The math is not QUITE that simple... I stumbled across this and wanted to correct it - just the pedantic engineer in me jumping out I guess.
The statement here is strictly true *if* all of the power drawn off the wall made it into the battery, which is absolutely not true. There are losses in the thermal management system, there are losses in the charging electronics, there are probably other losses as well. Let's motivate by example here:
we'll assume, for argument sake, that the thermal management of the battery (the pumps, fans, etc) require 300 watts of power while the car is charging. We can also assume that the charging itself is 90% efficient (which is about what a high quality AC->DC conversion gets you).
So, with that in mind - If you're plugged into a 5-15, 120v, 15A circuit you are limited to 12A of constant current, which means 12A*120v -> 1440W of power "off the wall". If we then lose 300W to the thermal management vampire, you have 1140W remaining. If *that* value is then converted at 90% efficiency into the battery, you're left with 1026W of power into the battery. I don't know what Tesla uses for the "miles per hour" computation, but if it says 3 miles of range per hour that would be ~342Wh/mi. If they say 4 miles of range per hour that's 256Wh/mile. In any case, these numbers pass the smell test.
Now let's look at the same setup but on a 5-20 outlet, still 120v but now 20A nominal and 16A constant. This yields an "off the wall" power of 1920W. Again we'll subtract the 300W for thermal management[1], leaving 1620. Subtract 10% loss, and we have 1458W.
Note that there is a meaningful difference between the "nominal" and "actual" power values that get into the battery...
1458/1026 is a ratio of 1.42 - meaning the 20A circuit actually provides you with a 42% increase in actual charging, rather than the 33% that you'd expect from just the increase in current from 15->20 (or from 12->16 in reality).
Please note that I'm *not* certain about the vampire draw of the thermal management system, nor am I absolutely certain about the charging efficiency. These are educated guesses - the main point is to illustrate that when you have *some* type of fixed, significant overhead it matters more than is obvious to increase the available power.
