3+3+3+3 =12 not good at all. Each LED drops close to 3 V.
If we add a diode for reverse polarity protection
3+3+3+3+0.6 >12 so won;t work.
we need (3+3+.6+R) * 20 mA to be less than or equal to 12V
R< (6-0.6)/20e-3
That turns out to be 270 ohms.
I'll pick 1 /2 W metal oxide.
https://www.digikey.com/products/en...9|0&quantity=&ColumnSort=0&page=1&pageSize=25
270, 240 and 220 ohms are available, so pick 270.
You can check the power dissipation P = I^2 * R is about 0.1 W. I would not use an 1/8W (0.125W) resistor
Use heat shrink over the diode and resistor bodies. heat shrink shrinks ny where from 1:2 to 1:4
I don;t know your logistics.
4 street lights?
4 houses
2 houses with 2 LEDs in each
That would change how I would do things.
So, two of these (1N4001+LED+LED+270) in parallel with (1N4001+LED+LED+270) combination will do what you ask.
You can reduce it to one diode if you want. The diode prevents damage if it's plugged in backwards.
and 12V is a common available power supply.
5, 12, 15 and 24 VDC are common available power supplies. 15 V is rarer.
+-15 was common for OP amp circuits.
1.8, 3,3 and 5V are common for logic supplies. 1.8V and lower reduces power requirements.
the first IBM power supply provided +5 and +12 and -5 and -12. The +12 for disk drives and the -5 and -12 for memory.
Cars and truck are 12 and 24. telco is 48 VDC.
Transistor radios ran off a multiple of 1.5V. Now 3.7 V is common for battery powered stuff.