Here's how it works... let's assume this circuit is for a single 100 watt landing light and the circuit is protected by a 10 amp fuse.  When you turn on the light virtually all of the voltage is dropped across the light bulb because of its very low resistance (maybe 1.5 ohms... 12v / 1.5 ohms = 8 amps).  At this point there's just about a direct short across the LED/resistor part of the circuit in the form of the fuse (something like .1 ohm) so the current takes the path of least resistance...  just about all of it flows through the fuse,  and next to none through the LED and resistor.  So the landing light is dropping 11.99 volts, and there's maybe a 8-9 millivolt drop across the LED/resistor, so there's nowhere near enough current there to light the LED.  Somewhere along the line we were working on our fuse panel and accidentally put a 5 amp fuse in the landing light circuit.  We go to turn on the landing light and bango!, we blow the fuse.  (100 watts / 12 volts = 8 amps... the fuse gives up the ghost at 5, maybe a little more.)  The current path is now through the LED and resistor instead of the fuse, so virtually all of the voltage is dropped by them and there's maybe 15 millivolts left over for the landing light, which is nowhere near enough for it to light up.  The LED, however, is merrily glowing away to let you know that the fuse has failed.  Pretty cool, eh?  There is a catch.  If the fuse is blown or removed _and_ the circuit is powered down, (ie, the switch to the landing light is off) the LED will not glow.  The path through the load must be complete for current to flow through the LED and resistor.  As soon as you flip that switch, though, the LED will light up telling you to chech that fuse and life is good once again.