Voltage drop with LED

So, to start out, you have to understand that the LED here has a voltage rating of 2V. This means that for anywhere from a few milliamps to enough current to destroy it, it'll have very close to 2V across it. Anything less than 2V and very little to no current flows. Anything more than about 2V and way too much current flows, destroying it. Now that you understand that, you can mathematically compute the current. You should probably know that the 10ohm and 20ohm resistor is series make a total resistance of 30ohms. And since the LED has 2V across it, the 30ohm resistor has 2V on one side and 3V on the other (since the battery/power source/whatever has a voltage of 3V). So, a total of 1V across it. Now, using ohm's law, that computes to 33.3 milliamps. This is a little higher than the actual value of 31.7 milliamps because you are overloading the LED and the voltage across it is slightly higher than 2V. But it's very close to 2V, so the current is only off by one and a half milliamps.

Thanks @jason9. I already understood this but thank you for the confirmation. I'm trying to find out why EveryCircuit computes the current to 31.7mA.

Place a voltmeter in parallel to led. It show 2.05 V and so following Jason explanation you jave to divide 0,95 / 30 = 0.03166666 ampere rounded to 31,7 mA

Thanks @giomix, that is clever. But why is it 2.05 volts? How does the led cause that.

Put your voltmeter across the LED and vary the resistor value. You will see that the current changes and, therefore, the volt drop across the LED changes (just as it does across the resistors).

With a total resistance of 50 Ohms you will have the desired 20mA current and 2V drop across the LED.

The other 1V will be shared between the two resistors according to their resistance values.

@wimjongman, check this hope it helps http://everycircuit.com/circuit/5939939360112640

Thank you @hurz. This is what I wanted to know. Now I can sleep again :)