Headlight replacement bulb

Sure it does. You say "unless you added something to reduce the voltage." It's the wire, man; it's the WIRE. You're regulating the battery voltage at the point the regulator senses, then there's cable drop to the bulb itself. Your chart above clearly shows at twice the power you pull twice the current. V=I*R, so at the bulb you have twice the voltage drop between the regulator and the bulb. That's one reason people care about wire gauge. It's a simple voltage divider comprised of the bulb's resistance and the wire resistance.

Since we're touting credentials. I'm an aerospace electrical power system engineer.

 

The bolded above is even more wrong and borders on insanity. With a battery supplying 12 VDC you cannot drop twice the voltage ( 24 VDC ). That dear sir is not possible. If that is what you are saying (and I read it that way from your words).

I feel sorry for our aerospace program.

You are talking about negligible voltage drops on a wire a few feet long. If the wire were dropping as much voltage as you are implying it would heat and melt the insulation.

And if you are adding and excessive load on undersized wire you are doing it wrong because of the increased load the wire will get hot and melt or burn.

That being said the Voltage Potential across the circuit will still be 12 VDC.

Kirchoff's Voltage Law:

Kirchoffs Voltage Law or KVL, states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero. In other words the algebraic sum of all voltages within the loop must be equal to zero. This idea by Kirchoff is known as the Conservation of Energy.

You can argue with me all day long, but you cannot break Kirchoff's law.

 

PLONK !

 

Good for you Nub!

 

V=I*R Well, the R hasn't come into the conversation yet. If it stayed the same, and I doubled, V would double.

However, we get greater power dissipation from an incandescent bulb filament by decreasing R (higher wattage bulbs), so yes, V mostly stays the same as I increases. However, a 100+ watt bulb inline with the average vehicle wiring tends to burn quite a bit more dim than a 60W bulb. That's because as I goes up, the wiring doesn't carry the extra current (I) demanded. We need wiring with a lower R in order to keep V the same. (I have seen 130W H4 bulbs added to 22~18-gauge-wired vehicles and the wires/switches have not even begun to melt, but very little light was produced.)

Vehicle electrical system designers value a balance between light output and system longevity, not to mention manufacturing costs and relay/switch reliability. This is why we get what most people consider a useable amount of light from our headlights with what one might argue is a "barely adequate" wire size. The voltage drop across the wiring is considered, in the drawing room, in order to increase bulb longevity, component reliability and affordability.

For those of us who know we'll be replacing bulbs more often but really want more light, we increase the wire size (lowering the R of the wires) to allow more I to reach our lower-R bulb filaments and less power to be dissipated by the wiring. P=I^2*R - which means power dissipated by wiring with higher R with be greater. Notice our "mostly-regulated" V doesn't even come in to play with that equation.

Long story short, all electrical power going into our system must be transduced into light and heat...larger wire sizes ensure that more of it exits the system as light, and consequently heat, from the bulb than anywhere else. The lower-R bulbs draw more power to begin with; and the lower R wires/switches/relays are there to draw even more system power into the whole system - plus "shift" it to the bulbs by consuming less of it themselves.
EDIT: This is a rewording of