It’s also a place where the theory of making accurate AC power measurements is freely discussed, and since that theory is based on the laws of physics, it transcends any particular implementation. I’ve never seen an IotaWatt or any of the other OEM products.
Here are two for your consideration:
The first is my microwave oven being a clock, as discussed above. The second is my pool pump when it’s sucking on air, i.e. it’s a 1.5hp synchronous AC motor with almost zero load on it.
Those are the actual signals on the wire, they’re not indicative of any measurement errors. In both cases the current signal is distorted (i.e. it’s not a sinewave), but they’re close enough that you could at least imagine what the 50Hz component would look like were you to extract it. In the Microwave case, the current leads the voltage by almost 90° and in the Pool case the current lags the voltage by about 50°.
Now imagine what happens if you introduce a 5° phase error (say) to both of those measurements. If your 5° error shifts the current to the right (i.e. delays it) it will increase the reading for the Microwave and decrease the reading for the Pool. If your 5° error shifts the current to the left (i.e. advances it) it will increase the reading for the Pool and decrease the reading for the Microwave (so much so it will likely make the Microwave appear as a generator rather than a consumer). So without knowing the nature of the load, it’s impossible to predict whether an introduced phase error will cause an increase or decrease in power reading.
When posting these examples it would be a lot easier to knock up something with a signal generator or a spreadsheet, but to avoid acsusations that I’m talking about some theoretical situation that will never happen in real life, I always try to use real measurements from real loads as found around my house: washing machines, hair dryers, microwave ovens, pool pumps, CFLs etc. These loads exist, you can’t just wish them away, and the more your monitor drills down to a per-circuit measurement the more likely you are to have such a load totally dominate the measurement you’re trying to make.