If you are using capacitors (like X2) in the supply mains (e.x. 230VAC/50Hz) or elsewhere, there might be a need for a resistor to make a controlled discharge the capacitors. This resistor is typically called a bleeder resistor.

The following formulas describe the Voltage over the capacitor (Vc) and the current through it (Ic).

$$Vc=V\times e^{-t/Tau}$$

$$Ic=-V\times R\times e^{-t/Tau}$$

Vc: Capacitor voltage.
V:   Start voltage on capacitor.
Ic:  Capacitor current.
R:   Resistor.
τ:   Tau is the time constant τ = R x C in seconds.
t:    time in seconds from V to Vc.

A = Charging curve for the capacitor and B = discharging curve for the capaciter.

By rewriting the formulas we get:

$$t=-ln (\frac{Vc}{V})\times R\times C$$

$$R=\frac{t}{-ln (\frac{Vc}{V})\times C}$$

Here is a example of calculating the bleeder resistor for discharging.

When disconnecting the main supply, the voltage over the capacitors can be very high. If disconnecting at the top of a of the 50 hertz sinus the max. voltage (starting voltage Vstart) will be 253V x  √2 = 360V. This voltage need to be discharged to about 60V (End voltage Vend) within max. 0,75 second. Let us say C = 1,32uF in this case.

We want to discharge the capacitor to 60V in max 0,75s. by using the formula above:

$$R=\frac{t}{-ln (\frac{Vc}{V})\times C}= \frac{0,75}{-ln (\frac{60}{360})\times 1,32u}=317Kohm$$

I choose 220K
Remember to calculate the power in the resister, when it’s not discharging:

$$P=U*I=\frac{U^{2}}{R} =\frac{230^{2}}{220Kohm}=0,24W$$

So use at least a 1W resistor.