by Klaas Robers » Mon Sep 27, 2010 9:13 pm
Three batteries in series gives 4.5 volts. That is more or less the voltage that white (= blue) LEDs need to illuminate at their full brightness. So I fear there is no resistor.
With 19 LEDs in parallel the nominal current (20 mA per LED) will be about 0.4 A. At that current the voltage of the small AAA batteries will drop due to the internal resistance. So the internal resistance of the batteries is used to limit the current. If the batteries get more and more exhausted the light output will dim.
In the older flashlights a 3.8 V incandescent lamp was used to run on a 4.5 V battery. I always was puzzeled where the remaining 0.7 volt was gone. Well, in the internal resistance of the batteries.