Electricity Labs

Setting up a breadboard

nothing to say here, literally plug it in

Simple LED circuit

1 The resistor is used to regulate how much current flows through the circuit. The LED itself has negligible resistance, so Ohm's law dictates that an insanely high current will run through the circuit. The resistor then causes the entire circuit to have a non-negligible resistance, so the current through the circuit isn't insanely high.
2 V is 5V, R is 560 ohms, so I is 5/560A, or 8.93 mA.
3 V is 5V, I is 15 mA, or 15/1000A, then R must be 333 ohms.
4 2.8V. The entire circuit drops 5V, and if the LED drops 2.2V, then the resistor must drop 2.8V.

Simple LED circuit with switch

(for this exercise, I had plugged in the switch in the wrong orientation, so it didn't function. I ended up recording my partner's circuit)

1 Nothing would change
2 Nothing would change
The reason nothing changes is because if the switch is on, then the entire circuit is connected and functional. if the switch is off, then the circuit is disconnected. You need the circuit to be a complete loop from and back to the power source.

Simple LED circuit with potentiometer

1 Same reason as the simple LED circuit. You might think that the potentiometer would be enough, but if the potentiometer is set to nearly 0 resistance (which it can easily be set to), then the same problem as the simple LED circuit would arise.
2 Use Ohm's law. V = 5V, R = 560+10k ohms, then I is 0.473 mA.
3 Photoresistors, thermoresistors, digital resistors, and rheostat.

Dueling LEDs circuit with potentiometer

1 The potentiometer has 2 sides, a top and bottom. Turning its dial changes on which side the majority of the resistance lies (and how much). As in, top and bottom must always add up to 10k ohms. So if the top is set to 1k resistance, then the bottom must be 9k resistance. So we can control which side is "dimmer" or "brighter" (or "equally as dim/bright"). Whichever side has the higher resistance has less current flowing through it, so the light on that side is dimmer.

Capacitor charging circuit

1 It will take longer to charge the capacitor. Adding the 2.2k ohm resistor implies that less current flows through the entire circuit (via Ohm's law), so the capacitor will take more time to build up charge (as current is charge per unit time).

Capacitor discharging circuit with LED decay

1 When the switch is off, there is a circuit within a circuit (the circuit made by the capacitor, 560 ohm resistor, and the LED). The capacitor here acts as a temporary power source (battery), and can light up the LED for a short amount of time (as in, discharges via the LED).