13.8v into 25 ohms is 0.552 amps but the resistor will only dissipate 7.6 watts. I don't know where you got your 12 watts from.
OOPS! Your correct...I was just working on another project which uses a 12.5 ohm solenoid...got my numbers mixed up
The idea idea came from how the computer knows whats a good or bad solenoid...it measures current draw...which for a 25 ohm solenoid would be anywhere from .48 to .552A (thanks again for the correction) at 12-13.8vdc (car voltage)
So instead of connecting 25 ohm resistors and having to dissipate 7.6 watts each...(keep in mind I might need to trick 1 other similar solenoids which makes it almost 23 watts total
) I would replace the solenoids with 2 of those voltage regulators in parallel to give me 2A output. Then use two (possibly three) 1/4 watt, .4 ohm resistors. When the computer grounds the new resistors it will still read a .5A current draw but each resistor will only dissipate .1 watts.
It makes perfect sense but while the resistors are not dissipating much I know that these voltage regulators are. I don't know enough to be able to take all the graphs in the data sheet and figure it out.
In operation all three solenoids (resistors) would never be on at the same time....either one,two or none but never all three. I could probably get away with a single voltage regulator but 2 in parallel helps spread the heat and if for some reason the computer malfunctioned and all three were on it can't hurt. So normally there would be a requirement for only .5A with one resistors or 1A for two to be shared between two regulators (or a max of about .5A each)
There must also be some heat involved with the switching of 12Vdc down to .2Vdc but again I dont know enough to be able to figure that out..any help appreciated!