mandulaj
commented
1 year ago When you take the simple product of Voltage and Current, you get something called Apparent Power (S). If you have a resistive load, the voltage and current are in phase and as a result the Real power (P) is equal to the Apparent Power (P=S=V*I).
However, in many cases, voltage and current are not in phase. This means that the Apparent Power contains two components, The Real power (P) and something called Reactive Power (Q). The Apparent Power S is complex with theta being the phase shift, but we often summarize the relationship using the power factor PF=cos(theta): P = |S| * cos(theta) = |S| * PF The Reactive Power Q manifests as this extra 'phantom' current. However don't be fooled, the current is really there! It just does not do any useful work and instead just moves the reactive power back and forth between the load and source.
In big power networks, this is a problem because you need to have wires sized sufficiently in order to handle both the Real and Reactive currents despite only the Real current doing useful work! So ideally your PF is 1 and your Apparent Power is equal to the Real power giving you maximum transmission efficiency. This is why Power Factor correction is a big business! However, you as a consumer don't have to worry about it. You will anyways only pay for the Real Power.
You can read more about this on Wikipedia: Power Factor.
If I am not mistaken, you are probably powering some kind of low-power switching device, something capacitive, or even just nothing (effectively capacitive). This results in a horrible power factor (0.01). If you multiply it by the Apparent power you get 231*0.212*0.01 = 0.5W which is close to the reported Real power (0.7W) and what your commercial power meter says (2.9W).
As to why all these numbers are not exactly equal, first of all, the low PF is usually not very accurate, I would like to one day measure its true resolution, but the discrepancy between the calculated 0.5W and reported 0.7W is likely just a numerical error of the resolution (0.01 vs 0.014). And second, the PZEM module is not great at measuring low currents overall, plus it has some power consumption on its own. So you cant trust it too much for the first few watts. It does get quite accurate for larger loads though.
So in conclusion, just use the Power: 0.7 value to get your Real Power and don't bother with reactive power.
mlaci001
vortigont
Hello When reading the device, I get the following values: Voltage: 231.900 Current: 0.212 Power: 0.700 Energy: 0.002 Freq: 50,000 PF: 0.010 The real consumption is 2.9 W (measured with a commercial power meter). According to the values, 231.9(Voltage)*0.212(Current)=49.16W would come out, which is impossible. Has anyone encountered such a problem? Thanks.