Answer :

24

Given:

Let's consider,

V₀ = Load voltage

I₀ = Load current

V_dc = Supply voltage

Duty cycle = D

Duty cycle (D) = 0.75

$V_0=V_{dc}\left[\frac{D}{1-D}\right]=20[\frac{0.75}{1-0.75}]=60V$

$I_0=\frac{V_0}{R}=\frac{60}{10}=6A$

$(I_L{)}_{avg}=\frac{I_0}{1-D}=\frac{6}{1-0.75}=24A$