LED Blink

Solving Approach

The Arduino gives 5 volts. The LED only needs about 2 volts. That means the extra 3 volts must drop across the resistor. If we want around 10 mA of current, we divide 3 volts by 0.01 amps. Since 300 Ω is not a common resistor value, we use 330 Ω, which is close. This makes the LED current about 10 mA, which is safe for the LED.

Code

void setup() {
  pinMode(12, OUTPUT);   
}

void loop() {
  digitalWrite(12, HIGH);  // LED ON
  delay(400);              

  digitalWrite(12, LOW);   // LED OFF
  delay(800);              
}

Output

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