Yea. Don't just chop the end off. My physics teacher in high school didn't even understand this concept so he taught it wrong, then told me I was wrong when I tried to correct him.
I'm leaving room for error because standard
voltage in the US is roughly 110v-125v, but each
outlet in a different area of a building and in a different part of the
grid will be different, so exact calculations are a best-cast scenario.
Each light
bulb on a standard string of 50 drops ~2.5v of the ~120v. Often you'll find multiples of 50 for the number in string based on the following diagram.
Each section of 50 bridges between a hot
wire and a
neutral wire. No matter what happens, 120v is going through that section when the lights are turned on. In a simple world, one light goes out in a section, all go out because the gap between the hot
wire and the
neutral wire is no longer bridged. Then, some excellent designers came up with a way for the filaments to fail such that no matter what, that entire section of the string wouldn't turn off. So if one 2.5v
bulb on a string fails, those 2.5v then get diffused across the remaining 49 bulbs because 120v still needs to get from the hot
wire to the
neutral.
For the purposes of this next paragraph, we'll assume the
voltage present is 125v which is divided equally among all light bulbs that work.
When one
bulb goes out, that's not a problem because that means the rest of the bulbs are still getting (125v/49bulbs) = 2.55v each. Usually this is fine. But let's say you leave that string out for a while and another seven bulbs fail. Now it's (125v/42bulbs) = 2.97v. That's almost 3v for filaments that are designed for only 2.5v. Eventually each
filament will
drop so many volts that they will become brighter and hotter. As that happens, more filaments fail from the heat.
This leads us to a really cool scenario. Eventually so many bulbs will fail that as each one fails, the rest of the string becomes noticeably brighter. Again, brighter = hotter, therefore as each one fails, the chance of another failing grows. So over the course of a few minutes you could watch almost an entire string of lights become very bright as more fail until eventually all of the filaments have burned out. (a lot of people don't believe this until they see it)
In your case, cutting the string down takes two steps. First, cut out two of the three sections of 50 lamps. Refer to the diagram to figure out what I mean.
Now just chopping the other 20 bulbs off would result in more
voltage being dropped across each of the filaments than they can handle. To prevent the bulbs from burning out, you would need to do one of two things -- you can
drop the source
voltage down in 2.5v decrements to be proportionate to the number of light bulbs that remain, or you can find a way to
drop the rest of the volts in the
system. In this case, the number of light bulbs removed is 20. At 2.5v each, you need to
drop 50v across a resistance.
However, if I'm not mistaken, we need more information to determine the resistances. Your 3amps is either a mistake or your idea of standard Christmas lights is different than mine. That would mean one
strand of 150 lights would produce (120v*3a) 360watts, and that is very, very high. A standard string of mini-bulbs would produce about 75w -- about 25w per 50 lamps. Assuming this information to be correct (source:
HowStuffWorks "Christmas Lights Power Consumption"), let's continue.
25w/120v = .2083a
125v/.2038a = 576.09ohms
576.09ohms/50bulbs = 11.52ohms/
bulb
11.52ohms*20bulbs = 230.4ohms
You would need a 230ohm resistance in
line with your 30 light bulbs, and it would need to be rated for at least 10watts. There's a number of ways you could do that, including just putting (20) 11.5ohm resistors (with a
rating of a minimum of 1/2w each, but you'd probably want them to be rated for 1w in this application) in series.
[Please don't actually use those numbers unless someone takes the time to check that they are accurate. It's entirely possible that somewhere along the way I made an error. I cross-referenced some data with that HWF article, so they look to be about accurate, but I make no guarantees.]
For all
practical purposes, you're best off using a 50-bulb
strand and finding a way to hide the other 20 lamps. Otherwise cut it down to 30 bulbs total and vary the source
voltage appropriately as Lotos described. Using resistors to do what you want to would be a PITA. It would work, but it would be tedious to find the right components and solder them in, plus then you have to pack the entire rig so that no one can come in contact with conductive materials.
Also remember if you go the route of varying the source
voltage, don't forget to
mark your string of lights for future reference NOT to be plugged into a 120v wall
receptacle. They'll be very bright for a short moment before every single
filament goes supernova and the entire string is dead.