Aboutcleggsan Expertise All technical areas of Electronics Engineering.
Experience BSEE, MBA, Design, R&D, University Research.
Senior Life Member of IEEE. Life Fellow of AES.
Organizations IEEE, Consumer Electronics Society, Audio Engineering Society.
Broad teaching experience; work experience mostly in consumer electronics and conversion from analog to digital technologies. Pioneer in digital audio at all levels.
Question I have a self-made wind turbine (daughter's project) with a 12VDC 10amp motor/generator attached to a resisitve load of 3V - 300mA. Have taken alot of readings to calculate mean power output at given speeds. For example, at 6 m/s mean v=3 and mean amp=3.5. Is the power it is producing simply p=v*i and how does changing the load effect determining the power output?
Answer Resistive load would be in ohms. Is the load a 10 ohm resistor? (E = I*R, or R = E/I = 3v/.3a = 10 ohms)
If the load is a fixed resistor, then increasing the rotational speed will increase the voltage (and the power) output and decreasing the speed will decrease the voltage (and power).
Power output it calculated by P = E*E/R or voltage squared divided by R. At 3v out, for example, the power into a 10 ohm load would be 3*3/10 = .9 watts. If speed increased to produce 4 volts out, then power out is 4*4/10 = 16/10 = 1.6 watts. Conversely, if voltage drops to 2 volts, power out is 2*2/10 = 4/10 = .4 watts.
Does this help? If I did not understand the scenario properly, please let me know.
