Electrical Losses (outside the motor)
Electrical losses scale as P = I^2 * R (
source). Note how current has an exponential contribution! And faster acceleration requires higher current.
Very happy to get an opinion from the resident physicists on this little point!
The best my non-physicist brain can come up with is as follows:
For the following, let's disregard all losses with the exception of a randomly chosen parasitic resistance of say 0.1 ohm in the power circuit. This also means that constant speed requires zero energy, so we are only looking at acceleration.
From the above observation (P = I^2 * R), it is valid to say that if you double the power, say from 100kW to 200kW, then for a constant voltage, you also double the current, so the resistive losses increase four-fold.
To pick an example with easy hypothetical numbers given P=EI, a 400V battery will supply
100kW at 250A, and
200kW at 500A.
If the total parasitic resistance is 0.1 ohm, then resistive loss (I^2 * R) is
250^2 *0.1 = 6.25kW at 100kW, and
500^2 *0.1 = 25kW at 200kW (a four-fold increase, as previously noted)
However! To get the car to the same desired speed (say 60 mph), we only need to supply 200kW for half the time (kinetic energy = power x time) - both cars end up with the same kinetic energy at 60 mph, so doubling the power will halve the time.
If 100kW gets us from 0 to 60 in say 10 seconds, then 200kW gets us from 0 to 60 in 5 seconds
Now we're actually interested in energy consumed (power x time), which we normally measure in watt-hours (Wh), but for simple maths, lets use units of kilowatt-seconds (kWs).
0-60 at 100kW consumes 100x10 = 1000kWs, with a resistive loss of 6.25x10 = 62.5kWs (6.25%)
0-60 at 200kW consumes 200x5 = 1000kWs, with a resistive loss of 25x5 = 125kWs (12.5%)
This is, of course, slightly inaccurate because not all our energy actually made it to the motor.
However, the approximate conclusion I'm trying to reach is that our resistive loss has actually doubled (rather than quadrupled), for the same energy delivery. Whether the actual value is significant to the topic is another question. Probably not, because my hypothetical 125kWs translates to a mere 0.035kWh.