Originally Posted by Carbon
If the lamps were to have the same resistance in each state, the power to the lamps would be 12.5% compared to powering the lamps in parallel. Since the bulb resistance is lower when the filaments are cooler, the power consumption will be more than 12.5%. Maybe it could be 50%-- if the cooler DRL-mode filaments have 25% as much resistance as the full-brightess filament.
So two 2 ohm filaments across 12 volts draw 72 watts. In series, two identical bulbs see 6 volts each. So to generate 18 watts each or 72 watts total, each filament would have its resistance down to 0.5 ohms.
After looking up the Voltage v Current curve for a typical incandescent bulb, I see that there was an error in my thinking. Using the curve for a typical bulb the current at 50% voltage is about 80%. With that in mind, at full voltage (12.8Vdc) 2 65W (9005) bulbs consume 130W total (about 5.08A each). At 50% voltage the total current is about 4.06A or 52W total.
So the DRLs consume about 40% (52/130) of full power, not the 50% I previously thought. I do not have a curve for the exact bulb so there are some assumptions being made.
If they were 2 purely resistive loads, 65W/12.8V=5.08A, 12.8V/5.08A=2.52 ohms in each branch. In series the total resistance is 2.52*2=5.04 ohms 12.8V/5.04ohms=2.54A, 12.8V*2.54A=32.5W or 25% of full power total.