The goal is to achieve consistency of brightness across the 2 colours. There are 6 blocks of LEDs and the switches are centre off. This means, there could be 0, 1, 2, 3, 4, 5 or 6 lights powered at any given time depending on switch positions.

To my way of thinking I'd need to offer resistance for the "worst case scenario" which would be ensuring "only 1 bulb on" doesn't receive too much juice. But this is where I get confused with series vs parallel... Below is my wiring setup. Those LED specs were all I got with the LEDs.

Ok, first Series v Parallel connections .... This applies when you connect a number of LED's using one resistor.

In your wiring diagram you are using single LED's with Single resistors so forget about series/parallel as it does not apply

When you use the resistor calculator I posted in my previous post you select the

**Single LED** page.

Next, as the two colours have a different LED voltage, you need to calculate the individual colours separately.

The Yellow Vf is 3.2Volt, assume 20mA current and the calculator comes up with a resistance required of 440 ohm. Use a 470 Ohm 1/4 Watt resistor.

The Blue Vf is 3.6Volt again assume 20mA and the calculator comes up with 420 ohm required again the next resistor size is 470 Ohm 1/4 Watt

Now to play with the difference in brightness. Try a larger ohm resistor on the brighter colour side, a larger resistance will lower the voltage across the LED and hence reduce it's luminosity.

The standard sizes in that area are 430, 470, 510, 560, 620.

Remember the resistors need to be rated at least 1/4Watt.

Now the rating of the Bi-colour LED

....... mcd means millicandela which is the units for the luminous intensity of the LED

so from the figures on your diagram (Yellow = 7000mcd, Blue = 8000mcd) the blue will appear "brighter" than the yellow.

As each of the six indicator circuits will only draw approximately 20mA each the total current draw on your power supply will be around 120mA with all circuits turned on. As the power supply is rated at 1.5Amp (1,500mA) the power supply will hardly 'know' there is any load and the individual LED's brilliance won't change with respect to how many are turned on