouldn't be creating energy from nowhere, just storing it. If a resistive load were connected, it would slow down and stop.Quote from: radiolistener on August 24, 2019, 08:51:53 pmWhere does the power go when it's applied back to the AC generator?To simplify things, suppose the alternator is being driven by a motor. We have an old fashioned motor generator set, which we could be using to change the voltage and frequency. The secondary side has a capacitor connected to it. The whole thing is cooled down to the point of superconductivity.Where are the losses, other than the tiny amount of electromagnetic radiation caused by the currents flowing in the loops of cable?How does connecting a capacitor which will draw 230mA of current cause the same amount of power dissipated as a resistor with the same impedance? The answer is it doesn't, the capacitor results in no extra losses, other than the tiny increase in electromagnetic radiation.Quote from: radiolistener on August 24, 2019, 09:05:58 pmNo one has ever implied any perpetual motion.You seem to be implying the reverse, i.e. energy can be destroyed. Energy doesn't simply disappear. It has to be converted into another form.

author: KaneTW
author_rank: Frequent Contributor
timestamp: August 24, 2019, 09:14:24 pm
content: Radiolistener, are you actually just trolling?Look at the power graph in your simulations. It has positive power (= power consumed) for one half-period, and negative power (= power returned to the source) for the other. Average power is 0. The energy loss is in resistance, and you can simulate that by adding ~100 milliohms of ESR to the cap. The theoretical circuit, as shown, has zero losses.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 09:23:31 pm
content: Quote from: Zero999 on August 24, 2019, 09:09:52 pmSo if you continue to apply mechanical power to AC generator, it will accumulate power and amplitude on the output will grow up just because we connected capacitor as the load?Do you talking about DC generator or AC?

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 09:49:52 pm
content: Quote from: KaneTW on August 24, 2019, 09:14:24 pmProbably you didn't understand what we're talking about. I didn't said that capacitor consumes power. I said that capacitor doesn't consume power. And power graph confirms that.The discussion is not about if capacitor consumes power or not. The capacitor doesn't consume power. OK?The discussion is about whether energy was wasted or not. I mean energy which was used for work to transform mechanical energy to capacitor charge and back.As I understand all agree that there is no power loss in circuit between AC source and capacitor and all agree that AC source consumes power back.So, the question is what happens with that power next? When it is consumed by AC generator.

author: soldar
author_rank: Super Contributor
timestamp: August 24, 2019, 09:55:50 pm
content: Quote from: radiolistener on August 24, 2019, 09:05:58 pmNo. Read again carefully. I said the problem is with you, not with the capacitor. We all understand what happens and you seem to be the only one who has a problem. The problem is with you. The capacitor knows what it does, we know what it does. You are the one who is confused. maybe we need to start talking about why you are having such a hard time understanding. What kind of feelings do capacitors generate in you? Happiness? Sadness? What is your oldest memory of capacitors? How old were you? Was it a happy memory? How does it relate with the rest of your childhood? etc. Tell us your thoughts. Maybe then we can understand your problem. Or maybe not.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 10:08:48 pm
content: Quote from: soldar on August 24, 2019, 09:55:50 pmIf you understand what happens, then tell what happens with power when it consumed by AC generator?

author: Circlotron
author_rank: Super Contributor
timestamp: August 24, 2019, 10:15:18 pm
content: Quote from: radiolistener on August 24, 2019, 08:51:53 pmI have never done the experiment, but I would expect that as the alternator is rotating and the capacitor voltage is increasing as energy is flowing into it, the alternator shaft gets harder to rotate. It is working as an alternator. Then later as the voltage is decreasing and the energy is flowing back to the alternator the alternator is operating like a motor and it would try to rotate ahead of whatever is driving it. It would actually return mechanical power to the gasoline engine flywheel or whatever.Think of it in familiar mechanical terms. Piston, connecting rod and crankshaft. For the first 90 degrees of crank rotation the piston mass is accelerated and it gains kinetic energy and it needs energy from the outside world to do this. Then in the next 90 degrees of rotation the piston is slowed down and while this is happening it returns the kinetic energy back to the crankshaft and whatever is connected to it. It tries to rotate the crankshaft faster than the crankshaft wants to go. The absolute fact that a connecting rod will stretch in a high speed race engine is proof of this chain of events. Physics being what it is, this comparison of electrical and mechanical is accurate enough for this discussion.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 10:48:15 pm
content: Quote from: Circlotron on August 24, 2019, 10:15:18 pmwhy ahead? I'm inclined to believe that it will try to rotate back (to slow down). It's hard to believe that capacitor will keep AC alternator rotating for infinite time (with no heat loss).

author: soldar
author_rank: Super Contributor
timestamp: August 24, 2019, 11:02:13 pm
content: Quote from: radiolistener on August 24, 2019, 10:48:15 pmYou are inclined to believe wrong. You are being told by everybody. You are wrong. Science is not based on your inclination. Please stop being wrong.

