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GATE 2015 EE - SET 2 - Question 31

Solution :

D = 0.6 , V_{0} = 36 V , f = 100kHz = 100 * 10^{3}Hz

I_{0} = \frac{DV_{in}}{R}

Inductor ripple current , \Delta I_{L}=\frac{D(1-DV_{in})}{2fL}

\Delta I_{L}=I_{0}

\frac{D(1-D)V_{in}}{2fL} = \frac{DV_{in}}{R}

R = \frac{2fL}{1-D}  \Rightarrow \frac{2*100*10^{3}*5*10^{-3}}{1-0.6}

R = \frac{1000}{0.4} = 2500 Ω

 

 

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