Recently I picked up some energy saver bulbs from Wickes. They are "advertised" as being 18 Watt (100W equivalent, but that's not relevant to this). However, on the top of the box, wher eit says 1100 lumen, it also says 130mA. Now, my current mains voltage is about 235V, so at that voltage, 130mA is just over 30 Watts, not the 18 W claimed.
Now this is my simplfied thinking - I haven't (and don't intend to) looked into the power factor of these puppies, but does anyone have any insights into the discrepancy between the figures offered?
I dimly remember from my C&G classes that the consumed power of a flourescent isn't V*I as it has in inductor in it using AC. It's something like P=V*I*COS(factor) where COS(factor)1 for non-resistive loads.
Yes, but the meter will record the current so the energy charged for will not take into account the reduction due to the phase angle. A very interesting point you've raised here.
As I understand it all domestic electricity meters measure the true power used. They record the product of instantaneous voltage and current so the phase angle is effectively accounted for. See
if the current waveform lags/leads the voltage then its not a true power reading, if the peak current happens at different time as the peak voltage the the overall power reading will be lower
'True power' is a vague and misleading term. There's KWH, KVArH, and KVAH. Now you could argue that KWH is 'True Power' coz it's the inphase component, but there again, you could argue that KVAH is true power, coz you're then taking KVArH into account......
OK for us innocents, what's the difference between VA and VAr?
BTW a little digging trying to find the answer suggests that watts means fuel at the power station, and VA means equipment to get it to you. But watts is power in my book...
It's been a while for me (20 years or so :-}), but...
Imagine two sine waves that should sit directly over one another - one for volts, one for amps.
Certain types of load can pull them a bit askew from each other, which reduces the overall effectiveness of the power "delivered" as against the power "sent" (phase angle).
A crap analogy might be a sports car that's meant to use Super Unleaded, but you only put Regular Unleaded in - you can have the same amount in a full tank of either, but one will have a bit more "go" in it.
In a purely resistive system the voltage is in phase with the current, so the calculation of wattage is simple -- volts x amps.
If the circuit is reactive (inductive or capacitive), the current lags or leads the voltage, so it's a bit more complicated multiplying the two sinusoidal values.
Your electricity meter 'multiplies' the voltage and current - if they're out of phase the measured power will be less than the effective power that you're using.
Try it on a bit of graph paper, drawing two sine waves (call them V & A) in phase and drawing the product, then put them out of phase and again multiplying the two... In an extreme (theoretical) case the voltage could be at a maximum whilst the current is zero. The power at that point is obviously zero.
If you imaging a resistive load as being like riding a bike along a constant plane. It takes a certain amount of push to make the pedals go round, and propel the mike at a given speed.
If one were to add a large spring that connected one pedal to the seat post - such that it was stretched as that pedal reached the bottom of its travel, then it would be rather like adding an additional reactive load. Pushing the pedal down on one side would now require more push than before. However, the actual power required to propel the bike at a given speed would stay the same (since the extra push on one pedal would be offset by the pulling effect of the spring reducing the push required on the other side).
The reactive element (the spring) is just acting as a temporary energy store - neither consuming or contributing energy to the system. However you need the capacity to exert more force on the pedals in order to be able to still ride the bike - so the bike has become harder to ride.
The same with the circuit - you may get the same power out, but you will need capacity to push more current through it than before.
Some interesting analogies, but no-one's yet hit on what's happening. There is more than one way to get a non-unity power factor, and what goes on in CFLs is that the mains input goes through a diode rectifier bridge and to a reservoir capacitor. This draws current around the voltage peaks, and nothing for the rest of the cycle. Thus what's happening is quite different to the waveform you get with a basic inductor ballast.
FWIW the phase shift occurring with an old fashioned switch start inductor ballast (linear fls) is only an approximation to what's happening there, as an electromagnetic fl circuit has 2 things in series, ballast and tube. The ballast operates at near 90 degrees out of phase, but the tube has a weird waveform specific to discharge tubes. The resulting mains current waveform is a hybrid of the 2, which is why its not possible to get such fittings to a power fator of
One or two others have talked about sine waves lagging and leading, I prefer to think in terms of vextors.
Take a look at
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It's a wikipedia article on AC power. I know you always need to approach Wikipedia articles with a touch of caution, but it presents the case accurately.
If the meter just multiplied the peak (or RMS) voltage and current to determine the power then this would be the case. But domestic electricity meters don't work that way. In the traditional mechanical meter the power indication is continuously derived from the fields induced by the voltage and current coils, at any instant this is the actual real power, the meter measures KW, not KVA. You are only charged for kilowatt hours, the out of phase component of the current does not represent power used by you, and you are not charged for it.
Modern electronic meters do it differently but end up with the same result of indicating Kilowatt hours.
Things are different for some heavy industrial users who are charged for power and maximum current.
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