From the youtube comments. Breaking down long questions and explanations is difficult on YouTube's comment system.
For context, the original poster was asking about a LED and Resistor found in an AC Mains extension cord. It's a surprisingly common way to add an LED indicator to an AC circuit. Just like in DC, the current limiting resistor drops all but the LED's forward voltage. A better design is to use a "capacitive dropper." That's where you use a series capacitor to couple the mains to a lower voltage DC circuit. It's more expensive and usually only used when driving more than a single LED.
Here was the follow-up to that response.
I've googled about this circuit and most of the responses in forums, did not believe that this setup is possible.
Not sure what to say. As I said in the original reply, this is common. I'd be about careful what you read on the Internet. 
As I understand in your video, the voltage will have no direct effect on the LED (even at 220v) because it is dropping or just using it's forward voltage (e.g 3.2 v) and the rest goes to the resistor.
Yes. It is the same concept whether DC or AC. The series resistor drops whatever voltage is left after the LED drops its forward voltage. In the case of AC two conditions apply:
- The voltage is always changing. So the LED is pulsing. You might notice this pulsing since LEDs respond to voltage change very quickly. However,
- The pulsing is swamped by the flickering. At a glance, you won't notice the flicker, but stare hard enough, and you'll see it. Since the LED is a diode, it only conducts current in one direction. So when the AC waveform 's polarity changes, the LED is off--meaning its only on "half" the time. Even at 50Hz, 25 flashes a second is probably fast enough to seem steady to most people.
Only the current has a direct effect on the LED, that's why there is a need for the current limiting resistor.
I don't understand your use of "direct effect." Current and Voltage are dependant on each other. The need for the limiting resistor is that LEDs do not safely self-limit their current when turned on. Also, the junctions have voltage limits. One concern is that the LED needs to have a reverse bias voltage rating of the line voltage. Otherwise, when the polarity goes negative (or opposite), LED is dropping the entire AC voltage.
I've also came across Kirchoff's Law while browsing that somehow explained the voltage drop across the resistor together with a load like an LED, based on this the voltage drop on the resistor will be 220V-3.2V (Vf of LEd)= 2168v (AC or DC?) is this correct?
Yes. In a series circuit, voltage drops add up, and current is the same through all components.
Forgive me for not trying this myself since I'm still new and a bit scared to test it out on live mains.
Please do not try. Unless you have had classes or training on working with AC mains voltages, I would not play around with them. There are many safety concerns that you can't just "pick up" from internet tutorials.
