That is correct Carl. Surface Mount Devices (SMD) are identified by their physical footprint.size. In US we use Imperial sizes. Looking at the designation is is quite easy to instantly know their size. Separate the 4-digit designation into two 2-digit numbers. Those are their footprint dimensions in hundreds of an inch. So 0603 component's footprint is 0.06" long and 0.03" wide. Or 1206 is 0.12" long and 0.06" wide. Once you know that, it is really easy to visualize their size. The thickness is not part of that nomenclature and it can vary. Here is a good page which shows you a chart of sizes and the power rating (for resistors):
https://en.wikipedia.org/wiki/Surface-mount_technology#PackagesAs far as power dissipation goes, quoting
http://www.electronics-tutorials.ws/resistor/res_7.html :
The Resistor Power Rating is sometimes called the Resistors Wattage Rating and is defined as the amount of heat that a resistive element can dissipate for an indefinite period of time without degrading its performance.But keep in mind that a resistor which is utilized at its 100% power dissipation rating will be very hot.
I recommend reading through the the above web page - there is lots of useful info (like several ways of calculating the power dissipation) and recommendations on how to choose the proper power resistor.
Basically I recommend to use a resistor which is rated at at least triple of the actual power it will dissipate. In enclosed space (like inside a locomotive) and when the resistor is not soldered to a circuit board which acts a a small heat sink, I would even recommend to triple the chosen power rating. Not because I worry that the resistor will burn up, but I worry that if the resistor is close to some plastic part, it might get hot enough to deform the plastic.
As far as the other question you have, I will assume few things:
1. The voltage between the common positive (blue wire) and a function output is 12V
2. You will be using a single white LED and a single resistor (in series) for each function.
One unknown is the current which will flow through the LED circuit. That will be chosen by you to get the desired brightness.
There are several ways to calculate resistor value and power dissipation. I will use the one I prefer.
White LEDs have forward voltage "Vf" (voltage across their terminal when they are illuminated) of around 3V. That fluctuates slightly depending on the specific chemistry of the LED and the current flowing through it, but for what we are doing 3V is a good value.
To calculate the resistor value we subtract the Vf from the input voltage of the LED/resistor circuit (which is 12V as assumed above). So that is 12-3=9(V) Voltage will be represented by the letter V later on.
Next, we need to pick the current we want to flow through the LED (and by definition through the series-connected resistor). With a small white LED in the headlight lens you might find that 3mA (or 0.003A) gives you a good brightness. Current will be represented by the letter "I" later on. Now we have enough information to calculate the current passing through the resistor.
To calculate the resistor value (R) we use the
Ohms Law Triangle. Specifically: V/I=R. So for this circuit it will be 9/0.003=3000(ohms). Or 3k ohm.
Since in a series-connected circuit (by definition) the same current passes through all the components in that circuit, if you feed 12V to a series-connected 3k ohm resistor and a white LED, 3mA will pass through both of them.
Power dissipated by that resistor (or "P") can be calculated using a Resistor Power Triangle (shown on the web page I initially mentioned. Here I'll use I*V=P or 9*0.003=0.027(Watt). If you triple that it is 0.081W. In this instance using an SMD 0603 resistor (rated at 0.1W) would be acceptable.
Here is the same circuit but you want 10mA passing though the LED (so it glows extra bright).
Resistor value will be 9/0.01=900(ohms).
Power dissipated by the resistor will be 9*0.01=0.09W Triple that is 0.27W. Here I would use a 1206 size SMD resistor rated at 0.25W (that is close enough to be triple the rating).
You don't have to worry about the power dissipated by a small LEDs. Some of the power is converted to light and the remaining power is dissipated as heat (but that is minuscule in our application).
Last thing: some resistor values obtained in the above calculations are not available as standard-value resistors which are available (as the least expensive 5% tolerance resistors). In that case select the next higher value. That will slightly reduce the current you wanted to pass through the LED but will be close enough. Google "standard resistor values" then find a table or a calculator showing 5% tolerance resistors.
I hope this helps.
You mentioned installing some capacitors. That is totally unrelated to calculating resistor values for LEDs. I suspect those caps are the ones used in sound decoders to prevent sound cutouts due to intermittent electric track pickup. I prefer using tantalum caps rated for 20 or 25V but if your DCC booster does not output more than 12V then 16V caps should be safe to use (few people here use 16V caps in their decoders).
