I need produce an 18 light strand of series connected light, based on a 100 light 2P50S system (2 parallel strands each of 50 series connected lights).

Based on the wiring, I believe the strand is engineered as two 50 light strands in parallel. The strand is rated for 120 V and consumes 40.8 Watts at 0.34 amps.
That is each strand of 50 lamps in series has a voltage per lamp of 120/50 = 2.4V / lamp, and each lamp draws 0.34A / 2 = 0.17A.

When I shorten the strand I need to adjust for the lost resistance from the bulbs that have been cut away.
ie I will now have a 18 x 2.4V = 43 volt string drawing 0.17A.

If I did my calculations right it seems I would need about 450 Ohms of resistance.

  • Resistance per lamp = V/I = 2.4/0.17 = 14 Ohms.
  • Lamps to substitute for = 50-18 = 32 lamps.
  • Equivalent resistor = 32 x 14 ohms = 448 Ohms.

My problem is I do not know what part to order to achieve this level of resistance and I am not sure if the heat put out by that resistor is realistic to deal with.

Question is, how should I best engineer this adjustment and what parts should I order to construct the solution?

  • \$\begingroup\$ But less lights will consume less power. An analogy: Do you add a resistor in your house each time you turn off a light? \$\endgroup\$
    – Bageletas
    Nov 29, 2016 at 4:52
  • 6
    \$\begingroup\$ The lights in his house are not connected in series. THey are connected in parallel. This is a completely different case. Don't confuse the issue. Mr. Stickley needs at least a 25-30 watt resistor minimum to do this. It is a terribly inefficient (and quite potentially dangerous) scheme. Much better to use a 20-bulb string and hide a couple of bulbs. Or operate the string on a 48v transformer, etc. We don't know what is the application, but I certainly would NOT put a 30-40 Watt heater anywhere near flammable decorations. Terrible idea. \$\endgroup\$ Nov 29, 2016 at 4:55
  • \$\begingroup\$ @Richard Crowley I agree with the general point of the safety. I misread it as 2 strands with 50 parallel bulbs and thought it could be reduced a lot easier. Suggest the wording change to "2 parallel strands each with 50 bulbs in series" if that's the case. \$\endgroup\$
    – Bageletas
    Nov 29, 2016 at 5:10
  • \$\begingroup\$ Whatever resistor you end up adding, I recommend using a couple in series to improve their max voltage rating. Costs close to nothing and is a bit more work, but its way safer. \$\endgroup\$
    – Wesley Lee
    Nov 29, 2016 at 12:54
  • \$\begingroup\$ It is NOT a repair question, so I am inclined to reopen. Given the time that has passed, not sure if that helps the OP at all. \$\endgroup\$
    – user105652
    Jul 12, 2020 at 2:09

2 Answers 2


I remember those lights from when I was a child, long ago. The obvious way to wire multiple lights is in parallel. This will allow them to be quite independent: if one blew then the others would not be affected and you could take out unwanted ones or shorten the chain easily. The problem with this obvious solution is that a small, low power bulb running at 100V would require a very thin filament and would be fragile with a short life. The next most obvious solution would be low voltage bulbs and a transformer. The third solution is to wire low voltage bulbs in series. This has disadvantages but it is cheap. One disadvantage is that if one bulb blows then all go out. So, the bulbs are usually designed to short when they fail. The others stay on but run a bit too hot and bright. As more fail, there is a danger of runaway failure, so there is usually a fuse bulb that will break the circuit when it fails. This is normally distinguished in some way; I remember a white coating around half the bulb and it is usually one nearest the supply. Do not remove this bulb.

Your 34A figure is very wrong. If you were drawing 34A at 100V then that would be 3400W which would be blinding and very hot in a typical room in a house. P = I V so the actual current should be 0.408A. Let's forget the .008. 100 identical in series connected to 100V will get 1V each. Since the current is 0.4A, we can deduce from V = I R that the resistance of each bulb is 2.5 Ohm. So, you want 2.5 Ohm per removed bulb. You are removing 82 bulbs so I calculate you need 205 Ohm. With 63 Ohm, they will run very bright and hot and blow soon. Note that even if you get it right, you will use the same power as the full set and generate as much heat. That resistor will get as hot as the 82 bulbs it replaced. Remember, don't omit the fuse bulb.

I have assumed a single strand of 100 bulbs. If it is two strands of 50 then the figures will be different. Obviously, just detach one strand entirely. I won't do those calculations as I don't suggest that you do this.

Although this is an interesting mental exercise, I don't suggest that you do it. Buy a new set of LED lights. These will use a lot less power and run a lot cooler. You could ask another question on how to shorten one of these sets but, since they use so much less power and run cooler, you may as well just hide the unwanted lights. If you run the set for long then it might pay itself in reduced utility bills. If it avoids a house fire then the saving will be even greater.


To use 18/50 lamps rated for 120V means you need a small power supply with 18/50*120V=36V or less.

The power supply level 18/100*40.8W = 8W or more.


The engineering content of this question is simply an application of Ohm's Law.

If 50 bulbs share 120V in series equally and draw 50% of ~40W then each bulb needs 2.4V and 1/3 A.

Thus 18 bulbs needs 43.2V * 1.3A = ~8W.

If you used a series resistor it must drop 120-43=77V or almost double and thus almost double the power rating. So it would be a poor design for efficiency and cost.

A better sol'n might be to find a transformer to match the requirements.

Some old printers from HP used 32 V supplies which would work ok using slightly less power.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Russell McMahon
    Jul 9, 2020 at 22:24

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