Then, for a 0.047uF cap at 60 hz , that would make

X_{C} = 1/(2*Pi*60Hz*0.000000047 F)

= 56,437.9 ohms

Xc = 56,437.9 ohms

P=I x V> I=P/V

I = 100W120V= 0.8333A

If my logic and math is correct, a capacitor with an

Xc of 56,437.9 would conduct 0.0000177A

I= V/R

= 120V / 56,437.9 Ohms

= 0.0000177A (17.7uA), which , conducting through the filament equivalent resistance of (120V/[(100W/120V)]A)

[(120V/0.833A) = 144 ohms)] V=I x R

= 0.0000177A x 144 Ohms

= 0.0025V. (2.5mV)

I think what I forgot was that for AC the cap has reactance which allows the application of ohm' law to calculate

what , if any ac passes through the RC snubber.