$V_{L}$(dc)=15 V

$I_{L}$(dc)=100 mA

r=0.01%=0.0001

The load resistance $R_{L}=\frac{V_{L}(dc)}{I_{L}(dc)}$

i)For $\pi$ section filter

$R_{L}=\frac{15}{100\times 10-3}=150\Omega$ ...............................(i)

ii) r=$\frac{5701}{C_{1}C_{2}R_{L}L_{1}}$

If $C_{1}=C_{2}=C$, then

r=$\frac{5701}{C^{2}R_{L}L_{1}}$

$C^{2}L_{i}=\frac{5701}{rR_{L}}=\frac{5701}{0.0001\times 150}$=380,066 ..............(ii)

We can choose commercially available values of inductor and then can calculate C.LEt us select $L_{1}$=20 H having DC-resistance 375$\Omega$

$C^{2}=\frac{3.80.066}{20H}$=190033

C=137.85$\mu f$

We can choose 170 $\mu f$ commercially available capacitor

iii)The dc voltage drop in the inductor will be

$375\times 100 mA$

=375 V ......................(iv)

Thus dc voltage across the capacitor will be

15+37.5V=52.5 V .....................(v)

The output DC voltage $V_{dc}=V_{m}-\frac{I_{dc}}{4fC_{1}}$

$V_{m}=V_{dc}+\frac{I_{dc}}{4fC_{1}}$

525+$\frac{100 mA}{4\times 50\times 170 \mu f}$=55.44 V ..............(vi)

The secondary rms voltage is

$V_{rms=\frac{55.44}{\sqrt 2}}$=39.2 V .........................(vii)

Hence, a transformer with rating 40-0-40,100 mA can be selected

iv) Diode selection : PIV rating of diode=2$V_{m}$

$2\times55.44$

=110.88V

Select the diode IN 4003 having PIV 200 V and current rating 1 A

v) To maintain good regulation if load is removed, we have to use bleeder resistance parallel to resistance $R_{L}$ So that,

$R_{b}\lt 3\omega L$ $\ \ \ \ \ $ $R_{b}\lt942L$

$R_{b}\lt 942\times 20$

$R_{b}\lt 18840$

Select $R_{b}=15 k \Omega$