Electrical – Wiring DPDT switch to add parallel resistor

Well, I'm studying electrical engineering right now and I can tell you that such jumps as you described take around two years of lectures at my university.

First thing which is important is to know which elements are passive and which are active. Then you need to know which elements are linear and which aren't. Next step is to get equivalent schematics for elements which you have and to see how they behave.

For example, let's take the switch. In off state, it functions as an open circuit, while in on state it functions as short circuit. Next, if you have sensitive equipment, you'll be able to notice that the switch isn't actually short circuit because it has some resistance, but that it's very low. Now let's take a look at the diode. Diode isn't linear component, so it doesn't have resistance in the classical sense in which for example resistors have. Instead there's the V-I curve of the diode. On a resistor, it's a linear function and we can use resistance as its characteristic, but on diode, it looks exponential.

As you can see from the image, certain voltage is needed for diode to start working properly and when you trigger the switch, that voltage disappears. That means, that the "resistance" of the diode just became huge. To get a feeling for this, use the parallel resistor calculation for say 1 mΩ resistor and 1MΩ resistor and take a look how much current goes through each of them. This is the way the circuit you mentioned behaves.

If these are 12V LEDs then they most likely have some sort of regulation built in, so you don't have to worry about putting in a resistor. If you have the documentation for them, I would put it in your post.

2) If you wanted to put resistors within each parallel circuit, it would also need to have the same resistance as the one you would put at the beginning or end node, not 1/5. The advantage of doing this is that you could have smaller resistors rated for a lower power than one big resistor rated for a high power.

3) Short answer, no, the resistor is generating heat that would have been made at the LED if the resistor wasn't there.

4) Those don't sound like regular LEDs, but if you mean LEDs like the through hole 3mm or 5mm most of them are indeed designed for 20mA current draw at a characteristic forward voltage.

LEDs are non-linear semiconductor devices, their current draw is non-linear with the voltage across it. Look at this graph to see what I mean. http://en.wikipedia.org/wiki/File:Diode-IV-Curve.svg