DJ_Dexinter wrote: Install the filtration condenser. Should be OK. 100nF or more at >300V
Firstly, the voltage of the capacitor must be at least 400V (if there is no abbreviation AC, i.e. the given value of 400V is the instantaneous value). This follows from a simple calculation:
230V AC in our mains is rms value and the instantaneous value is 230 x 1.41 (root of 2) = 330V. Taking into account mains voltage fluctuations, 400V is assumed for safety reasons.
If there is an addition AC (e.g. 250V AC), it means that the capacitor withstands 250V rms value (i.e. about 400V instantaneous value). Such markings are most often used for capacitors of the X2 type (so-called across), which are constructed as typical filtering systems at the network input.
Added after 2 [minutes]: By the way, a helpful post in solving problems with LED flashing after power off.
Added after 12 [minutes]: tomsurm wrote: I have a problem. As you wrote with an ash capacitor (10 micro F, 400 V) parallel to the bulb. Before turning on the power, the LED bulbs actually stopped glowing. However, after turning on the power, after a few seconds, the capacitor burned, emitting a lot of smoke. I immediately say that I have kept the polarity (the longer leg of the capacitor is plus, the shorter one is minus). I connected the capacitor in the switch box (about 10 meters before the light bulb).
Dales 10uF parallel to the grid? No wonder you boiled it in 3 seconds. Do you know what the current flowed?
The guys wrote about 100nF or 0.1uF and you were 100 times bigger. For such a capacity, the lights must have dimmed throughout the house before the capacitor exploded :)
Second thing, what does "I kept polarity" mean? Plus to phase and minus to zero or vice versa? Anyway, it doesn't matter. in both cases, the electrolyte will not last long. It will evaporate in an instant.
Don't do any more experiments like this because it will end badly. And so much detail...