Capacitor drop power suuply
Simplified tremendously a capacitor looks like a resistance to AC. (actually its an impedance but i'm simplifying). The conversion is 1/(2*pifc). In your case the capacitor looks like a 16k resistor. Then it's a simple voltage divider 

230*(800/(470+800+16000)) = 10.5v

There are going to be some losses unaccounted for and the diode will drop something. 8v sounds about right. 

You can not replace the 800 ohm resistor. This would be dangerous.







simulate this circuit &ndash; Schematic created using CircuitLab

The voltage regulator would look like a variable resistor depending on the load. If there is every a condition where the is no load the regulator would look like a near open circuit. This would expose the regulator to extremely high voltage. 







simulate this circuit

The two circuits pictured above are safer. They both limit the voltage to the regulator even if the regulator is unloaded. They also place two 100k resistors across the big capacitor that drain any residual voltage. My favorite is the circuit with the zener diode.