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So I have learned that in order to shorten christmas lights, you need a resistor. And unless you get a very very big resistor (which is expensive and hard to find, or use a heatsink, or string many together to up the wattage a lot), it will get very hot and you can't let it touch anything.

I don't know what I am missing why I needed a resistor for shortening a christmas string unlike the shorter string. For example, someone could manufacture a string with 78 bulbs instead of 100, and I am sure they would not have a hot resistor on there.

But why does a shorter string of lights not need a resistor?

I'm not sure why I am needing to add a hot resistor, I wondered what they do in the manufacturing process to determine how to make the string without a resistor? Is it the type of bulb? Or is there something else involved?

Is there a way to emulate the shorter string of lights without the resistor?

Note: these are strings which plug into a wall outlet of 120V. The voltage therefore cannot be adjusted since the power outlet cannot be adjusted.

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  • \$\begingroup\$ 0.5W bulb? or what \$\endgroup\$ – Sunnyskyguy EE75 Nov 1 '16 at 2:26
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    \$\begingroup\$ I think, IIRC, that the bulbs are generally in chains of 25 or so. As such, a string of "100" bulbs is really four smaller strings, whereas the 78 bulb string is three sub-strings, etc... \$\endgroup\$ – Connor Wolf Nov 1 '16 at 3:27
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    \$\begingroup\$ dollar tree string is 20 bulbs. the string of 100 is only two strings. Trust me, check my other questions, I unwound the whole set of 100 and it is 2x 50 in parallel. \$\endgroup\$ – hbsrnbnt Nov 1 '16 at 3:27
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    \$\begingroup\$ @J... Clever use of shunts that only kick in when a bulb fails means they keep working unlike the really old series sets where you had to test every bulb in sequence. energy.gov/articles/how-do-holiday-lights-work Of course now LED lights are common and they're normally a series-parallel DC circuit \$\endgroup\$ – Chris H Nov 1 '16 at 10:08
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    \$\begingroup\$ @ChrisH Yes, but clearly OP is not so familiar. \$\endgroup\$ – J... Nov 1 '16 at 12:12
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The strings are designed to use bulbs whose voltages sum to equal the supply voltage. So a string that uses 20 bulbs for a 120V power source will use bulbs designed to operate at 6 volts. And a string that uses 50 bulbs for a 120V source will use bulbs designed to operate at 2.4V. When you are making hundreds of thousands (or millions) of strings you can have custom bulbs made for whatever voltage you wish.

If you want to remove some of the bulbs and shorten the string, then you must compensate for the power the bulbs were using or run the risk of premature failure of the remaining bulbs which you are operating over-voltage.

A 2.4V bulb designed for a 50-bulb string is NOT "interchangeable" with a 6V bulb designed for a 20-bulb string. No matter how similar they may appear to the naked eye.

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    \$\begingroup\$ Thank you, this is very helpful. I think this solves my problem - it is the bulbs which are the variable in question. By removing bulbs, it reduces the total resistance, so the resistor is needed. A resistor will get hot because it is the equivalent of a bunch of bulbs all in one object, the resistor. \$\endgroup\$ – hbsrnbnt Nov 1 '16 at 4:05
  • \$\begingroup\$ if a 6V 80mA (480mW, 75Ohm) bulb is used in a string that drops 2.4V *170mA (400mW, 14.1Ohm) it will be adding only 61 Ohm to 640 Ohm so current may drop 10% but running twice the current of 80mA it was before thus it will be drawing twice the power 155mA*6V(1000mW) that should be obvious \$\endgroup\$ – Sunnyskyguy EE75 Nov 1 '16 at 4:41
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    \$\begingroup\$ @User91232 Yes, exactly. The resistor will release about as much energy as the missing bulbs would have - it's just concentrated in one small component, and all in heat (not that typical incadescent bulbs are very efficient at converting electricity to light :)). Light bulbs are rather large heatsinks, and get quite hot regardless - imagine all that in a tiny resistor, and summed up from multiple bulbs. \$\endgroup\$ – Luaan Nov 1 '16 at 9:45
  • \$\begingroup\$ The string2P50S acts as a current source to any resistance added with 120V applied anything replacing a dead bulb so the 6V bulb sees twice it's normal current, then what? \$\endgroup\$ – Sunnyskyguy EE75 Nov 1 '16 at 12:39
  • \$\begingroup\$ @Luaan Actually, filament of a classic bulb create mostly radiant heat that doesn't stay in the system. Resistors and even LEDs have much bigger problem with cooling because they must rely on convection or conduction, otherwise the heat accumulates and temperature rises. This is why 5W bulb can be held in a hand, while 5W resistor causes instant burns. \$\endgroup\$ – Agent_L Nov 1 '16 at 13:32
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If you want to shorten a string of 100 bulbs to 78 (actually two parallel strings of 39 for 120VAC) you can wire a 1N4007 rectifier diode in series with the strings. It will reduce the RMS voltage by about 30%, which is about right.

