shaneshane1
New Member
I have a simple LED voltage drop question
This is a really stupid question that i cant seem to get my head around
here goes
I have a LED running of 12V
so 12V - 2.1V(voltage drop) = 9.9V
then 9.9V / 0.020(Amps) = 495ohms
so my question is do i NEED to include the voltage drop in the math?
when i put a multimeter on the anode(+) of the LED it shows 2.1V
and when i do the math without the voltage drop included
so 12V / 0.020 = 600ohms
I still get 2.1V on the anode of the LED?
is the current to the LED decreasing if i dont add the voltage to the math?
and also if im reading 2.1V on the LED anode does that mean there is a voltage loss of 9.9V through the resistor?
This is a really stupid question that i cant seem to get my head around
here goes
I have a LED running of 12V
so 12V - 2.1V(voltage drop) = 9.9V
then 9.9V / 0.020(Amps) = 495ohms
so my question is do i NEED to include the voltage drop in the math?
when i put a multimeter on the anode(+) of the LED it shows 2.1V
and when i do the math without the voltage drop included
so 12V / 0.020 = 600ohms
I still get 2.1V on the anode of the LED?
is the current to the LED decreasing if i dont add the voltage to the math?
and also if im reading 2.1V on the LED anode does that mean there is a voltage loss of 9.9V through the resistor?