That's for phase shift due to inductive or capacitive load. There's almost no phase shift with a CFL, but the power factor is still very low due to the spikey current draw (only at the mains peaks).
Nope -- a domestic meter meter will only charge for the power consumed, which is 18W.
They do, and have for a very long time.
Thomas Prufer
Interesting, so the Voltage is the mains sinewave and the current only runs when voltage exceeds ??. Effectively a high peak current clipped either side of these voltages, a sort of pulse width modulated current. Is this why dimmeres won't work with CFLs?
AJH
go on then what do you base that theory on??
from the web
Modern electricity meters operate by continuously measuring the instantaneous voltage (volts) and current (amperes) and finding the product of these to give instantaneous electrical power (watts) which is then integrated against time to give energy used (joules, kilowatt-hours etc). The meters fall into two basic categories, electromechanical and electronic.
no way can a meter that measures volts and amps instantaneously take into account a leading or lagging current
I think you've got the logic back to front there
how?
No, meters cope with phase angles.
don't confuse peoples strongly held beliefs with actual facts.
They wont thank you for it.
meters measure the integral of VI over the cycle: thats what POWER you are drawing.
The fact that a rotten power factor costs the generating company money, is essentially ignored, as by and large the main users of electricity are resistive heating.
You can of course in industrial situations where a lot of inductive loads are running, correct the power factor with capacitors. There is some rule about that, but I dont know what it is.
SMPS are buggers, as they draw all their current in short bursts at peak voltages mostly.
Hard to correct without L and C filters..
shows how little you understand electrical theory.
in your usual way. By thinking you understand something when you don't.
I suspect that is exactly it. Unless they have a detector to work out what the dimmer was trying to do, and behave accordingly.
As it says. If it takes instantanious (>than 20ms it will give a true reading of energy used, whatever the waveform.
Because by measuring instantaneous values it's following the waveform and doing exactly what you say it can't do.
There are some graphs at which show the effect of taking samples of voltage and currents and calculating the power curve. This is effectively what happens in the electricity meters.
Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power can be greater than the real power
and as a domestic meter reads both current and voltage at the same time IE instantaneous how is that taking into account the power factor??
Depends on what you mean by "taking into account". The instantaneous product sum will arrive at the real power consumption. The meter does not have the capability to calculate the apparent or reactive powers though. Since you are not charged for apparent power usage in a domestic situation, that does not matter.
It doesn't need to know anything about the power factor.
Thats simply way of looking at how out of phase the V/I waveforms are, if both are sinusoidal. Whih in many case they are not anyway.
It doesn't need to. The meter measures the *real* power, i.e. the product of instantaneous voltage and current.
Take another look at the graphs at . The dark blue sine wave is the result of multiplying instantaneous voltage and current, that's the real power. The total power consumption is equal to the area between this curve (or the pale blue average line) and the X-axis. This is what is measured by your electric meter.
You would only need to take the power factor into account if you wanted to know the apparent power (i.e. KVA). Your electricity meter neither knows nor cares about power factors, the electric companies don't charge domestic customers for the reactive component so that's no problem anyway.
so why did you say and I quote "they record the product of instantaneous voltage and current so the phase angle is effectively accounted for"
now you are saying "Your electricity meter neither knows nor cares about power factors"
so which is it going to be?
the point I was trying to make is domestic meters don't allow for power factor
Yes, I follow that the product of RMS current and RMS voltage is not the same as the RMS product of instantaneous voltage and current.
The links I am seeing suggest that VAr and VA are the same. The Wanderer implies that they are not.
Andy
Have something to add? Share your thoughts — no account required.
Ask the community — no account required