[morzh]

Terry

Here's how you do it.

You have 120V.
In order to drop what you need you do not care how much you drop, you care to keep the current 300mA, this way you keep your voltage 18V and drop the rest.

So

1. your current is your V over Z, where Z is total impedance.
I=V/Z=120/Z
I= 300mA= 0.3A
Z=V/I=120/0.3=400 Ohm


2. Express Z via Rc (capacitive impedance) and the R active (filament).
Z=SQRT(R*R + Rc*Rc) =400

3. Your R active
R= Vf/0.3= 18V/0.3A = 60 Ohm

4. Find cap impedance
R*R + Rc*Rc= Z*Z= 400*400= 160000
Rc*Rc=160000-3600=156400
Rc=395

5. Find Capacitance
Rc=1/(6.28*f*C)
C=1/(Rc*60*6.28 )=1/(395*60*6.28 )=6.71*e-6 F= 6.71 uF