paternoster2012
New Member
Keep the horizontal ones 68K - assuming the previous example worked OK on 110V? (and I presume it did?) then all you're doing is duplicating it with slightly higher resistors (but the same ratio).
If I did that, and given a reverse polarity outlet. Wouldn't L1 remain unlit?
(R2 / (R1 + R2)) * V1 = V2
(68k / (150k + 68k)) * 230v = 72v
It was stated earlier that neons require 90v to strike.
With 150k resistors...
(R2 / (R1 + R2)) * V1 = V2
(150k / (150k + 150k)) * 230v = 115v
This would enable the neon bulb to strike, thus signifying a reverse polarity?
It's quite the balancing act it seems