The relation between V, I and R is called Ohm's Law, in honor of good ol' George. The law should be familiar in all three of its possible forms:

You may want to write V = IR on the back of your hand (note that in the product "IR," the "times" sign is implied). From V = IR, you can quickly factor the other expressions -- they'll serve you well. And don't wash that hand.

A MODEL RR EXAMPLE

1. choose a higher-voltage and/or lower-current bulb;
2. cheat the system a bit -- you know that you won't have the full 12 volts on the rails at all times (unless you're into train racing), so you might use a 1/2 watt resistor; a liitle calculating shows that this is just fine up to 10.6 volts on the rails, and feel comfortable with this; only experience will help you make these determinations (and don't forget that the resistor's power rating is always double the calculated dissipation);
3. use two or more lower-wattage (and hence smaller) resistors to yield the desired value -- by connecting two 160 or 180 ohm 1/2 watt resistors in series (see "Series & Parallel Connections," below), we get the equivalent of a 320 or 360 ohm, 1 watt device...and either combination would be OK here; or, we might connect two 680 ohm, 1/2 watt resistors in parallel, yielding 340 ohms at 1 watt. Perfect.

SERIES & PARALLEL RESISTORS

RESISTANCE OF COPPER WIRE

For the sake of completeness, you'll surely want to know that #22 wire is 0.016 ohms/foot; #24 is 0.026 ohms/foot; #14 is 0.0025 ohms/foot; and #12 is 0.0016 ohms/foot. All this at 25° Centigrade...close enough for our purposes.

Almost. PC supplies accomplish lighter weight and lower power dissipation via a design technique called the “switching regulator.” Switching regulators depend on sophisticated closed-loop control circuits to maintain tightly regulated outputs; what you need to know is that these circuits depend on each output having a minimum load. At home in that old PC, each output was properly loaded; your layout may or may not provide adequate loading -- if it doesn't, you'll need to lend a helping hand. Keep reading.

Whatever you do, DON'T OPEN THE METAL CASE that the supply lives in! There's nothing inside you need be concerned with...and careless poking around could be fatal. If it has a fan, leave space around the case for it to breathe; if it doesn't, don't worry about it. Newer supplies are usually smaller than older supplies; they may have a different set of output voltages, but you'll likely not care...so long as it does have the ones you want.

ADEQUATE LOADING?

AN EXAMPLE

Let's calculate a 10% loading for each of the outputs:
     from above, the 5V supply needs a 5 ohm, 10 watt resistor
     for -5V, R = V/I = 5/0.05 = 100 ohms @ P = V²/R = 25/100 = 0.25W (use 1/2)
     for +12V, R = 12/0.18 = 67 ohms (use 62) @ P = 144/62 = 2.3W (use 5)
     for -12V, R = 12/.02 = 600 ohms (use 560) @ P = 144/560 = 0.26W (use 1/2)

If you've been paying attention, you're probably muttering something like, "Where am I going to get a 62 ohm, 5 watt resistor?" It's a good question. Unless you have a very well stocked parts bin, you're going to have to make a few "adjustments." Good ol' Radio Shack offers several "wirewound" resistors with higher power ratings. [Note: wirewound resistors are inductive, so never use them in AC circuits.] 50 ohm, 10 watt units are available (271-133), so what if we tried a 50 ohm unit in place of the 62 ohm critter we calculated? Well, we'd draw a bit more current (which is fine), and we'd dissipate a bit more power (more current) -- how MUCH more power? P = V²/R = 144/50 = 2.88 watts (use 6W, minimum), so the 10 watt rating would be plenty. Problem solved.

But we're not done yet -- the two 1/2 watt units are easy enough to come by, but what about that 5 ohm, 10 watt thing we just breezed by so quickly? A quick perusal of RS's stock will reveal no such animal; but, recalling the formulas for series and parallel resistors,

On the above schematic, you'll notice a few things we've not discussed. You do need to connect your supply to the AC mains, thereby requiring a line cord (I strongly urge you to use the 3-wire type), an on/off switch, and a fuse (with holder). I also suggest a "power on" indicator of some sort; I prefer using LED indicators on the DC side of things, but you could also use a pilot lamp designed for the 115 volt AC side of things (or even a power switch with a built-in indicator). Be careful when connecting things to the AC mains -- shocks can hurt...and in some cases might be fatal! If in doubt, have a knowledgeable friend do this for you.

A special word about using two or more PC supplies on the same layout:
Many of these critters have transformers which are not used in the "traditional" way, and may therefore lack isolation from the AC mains; if you use two or more, it's critical that they each use 3-wire power cords to ensure that the cases are at earth ground (and NO cheater plugs). Note that it's a very good idea to use the 3-wire cord even if you're using only one supply. Now...if your home has only the older 2-wire outlets and you want to use multiple supplies, you MUST provide a reliable earth ground for that third wire; if you're unsure how to do this, please consult a qualified electrician.

