Hi Sonam,
Basically as the frequency of the current flow through the capacitor decrease, the voltage drop across it increases, and it is dependend on the capacitance value. I think Mr. Mohankumar explained all the details clearly using design equations in the article which you are talking about.
In the equation I = V / X
find out X, you know the required volatge drop V and the required current I. X can be found uisng another formula X = 1 / (2 ΒΆ fC ).