5 ms·

Watts = Amps * Volts (or, more scientifically, P = I * V) If something draws 48 Watts of power, you can either supply it with 4 Amps @ 12 Volt, or 1 Amp @ 48 V

Watts = Amps * Volts (or, more scientifically, P = I * V) If something draws 48 Watts of power, you can either supply it with 4 Amps @ 12 Volt, or 1 Amp @ 48 Volt. It's the amps that determine how thick the wiring needs to be, so by lowering the amperage, you lower the amount of wiring needed. I believe the only reason cars are 12 V is because that's a practical voltage to build lead-acid batteries at (which is actually closer to 14 V in practice). Early car electric systems were only 6 V back when the only accessories were the front lights and the horn, but as more stuff was added on cars moved to 12 V.

ace2358 3y ago
Things like stator motors were also probably quite difficult to do at 6v. I know the older Honda CT110 postie bikes came in 6v in the old days. That was a kick start.
grepfru_it 3y ago
Car electrical systems typically run at 13.8-14.4V while the alternator is charging and drops slightly after the battery is topped off. When the engine is off the battery regulates a 12v output
loufe 3y ago
Resistance is proportional to the square of current. This would mean 1/16th the loss instead of 1/4th.
- 05 3y ago
  Nope. Consumed power is voltage*amperage, and it’s constant, so 4x more voltage means 1/4 the amps. Wire heating is I squared R, so with both I reduced and R increased the losses are just 1/4, not 1/16 of the 12V system. Because you wouldn’t keep the same wire gauges you’ve had in the 12V system, you’d reduce them to actually have some benefits from conversion.
- ahepp 3y ago
  Isn't the resistance of a wire usually modeled as a constant value? Power dissipated in the wire as heat is proportional to the square of the current. It's equal to the voltage drop times current (P = I * V). Voltage drop is current times resistance (V = I * R). So P = I^2 * R.
- deleted 3y ago
  [deleted]
LoganDark 3y ago
This makes such little sense to me. I'm probably just stupid, but if the same amount of power is flowing through the wire, don't you need the same size of wire? A higher voltage system will push more energy through the same component, so you need thicker wires. If you up the resistance of the component to consume the same amount of energy as before, you arrive at the same size of wire you were using before, don't you? What exactly is magical about higher voltages that makes them suddenly able to carry more power across the exact same wire? I know you just said it's "lower amperage" but I don't understand how amperage can be the only thing dictating the required size of a wire.
- hansvm 3y ago
  A higher voltage system enables the same energy at a lower amperage, so constrained to wanting to do a particular thing (loud stereo, bright headlights, ...), higher voltages result in lower amperages. I'll opt to not try to explain that intuitively, hoping that somebody else chimes in. For the other half, "why do amps determine wire size?" Amps are coulombs per second -- how much electrical stuff moves through a cross section in a second. The amount of heat produced is directly proportional to amperage (every bit of movement has a chance to whack into something and produce heat, and more movement per second or more seconds generates more heat). Wire size is ultimately constrained by being able to release that heat while still acting like a wire (not melting, not sublimating, still conducting, not setting anything else on fire, ...). The amount of heat released is proportional to surface area (proportional to the square root of diameter, but that difference doesn't matter a ton right now), and the amount of heat produced is proportional to amperage and approximately nothing else. The point where those two terms are equal is the limit for the system (engineers add in huge safety factors to account for conduit and insulation and other imperfections), above which your wire melts and below which it behaves reasonably. Higher amperages require higher surface area to release the extra heat (so much higher thicknesses), so a given amount of power (stuff you want your electrical system to do) requires lower amps or thicker wires to work appropriately, because you can achieve the same power by increasing the voltage on the same amperage, thereby solving your power problems without extra heat.
  - hansvm 3y ago

ace2358 3y ago

Things like stator motors were also probably quite difficult to do at 6v. I know the older Honda CT110 postie bikes came in 6v in the old days. That was a kick start.

grepfru_it 3y ago

Car electrical systems typically run at 13.8-14.4V while the alternator is charging and drops slightly after the battery is topped off. When the engine is off the battery regulates a 12v output

loufe 3y ago

Resistance is proportional to the square of current. This would mean 1/16th the loss instead of 1/4th.

05 3y ago
Nope. Consumed power is voltage*amperage, and it’s constant, so 4x more voltage means 1/4 the amps. Wire heating is I squared R, so with both I reduced and R increased the losses are just 1/4, not 1/16 of the 12V system. Because you wouldn’t keep the same wire gauges you’ve had in the 12V system, you’d reduce them to actually have some benefits from conversion.
ahepp 3y ago
Isn't the resistance of a wire usually modeled as a constant value? Power dissipated in the wire as heat is proportional to the square of the current. It's equal to the voltage drop times current (P = I * V). Voltage drop is current times resistance (V = I * R). So P = I^2 * R.
deleted 3y ago
[deleted]

LoganDark 3y ago

This makes such little sense to me. I'm probably just stupid, but if the same amount of power is flowing through the wire, don't you need the same size of wire? A higher voltage system will push more energy through the same component, so you need thicker wires. If you up the resistance of the component to consume the same amount of energy as before, you arrive at the same size of wire you were using before, don't you? What exactly is magical about higher voltages that makes them suddenly able to carry more power across the exact same wire? I know you just said it's "lower amperage" but I don't understand how amperage can be the only thing dictating the required size of a wire.

hansvm 3y ago
A higher voltage system enables the same energy at a lower amperage, so constrained to wanting to do a particular thing (loud stereo, bright headlights, ...), higher voltages result in lower amperages. I'll opt to not try to explain that intuitively, hoping that somebody else chimes in. For the other half, "why do amps determine wire size?" Amps are coulombs per second -- how much electrical stuff moves through a cross section in a second. The amount of heat produced is directly proportional to amperage (every bit of movement has a chance to whack into something and produce heat, and more movement per second or more seconds generates more heat). Wire size is ultimately constrained by being able to release that heat while still acting like a wire (not melting, not sublimating, still conducting, not setting anything else on fire, ...). The amount of heat released is proportional to surface area (proportional to the square root of diameter, but that difference doesn't matter a ton right now), and the amount of heat produced is proportional to amperage and approximately nothing else. The point where those two terms are equal is the limit for the system (engineers add in huge safety factors to account for conduit and insulation and other imperfections), above which your wire melts and below which it behaves reasonably. Higher amperages require higher surface area to release the extra heat (so much higher thicknesses), so a given amount of power (stuff you want your electrical system to do) requires lower amps or thicker wires to work appropriately, because you can achieve the same power by increasing the voltage on the same amperage, thereby solving your power problems without extra heat.
- hansvm 3y ago