prateeksikka Posted May 17, 2005 Report Share Posted May 17, 2005 hey guys!the electricity meters in our homes show the monthly usage by us of active power .i.e V*I *cos(Q).i.e voltage*current*power factor.as power factor is less than 1,this power is less than voltage *current actual power used by us. So do we pay less than actual power used by us. i mean if we have only choke coils at our home we need to pay nothing. Use tubelights only !say no to bulbs.Could anyone explain the fallacy?prateek ;D Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 17, 2005 Report Share Posted May 17, 2005 hey guys!the electricity meters in our homes show the monthly usage by us of active power .i.e V*I *cos(Q).i.e voltage*current*power factor.as power factor is less than 1,this power is less than voltage *current actual power used by us. So do we pay less than actual power used by us. i mean if we have only choke coils at our home we need to pay nothing. Use tubelights only !say no to bulbs.Could anyone explain the fallacy?prateek ;DIn the UK we pay for VA or apparent power - we are being charged for more power than we use!A pure resistive circuit like a light bulb will have a power factor of 1 (not quite as there will be some parasetic L or C that dominates) A perfect inductor or capacitor will have a power factor of 0 even though it's drawing a current.It is very important to understand that capactors and inductors don't disipate any power they only store energy. The only element in any circuit that dissipates power is resistance.For example a 100mH inductor at 50Hz will have an impedance of:2pi*50*0.1 = 31.4ohmsConnected to 240V the current will be:240/31.4ohm = 7.64AThe apparernt power (VA) will be:240*7.64 = 1833VABut no energy is disipated in the inductor it is all returned to the mains, the only power loss is due to the I2R losses in the cable.There aye many articles on the web about power factor, here are some links:Tutorials:http://users.telenet.be/educypedia/electronics/electricitybasic.htmhttp://www.du.edu/~jcalvert/tech/acessent.htmhttp://en.wikipedia.org/wiki/Real_powerhttp://www.lmphotonics.com/pwrfact.htmhttp://www.ibiblio.org/obp/electricCircuits/AC/AC_11.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/powfac.htmlhttp://www.microconsultants.com/tips/pwrfact/pfarticl.htmhttp://www.esru.strath.ac.uk/Reference/concepts/elec_cepts.htmOther information and calculators ect.http://www.oit.doe.gov/bestpractices/motors/factsheets/mc60405.pdfhttp://www.pcnet.com/~jhg/calc2.html Quote Link to comment Share on other sites More sharing options...
prateeksikka Posted May 18, 2005 Author Report Share Posted May 18, 2005 HI ALUN! thats what i said ,the power meters only measure the power used by us i.e the power dissipated in the resistive elements at our homes.If we have pure inductances we need to pay nothing as we are dissipating nothing!any explanation for the fallacy? :( Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 18, 2005 Report Share Posted May 18, 2005 I don't know about where you live but in the UK we pay for apparent power, not just the power dissipated in the resistive elements. So if I connected a big inductor to the mains a large current would flow and even though I'm not cunsuming any power I would still be paying for it. Quote Link to comment Share on other sites More sharing options...
prateeksikka Posted May 18, 2005 Author Report Share Posted May 18, 2005 hi alun!in your country how do the electric companies measure the power used by you when you are feeding back the energy recieved in form of magnetic field.I mean how does your meter come to know the amount of power which u have apparently used and not actually used!the meter must be reading V*I *Cos(Q).no? :o Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 18, 2005 Report Share Posted May 18, 2005 The answer is, it doesn't.All the meter measures is the voltage and current, this will give VA (the apparent power) the don't measure the power factor, so they don't know how much power I'm really using. Quote Link to comment Share on other sites More sharing options...
