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My power supply requires 2Amp' load at 7.5 volt..Transformer--230V-->9V or 12V AC---with 27VA...
how to calculate value of IL,ΔV and k for 50 Hz.
If you are using either a full-wave bridge rectifier, or a Center-Tapped transformer with two diodes, the ripple frequency is twice the line freq, hence 100Hz. That's why the formula is for either 100Hz or 120Hz. No one uses a half-wave rectifier...
The key point is that C1 must be big enough so that its voltage doesn't sag below the sum of the supply's output voltage plus the regulator's dropout voltage. C1 gets a pulse of current from the rectifier twice per 50Hz cycle, i.e. every 10msec. It charges to a peak determined by the transformer secondary voltage minus some IR drop and the forward drop of either one or two rectifier diodes. For the next 9msec or so, it is discharged by the load current on the supply. Charge is Q=C×ΔV= i×Δt. Rearranging gives C=i×Δt/ΔV.
You know i (supply load current) and Δt (about 9msec at 50Hz), so if you know ΔV (how much C1 can sag) you can solve for C.
Guessing that the dropout voltage of your reg is about 2V, and you want 7.5V out, that means that the minimum input voltage to the reg is 9.5V. If you start with a 12Vrms tranny, the capacitor will charge to 1.4*12-2V (peak is 1.414×rms, 2V drop in two diodes)= 15V. ΔV=15-9.5=5.5
Finally, C=2×0.009/5.5 = 0.0033 FARADs!!! = 3300uF
Left as an exercise for the student, figure C1 for a 9V secondary on your tranny.