Question about time constant of Voltage Multiplier
You have a pair of half-wave rectifier DC power supplies connected back to back to double the output voltage.



Figure 1. A simple half-wave rectifier with DC smoothing capacitor. Source: Electronics Tutorials.


Without the capacitor the output voltage from this circuit would just be the positive half of the sine-wave.
By adding a capacitor we can charge it up when the AC voltage > capacitor voltage.
The capacitor then acts as a reservoir during the dips in the AC and supplies the load.
The slope of the discharge of the capacitor will be inversely proportional to the \$ RC \$: the higher the resistance of the load, the slower the discharge; the higher the value of the capacitor, the slower the discharge.
The variation between the peak voltage and the lowest dip between half-cycles is known as "ripple".


All that \$R_LC_{1,2}&gt;&gt;T=1/f\$ is telling you is to make sure you have enough capacitance in there to keep the ripple small for the mains frequency you are on.

Example:


100 Ω load.
f = 50 Hz so T = 20 ms.


Rearranging the formula to find C: \$C_{1,2}&gt;&gt; \frac {T}{R_L} = \frac {20m}{100} = 0.2 \ \mathrm {mF} \$ 

So if we consider that a factor of ten will satisfy the '>>' requirement we would choose 2 mF or 2000 µF capacitors.



Note that the calculation in the original circuit is complicated by the fact that the load resistance is connected across both supplies. For this reason when calculating the RC values we should consider half the load resistor for the R value. (The mid-point of the resistor is approximately at midpoint of the voltages.)