a. What resistance do you need to limit current to 30 mA (if using red LED) or 25 mA (if using yellow or green)? Be sure to state which color LED you are using. This resistance refers to the total resistance in series with the LED.

Hint: Make sure that you account for the forward voltage drop (Vf) of the LED that you're using.
A: Using 5V and a current of 30mA

V rest = V bat - V led. =5-1.8
V res = 5 - 1.85
V res = 3.15 V

V= R*i
3.15 = R * 0.0030
R = 105 Ohm's
✓

b. Is the resistance from question a) a maximum or minimum resistance? That is, in which direction if you change the resistance (higher or lower) would the LED likely fail.
A: is the minimum resistance that you need... Lower than 105 Ohms will make the led fail
✓

c. What is the resistance range of the potentiometer?
A: It is 10K Ohm for the 302B potentiometer 
✓ ~0 to... 

a. Does it matter what order the components of your circuit are arranged between power and ground? Why or why not?
A: no, because they are connected in parallel
✓ They're actually not connected in parallel... What does it mean for components to be connected in series/ in parallel?

Now, let's make things a little more portable. Disconnect the breadboard from the power supply. Use the battery clip and a 9V battery to power your LED circuit. You'll need to replace your 150 Ohm resistor for the higher voltage. Be sure to do the calculation below so you know what resistance to use. You can choose a conservative resistance so you are not at the edge of failure.

a. Using this battery, what is the minimum resistance required for use with your LED?

V = R * I

V res = V bat - V led
V res = 9 - 1.85
V res = 7.15 V

V = r* I
7.15 = r * 0.0030
R = 2,373.33 Ohm's

-0.5 Look at the value you're using for your current!