Inrush current calculation in capacitive circuite.............?
If you assume that AC is 240V,RMS and can be turned ON instantaneously at any angle, that C2 is completely discharged before AC power-up, and that the the voltage drop across each of the diodes in the bridge is 1 volt, then if the mains  are enerizgized  at at either  \$90 ^{\circ}\$ degrees or \$270^{\circ}\$  the instantaneous initial current out of the mains, on power-up  will be:

$$ I = \frac{340V - 2V}{100\Omega} = 3.38 amperes $$