In this robotics challenge, we built breadboard circuits with multiple LEDs in series or parallel, and tested the voltage drop between components to learn about parallel and series circuits. These are the notes we took during the challenge:
Voltage Calculation
9V source
R = (Vbat - Vled) / (Iled + Iled)
R = (9V - 2V) / (20ma + 20ma)
R = (7V) / 0.04A
R = 175Ω
We chose a 220Ω resistor because it seemed to be the closest we could get to 175Ω without going below, which could harm the LED’s by giving them too much current.
Breadboard circuit with 2 LEDs in parallel:
Voltage Drop Across Components (2 parallel LEDs):
9V Battery: 7.9V
Resistor: 5.0V
LED1: 2.9V
LED2: 2.9V
Breadboard circuit with 2 LEDs in series:
Voltage Drop Across Components (2 series LEDs):
9V Battery: 8.0V
Resistor: 2.3V
LED1: 2.9V
LED2: 2.9V
Breadboard circuit with 3 LEDs in series:
Voltage Drop Across Components (3 series LEDs):
9V Battery: 8.1V
Resistor: 0.3V
LED1: 2.6V
LED2: 2.6V
LED3: 2.6V
I believe that the resistor consumes less voltage depending on how many LEDs there are in series because each LED in series consumes the same amount of voltage to itself; If there are 2 LEDs in parallel, the voltage across both is the same and therefore the voltage required for both is the same as is consumed by one LED. When LEDs are in series, each one takes the same amount of voltage from the circuit, so 3 times the voltage is needed to power all three compared to a parallel circuit.
The easiest aspect of this challenge was assembling the circuits, because I already have some experience with breadboards. The hardest part of this challenge was calculating the resistance required for the LEDs, because I’ve never used the formula before and didn’t know it. In the process of this challenge, I learned how to calculate the required resistance for components, and learned about how voltage drop varies between parallel and series circuits, and how resistors take less of the voltage drop depending on the other components requirements.
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