Originally Posted by egads
Ignore this advise.
LED lamps have a high in rush of current when they first light. It might be possible to use a fuse the right size that is slow to blow, allowing an in rush but still will blow if the continuous load exceeds 10 amps.
With all due respect, LEDs do NOT have any higher than the normal running current when starting to light.
Wire a LED bulb and a dropping resistor in series with the resistor going to ground.
Connect a dual trace oscilloscope as follows;
1. Connect the ground lead to the resistor ground lead
2. Connect one scope probe to the positive lead of the diode
3. Connect the other scope probe to the negative lead of the diode.
Now apply a positive low frequency square wave to the circuit and compare the two square wave traces. You will see that the negative diode connected trace follows the input square wave trace. The only difference you will see is the voltage drop across the LED.
The LED and resistor have no capacitive or inductive loading of the input signal and therefore can't cause any change in the voltage rise (causing current changes and/or phase shift).
Now if you were to connect a high value capacitor across the resistance you would see an inrush of current as the capacitor charges. And it might be enough current to blow the LED too.
A normal incandescent light bub does have a start up change in current but that is due to the bulb filament changing resistance as it heats up. The resistance is much lower when cold compared to being lit for a few seconds.
I suspect Farmerjosh just overload the 10A fuse with the final installation of more lighting. Try measuring the current without the last addition across the blown fuse to see what the current is with all the other changes are drawing.