On Nov 15, 7:43*am, klem kedidelhopper
wrote:
I think I know this but I'd like to be sure. I need to affix a small
neon indicator on to a 240V line. Would 80K be an appropriate resistor
to use? Thanks, Lenny

bare bulb?

assume neon turns on at 90V and extinguishes around 60V.

Sinewave are 'so fast' just assume the neon is like a bidirectional
zener of 60V:

First what power idssipates if there were NO neon bulb:
to find R try 240*240/R=Power
therefore, R=240*240/Power
rule of thumb for resistors is use half their rating, so change
formula to
R=2*240*240/Power
so a 1/4W means R = 460k, close value 470k 5%
and 1/2W means R = 230k, close value 220k 5%

Now let's 'clip' the voltage that goes to the neon:
as a simplistic estimate, 240-60=190, and try again
R=2*190*190/Power
1/4W R = 289k
1/2W R = 144k

use 1/2W 220k, looks like won't hurt anything.

just as a check, how much power is going into the neon?
current is (240-60)/220k = 0.86 mA
power into the neon bulb is 60*0.86mA = approx 50 mW, so neon is not
likely to burn up.

Plus, dopn't those GFI outlets limit AC current to something like less
than 1 mA, so even if you get your fingers in there you're not likely
to be killed.