What I am doing is I have cut off some bulbs off a string of incandescent Christmas lights. However, now I need a resistor for it so it doesn't blow the bulbs out.

Without a resistor, within a few minutes it blew all the bulbs on that half of the string, and now I understand the reason is because the bulbs I removed caused a greater load to be placed on the remaining bulbs.

I am almost done, and I even soldered the resistor onto the string properly.

However, because of the following two factors, I believe it is the wrong resistor, perhaps because I need a resistor to handle much higher wattage, per my speculations. * resistor got too hot immediately * bulbs were too bright like without any resistor, indicating the resistor was not taking the load as expected.

Now, keep in mind that not only am I just trying to have good Christmas lights but also I am doing this as a learning experience. Obviously I could just take it down and use an unmodified string, but I want to learn so that is why I want to use the resistor.

Here is my diagram:

enter image description here

Project details:

I removed 27 bulbs from a 100 bulb set, from one side.

You can see in my diagram how these lights are wired.

The Math (my math is wrong, see below*):
So I did all the math, and I know all the variables I need for the project.

1.2v x 27 bulbs = 32.4 v
.408w x 27 bulbs = 11.016 w
amps = 0.34 a
ohms (Ω) = 95.29412 Ω*

>>>> *NOTE: Math is wrong, see update below. Keeping this here for reference.*

(NOTE: I calculated ohms using online calculators. I used this one: http://www.ohmslawcalculator.com/ohms-law-calculator. I also double-checked using other online calculators, with the same result.)

Ok, so according to the radioshack guy, all I needed was a 100 ohm, 1/8 w resistor, he said there is no need for a bigger wattage.

So I went and soldered on a 100 ohm, 1/8 watt resistor, but I noticed that first the bulbs were too bright like before, so it seemed like the resistor wasn't taking the load as expected. Second, the resistor got immediately so hot it burned a hole through the electrical tape within seconds.

So I unplugged immediately and now I am researching the correct solution.

I am wondering is if he had no idea what he was talking about and if in reality, I needed a 100 ohm, 11+ watt resistor?

Also from searching the web I read that a standard is to get twice the wattage of your needs for the resistor.

I have also already done all the connecting, testing and soldering. I just need to know which is the correct resistor to use for my project. Once I have that I can just attach it in place of the wrong resistor and then everything should work!

Please let me know if I need the 100 ohm, 20 watt resistor, or if not which is the correct resistor.

Thank you in advance!


Ok, so commentors are saying that due to the KVL law of physics, actually it is not 120v split evenly into both strings.

Instead, it is 120 v into EACH string. This changes things.

Also another answerer said that amps is split among the parallel circuits. So looks like I had it backwards originally.

New formula:

120v / 50 bulbs = 2.4v per bulb
2.4v x 27 bulbs = 64.8v for the 27 bulbs.

2.4v x 27 bulbs = 64.8v v
0.34a/2 parallel circuits = 0.17 a
watts = 11.016 w
ohms (Ω) = 381.17647 Ω

So in this case, does this mean that I actually need:

~381 ohm, ~20 watt resistor?

(~ means here "around" since an exactly 381 ohm resistor doesn't exist.)

(20W because 11W is needed and due to research good rule of thumb is double wattage needed)

  • \$\begingroup\$ How did you calculate the resistor value? \$\endgroup\$ Commented Oct 27, 2016 at 19:26
  • \$\begingroup\$ I used an online calculator. I tried several calculators and they all came up with the same result. Here is one I used: ohmslawcalculator.com/ohms-law-calculator \$\endgroup\$
    – hbsrnbnt
    Commented Oct 27, 2016 at 19:27
  • \$\begingroup\$ What was the formula you used - what voltage and current? \$\endgroup\$ Commented Oct 27, 2016 at 19:29
  • \$\begingroup\$ I'm not sure what you mean? I thought you can see that from the diagram - 32.4v (voltage), 0.34a (current). Please let me know if you need me to clarify. \$\endgroup\$
    – hbsrnbnt
    Commented Oct 27, 2016 at 19:30
  • \$\begingroup\$ Two parallel strings of 50 means 2.4V per bulb, not 1.2V. \$\endgroup\$ Commented Oct 27, 2016 at 19:30

