# Electricity Labs

## 1. Setting Up A Breadboard ## 2. Simple LED Circuit  Questions:

1. The purpose of the resistor is to limit the amount of current running through the circuit. An LED light is a semi-conductor and can only accept a certain amount of voltage (~2V); thus, the resistor acts to make sure the LED does not overheat or explode.
2. Ohm’s Law: V=IR. V=5V, R=560Ω. The current running through this circuit is I=V/R=(5/560)=8.9mA.
3. In order to have only 15mA of current in this circuit, the resistor value must be 333.333 ohms. I=15mA, V=5V, R=V/I=(5/0.015)=333Ω.
4. Kirchhoff’s voltage law states that the sum of all voltage added and all voltage dropped in a closed loop circuit is equal to 0. Thus, assuming the LED drops 2.2V, the resistor drops 2.8V (5V-2.2V=2.8V).

## 3. Simple LED Circuit with Switch

Questions:

1. If you moved the switch so that it was connected between the resistor and the LED, the circuit’s behavior would not change. As long as the pushbutton switch disconnects the power supply from ground, the circuit remains open and would still act in the same way.
2. If you moved the switch so that it was connected between the LED and ground, the circuit’s behavior would not change.

## 4. Simple LED Circuit with Potentiometer

Questions:

1.  The 560Ω resistor is necessary to limit the amount of current running through the circuit. Since the potentiometer is a variable resistor, when it is turned all the down (i.e., at its lowest resistance) there may be an excess amount of current which can burn out the LED light.
2. When the 10kΩ potentiometer is turned all the way up (maximum resistance), there is 0.47mA of current flowing through the circuit. Ohm’s Law: V=IR. V=5V, R=560Ω+10000Ω=10560Ω. I=V/R=(5/10560)=.47mA.
3. Variable resistors: potentiometers, rheostats, thermistors (temperature sensitive), photoresistors (light sensitive).

## 5. Dueling LEDs Circuit with Potentiometer

Questions:

1. This circuit behaves this way because both LEDs are connected in parallel with the resistor in between – therefore; turning the potentiometer increases resistance on one end whilst decreasing resistance on the other. The LED with higher resistance is dimmer than the opposite LED with a lower resistance.

## 6. Capacitor Charging Circuit

Discharging the Capacitor:

Questions:

1. Adding a 2.2kΩ resistor between the pushbutton and the capacitor affects the capacitor’s charging behavior by slowing down the charging time of the capacitor. When the button is pressed, the current is limited as it flows through the resistor and thus slows down the time it takes to charge the capacitor.

## 7. Capacitor Discharging Circuit with LED Decay

Questions:

1. The capacitor discharges up through the LED part of the circuit and not directly from the capacitor down to ground because when the pushbutton switch is let go (open), the capacitor becomes the power source for the LED. Since charge has to flow from power to ground, this can only be accomplished by discharging through the LED part of the circuit, not directly from the capacitor to ground.
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