The current or LED can be varied. In my case, I assume 10mA.
For 240V AC, to get around 20mA -> 240 / 10m = 22K ohm and
power dissipated on resistor will be around 3W. The resistor will run
hot. Usually a capacitor can be used as a resistor.
150 nF/400V will work fine. The capacitor should be non-polarised.
LED needs to be protected from large reverse voltages, another LED
in the reverse direction might just work. Or just put a diode like 1n4007.
Using capacitor only, will have a problem with inrush current during
switching and it is preferable to have a resistor in series with the capacitor.
A 1K or 2K resistor can be connected in series to the capacitor to reduce
the maximum inrush current to 240mA with 1K ohm or 120mA with 2K ohm.
After turning off, we also like to have the capacitor to discharge quickly. Any
voltage residue on the capacitor may cause shock. Therefore, a resistor is
usually installed in parallel with the capacitor. A 220k ohm resistor in parallel
with the capacitor should remove the 70% of residue in 35msec.
Some links to such a designs:
I downloaded their schematic and put here for access in case their site goes offline:
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