Those blue led gets very bright. I suggest to start with a 200K resistor in front of the led. I remember someone on this forum install two leds - one on each rail - just so that it "looks balanced electrically".
This helps you when you play with resistor values.
Is a protection if you decide to put the led in parallel with a resistor in a PI filter. There , the the turn on voltage values may go as high as 6 or 7 volts
and so the led is protected.
I have the diode bridges between the transformers and
the PS capacitors for my BOZ and Zen V2.
I just read the reference AC supply voltage at the bridge position,
e.g. 24V AC in case of my Zen V2. Meanwhile, my blue
LED lightens lovely when about 5mA currents flow thru it.
Therefore, I calculate the necessary resistor value in this way:
24 x 1.414 / 0.005 = 6787, i.e. about 7.5K ohms (watts?
24 x 1.414 x 0.005 = 0.17w).
In the same way,
if the AC supply is 30V, R of about 8.5K and
if 40V, R of about 11.5K and
if 50V, R of about 14K and
if 60V, R of about 17K...
If we know the AC supply value and the necessary current thru
LED for a lovely light, we can calculate the necessary resistor
easily. Try it.
Calculate power of the resistor =
[(power supply voltage) - (LED voltage)] * (LED current)
Double this (at least) to get a power rating for the resistor. You can put a bunch of resistors in series to use smaller-power resistors.
LED voltage, current are about 3V, 5 mA (look up the specs of the LED). Tweak resistor value to adjust brightness.