gate 2017 EE(Set-B) q39-solution

Given is Buckboost converter
$$V_0=\frac{DV_s}{1-D}$$
Given$$V_s=50V,D=0.6,V_0=75V$$
$$\frac{V_0}{V_s}=\frac{I_s}{I_o}=\frac{D}{1-D}=\frac{0.6}{1-0.6}=1.5$$
$$I_0=\frac{V_0}{R}=\frac{75}{5}=15A$$
$$I_s=\frac{D}{1-D}.I_0=\frac{3}{2}\times 15=22.5A$$
Since capacitor is very large, $$i_c=0$$
$$i_{L avg}=i_{s avg}+i_{o avg }$$
$$I_L=I_s+I_o=22.5+15=37.5A$$
$$\Delta I_L=\frac{DV_s}{f_L}=\frac{0.6\times 50}{10,000(0.6\times 10^{-3})}=5A$$
therefore $$i_{L peak}=I_L+\frac{\Delta I_L}{2}=37.5+\frac{5}{2}=40A$$
Peak current drawn from source is 40A