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 11:16:06 pm
content: ok. I'm not familiar with AC alternators. May be that's true.So, if I understand correctly, capacitive load leads AC alternator to keep rotating. And inductive load leads to slow down. Is it correct?

author: Circlotron
author_rank: Super Contributor
timestamp: August 24, 2019, 11:21:37 pm
content: An inductor would have the same effect on an alternator.Ideally, a voltage source alternator with an inductor or a current source alternator with a capacitor, but that is complicating things even further.

author: emece67
author_rank: Frequent Contributor
timestamp: August 24, 2019, 11:33:11 pm
content: .

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 11:40:36 pm
content: Quote from: emece67 on August 24, 2019, 11:33:11 pmAC source can consume external power the same as DC source.

author: emece67
author_rank: Frequent Contributor
timestamp: August 24, 2019, 11:42:32 pm
content: .

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2019, 11:45:08 pm
content: Quote from: emece67 on August 24, 2019, 11:42:32 pmIf capacitor voltage is higher than AC or DC source, why not?Above you can find some screenshots from LTspice with example. AC source consumes power from capacitor. You can see it on the graph.

author: emece67
author_rank: Frequent Contributor
timestamp: August 24, 2019, 11:48:43 pm
content: .

author: radiolistener
author_rank: Super Contributor
timestamp: August 25, 2019, 12:34:34 am
content: Quote from: emece67 on August 24, 2019, 11:48:43 pmfor AC it matters, because power will be returned with some time lag. AC source will be changed after this lag, so the result will depends on how new AC value will be different from a lagged value. Lagged value will change load impedance on AC source and the new impedance will depends on this time lag and AC frequency.BTW, you can connect 1000 Farad capacitor to AC alternator and it will be almost the same as short circuit. I'm suspect that idea that AC alternator with such capacitor will rotate almost forever (as claimed by soldar&co) looks not realistic.

author: Brumby
author_rank: Unknown
timestamp: August 25, 2019, 01:22:33 am
content: 1. Whether the capacitor is a microfarad or a megafarad makes no difference to the principles involved. It just changes the number in the arithmetic. Please don't try and obfuscate the discussion with such irrelevant extremes.2. Any talk - or perceived talk - of concepts that might lean towards "perpetual motion" here is based on IDEAL components. In the discussion I have read, such statementsare, indeed, correct. These must also include completely isolated systems and every other requirement for an ideal system - including not making any measurements. IN PRACTICE, we DO have losses, so the "perpetual motion" viewpoint is completely irrelevant within this discussion. However, IN THEORY, such discussion with the assumption of ideal components, is correct when we wish to consider the fundamental processes in play. In fact, we really need to consider the system with ideal components in order to get a grasp on those principles.3. I am concerned about the use of the word "consume" byradiolistener. It infers a loss, whereas I get the feeling it needs to be interpreted differently - more along the lines of an energy transfer function.Just my 2 cents.

author: radiolistener
author_rank: Super Contributor
timestamp: August 25, 2019, 01:34:58 am
content: Quote from: Brumby on August 25, 2019, 01:22:33 amUpdated: I have thought and it looks that I was not right, we cannot compare short circuit with large capacitor. These things with AC alternator looks very strange, needs to learn it more deep and make some experimentsQuote from: Brumby on August 25, 2019, 01:22:33 amI understand that. We assume that there is no heat loss for such "perpetual motion". But I still don't believe that it will keep rotation with any capacitor.

author: james_s
author_rank: Super Contributor
timestamp: August 25, 2019, 02:54:37 am
content: You've got it backwards, returning energy to the generator makes it turn more easily because it acts as a motor, not harder to turn.At this point I'm starting to think you're just trolling because any EE or textbook will tell you that you're wrong and yet you continue to argue against it. Just what is it you think you know that engineers do not?If you're not trolling then you're wrong, you don't understand it and you're not trying to understand it, do you just enjoy arguing?

author: james_s
author_rank: Super Contributor
timestamp: August 25, 2019, 03:07:57 am
content: Perpetual motion cannot happen because even 0.00000000000000000000001% loss will eventually deplete the system. When talking about a small increase in I2R losses it's not really significant until you get to a large scale like industrial settings and those historically have been heavy on inductive loads, hence the capacitors added to correct the power factor.Nothing is perfect and there always are some losses but they are small. A capacitor that caused 500W of apparent power to flow might result in a few watts of energy wasted in heating the wires. It is this I2R loss that will result in a very slightly higher mechanical load turning the generator but this is negligible, nowhere near what you would calculate from apparent power.

