In parallel 1/R = 1/R1 + 1/R2. If the LED has a lot less resistance than the tungsten, you will draw much more current this way.
I'm a Signal Engineer, not a Shunter!
Correct formula but not LED characteristic. You need to consider this Bob. For the common electro-mechanical indicator relay to flash normally it needs to see the current draw of two 21 watt bulbs = 42 watts - they often have this wattage marked on them.
Thank you, Richard. What I didn't realise was that LEDs have a high resistance. I assumed that it was very low resistance in one direction and high in the other. Obviously, with this high resistance, you need a bypass resistor, not a series one. Cheers.
