your calcul is good but if you are after a full bridge rectifier (diode bridge I think  ?), your frequency is the double of the input, so here 120Hz.

Moreover, there is the drop voltage of diodes, so with a full bridge, it's roughly 1.2V => Vp = 9-1.2 = 7.8V

Notice that a ripple of 10% is very good. Usually we add a regulator after the capacitor to have a better constant voltage supply.

Your calculations are good but if you are using a full bridge rectifier (diode bridge, I think?), your frequency is the double of the input, so here 120Hz.

Moreover, there is the voltage drop of the diodes, so with a full bridge it's roughly 1.2V. Hence Vp = 9-1.2 = 7.8V.

Notice that a ripple of 10% is very good. Usually we add a regulator after the capacitor to have a better constant voltage supply.

your calcul is good but if you are after a full bridge rectifier (diode bridge I think ?), your frequency is the double of the input, so here 120Hz.

Moreover, there is the drop voltage of diodes, so with a full bridge, it's roughly 1.2V => Vp = 9-1.2 = 7.8V

your calcul is good but if you are after a full bridge rectifier (diode bridge I think ?), your frequency is the double of the input, so here 120Hz.

Moreover, there is the drop voltage of diodes, so with a full bridge, it's roughly 1.2V => Vp = 9-1.2 = 7.8V

Notice that a ripple of 10% is very good. Usually we add a regulator after the capacitor to have a better constant voltage supply.