"Bill Jeffrey" wrote in message ...
Chuck wrote:

Just count the number of bulbs that go out when you remove a bulb from a
working string...
then divide 120 volts (if in North America) by the number of bulbs in
each section,

10 bulbs would be 12 volt bulbs
20 bulbs would be 6 volt bulbs
50 bulbs or so would be 2.5 volt bulbs
These are nominal voltages, Some strings use a few more bulbs and run not
as bright.

Now that was a good common-sense troubleshooting response. Right to the
point. Excellent!

Good answer, but that wasn't his question. If I read the OP correctly,
he has a bulb in his hand (not in a string) and doesn't know which
string it goes into. In other words, he doesn't know how many volts it
takes to light this particular bulb.

Bill

Bill, you're right, and thanks: I have a bulb in my hand, and want to know the proper voltage it requires.

I have measured resistance, and get a reading of 1 (full resistance) for a burnt bulb.

Working bulbs have varying resistance, depending on voltage, I guess. My meter is set at 200 on the OHM scale (selections from 2000K to 200).
Results:
1.7 or 1.6 for bulb from the 2.5v set
3.5 for unknown volt bulb
2.2 " " " "
and
20.2 for a motorized ornament of unmarked voltage which plugs into a socket of the 2.5v set.
This last may be either bad or good for the other bulbs; if the motor draws more or less voltage. My guess is it draws less voltage, thus is bad for the other bulbs.

The other info is interesting, too, i.e. Ray's idea. And I now understand why not replacing burnt bulbs is hard on the working bulbs - they receive more voltage, burn brighter, and thus burn out faster... if I infer correctly.

Thank you all. This group is great, and has also helped me before.

Dugie