i) The transformer secondary voltage

$V_{m}=\frac{\pi V_{dc}}{2}=\frac{\pi \times 50}{2}$=78.53 .......................(i)

Therefore the secondary voltage =$2 V_{m}$=157.07 V

ii)Transforming rating

The rms secondary voltage =$\frac{2V_{m}}{\sqrt 2}=\frac{157.07}{\sqrt 2}$=111.06

Allowing diode voltage drop of 0.7 V

The transformer secondary voltage =111.06+0.7=111.76V

Select center tapped transformer secondary having voltage of 55-0-55 V at 4.9 mA .........................(ii)

iii) Selection of diode

The average current through diode is calculated as

$I_{dc}=\frac{2I_{m}}{\pi}=\frac{2}{\pi}\times \frac{V_{m}}{R_{L}}$

$I_{dc}=\frac{2\times 78.53}{\pi \times 10\times 10^{3}}$=49. mA

The diode will be selected with the average of dc current greater than 25% of ull load current

Average current through each diode

$(I_{dc}/2)+25%of I_{dc}$

=2.45 mA+1.23=3.68 mA

Select diode having PIV =2$V_{m}=2\times 78.03=$156.6=157 V and average current of 2.45 mA

Hence select diode IN4003 which has

PIV= 200 V, $I_{avg}=750 mA and V_{F}=1.1 V$ .............................(iii)