2 Answers 2


You're on the right track in two regards; your calculations headed in the right direction, and the Radio Shack guy doesn't know what he's talking about. A general good start in thinking about electrics is to consider vague magnitude. You have a project dealing with 40W here. Does 1/8 of a Watt sound like the right sort of answer?

The bit of calculation you need here is the power dissipation in the resistor. A series resistance circut (when resistors are connected end to end, "in series") is pretty simple. In this case, we know that the current flowing in the circuit is 0.335A or so. We also know that the resistor you are using is 100 Ohms.

The power dissipation in a component is Volts X Amps. Combining this with Ohms Law, we can show that this power is also Amps squared X the resistance in Ohms.

This gives us 0.335 X 0.335 X 100. Which gives an answer of 10.77 Watts.

So the vague magnitude speculation actually gave you a pretty good answer, calculating it we find a 20W rated resistor will comfortably handle this load. And not to trust the guy in Radio Shack.

edit Oh poo, I didn't realise you have two parallel strings. And now I'm typing this retraction I can't see your picture. BRB.

edit 2.

Okay, we now have 50 lamps across 120V. The total resistance of all the lamps is V*V/P, which is 120*120/20, or 120*6, which is 720 Ohms. Divide that by 50 to get the resistance of 1 lamp, which is 14.4 Ohms.

Now the resistor needs to be the same resistance as 27 lamps, so it's 27*14.4 which is about 388.8 Ohms. So that's our resistor.

The current flowing is I/R, which is 120/720, about 0.166 Amps. And the power dissipated by the resistor is 0.166*0.166*388. Which comes back to 10.77 Watts again; which it ought to because we're dissipating about (rule of thumb) 1/4 of the entire string's rated power as useless heat in the resistor.

  • \$\begingroup\$ Thank you for confirming my suspicions about the radioshack guy. Also, another commentor noticed that I calculated volts wrong since this is a parallel circuit of 2 x 50 bulbs, meaning that 120v is going into each string. I have updated my question with the new calculations at the bottom. p.s. it's actually 40.8W not 40W, does that make a difference? \$\endgroup\$
    – hbsrnbnt
    Commented Oct 27, 2016 at 19:39
  • \$\begingroup\$ Where are you getting the wattage from? Is the whole string specified as 40.8W? \$\endgroup\$
    – Ian Bland
    Commented Oct 27, 2016 at 19:42
  • \$\begingroup\$ I got the wattage from the box, although it can also be calculated using the voltage: 120v * 0.34a = 40.8w. (Amps are also given on the box). \$\endgroup\$
    – hbsrnbnt
    Commented Oct 27, 2016 at 19:44
  • \$\begingroup\$ Also I wonder where in the world can I find a 200Ω 40W resistor (if that is what I need)? \$\endgroup\$
    – hbsrnbnt
    Commented Oct 27, 2016 at 19:50
  • \$\begingroup\$ @User91232 I've updated the answer, it comes to the nearest to 388 Ohms you can find, rated comfortably over 10W. \$\endgroup\$
    – Ian Bland
    Commented Oct 27, 2016 at 19:55

If your schematic is right, and I believe it is, you need 330 Ohms at 10W.
Use 20W so you're safe.
Keep in mind you're dealing with 10W of HEAT, so be sure it doesn't get too hot.
Don't wrap the resistor(s) in anything too insulating.
Protect the wiring so no one gets lit up though. Keep the resistors on the N side of the plug if you can.

If you put all those bulbs in a coffee can and lit 'em up, how much heat would you generate? You need to get rid of about 1/4 of that through your resistor, or 10W.


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