As a first cut, power loss in the charger goes as I^2 R, where I is the current and R is the wire resistance.
The wire resistance is not a function of current in general. The power the battery gets is IV, where I is the current and V is the voltage.
For a given charging rate, higher voltage requires lower current. So if you’re charging at 11 kw/120 V, the current required is twice as much as 11 kw/240 V. So the I^2 R losses go down by a factor of 4.
Without knowing voltages/currents on the 11 kw and 6 kw chargers, I couldn’t answer for certain. But it’s certainly possible for the 11 kw to be more efficient.
As a first cut, power loss in the charger goes as I^2 R, where I is the current and R is the wire resistance.
The wire resistance is not a function of current in general. The power the battery gets is IV, where I is the current and V is the voltage.
For a given charging rate, higher voltage requires lower current. So if you’re charging at 11 kw/120 V, the current required is twice as much as 11 kw/240 V. So the I^2 R losses go down by a factor of 4.
Without knowing voltages/currents on the 11 kw and 6 kw chargers, I couldn’t answer for certain. But it’s certainly possible for the 11 kw to be more efficient.