Power supply = 5.15V / 2.5A

LED = 3.0Vf / 20mA

As I am trying to fully grasp Ohms law, I came up with the following:

V=IR

I=0.02A ( from the LED spec)

V=2.15V (5.15 from the supply - 3.0 drop from LED spec)

So R=0.02/2.15

R=107.5 Ohms

Common sizes of resistors near this seem to be 110 and 120 Ohm, and I'd probably go with the 120 to give a little safety margin given its an unregulated supply and may fluctuate.

Question 1 - is my thought process correct so far?

Now, I don't have any resistors of that size on hand but I do have a 210 Ohm on hand. I confused myself trying to figure out what this would do. If the resistance is now fixed, but I am not sure which would change - voltage or current. My gut says voltage would stay at 2.15 and it is current that would change. So that means the I would now solve for I:

I=V/R

I=2.15/210

I=10.24mA

Running almost half the current would result in a much dimmer led is my assumption and the voltage would not be changed at all due to the different resistor size.

Question 2 - do I have this part right?

Thanks for putting up with some very basic questions.