- 9 Volt Battery
- X-Mas light

Step 1: Putting the Lightbulb on the Battery

'Take the wires on the bulb and spread them apart and Then put the wires onto the battery +,-

Step 2: DONE!!!!!!!

That was easy....
You can also take the wire with the holders for the christmas and cut off two of the holders and hock it up to the battery and it wont burn out the bulbs so fast
it works but can fry the lights due to over voltage
according to Howstuffworks.com (which is probably the origin of the project) the voltage rating on the bulbs is 2.5v. so if you connected 4 in series it would work just fine.or using some simple calculations you could figure out that for one bulb you could connect a 30 ohm resistor in series with the Christmas light.<br/><br/>here's the math:<br/>25w per 50 bulb strand connected to 120v ac source.<br/>P=25w<br/>V1=120v<br/>V2(voltage drop across one bulb) = 120 / 50 = 2.4v<br/><br/>P = V1 X I<br/>25w = 120v X I<br/>I = .208333<br/><br/>V2 = I X R1<br/>2.4v = .208333A X R1<br/>R1 = 11.52 ohms per bulb<br/><br/>V3=9v<br/><br/>V3 = I X (R1 + R2)<br/>9v = .208333 X (11.52 + R2)(we will neglect the internal resistance of the 9v battery)<br/>R2 = 31.68 ohms<br/>
It's been 10 years since high school physics class, but I do remember watching another 2 man team fry a long chain of lights. So pyro I believe your 4 light solution would fry them 1 by 1, It may, however, work if you ran them in parallel. P.S. the 4 bulbs is probably a cheaper solution, especially if you bought them on after Christmas clearance
NO my solution would work go look up the physics! I calculated the correct number to keep them from frying. if you used just 2 then that would probably be the case or plugged only 40 in to the wall socket instead of 50 then it might pop one by one. IF you decide to put it in parallel then you are in fact increasing the likely hood of blowing the bulbs unless you add more because the total resistance is less in parallel than in series. Basic Electronic knowledge. take your battery and light bulbs run once in series then in parallel. the ones in parallel will burn brighter because they receive the full voltage no matter how many you connect but will drain the battery faster. Also the bulbs are rated for AC current not DC so are actually higher rated for DC but I have not covered that in my classes yet. By the way I am an Electrical Engineering Student at University of North Texas. go read the article at <a rel="nofollow" href="http://christmas.howstuffworks.com/christmas-lights1.htm">Howstuffworks.com</a><br/>
the 9 V battery is pretty ok. the lights get way more when theyre stuck in the xmas tree wire the resistance of the battery is not to be ignored. its quite big in 9 V batteries. on cheaper ones it may be > the resistance of the light itself most that lights make short and not open when they fail. if you put them in parallel (on better source than 9 V battery) one failed will shut em all and kill the battery. they sometimes fail without visible blackening or void in filament
I don't think thats correct. however a way to find out would be stick your bulb in the Christmas tree socket and look at its brightness. then stick it across a fresh battery. If what you say is correct then it should be dimmer if what I say is correct it should be brighter. If you want to be precise try and isolate the light of the single bulb by sticking through a hole in a box so its the only light glowing within. I don't know much about the internal resistance of a fully charged 9v but given that "In practical batteries, the internal resistance will increase as it is discharged" (wikipedia.) I could see using a depleted battery that would no longer be yielding the full 9v. I know in my classes we usually neglect the internal resistance due to the fact that it is usually minimal and the tolerances of our resistors is larger. agreed.
i got this from the following calculation<br/><br/>take a 50 light wire on 240 V. it actually is made of 3 series strips of 1x16 + 2x17 lamps. divide 240 / 17 and you get 14+ V. the resistace of the 240 V source is really small and i think of the wires and stuff in the lamp wire too<br/><br/>take a 20 light wire on 240 V. its a single strip of 20 lights and i think there are the same lights as in the 50. still each light gets 11+ V<br/><br/>to see visually the effects of the 9 V battery resistance try :<br/><br/>add a battery in parallel with the light (that reduces the resistance to 1/2)<br/><br/>add another light in parallel (thats close to the effect of multiplying the resistance by 2)<br/><br/>to find it exactly use multimeter and calculator<br/><br/>measure battery voltage with light connected and without (i'd measure short time after light disconnected and the battery did not 'recover' yet)<br/><br/>V with light / V without light = R light / (R light + R battery)<br/><br/>measure current light takes (in mA)<br/><br/>V without light / current = (R light + R battery)<br/><br/>V with light / current = R light<br/><br/>the resistance of light depends on the temperature (brightness of light). its resistance much higher when its hotter<br/><br/>i liked to make and hack stuff since i remember myself. especially electrical stuff. and i study electronics in school<br/>
by the way I just wanted to say you seem pretty inteligent for a 17yr old. Did you take an electronics class in school or just learned on your own?
a cool thing you can do is to crack the glass case but be careful not to harm the filament. attach that to you 9v battery. It pops as it burns in the oxygen rich environment.

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