How do I calculate resistor size

You say 50 mA per lamp.  I thought you were using 12V LED strips?  Anyway, if one 150 Ohm resistor successfully dims 2 lamps in parallel, then why wouldn't you just use a 2nd 150 Ohm resistor to dim the other 2 lamps (also driven in parallel)?

I only had two resistors that size. I made four highway lights with the two sets of three leds  from the strips. I also had two double lamp highway light poles that I used one set of three for each side. I have them hooked in parallel to the other resistor. I still have two poles with the six leds from the strip to hook up, and would like to hook them into the same resistor with the other two similar poles, if it's big enough. Waters muddy enough?

My head is spinning.  I'll just say this.  You have a 12V power source.  You have a lamp/LED assembly that lights up to your satisfaction thru a 150 Ohm resistor.

If you add a 2nd equivalent lamp/LED assembly in parallel, to achieve the same dimming use 1/2 the resistance or 75 Ohms.  The power going thru that single resistor would be double. 

If you have a DC Voltmeter, measure the voltage across the 150 Ohm resistor in first configuration.  The power in Watts is Volts * Volts / Ohms.  

I'm a little surprised that 50 mils across 150 ohms doesn't at least warm up your resistor, that's .375 watts.  Generally, I try to have the rated value of the resistor double the load.

I don't think that 50 mA is thru 150 Ohm.  That would be a voltage drop of 7.5V.  Since he has a 12V DC source, that would only leave 4.5V DC to drive a 12V strip(s).  That would be a very dim strip. 

That's why I simply suggest measuring the Voltage across the 150 Ohm which of course does not require inserting a DC Ammeter in series with the wiring - can be a nuisance.  Then calculate power in Watts = V * V / R.  Or calculate the current in Amps = V / R.

I have one single lamp dropping 2.7 volts across the resistor. The pair of double lamps drops 2.25 volts. The picture is one single with 2 sets of three leds and no resistor on the bench. The double lamps are essentially the same thing with a pair of 3 led sets.

single lamp. 2.7V across 150 Ohms = 0.05 Watts, 18 mA.

double lamp, 2.25V across 150 Ohms = 0.03 Watts, 15 mA.  If split 2-ways, that's 7.5 mA per lamp. 

So in either case, the 150 Ohm 0.5W resistor would be cool to the touch...

I take it that I should have a resistor for each of  the other three single lamps, or at least for two single lamps, as  that would be the same number of leds as the pair of double fixtures. Maybe I'm better off just using a DC to DC converter to drop the voltage to all four. I'm still lost as to how to figure the size of the resistor.

The DC-DC converter isn't going to solve your problem unless you are accepting that the single lamps will not output as much light as the double lamps.  Obviously, the lamps you only used one set of three LEDs in won't be as bright as the the ones with two sets of three LEDs.

The problem here is this is all subjective based on what you are seeing as far as acceptable intensity.  It's not a pure analytical issue of calculating the resistors.  We can't view this problem through your eyes.

I'm sorry, I mean figuring the wattage for the resistor. Is that just using Ohm's Law to figure the watts and keep the resistor larger than that? Except that doesn't compute for the two double lights drawing 40 ma each at 11.5 volts. Isn't that .9 watts?

0.9 Watts is power going into the lights.

For the matter at hand, you're trying to calculate the power in the resistor.  Measure the voltage across the resistor, square it, and divide by the Ohms.  That gives you Watts.  For example, say you measure 2 Volts across the resistor.  Power in the resistor is 2 * 2 / 150 = 0.03 Watts.

End of story!

John and Stan, thank you both. With your help, I think I can keep the magic smoke contained.

John, I assume these are the LED strips from your earlier post.

First the shortcut answer: each 3 LED segment has three LEDs and a 150 ohm resistor in series.  The LEDs are 3.5mm x 2.8mm (hence the 3528 number).  From this, estimating the physical size of the 150 Ohm SMD resistor, it is either a 1/4 Watt or 1/2 Watt.  Your added external 150 Ohm resistor will lower the current so 1/2 Watt we be fine.

OPTIONAL READING:

Next I did a more detailed analysis, just for fun.

Looking at the datasheet for a typical 3528 LED, I compared the typical current and voltage spec. (20 mA, 3.2V) to the maximum current and voltage spec (25 mA, 3.6V).  From these values I determined a rough model for the LED is an 80 Ohm resistance with a 1.6 Volt offset.  This model is a linear approximation of the Current vs. Voltage curve of the LED around the typical operating point.  Note, all these specs. have a 10% tolerance so all the numbers are approximate.

Three of these LEDs in series yield an 240 Ohm resistance with a 4.8 Volt offset.  The 150 Ohm resistor included in the strip gives you 390 Ohms total.  When you apply 12 Volts the current will be:

(12 - 4.8) / 390 = 18 mA

If you add an external 150 Ohm resistor you get:

(12 - 4.8) / (390 + 150) = 13 mA.

The power in the added resistor in 0.013 x 0.013 x 150 = 25 mW.  Double this for 100% derating factor- a 1/20 Watt resistor would do.

If you have two strips of 3 in parallel the 390 Ohm resistance is cut in two.  (For N strips in parallel the resistance is 390/N)  Applying 12 Volts the current will be:

(12 - 4.8) / 195 = 37 mA   This means 18.5 mA through each set of three.

If you drive these through a single external 150 Ohm resistor the current is:

(12 - 4.8) / (195 + 150) = 20 mA  This means 10 mA through each set of three.

The power in the added 150 Ohm resistor is 0.02 x 0.02 x 150 = 60 mW   In this case a 1/8 Watt resistor would be OK.

Bottom line: a half watt resistor is fine.

Thanks Cam. I'll work on digesting that for the rest of the day. That helps me figure something out before it gets too hot under the layout.