Reactance

Presumably you are asking what current will flow.  You can work that out
simply using Z = 1/ (2 * pi * f * C), then I = V / Z

f in Hz, and C in farads, V in volts, I in amps.

Assume that the LED will drop only about 2 volts forward, negligible in
comparison to the 300 odd volts peak from the supply.
The LED will also conduct in reverse, which may not be a good thing.

But the major thing that may happen is that you use a capacitor not
rated to withstand the high voltage spikes ( several kv ) which exist on
most raw mains supplies.  These can cause catastrophic failure of the
cap, and subsequently any components downstream.

"Adrian Jansen"
Roger



** You need to consider the situation at the moment of switch on.

     Big current spike -  then maybe LED no workee.






** The cap needs to be rated for continuous operation on the AC supply.

Only class X or class Y caps are usually suitable.

The problem is not simply from transient voltages because continuous
application of high AC voltages results in internal damage to most film caps
by a process called " corona discharge ".



....   Phil

Thanks all.

If it didn't the capacitor would charge up once, producing a brief flash
from the diode, and then that would be that.

Sylvia.

And blow up the LED. A normal diode put anti parallel to the LED would
probably help. Nice way to find out is quickly running a simulation, like on
http://www.falstad.com/circuit /

Tony