When you dim the bulb you burn off extra heat through a resistor. Not too many ways to do it really... High power potentiometer and resistive load. Kind of like two garden hoses on one spigget when you open the other hose the more you open it the less the water flows out the other side....
Agreed, the resistor burns off power, and creates heat. Heat or work must be produced, the energy cannot dissappear (conservation of energy. - Wiki "The only thing that can happen to energy in a closed system is that it can change form, for instance chemical energy can become kinetic energy.) Translation - It either produces heat or does work. It cannot dissappear.
But the lumateks do not get hotter when dimmed, I have 2, I know. So if there is not extra heat, then they are stepping down the power. I will look for the post where someone took a killawatt to a lumatek and shows that it decreases the amount of power drawn from the circut.
Think about it, the electronic ballasts already have a complex circut to use power inputs from 90 to over 240. Not much harder to put in a circut that after igniting simply lowers the output power. Im no electrical engineer and if im incorrect in my knowledge please correct me. :)
