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.
The LED I used is a green LED. According to the data sheet, the forward voltage is 2.2V， which is the voltage on the LED. So,
V(resistor) = V(power) - V(LED) = 5.0V - 2.2V = 2.8V.
Since the resistor and the LED is in series, the current is the same, which is 25mA. So,
R = V(resistor) / I(resistor) = 2.8V / 25mA = 112 Ohm.
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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.
This is the minimum resistance. If the resistance is lowered, then the current will become higher, which leads the LED fail.
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c. What is the resistance range of the potentiometer?
To be precise, according to the multimeter, the range is 0.3 Ohm up to 9.9 KOhm.
So it should be 0 Ohm to 10 KOhm.
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a. Does it matter what order the components of your circuit are arranged between power and ground? Why or why not?
No. The components is all in series, which means the current of every component is the same. So the components will do their jobs regardless of the order.
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a. Using this battery, what is the minimum resistance required for use with your LED?
For the new resistor,
V(resistor) = V(power) - V(LED) = 9.0V - 2.2V = 6.8V.
Since the resistor and the LED is in series, the current is the same, which is 25mA. So,
R = V(resistor) / I(resistor) = 6.8V / 25mA = 272 Ohm.
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Link: https://www.youtube.com/watch?v=4l4ydhmqI7Q&feature=youtu.be