Bhuvanesh123
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analogWrite(pin num,255)//100% duty cycle so it completely on
analogWrite(pin num,125)/50% dutycycle so its 50% of time on and 50 percent of time off)
my question is in the above image whatever may be the duty cycle it on (5v) and off(nearly o v)
then how do we getting half of that voltage i mean around 2.5 from the actual 5v(change in value between 0 to 255 only says on time of time )
.thank you in advance