8. LED Blink

How do you plan to solve it?

 1->>Set a digital pin as OUTPUT

2->>We need to limit current 40mA to 10mA so we should design a resistor value,Let us assume RED LED forward voltage drop is 1.1V and given that forward current is10mA

R=(Vin-Vf)/(If)=(5-1.1)/(10*10-3)=390 ohms

for better saftey as well as refering standard resistance value we take 470ohms.

Since while calculating the power we get P=VI=1.1V*10mA=0.011W

so for better saftey we take 0.25W resistor

so finally we take 470 ohms 1/4 W resistor
 

const int ledPin = 8; 
void setup() {
  pinMode(ledPin, OUTPUT);
}
void loop() 
{
  digitalWrite(ledPin, HIGH);
  delay(800);
  digitalWrite(ledPin, LOW);
  delay(400);
}

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