How to power a homemade lightbulb safely

I have created a homemade lightbulb as part of a history project and am wondering how to power it in a manner that is safe enough for school.

The lightbulb consists of a piece of .9 mechanical pencil graphite as a carbon filament in a sealed container (http://www.instructables.com/id/Homemade-Lightbulb/?ALLSTEPS)

I currently have a 12 volt drill battery that successfully powers the bulb to an adequate level for a short time before the alligator clips heat up too much.

I also have a step-down transformer I pulled from somewhere that (according to my measurements) steps down the 120v wall power to two secondary windings that each output 12 volts.

After some research I found out that shorting the secondary coil (which is basically what happens when I connect the light) creates lots of heat and can/should trip a circuit breaker on the primary side. So in other words inadequate for school.

My question is how I can power this lightbulb with minimal heat generated outside the jar in which it is contained. the project is due this Friday and I could turn it in as-is but I'm just trying to optimize it so it's safer. My theory of creating another bulb to increase the resistance wouldn't work due to time constraints

I'm competent with basic electrical engineering tools and terms

• Use a 12V fan to cool the alligator clips whole powering the graphite? There will be a 12V fan in any desktop computer you find. Feb 16, 2017 at 2:53
• Can you measure what power levels are being dissipated now? Current and voltage out of your transformer with the load also. p.s.: interesting question, but the title needs some work.. Feb 16, 2017 at 2:55
• And what do you mean the alligator clips heat up too much? Feb 16, 2017 at 2:55
• Have you thought of adding a series resistor? Right now your circuit's resistance is only the filament plus wiring (so ~0). Adding a small resistor (Rs >> Rload) would limit the current to approximately 12V/Rs. Feb 16, 2017 at 2:57
• @Passerby the entire thing is basically a short circuit so all the wiring is going to heat up Feb 16, 2017 at 5:08

The main issue here is really safety. You are concerned about not being able to unplug the bulb due to how hot the alligator clips get. This is a valid concern.

Depending on the current used, a typical automotive 12V switch may help. Most have multi amp ratings at 12V. This 8 dollar switch at Orielly Auto Parts (common auto store) claims 30 Amps.

Wire this in series, and flip the switch when you want to turn your project on and off. RadioShack has similar ones.

Alternatively, you could use a simple switch and a relay. Just as easy.

To address the actual heat issues, a 12V computer fan, in parallel, could be used for some active cooling.

My theory of creating another bulb to increase the resistance wouldn't work due to time constraints.

Your Theory is correct. Two of these bulbs in series would increase the resistance, and thus decrease the current draw. You would be redistributing the heat created as well, half on each bulb.

• Being a cheapskate, wouldn't holding an alligator clip to the battery with pliers work? Great answer either way Feb 16, 2017 at 5:45
• @DeveloperACE why not? I have like 15 auto stores in a 10 mile radius, and two radioshacks left. The same switches can be found at Target or Walmart or Kmart, even Home Depot (they have a little auto section for a year or two now). You only need one, they are really common and cheap. Feb 16, 2017 at 5:47
• Is there a word missing? "Two in series would increase the resistance and the current draw." Seems like it ought to reduce the current draw.
– JRE
Feb 16, 2017 at 6:20
• @JRE nice catch. up Ω, down A. Feb 16, 2017 at 6:24

So I guess all of my ideas are super dangerous and impractical. I'm just going to hold one of the wires with (insulated) pliers or something just to demo the bulb for a few seconds in class and have that be it.

• That's a possible solution, but look at my answer. Feb 16, 2017 at 5:42