I found the following circuit in a random video (don't know what it's worth) and have two questions about it. I get that it's a full wave rectifier, and I think I have a decent idea on how it works. However:
That is a de facto circuit for cheap generic mains powered LED lamps.
The capacitor is an essential part, the circuit is called a "capacitive dropper" which is used to bring down the voltage at modest current level to power the LEDs.
The 220k resistor is necessary to keep the LEDs turned off, so that they do not glow faintly due to capacitive coupling in the mains wiring powering up the LEDs ever so slightly.
Not sure why they put the 220K there. It's not essential to the operation- it will make the LEDs go completely dark a bit faster when power is removed. As others have mentioned, it will help prevent a faint glow from the LEDs caused by capacitive coupling to the wires (say there is a long run of wires to the on/off switch, a tiny bit of current will flow even if the switch is off- and modern LEDs are efficient, and the eye is very sensitive when the ambient lighting is low).
The 105JF (1uF) capacitor is the dropping element to power the LEDs. It has an impedance of a few k\$\Omega\$ at 50Hz. It is crucial to the operation. The 470K is just to bleed off the charge when the power is removed so you won't get too much of a tingle from pulling out the male plug and touching the prongs. You could use a ~3K resistor instead, but it would get quite hot and waste power.
This is a pretty crappy circuit. At a minimum it should have a fusible series resistor or fuse + series resistor to protect the connections and to limit current if powered near the AC waveform peak.
That's a cheap and nasty "capacitive-dropper" low-voltage PSU.
I like the other ideas suggested, here, but just because it isn't stressed much in the answers, bleed resistors seem quite likely. The one in parallel with the LEDs makes sense to me aside from the turnoff and other reasons mentioned could be because resistors are generally more reliable than LEDs and also the most likely time someone will cut the power to actually mess with the circuit, potentially leaving 105J charged, is after an LED fails open (which in this case could be inevitable with the crappy nature of the driver), potentially leaving 470U charged to high voltage. The resistor can also be isolated and potted, whereas the LEDs can't shine light if they're not exposed. Importantly the most likely time someone will endanger themselves by messing with those caps is right after that happens.
Worse yet, the damage could have been mechanical. Someone mentioned this is a cheap driver for an LED bulb. Those bulbs are durable but far from indestructable. It would be foolish to assume a layman will think "There could be a capacitor in there. I should give the bulb time to discharge after I power it down and be careful not to touch contacts.". Instead, the device should simply be safe within a reasonable interval of time after power down.
Well one of the answers above was mostly right. It is the most basic of a single phase ac to dc full wave rectifier. The cap is in there to smooth the ripple and the res is to discharge the cap when power is disconnected. And the above answer is spot on about never assuming someone would never put their little piggies in where they don't blong and 470uF is enough go knock the daylights outta 300 pound man if its anything rated above 18 volts so it would arc thru you.
Ingenious simplicity. I am sorry but I can not agree with such offensive classifications regarding this circuit - "crappy", "nasty", etc. If it really was, it would not exist perhaps more than a century. Instead, I would say this is another example of what Don Lancaster has called "elegant simplicity." Moreover, I would say "ingenious simplicity" because it is essentially just a capacitor in series to the load. That is why I propose, first at all, to explain its idea in simple words... and not in learned phrases worn out by repetition...
The problem is to reduce an (AC) voltage. For this purpose, we can connect an element in series to the power supply with the idea of subtracting ("stealing") a part of the voltage.
The "stupid" solution is to connect a simple resistor or the more sophisticated transistor. Both will dissipate colossal power in this case especially if the load requires very low voltage. To show this, a few days ago, I suggested to my students to power an LED (2 V) from the mains (310 V peak value here) in this way - by a resistor in series. Then I reminded them to calculate the power dissipated by it... and they were very surprised by the result. Thus, they became convinced of the futility of losing a voltage of 318 V across the "ballast" element to obtain a voltage of only 2 V across the LED.
The clever idea is to connect an opposing voltage source in series. Its voltage will be subtracted from the supply voltage and only the residue will be applied across the load. What should be this source? It should be time-varying and rechargeable source. Thus it will alternatively absorb and return power to the circuit… and there will be no loss of power. The capacitor acts as such a source here...
Capacitor vs resistor. So, the capacitor's clever trick to decrease voltage is to absorb power by storing it while the resistor's "stupid" trick is to absorb power by wasting it. Another clever trick to reduce voltage is "by switching"... but this is another story...
The load. However, the load must pass current in both directions. Such a bilateral load is, for example, an incandescent lamp. In addition, it is slow-acting; so there is no need for a filtering capacitor. Since the LED is a one-way load, we connect a diode bridge before it… and since it is fast-acting, we connect a filtering capacitor in parallel to it.
