Using Ohm's law, 55 watts at 14.8volts is 3.716 amps. If you disconnect everything from the alternator except one headlight and you spin the engine up enough to get to 14.8v, somewhere between 600-1200 depending on alternator and pulley, the load would be 55 watts. The amount of current is totally dependent on the load. The alternator does not supply current (Amps), it puts out voltage, around 14.8v in GMs case. seeing how you would be putting less amperage through it. On a side note, you would see less wear/tear through the electrical system. Now if you wanted to decrease the parasitic loss by putting on a bigger pulley on the alternator, you would decrease the output of the alternator, and spin the alternator at a slower speed, putting less drag on the pulley system, but you would also be increasing the rotational weight of the pulley. The only way you can change the amp output of your alternator is either by overdriving it (by putting a smaller pulley on it, which will greatly increase wear/tear), or by re-winding the coils with thicker wire. whether or not you use all 130 amps won't vary the speed or how much voltage gets put into/out of the alternator. It's not like the truck's computer is telling your alternator it needs X amount of amps and the alternator puts that out. You can't change the amperage at a flick of a switch. varying on what package/options you got). When your engine reaches ~1000 rpms, it should be putting out it's maximum amperage (IIRC most dmax's come with 130 amp alternators.

The armature (attached to the belt pulley) spins inside the coils (IIRC there are 3 in most alternators), and when the coils are energized you get electricity. You have two basic working parts in the alternator (I'm generalizing this A LOT, there are actually more than two parts, but there are two main parts). You have a good general idea of how it would up until the most important part, the alternator. Click to expand.Close but no, it will not.

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