Let you know what? The answers to my questions? It's math, using expertise relating to the topic, which I was seeking. Others understood, and answered effectively.U have it in ur inventory right?
so why dont U try it, and let us know, U got the bulb![]()
Got it. Thanks, Chief.ALNims will be correct. Realize that you are dealing with an impedance not fixed resistive # of ohms. The filament exhibits less glow at lower current and voltage and it is not linear. Lower temperature means higher impedance with less thermal activity. Chief
Based on a constant resistance the power would be 1/4 with 1/2 voltage, but the resistance of an incandescent bulb changes with current due to temperature changes in the bulb. That makes calculations more difficult.1/2 the voltage will be 1/4 the wattage. In addition the amount of light from an incandescent bulb is not linear. I don't think you can easily calculate the amount of light with reduced voltage. Just get the appropriate bulb unless you are using it for a tiny heater.