BUT...I DON’T HAVE AN OLD PC SUPPLY…

OK...HOW DO I DO IT?

Enter small gauge solid copper wire. Remember the discussion about such things in the Ohm's Law article above? Well, here's a excuse to apply the theory. #30 wire, commonly used for wire-wrapping, is very useful (and available at RS as 278-501, -502, or -503...in colors). The resistance of #30 solid copper wire is only 10.5 ohms per 100 feet (at 25° C); that's 0.105 ohms per foot, so 12 inches of this stuff would make a great 0.1 ohm resistor (actually, 11.43 inches is exactly one-tenth of an ohm, but 12 inches in within 5%, and that's pretty close). How about that 0.17 ohm beastie I postulated above? Think 19.43 inches of #30; I'd probably settle for 19½", and be very satisfied. Here's a table of the resistances of 100 feet of some common solid copper wire sizes (@ 25°C):

Radio Shack (and most other electronics suppliers) have spools of #18 thru #22; category 5 (Cat5) networking cable, telephone cable and most intercom cables use #24; wire-wrapping is #30 -- it's all out there, but you may have to hunt a bit. Finally, here's the simple formula for finding the exact length of wire you need given the resistance/100 ft from the table above:

Let's try the formula for the length of #30 wire (@ 10.5 ohms per 100 ft) that would give 0.1 ohms:

Easy, huh? Just remember that the resistance is measured between the points where you actually attach a larger wire to the smaller wire from which we're making that resistor; in other words, don't cut that #30 wire exactly 11.43" long -- you gotta' leave room to solder your #20 wire (or whatever) to it...or to attach you alligator clips. Power dissipation should not normally be an issue, so long as you're not over 10 amps; to make sure you're not overdoing it, just feel the wire -- if it's too hot to touch or melting the insulation, you may be overdoing it. If the insulation IS melting, I'd suggest using two paralleled pieces of wire, each twice the length calculated (or 3 pieces of 3X the length, etc.).

WHY?

I'll admit that we're getting pretty esoteric here, but bear with me -- constant-current sources do have their place in the world of model trains. If you think back to the discussion on Ohm's Law, you'll know that the current through a resistor varies according to (a) the voltage, and (b) the resistance. But what if you want the current to be constant regardless of typical voltage variations -- such as the headlamp in a locomotive? Enter the "Constant-Current Source" (CCS), a handy way to sorta' "amend" Ohm's Law on a case-by-case basis.

For power supply voltages greater than 12 volts and output loads less than 1mA, R1 = 10K is a good starting point. Then, if we assume the transistor is OFF, the output current will all flow thru R2, producing a voltage equal to I(out) x R2. If I(out) increases such that the voltage drop across it exceeds the voltage required to forward-bias the Base-Emitter junction of the transistor (about 0.65V), then the transistor will begin to turn ON, causing current to flow from its Emitter to its Collector, and hence to ground. This "shunts" current away from the output, thereby causing the output current to remain virtually constant right at the value where conduction begins. If we let R2 = 6.2K, then the magical value of output current is given by I(out) = V/R =
0.65/6200 = 0.000105A, or about 0.1mA.

DO I REALLY CARE?

You're probably wondering, "What am I gonna' do with a constant current of only 100uA?" Practically speaking, not too much...although it is a good value to produce a linear voltage ramp on a capacitor (remember: if you charge a cap thru a resistor, the voltage will increase exponentially, not linearly; sometimes it matters). Let's try to think of something a bit more useful here: suppose we wanted to illuminate an LED with a constant 15mA (0.015A); what values of R1 and R2 would be called for? Using the formula R = V/I, we'd calculate R2 = 0.65/0.015 = 43.3 ohms, so we'd hunt for a 43 ohm resistor, and probably end up setting for 39 or 47 ohms (or even paralleling two 82 ohm units to produce a closer fit of 41 ohms). If we're using a 12V supply, then we'll want R1 to drop the "leftover" voltage; we've got 2.2V on the LED and 0.65V across the transistor, so the remaining 12-2.2-0.65 = 9.15V would indicate R1 = 9.15/0.015 = 610 ohms. Did you spot the fly in the ointment? It's the assumption of a 12V supply; if it were constant, then we wouldn't need the CCS... resistor would work just fine. But now imagine that the supply varies between 6 and 12 volts, but we want 15mA thru the LED at all times (ie, constant brightness) -- if we recalculate the minimum voltage across R1, we get V = 6-2.2-0.65 = 3.15; so R1 needs to be R1 = 3.15/0.015 = 210 ohms (use 220 or 200). Give this a try in the old laboratory just to convince yourself it really works: vary the supply voltage from 6 to 12 and observe the constant brightness of the LED. Once you're convinced, you can amaze your friends (or in my case, our Basset Hounds, who watch everything I do on the bench very closely).