4 ms·
Power is volts x amps. The more amps, the thicker the wires need to be (volts are irrelvant). So to minise weight/wire size, use the highest voltage possible (
by zh3 5y ago
Power is volts x amps. The more amps, the thicker the wires need to be (volts are irrelvant).
So to minise weight/wire size, use the highest voltage possible (and thus the lowest current). Losses are purely related to current (I^2*R) so the incentive is to squeezee the current (so needing to increases the voltage).
There's a reason the high-voltage overheads are 432 kilovolts (or more); 10 amps at 432kV = 4.3MW (MegaWatts) while 10amps from a stock AC outlet is 1.2KW (KiloWatts). The wire thickness required for both is the same (though the HV wires need to be better insulated, out of the way of crazy fools etc).
So a 60V system for a car carries 1/5 the current of a 12v system, and the wires can have 1/5 the cross-sectional area.
(yes, I know there are caveats when it comes to AC).
- a9h74j 5y agoRecent TIL: Wire guage follows a 10log10 of ohms per 1000ft, relative to 0.1 ohms per 1000ft: 0ga .. 0.1 ohms per 1000 ft 10ga .. 1.0 ohms per 1000 ft 20ga .. 10 ohms per 1000 ft 30ga .. 100 ohms per 1000 ft http://www.interfacebus.com/AWG-table-of-different-wire-gauge-resistance.html http://www.interfacebus.com/AWG-table-of-different-wire-gaug...
- a1369209993 5y ago> The more amps, the thicker the wires need to be (volts are irrelvant). Nitpick: the wire as a whole includes insulation, which technically does need to be thicker at high voltage (though at 40-60V, and maybe even at 40-60kV, it's probably dominated by tolerances for erosion and abrasion and such). High voltage power lines use air, which is actually a fairly good insulator per se, but has the problem that wires tend to pass through it on their way to a short circuit if not well-restrained.