LED Blink

Solving Approach

To solve the LED blinking task, you need to write a code that turns the LED on for 400 milliseconds and off for 800 milliseconds, and you must calculate the correct resistor value to limit the current to 10 mA with a 5V power supply. Resistor Value Calculation To determine the value of the current-limiting resistor, we can use Ohm's Law: R = \frac{V}{I} However, in a circuit with an LED, we must account for the forward voltage drop (V_f) of the LED itself. The voltage across the resistor (V_R) is the supply voltage (V_s) minus the LED's forward voltage drop. V_R = V_s - V_f The current (I) is the same through both the resistor and the LED since they are in series. Therefore, the formula for the resistor value is: R = \frac{V_s - V_f}{I} Based on the image, the supply voltage (V_s) is 5V and the desired current (I) is 10 mA (or 0.010 A). The forward voltage drop (V_f) for a typical red LED is about 2V. Plugging in the values: R = \frac{5V - 2V}{0.010A} = \frac{3V}{0.010A} = 300\Omega So, a resistor with a value of approximately 300 Ω is required. Since 300 Ω is not a standard value, you would typically use the next highest standard value, such as a 330 Ω resistor. Arduino Code for LED Blinking The code for the blinking task needs to be written inside the loop() function and use the digitalWrite() and delay() commands.

Code

const int ledPin = 13; // Use the pin number the LED is connected to
void setup() {
  pinMode(ledPin, OUTPUT); // Set the LED pin as an output
  }
  void loop() {
    digitalWrite(ledPin, HIGH);  // Turn the LED on
      delay(400);                  // Wait for 400 milliseconds
        digitalWrite(ledPin, LOW);   // Turn the LED off
          delay(800);                  // Wait for 800 milliseconds
          }