Lab: Feb 4 2015

Part 1: Simple LED Circuit

1. The purpose of the resistor is to weaken the current running through the circuit, preventing the LED (or anything else) from burning out.
2. Since V = 5V and R = 560 Ohms, then I = 5/560, or 8.9 mA.
3. If V = 5 and I = 15mA, then R = 5/0.015 =  330 Ohms
4. 5 - 2.2 = 2.8 V

Part 2: Simple LED Circuit with Switch

1. The circuit's behavior would not change. It'd just have the resistor dampen the current before it gets to the switch.
2. The circuit's behavior still would not change.

Part 3: Simple LED Circuit with Potentiometer

1. The resistor is necessary because without it, if the potentiometer was at its lowest setting (allowing the most current to pass through), the LED would burn out.
2. V=IR ; V = 5 and R = (10,000 + 560), so I = 5/10560 = 0.047 mA.
3. Rheostat, Thermistor, Humistor, and Photoresistor.

Part 4: Dueling LED's Circuit with Potentiometer

1. When the potentiometer is turned, it increases the amount of current going through one LED (and therefore brightening it), and decreases the amount of current going through the other LED, dimming it.

Part 5: Capacitor Charging Circuit

1. The capacitor will charge more slowly than before, since there will be less current flowing into it and charging it. As a result, the LED will stay lit longer.

Part 6: Capacitor Discharging Circuit with LED Delay

1. The capacitor discharges up through the LED part of the circuit because that's where the positive charge is held, so in order to discharge, the capacitor has to go from the positive side (facing the LED part of the circuit) and travel through that section of the circuit in order to reach the ground.