I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.
0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.
Think about it again, and how I did the calcs in the video for a home, and then get back to me.
Sorry for my critical thingking but the calculations shown in the text of the video make no sense to me because you keep mixing Watts and VA as if they are equal.
They are not equal. BUT if a product takes 80mA in measured RMS current, and that is NOT compensated for in the grid by filtering and/or phase correction, then to maintain the same mains voltage that power MUST ultimately be generated at some point, there is no free lunch there.
Thought experiment for you: Assume your grid is at almost maximum capacity (minus the 1.2MW real device power (1.2W * 1,000,000 devices)), and there is no more harmonic filtering or phase correction left to give, and you suddenly connect a million of these devices to the grid, where do you think the almost 100MVA magically comes from to maintain that 240V at the home? The current fairy?
So please enlighten us with a real calculation including a good estimate of the actual grid losses. Hint: if small devices with low PF are deemed a problem then they would be subject to regulations regarding power factor.
Just because they didn't bother to do <75W devices in EN61000-3-2 and EnergyStar doesn't mean it doesn't matter. The entire point of the video is to consider this in product design.
If it's a problem at 75W for a single device, why shouldn't it be a problem with 75 devices of 1W each (localised EMC excluded).
My simple, off the cuff, calculation already shows that the I2R losses for the smoke alarm are neglectible even if my resulting number is 10 times too low.
Seeing as that you couldn't figure out what was wrong with your own calculation, let me oblige.
5 units per house (assume 0ohms in-house wiring) = 0.4A² * 0.5ohms grid impedance = 80mW, not 3mW. Be careful of your assumptions before you square stuff.
Again, I'm done on this 100MW thing, I will ignore all further questions on it. If you want to talk harmonic PF, filtering, and low power design, please do so.