rbewley

Well, thanks for all your help with this, I've actually made an induction coil and did some pretty cool things, but unfortunately, my lightbulb still does not work. Since a transformer would be for AC, I'd have to use the mains, and using the mains to power my lightbulb would defeat the point, as the lightbulb needs 120V and it's already at that. I really wanted to get my DC batteries to power my lightbulb, so would I need to construct a transformer with a pulse DC current? If so, can anyone point me in the right direction of doing so? And, actually it makes sense with the pulse, because if the magnetic field isn't changing, then it's not going to produce a current on the other side, just like putting a magnet that's sitting completely still inside of some coil, it's not going to generate a current either without some movement... Thanks, Richard

Also as a side note, the wire I would be using is 18 gauge wiring. Thanks, Richard

Well, thanks for everyone's help with this. Last question, assuming I'm using insulated wire to make my transformer, if I would like to make a step up transformer, how many loops do you think I would need for the following, and is there an easy formula for this? I understand it is ratios, but I'm not sure how to figure the primary side, any help would be appreciated. Currently: 10 volts 8 amps Thanks, Richard

Is it still ok to wrap the coils around themselves if my wire is uninsulated? The current won't "circumvent" the loop will it? Is it important that it actually goes through all the loops? I realize I may be rather vague in my question, but what I'm asking is... if all the uninsulated wires are wrapped around each other, will this cause a problem? Thanks, Richard

Should the wire be insulated? Especially if it is coiled around itself...

Hi, I do understand the concept behind building a transformer (more coils on one side than the other), but if you could explain to me an easy way to build one, I'd appreciate it. Thanks, Richard

Well, the whole idea behind a filament is, since it just gets hot from the current going through and creates light and heat, I don't see how AC or DC would make a difference. Let me just try pumping up the DC juice a bit and see if I can get the thing to light... Richard

I have no actual reason to power the bulb, and I disassembled my flashlight. What makes a bulb "AC", and will any amount of DC power work to power the thing? Thanks, Richard

Hi, Thanks for the information, it was very helpful. So, are you saying my lightbulb will only pull 170mA no matter how much is available?

Oh sorry, series.

Hi, I'm trying to power my 40 watt lightbulb, and I've been at this for a bit, and was told this morning I would need 22,000 "C" batteries, which seems like it would make my lightbulb explode... Anyway, my standard C battery I am using has 1.5 volts and 7.5 amps when I tested it. Now, my first possible pitful of not being sure is, if I stack these batteries together, I know the volts will just add, but will the amperage stay constant? Now, assuming the amps did stay relatively constant, I figure it would take 5.3 volts to power my 40 watt lightbulb. This is using the following equation: P = iv 40 = 7.5x x = ~5.3 Now, assuming 1.5 volts per battery, i figure that 4 C batteries should more than do the job. So, with my 4 C batteries hooked up, I test again the current and voltage. I am now getting 6.1 volts (About what I expected to see) and also 9.5 amps. This is about 58.0 watts, which I figure should power my 40 watt lightbulb. But, now that I have my lightbulb hooked in, it doesn't even glow. And, when I re-measure everything, I find that I get 104 mA and 5.9 Volts, which of course is only 0.6 Watts, not nearly enough to power my lightbulb. My question is, how can this be? Is it the extra ohms of resistance in my lightbulb? Does my lightbulb really pull only 40 watts of power? If so, how come it takes that much... Thank you, Richard Bewley