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  • \$\begingroup\$ I did not downvote you, someone else did, fyi. However I really don't understand your answer. \$\endgroup\$ – hbsrnbnt Nov 1 '16 at 4:03
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    \$\begingroup\$ Using a rectifier will reduce the power by 50%, will it not? \$\endgroup\$ – Richard Crowley Nov 1 '16 at 4:03
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    \$\begingroup\$ @RichardCrowley it will reduce the RMS voltage to 0.707 so if the resistance was constant it would reduce the power to 50%, but we are reducing the resistance to 0.78 so the power per bulb is just a bit less. \$\endgroup\$ – Spehro Pefhany Nov 1 '16 at 4:10
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    \$\begingroup\$ @User91232 yes the diode will not get hot \$\endgroup\$ – Spehro Pefhany Nov 1 '16 at 4:11
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    \$\begingroup\$ @NickGammon Half on half off means the heating power is reduced to 50% for the same resistance load. Since P=V^2/R RMS voltage must be 1/sqrt(2)~=0.707. Average is something else, I could give you the formula for it for a half-wave rectified sine wave, but it does not predict the heating effect- for that you need RMS. \$\endgroup\$ – Spehro Pefhany Nov 1 '16 at 10:45
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Bulbs can be manufactured to any desired voltage.

I'd assume the strings use different voltage bulbs. Did you measure the voltage across each bulb when it's in it's string and lit?

Incandescent bulbs also don't have a rigid threshold of "working" vs "not working". The bulbs in one string might be overdriven with a voltage higher then they're really supposed to be run at, and the other string might be under driving the bulbs. Overdriving will produce more light, but shorten the lifetime. Underdriving will do the opposite, but you'll still get some light unless it's very, very underdriven.

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  • \$\begingroup\$ upvote because helpful and accurate. p.s. i did not measure the voltage across the string, only calculated it based on the 120v power source. \$\endgroup\$ – hbsrnbnt Nov 1 '16 at 4:07
  • \$\begingroup\$ Small bulbs can not be manufactured for less than a few millivolts or more than some kilovolts, but for voltages between some 0.5 and 300 volts. A bulb for a very high voltage and low power would require a filament wire too thin to be manufactured. A bulb for a very low voltage would require a very short and thick filament wire, the heat loss to the connecting wires would be too large. \$\endgroup\$ – Uwe Nov 2 '16 at 12:05
  • \$\begingroup\$ @Uwe - Point taken. \$\endgroup\$ – Connor Wolf Nov 2 '16 at 17:49
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If the bulbs are connected in series then their voltages will add up to whatever the power supply outputs - this means any bulbs removed must have their voltage drop made up for with resistors.

If you remove 2 bulbs at 0.5W each you will need to dissipate 1W of power if you want to maintain the same brightness in the other bulbs, for example. Add up the combined wattage of the removed bulbs and that's how much extra power you have to dissipate.

In a shorter string you would either use a lower voltage power supply (if not running the bulbs directly on the mains) or you would use higher resistance bulbs to bring the current down (with more-or-less even voltage drop across all bulbs).

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  • \$\begingroup\$ both strings plug into the wall which is 120V. \$\endgroup\$ – hbsrnbnt Nov 1 '16 at 3:09
  • \$\begingroup\$ thanks for your answer, upvote because helpful and accurate. \$\endgroup\$ – hbsrnbnt Nov 1 '16 at 4:08
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The lights were designed to 120 VAC input. The power output (light and heat) can be estimated by simple Ohm's law (supposing ideal circuit): P=VI=V^2/R.

By leaving some lightbulbs out, you are reducing the overall resistance and thus increase the power drained from the supply. With constant voltage supply, that means you are drawing higher current. Every element of the circuit dissipates power P=RI^2, where R is resistance of the element and I is the current flowing through it. That's why the resistor gets hot and bulbs are brighter.

Your note is false, though. You can adjust the voltage applied to the string. Add (auto)transformer between wall plug and the string. This way you can reduce the overall voltage to fit the 78 bulbs instead of 100.

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  • \$\begingroup\$ upvote can you tell me more about this transformer and if it is cheap and easy to use? Where do i find one and how do i calculate what kind/size I need? \$\endgroup\$ – hbsrnbnt Dec 31 '17 at 22:53

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