prateeksikka Posted May 19, 2005 Author Report Share Posted May 19, 2005 i guess here in india the meters measure active power.i.e the power actually dissipated.perhaps that is the reason why the government electricity boards and companies connect power factor boosters at some intervals along the line these boosters push up the power factor,otherwise companies would be bankrupt!even for a power factor of 0.5,they will be losing half their income ;Dprateek Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 19, 2005 Report Share Posted May 19, 2005 Power companies in the UK use power factor correctors too as they want to make as much money as possible, companies can be fined if their power factor is too low so. many companies have power factor correction systems often they just have capacitors on their induction motors and most flouresent tube fittings have power factor correction capacitors. Quote Link to comment Share on other sites More sharing options...
prateeksikka Posted May 19, 2005 Author Report Share Posted May 19, 2005 hi alun!but why do they require it in ur country,when u r already paying maximum. i.e power factor=1 at ur end and so at the companies end.they cant increase beyond 1.so why do they use these capacitors.i guess similar is the case in india.plz reply ;D Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 19, 2005 Report Share Posted May 19, 2005 A poor power factor means that the current is a lot higher than it needs to be, this means more power is disipated than necessary as inceased current means increased I2R losses. I suppose some companies can take higher electricity bills but a fine is just and extra kick up the arse. ;D Quote Link to comment Share on other sites More sharing options...
prateeksikka Posted May 20, 2005 Author Report Share Posted May 20, 2005 hi there alun!do u mean to say that actually we dont require these power factor boosters but the companies connect it to earn more?Why do the companies in UK boost up the power factor when they are already charging u for V*I? and it will in adverse increase the losses too!! :o Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 20, 2005 Report Share Posted May 20, 2005 hi there alun!do u mean to say that actually we dont require these power factor boosters but the companies connect it to earn more?Why do the companies in UK boost up the power factor when they are already charging u for V*I? and it will in adverse increase the losses too!! :oNo boosting the power factor reduces the losses, and power companies can make more money if they charge people for V*I and then correct the power factor them selves to reduce the lI2R losses in their transmition lines. They're quite cheeky realy charging you for VA and fining you if your power factor is too low but they boost the power at the local transformer anyway. I suppose it still makes sence to havave a high power factor because there are still losses in the cable from your house or company to the local sub station. Quote Link to comment Share on other sites More sharing options...
prateeksikka Posted May 21, 2005 Author Report Share Posted May 21, 2005 so u mean to say that companies reduce the power factor to reduce the transmission losses bu still charge you for V*I?very clever! ;D Quote Link to comment Share on other sites More sharing options...
jaganp Posted May 24, 2005 Report Share Posted May 24, 2005 can u explain how a single phase two wire static wattmeter (electronic with blinking LEDs) work? i just saw the connection diagram on the body of the meter. two wires go in(L,N) and two come out and they are internally shorted respectively. Infact the measurement device internally is hooked in parallel across the two wires ( Live and neutral). how do they measure the current or voltage or watts effectively out of this parallel connection across L & N ??? jagan from india. Quote Link to comment Share on other sites More sharing options...
prateeksikka Posted May 24, 2005 Author Report Share Posted May 24, 2005 hi jagan!i can tell u the general functioning of a wattmeter.the 2 coils which u see inside are current coil and voltage coil.voltage coil is in parallel (less no of turns)and current coil with more number of turns.the resultant field produced is such that it is proportional to product of currents in 2 coils.this resultant field is responsible for moving a needle .movement of needle is then calibrated in terms of power.V*I in otherwords i1*i2. ;D Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 24, 2005 Report Share Posted May 24, 2005 You mean like this. Quote Link to comment Share on other sites More sharing options...
prateeksikka Posted May 25, 2005 Author Report Share Posted May 25, 2005 yeah!this is what i mean the link given by you says the torque is proportional to V*I.but actually it is proportional to the product of 2 currents in coils I1*I2. but since potential coil has high resistance,the voltage drop is what we consider(simple ohms law) so it becomes I1*V2.i.e power in the load ;D ;D Quote Link to comment Share on other sites More sharing options...
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