author: soldar
author_rank: Super Contributor
timestamp: August 25, 2019, 08:30:15 am
content: There is so much confusion in this thread. radiolistener, you are not listening.Assume an AC generator directly connected to a resistor. The power supplied is in phase with the voltage.Now we connect a capacitor in parallel with the resistor.During the first quarter cycle the generator has to work harder because, besides the power being supplied to the resistor it is also charging the capacitor storing energy in it. It is working double time.During the second quarter cycle the capacitor is returning the energy which goes into the resistor and this means the generator has to work less. Whatever current is supplied by the capacitor is less current the generator needs to supply.If you do not understand this you should not continue to argue. You should go back and study some more until you understand it.

author: capt bullshot
author_rank: Super Contributor
timestamp: August 25, 2019, 08:55:11 am
content: Quote from: radiolistener on August 25, 2019, 01:34:58 amIn terse: Your general AC generator / motor (in German EE we don't use these terms, we just say AC Machine or Engine, regardless of whether it is used as Generator or Motor), will act as an inductor if you load it with an capacitor and vice versa. So the energy that is "consumed" back by the generator from the capacitor isn't consumed but stored in that inductor and than delivered back to the capacitor in the next cycle. Think of an LC tank, works the same way as one compensates inductive load by using capacitors. The amount of inductance or capacitance that the generators shows to deal with its load is determined by that nice term "Polradwinkel" (angular displacement in synchronous generators) - one can set this either by changing the excitation of the generator (think of power plants) or it just works implicit (think of small generators / motors).And yes, if there weren't losses, a generator loaded by a capacitor would run infinitely, same as an unloaded generator would do. The capacitor doesn't make the generator run forever, slower or faster, indeed it wouldn't change anything (in the ideal lossless model).

author: soldar
author_rank: Super Contributor
timestamp: August 25, 2019, 09:34:02 am
content: People with a poor grasp of things tend to confuse concepts and there is where we end up with people saying they think they have found a free energy source.Saying perpetual motion is not possible is only a way of saying any machine we build in reality is going to have losses. But ideally perpetual motion is possible and we can see examples which get pretty close. The planets have been revolving around the Sun for some time now and will continue to do it for the foreseeable future.What is impossible is not "perpetual motion", what is impossible is any system which yields more energy than it has stored and is receiving. Energy cannot be created out of nothing. An ideal machine can only yield as much energy as it is receiving. It can store it, it can transform it but it cannot create it. You cannot get energy out of nowhere.With an ideal inductor and an ideal capacitor we could build a resonant LC circuit which would resonate forever but we could not extract energy from that system without slowing it down.You can imagine an ideal pendulum which will oscillate forever but you cannot extract any energy without slowing it down. The amount of energy you can extract rom any system is what was put into it in the first place. No more.

author: emece67
author_rank: Frequent Contributor
timestamp: August 25, 2019, 10:46:07 am
content: .

author: radiolistener
author_rank: Super Contributor
timestamp: August 25, 2019, 11:18:26 am
content: Quote from: emece67 on August 25, 2019, 10:46:07 amDon't confuse "time delay" with "time". This is different things.

author: soldar
author_rank: Super Contributor
timestamp: August 25, 2019, 11:42:20 am
content: Quote from: radiolistener on August 25, 2019, 11:18:26 amIf there is anyone in this thread who is confused it is you and you would do well to stop posting and go do some studying and some learning and stop wasting our collective time.

author: radiolistener
author_rank: Super Contributor
timestamp: August 25, 2019, 12:00:06 pm
content: Quote from: soldar on August 25, 2019, 11:42:20 amAs I see, all what you can is just insulting and talking that other peoples don't know anything. Probably many people insulted you and talked that you don't know nothing, and it seems that they said truth, because it hooked you, and now you're trying to insult other in the same way for revenge.I'm really sorry you were so unlucky in life, but I cannot help you with that.Your words is nothing for me, because it cannot be applied and do not matter. So, I don't care what is your thoughts. Is it possible to take offense at the donkey who kicked you?May be I'm interested to discuss about AC alternators, because I really do not familiar with it, but definitely I will not do it with you, because you cannot provide any useful information, except stupid talking about "you're wrong", "go do some studying and some learning", etc...I believe that the topic has exhausted itself. Sorry, but I don't see the reason to discuss with you about personalities. I’m not at all interested in this. This is a boring conversation.

author: Zero999
author_rank: Super Contributor
timestamp: August 25, 2019, 02:27:00 pm
content: Quote from: radiolistener on August 24, 2019, 11:16:06 pmGood, it seems like you've finally realised there's a gap in your understanding. Please go back and do some research on alternators.QuoteNo, purely inductive or capacitive loads have little effect on the mechanical loading of the alternator.Quote from: radiolistener on August 25, 2019, 12:00:06 pmNo one is insulting you. Please don't take it personally. It's understandable that you feel everyone is ganging up on you.In reality people are frustrated because they feel you're not listening. If you do some research, come back with more questions or admitting you were mistaken, then people will be happy to help.

