how do you get 3 light bulbs to light up and give off the same amount of light as one light bulb?
929481
You need to wire them in parallel, with wire heavy enough for the total current and also have a battery that can deliver the current required at the voltage required without voltage drop.  You don't state the battery voltage or the lamp wattage, so I cant work it out for you!\n\n(I take it that you want EACH of the three bulbs to be as bright as one single one?)
518406
equivalent to 4   100 Watt  Bulbs
1088715
3 groups in series with all the lights in each group in parallel.
261216
The classic answer is "one, but the light bulb must want to change."
943748
use voltage dividers with a transistorized switch.  this method lights  a number of lights matching the input voltage in this case, all the way up the tree to it, rather than just one, so note that.  (7V in gives 7 lamps on).\n\nthe first transistor comes on if the base reaches .7V (2N2222 or so will work).  the next transistor is attached to the base of the first transistor through a voltage divider, the third ....  From the collector stick in a 1V Zener for regulation (make sure you get one that withstands 12v in), and the lights then go from there straight to ground (include a resistor here to limit current, try 15 ohms to start, experiment!).\n\ndecide how many lights you want, divide your final voltage by that number, and that is the voltage per step (here, 1V).  the voltage divider then comes from the voltage per step versus the transistor base ON voltage, across a good sized resistor (say 470 ohms or 1k, whatever you want (watch the current and the power drain here).  in your case, if you use a 470 load resistor on the voltage divider, then the second leg of the divider is (470 / (1V/1V-.7V)) = 141 is about a 130 (less is more on the base!).\n\nrecap:\n\nso the voltage divider goes from the input into the 130, into both the base and the 470, the 470 goes to ground.  the emitter can tie to the input voltage, the collector goes to a 1V Zener, to a 1V lamp, a small resistor, and ground.\n\nthe other stages stack right onto this.  the input still goes to the 130, then to the base and the 470, the 470 ties to the base of the previous stage.  the emitter again goes to the input voltage, and the collector goes to a 1V Zener, a 1V lamp, and a small resistor to ground.  sorry, i can't attach a picture here.
1186464
I hear that they have trouble with light bulbs. For some reason it takes three of you guys to screw one in.
233283
a regular light bulb \nas apposed to a floresent
883981
If you really want to be technical about this then here you go:\n\nHere are the three main consideration you have to follow when making a flashlight:\n\n1.) Amount of current the bulb will draw:\nI standby this clearly to show that current draw will affect the size of the wire that you will use to power up the bulb.  If your bulb draws lots of current from the source, then you will want to use a bigger wire.\n\n2.) Safe wire colors\nSecondly you have to consider safety standards as well.  You can't simply start to wire up the flash light and put the red wire on the negative end of the battery / power source.   You can possibly confuse someone, if they intend to repair it. \nThus in DC CIRCUITS ONLY\nRed = Positive Voltage w/ respect to GND\nBlack = Ground
642130
The best way to learn is by doing, And it is a difficult concept to understand that light can be broken.\n\nI would reccommend that you use prisms to break up the light into spectrums. you can use chandelier crystals or something like that.\n\nThen you can have them put the light back together: Heres how: get three slide projectors, and get some colored acetate in cyan (turquoise). magenta, and yellow to place in front of the projection lenses OR Better Yet take slides of peices of paper that are these colors and place one in each of the slide projectors. If you angle the projectors so that the projections overlap, it should form white light.
403379
Standard light bulbs contain a tungsten filament connected to electrical contacts inside a vacuum so that there is no (or very little oxygen).  When electricity flows through the high-resistance filament, the filament glows white hot, giving off light and heat.\n\nOther types of light bulbs (fluorescent, halogen, etc.) work a little differently.  If you want information on those, just say the word.
137750
First, take the bulb out of a flashlight.  Carefully remove about an inch of the plastic covering on both ends of two one foot lengths of wire.  Attach one bare end to the bottom of a D-sized battery (good tape might work).  Attach the other end of this wire to the SIDE of the metal part of the bulb.  \nAttach one end of the other wire to the top of the battery.  And attach the end of this wire to the bottom of the bulb.\nThat should do it!
1072750
turn on the first 2 switches  let them sit for a little while.  then turn on the 1st and 3rd switch.    go into the room right away, if the bulb is hot but off that means it was the 2nd one (because the switch is now off but it had to be on)  if the bulb is cold still (not really hot) it was the third one because the light just turned on and the 1st one didn't turn it on at all  if it is not on and the bulb is not hot at all it was the fourth because you never touched that switch  it was on and hot then it was the first because it was on long enough for the bulb to get hot.  \n\nso... when only the 1st and 3rd swith was on (after turning off 2)\n1=if it was on and hot\n2=if it was off and hot\n3=if it was on but cold\n4=if it was off and cold
43511
I think you are asking, why do light bulbs often burn out just when you turn them on, not a few minutes or hours later.\n\nThat's because of what happens to the filament of a light bulb as it ages, as explained here:\nhttp://members.misty.com/don/bulb1.html#wbt
1020503
1. A battery lighting up a filament bulb.\n\n2. Use LM317 IC to build a constant current source.\n(http://www.uoguelph.ca/~antoon/gadgets/sccs.htm)\n\n3. Battery Charger
1011849
a) probability of finding exacttly 2 working bulbs in first three =  3*( 9/12*8/12*3/10) = 27/55.\n\n\nb)\nprobability of not finding even one = 3/12 * 2/11 * 1/10 = 1/220\nprobability of finding exactly 1 = 3 * ( 9/12 * 3/11*2/10)  = 9/220 \n \nthus required probability = 10/220 = 1/22.
1170607
here is a link which I hope will help. http://www.hgtv.com/hgtv/dc_lighting_other/article/0,1793,HGTV_3466_1372885,00.html \n\nyou may have to adapt the info to the flower pots you want to use.  it has clear pictures to illustrate the process.   have fun!
1150061
Just one. But the light bulb really has to WANT to change...