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GATE 2015 EE - SET 2 - Question 32

Solution :

Continuous condcution , the average inductor voltage ,V_{L} = 0

V_{L} = V_{P}\frac{T}{T_{s}}

V_{L} = 15 * \frac{T_{ON}}{T_{s}}+(-45)*\frac{T_{OFF}}{T_{s}}= 0

V_{L} = 15 * \frac{T_{ON}}{T_{s}}-45*\frac{T_{OFF}}{T_{s}}= 0

15 T_{ON} - 45 T_{OFF} = 0

15 T_{ON} †= †45 T_{OFF}

\frac{T_{OFF}}{T_{ON}} = \frac{1}{3}
\frac{T_{ON}}{T_{OFF}} =  \frac{3}{1}

\delta = \frac{T_{ON}}{T_{ON}+T_{OFF}} = \frac{3}{1+3} = \frac{3}{4} = 0.75

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