author: ejeffrey
author_rank: Super Contributor
timestamp: August 26, 2019, 02:58:55 am
content: No, the energy returned back will make thing easier for the generator, i.e., decrease torque and fuel consumption aside from the extra transmission losses.Very large reactivate/reflected power is hard on linear amplifiers and rf power supplies because the reflected power is dissipated in the pass transistors instead of the load and can cause them to overheat. But generators and switching/class D supplies don't have that problem.It certainly isn't ideal. There is a reason for the push for better power factor. But reactive power is (mostly) not lost.

author: gbaddeley
author_rank: Regular Contributor
timestamp: August 26, 2019, 09:02:52 am
content: If the capacitors were really wasting power, it could only dissipated as heat. A component dissipating a few watts will get quite hot, unless it is heatsinked or large.

author: GerryR
author_rank: Frequent Contributor
timestamp: August 26, 2019, 11:28:34 am
content: Quote from: james_s on August 23, 2019, 04:16:56 pmIt isn't indicated on your bill as "reactive" power, but believe me, we are paying for it!The purpose of pf correction is to put the current in phase with the voltage; pf is the cos of the angle. That is why they have huge capacitor banks in motor-rich industrial environments. If you have ever been inside a pf-correction cabinet, you will see that there is some loss by the amount of heat generated inside the cabinet, but not as much loss as what is being saved.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 26, 2019, 11:34:49 am
content: Quote from: GerryR on August 26, 2019, 11:28:34 amI'm curious.. how exactly are you paying for it?

author: GerryR
author_rank: Frequent Contributor
timestamp: August 26, 2019, 12:31:57 pm
content: I have no way of knowing "exactly," but I'm sure the power companies have a good idea of the total reactive power being used and have that factored into the rates we pay each month.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 26, 2019, 12:40:13 pm
content: Oh, sure, no doubt. But you can use 0 VAR or 10kVAR at your house and not pay a dime more.

author: Circlotron
author_rank: Super Contributor
timestamp: August 26, 2019, 01:15:41 pm
content: I suppose we all pay from the point of view that the electricity supply companies install larger lines, transformer and switch gear than would be necessary if all loads were unity power factor. The extra cost of that larger equipment is passed on the consumer in one way or another.

author: radiolistener
author_rank: Super Contributor
timestamp: August 26, 2019, 02:53:01 pm
content: Quote from: Ice-Tea on August 26, 2019, 11:34:49 amAll expenses are included in the price. If your equipment has good power factor, the price for kVA will be included in the price for active kWh, the same as other expenses. This is what happens for usual private person with low power consumption.But if your equipment has bad power factor it requires an increase in the generation and transmission capacity of the electric utility to handle the reactive power component. If reactive power component is high enough then you will be charged for each kVA separately.Your equipment don't consume this energy, it just get it and then put back, but electricity company has additional expenses due to reactive power in your equipment (for exampel higher energy loss in transmission line due to heat), so they will include their expenses in your bill. If you're private person with low power consumption, it will be included in price for kWh. If you own some factory which has high power consumption, you will be charged for reactive power separately.

author: stefon
author_rank: Newbie
timestamp: August 28, 2019, 06:06:33 am
content: OK... now I understand power is lost only because of impedance in lines and ac source. Can anybody just give more or less how much is lost for example 1uF conencted on home power, because have no idea what is typical "power outlet" impedance...

author: Circlotron
author_rank: Super Contributor
timestamp: August 28, 2019, 06:20:12 am
content: Never bothered measuring, but if you have 240V available and you plug in your fan heater that pulls 10 amps the voltage might drop to 235V. This 5 volts drop for 10 amps = 0.5 ohms power outlet impedance. Might be more, might be less in any particular case. 1uF has a reactance of 3183 ohms at 50Hz so it would pull 0.0754 amps at 240V so the I2R losses in 0.5 ohms would be about 0.00284 watts. 2.8 milliwatts. Not a lot in the big scheme of things.

author: Simon
author_rank: Unknown
timestamp: August 28, 2019, 07:06:08 am
content: I think power companies will accept a minimum of power factor and after that start charging. Most large devices are now controlled by legislation about power factor. Also, the energy companies are supplying the whole grid, not just you so you capacitance will cancel someone else's inductance if they are close enough that the wiring between you is negligible.If they want to charge me for reactive power they better pay me for reactive power too as I am also a generator. Currently, I am supplying at a power factor of 1.

author: Ysjoelfir
author_rank: Frequent Contributor
timestamp: August 28, 2019, 07:13:31 am
content: Quote from: stefon on August 28, 2019, 06:06:33 amThe typical impedance of the power grid is specified in IEC61000-3-3 with 0.24 ohms + j0.15 ohms.