https://www.youtube.com/watch?v=bHIhgxav9LY
Cheat: Assuming a 12 V & 3 W LED bulb which draws 0.25 A,
and a 1.5 sq mm/ AWG 16 lead (having an inherent voltage drop of 6 V /km),
the applied battery voltage would only have to be 3 600 012 V [or 3.600012 MV].
Have a nice weekend, all! :idea:
Johan

