I have made several portable "light boxes" for a specific application in our manufacturing plant. What this consists of is 12 volt 35 watt automotive lamps mounted in a row in a box. I have done several with using transformers to drop from 120 volts to 12 volts. Also I have done several using a total of 10 lamps wired in series to 120 volts. (120 volts divided by 10 = 12volts. this works great except when a lamps burns out they all go out but that is not a problem for us.
I need to build one with 5 - 12 volt 35 watt lamps. Rather then using a transformer to drop down to 12 volts I would like to use my series method. Of course that would give each lamp 24 volts which would not work.
My question is, can I put resisters of the proper resistance in series with the lamps to get down to 12 volts? If so, how do I size them for wattage? I know how to size them for resistance, but I know they also have a watt rating and I would need to know how to size them that way

If each lamp is 35W @ 12V , then the lamp current is:

I=P/E

I=35/12

I=2.92 Amps

Assuming a series circuit, the current is the same through all the lamps.

If you have 5 lamps, each dropping 12 V, the total dropped across the lamps is 12V x 5, or 60V.

If you want to run this lamp bank from a 120V source, you will need a series resistor that will drop 60V at a current of 2.92A. According to Ohm's Law:

R=E/I

R=60/2.92

R=20.5 ohms

The power dissipated in this resistor is given by:

P=ExI

P=60x2.92

P=175.2W

The nearest standard resistor would be a 200W wirewound type, which is a pretty large device. Imagine a ceramic tube about a foot long and 1.5" in diameter.

You may have a problem getting the lamps to start well with a simple resistive ballast like this, because the resistance of a filament-type lamp is much lower when cold, resulting in a much lower drop across the lamp and a consequently higher drop across the series resistor. The lamps may remain cold while the series resistor burns up from excessive dissipation. Even if the lamps eventually heat up and run properly, you need to take this startup current surge into account when sizing the series resistor.

All in all, the use of a transformer is a better way to go. You save 175W of wasted power!

[This message has been edited by NJwirenut (edited 09-25-2001).]

I agree with "wirenut".
The Resistor would be really large and waste a lot of power [true power, in watts or Kilowatts]. The wasted true power in the Resistor will result in a lot of heat [which is the wattage dissipated in the resistor].

As wirenut said, look for a Transformer with the output voltage required, along with the total KVA required.
You could use a 250 VA [0.25 KVA] 120 x 12 VAC Transformer to drive all 5 of the lamps in Parallel. This would be a much safer method, as to eliminate 120 VAC from being present at the low voltage lamps.

Scott SET

I agree with all the previous comments. The dropper resistor would be huge & run VERY hot.
You may want to consider using a single-winding autotransformer, if you can find one with a suitable tap. It will be slightly smaller & lighter than the equivalent double-winding type.

[This message has been edited by pauluk (edited 09-26-2001).]

Thank you all for your replies, I was envisioning having 5 resisters in series between the lamps. But I see now that would not even be the practicel way to go. I would still be dealing with the same heat and wasted power only differance would be I would now be dividing it among 5 resistors instead of one. Also, That would still require each of the 5 resisters to be 4.1 ohms / 35 watt size. I'll just use a transformer. I can get those for about $40 to $50 for a 200va

This reminds me of a few years ago whe I was regularly asked to do the lights & sound for my local amateur drama group.

No fancy modern triac dimmers there. We had old rheostat lamp dimmers to control about 8kW of lighting, in a control room about 10 x 6 ft. Sure got hot in there during dimly lit scenes!