Just to add to Keith's reply, you can think of it either way as he says, you just need to think of it the same way all the time so you don't get confused. If you are comfortable with conventional current flow, positive to negative, use that. The actual electrons that are the physical current are moving the other way, but it really doesn't matter at this level. It is good to know, but it doesn't matter unless you are working at a molecular level.

Regarding resistor position:

If you are creating a single path for current, e.g voltage source>LED>resistor>ground, the position is not important because of WHY you are using the resistor. You are using it to limit the current through the LED, which is actually a diode. A diode will destroy itself if you let unlimited current flow through it. In this case, you want to keep the current to a safe level, and since all the current is the same in all points of a series circuit, it doesn't matter if the resistor or LED are connected to the positive terminal (in this case your arduino pin).

For someone starting out in electronics, I would highly suggest mastering Ohm's Law. It is the key to understanding the whole of electronics. Your Arduino will help with that. If you actually take the time to figure out how the resistor protects the LED, (and your arduino pin), then do some very simple calculations, you can start to understand.

For example, the very most current an arduino pin should be required to sink (when it is the negative terminal) or source (when it is the positive terminal) is .040A or 40mA. In practice, you should always use less current, but that is the maximum. So, if you just hook up a regular garden variety red LED, it will likely drop about 2.0V across it, and it needs 10-20mA to make a pretty bright red light. So, let's just arbitrarily pick 15mA of current through the LED. Since a diode (LED) will conduct in a forward biased mode as much current as you allow it to have, but it is a small device and cannot get rid of a lot of heat quickly, it also has a maximum power rating. Let's say 100mW. (.1 Watt). If you check the power the LED will have to disburse as heat, by Ohm's Law, 2V*.015A = 30mW (.03 Watts). So, we know we can safely put that current through the LED. Now, what about the resistor? How much resistance to supply only that much current?

The LED drops about 2V, the resistor has to drop the other 3V, because your arduino pin is going to put out 5V. Since we need to use all the voltage, we want to know what resistance will use up (drop) 3V across it at our desired 15mA of current. Ohm's Law...3V/.015A = 200 ohms. So we know the current in all points in a series circuit is the same, and we know that 200 ohms will drop 3v at our desired current. We're good....except for one thing, the resistor gets drops that voltage the same way the LED does, by converting to heat, so how big (physically) of a resistor do we need to get rid of the heat? Ohm's Law 3V*.015A = .045 W (45mW) which is way less than a 1/4W, less than half of an 1/8W, so thought there are 1/16W resistors with wire leads, it is pretty safe to say anything you would have that is easy to see the colors on would work just fine! (most common size in kits is 1/4W) And everything works great!

Parallel circuits are different. The current through each of the paths for current are different according to the resistance in that path. Get the series thing down, move to parallel after you can do what I did for any series circuit. In a series circuit, the voltages are different, in a parallel circuit, the currents are different.

It all still comes down to Ohm's Law!

HTH,

-fab