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daPoseidonGuy

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  1. Hi, i was doing a science project on LEDs, and i needed to calculate optical output power versus input electrical power to find relative wall plug efficiency. This is the only good formula i found Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts) Pout = N × Vres What is N exactly? I do not know how to calculate optical output power as I dont know what help is. Id appreciate help a lot, thanks. All the values I have are experimental. Im trying to calculate the relative wall plug efficiency of an LED. heres my data right now. I do not have a data sheet and the steps Im following at this point are somewhat modeled of those shown here: http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p003.shtml#procedure in the testing and data collection section. Voltage across resistor (V) ± .01 = 2.49 Distance from photocell to light (cm) ± .05 = 4.00 Voltage across light (V) ± .01 = 5.75 Current intensity (mA) ± .01 = 360 If N is just something I have to leave as a variable, then what is the point of measuring the distance between the photocell and the light bulb? In the experiment I moved the breadboard closer or further to get 2.5v across the resistor, as thats what I understood from the procedure. Was I supposed to do that? What do I do with the value for the distance between the photocell and light bulb? This project comes from here btw: http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p003.shtml#procedure
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