here's what I want to do

by the way, space is a concern too....as small as possible

Hello "Guest",

You have far too many variables.

Advise the following:

1. Intended battery voltage.
2. Intended lamp voltage/s.
3. Intended lamp wattage/s.
4. With the lamps, ensure each lamp voltage + wattage is listed separately.
5. Are some lamps closer to the batter than others - advise please.
6. Other relevant information - Advise please.
   

Reply with

Kind Regards....

Like Spark station said not enough info, what are the voltage and the current requirements of the lamps(Wats)? what DC battery power source are you planning tto use?

More info please.

there are diffrent ways of doing this

Check out LED possibilities.

TRIAL AND ERROR DUDE ~

Assumption; bulb filament voltage and resistance are directly proportional. Put all three in series with a series resistor to compensate for voltage drops accordingly. For instance-2, 1 and 9 volt bulbs may work connected in series and attached to a 12 V battery. If using a 24V battery, add a resistor to drop the additional 12 V. Matching the battery to the bulbs is the tipping point for heat and power.

T

As szwasta said you could put them in series how ever if the 3 bulbs you want do not total up to the voltage the battery supplies then you need to use voltage regulation. In series if one stop working they all stop, use a seperate voltage regulater throw off some current, off the top of my head I can't tell you how much as Sparkstaion said I don't have enough information. something that may help is Ohms Law.

Volts = Resistance * Current

Resistance = Volts / Current

Current = Volts / Resistance

Good luck

Hello Friend,

Perhaps a lower tech approach might be in order.

Its 12 volt your working with so, go by the local junk yard and ask about the purchase of a truck or automobile wiring harness and the attached fuse box with leads from an older model vehicle, as they will be much less expensive.

Remove the fuse box with at least about a foot of wire extending from the box so you can reconnect your wires to it. Get all the wire they will give you , certainly the battery wire and ground leads , over to the battery , as you will need them soon. Get your input connector leads to the fuse box , ( generally they are the 2 center wires but trace on the back side of the fuse box to make sure you got them ) with 1 foot extra as well , so you can power the box with your 12 volt battery supply charge .

Next get yourself, 2 knife switches or some suitable DC fuses for the Positive and Negative to be protected by and cut off by , when needed. Then mount all to the wall and the fused wire leads to each 12 volt light positive lead side by way of your individual supply wire that is fused according to each lights needs . Always watch your wire sizes, so as not to over load the draw on a too small wire , as it may cause a fire.

Make a ground bus bar from a piece of 3/4 hard copper tubing about 14 inches long pounded flat and drilled for small terminal bolts to go through for connection to your lights. Connect this to a block of 2 x4 , maybe 1 foot long , with a couple of wooden & tire tube rubber covered wooden blocks at each end, as the Negative Bus Bar device, mounted to the copper tubing suspended between the two blocks of rubber covered ends.

Then mount that on the wall in a safe place near your fuse box and make some sort of shroud for all to be protected later under. Then your off to set up the 12 volts lights , that you needed powered, one to each of the fused supply lines and the other to the Neg. Bus Bar.

This is not rocket science but, do watch yourself , be safe, and if there is something you don't know or understand then educate yourself with an automobile repair manual and/or electrical tutorials from your public library until your confident of your own abilities; After all, this is real electricity your playing with now. One thing you don't want to do is have on jewelry while messing with this stuff. To give one bad example, I once had a cheap metal watch to ground out on a battery and sweaty old me , while I worked on a installation. Burned my arm and belly badly and torched the watch band. Be safe !

Well hope that was some help or at least gives you an idea of how you might proceed.

Best Regards,

Joe Woodall, Managing Partner

Georgia Adobe LP Rammed Earth & Renewable Energy

2395 Bowman Hwy. NW.

Dewy Rose GA 30634

706-213-7693

http://www.georgiaadobe.com

They are all right in asking for the deficient information. Provided with what you've said, you may use the following:

Let V be your DC Supply voltage, assume it provides the necessary current;

and V1, P1 be the voltage and power of the bulb1 would take;

V2, P2 for that of bulb 2,

V3, P3 for that of bulb 3.

Then, for each of the bulbs, connect in series with a resistor, R1, R2 and R3 respectively. The value of these Rs would then be given by:

R1 = (V - V1)/(P1/V1) or [(V - V1)V1]/P1;

and the power, Pr1 of the resistor can stand should be given by:

Pr1 = [(P1/V1)^2]*R1*1.2 or [(V - V1)P1*1.2]/V1, to give it 20% power margin to avoid over heating.

Similiarly,

R2 = [(V - V2)V2]/P2 of Pr2 = [(V -V2)P2*1.2]/V2

R3 = [(V - V3)V3]/P3 of Pr3 = [(V -V3)P3*1.2]/V3

Connect these Resistor+bulb combination in parallel and then to the electrodes/terminals of your DC supply. These would work provided if the DC Supply is good (of high enough voltage and delivers high enough current to) enough to power up the bulbs.

You surely cannot light up a 5V bulb using a 3V battery. If so, you have to using more complicated stepping up circuit!

If you cannot find an exact match of resistances, always get one with resistance next step higher, that would allow you to light up the bulbs but the larger resistances may dim the light bulbs a bit.

Same thing apply to the power dissipation ability of the resistor as well. Look for next step higher rather than anyone lower in value! In case of power dissipation, the higher the value of the resistor power dissipation, the better it would be but it would be more expensive!

If any of the bulb has exact match of voltage requirement to that of the DC supply you have, then you can skip the concerned resistor!

These circuit would be 3 resistors more (maximum) than what szwasta suggested but it would work independently, i.e. one of the light dead, the others would still work!

Hope that would help!

Cheers,

AC Wing.

TRICK QUESTION?

To "run 3 lights 'OFF'", the best way to accomplish this is to do nothing. No heat, no power, no problem.

Hello szwasta

Yes, you don't even need a battery, wiring, lampholders etc, just the 3 bulbs.

Kind Regards....

So, series or parallel connections are equally stup.., meanlingless!

Mr. Guest, was that what you want? Why is there still no response from you?

Or, was it actually posted by szwasta?

AC Wing.

I cannot take credit as originator for this posting.

I too am curious about the intent of this posting

T

Here is what I want to do. It is 9:55 pm, I am starved, and I want to eat dinner. If you want to power up 3 DC lights requiring different voltages, from the same DC power source, then get your self some resistors to reduce the DC power to the voltage required for the particular light you wish to light up, and place them in line a head of the light.

TMF