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.
Current draw of one bulb is nominally 21/12 = 1.75 amps. Ohms law gives resistance of bulb as 12/1.75 = 6.86 ohms. This is the resistance (รท2 as two indicator bulbs) the circuit needs to see to give correct current draw for indicator relay to flash at correct speed. We can ignore any extra low wattage repeaters.
LED bulbs present a
high resistance, leaving a current draw so low as to be almost insignificant in this case. By placing a nominal 7 ohm resistor across the LED (shunting or in parallel with it) most of the current flows through this resistor (it needs to have a suitable power rating typically 50 watts) so the indicator relay is happy as it see's no difference to the presence of filament bulb and the LED is happy as the full 12 volts is across it. Any canbus monitoring is happy as well although not relevant to most classic Morgans.
BTW if you placed the LED directly in the circuit (with or without an extra resistor in series??) it's high resistance would provide a current draw so low the indicator relay would flash very quickly. Similar to the situation when one indicator bulb blows.
See
here for further explanation.
Note most 12 volt LED's do have an internal series resistor to present correct voltage and limit current to LED itself but this is not relevant to my argument.
Phew!!!
