$V_{L}$(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 …