GATE 2026 EE Q.48: solution and answer

Official answer: CSource: Official answer key, GATE 2026 EE (IIT Guwahati)

1. Read the waveform: the current rises for $T_{on}=10\mu \text{s}$ and falls for $T_{off}=10\mu \text{s}$, so $T=20\mu \text{s}$ and the duty ratio is $D=\frac{T_{on}}{T}=0.5$. The current just reaches $0\text{A}$ at the end of each cycle, so the converter is at the boundary of continuous and discontinuous conduction. The peak current is $I_{pk}=6\text{A}$.

2. The diode conducts only while S is off, as a ramp falling from $6\text{A}$ to $0\text{A}$. Its average current over the full cycle is $$I_{D,avg}=\frac{1}{T}\cdot \frac{1}{2}\cdot I_{pk}\cdot T_{off}=\frac{1}{2}\times 6\times \frac{10}{20}=1.5\text{A}.$$

3. With negligible output ripple, the average capacitor current over a cycle is zero. So the average diode current equals the load current: $I_o=1.5\text{A}$.

4. Output voltage: $V_o=I_{o}R=1.5\times 10=15\text{V}$.

5. Find the input voltage from the inductor. While S is on, the inductor voltage is $V_{dc}$, so $V_{dc}{T}_{on}=L{I}_{pk}$. While S is off, the inductor voltage is $V_{dc}-V_o$, and the current falls from $I_{pk}$ to $0$, so $(V_o-V_{dc})T_{off}=L{I}_{pk}$. Equating the two gives $V_{dc}\cdot 10=(V_o-V_{dc})\cdot 10$, so $$V_{dc}=\frac{V_o}{2}=\frac{V_o}{1+T_{on}/T_{off}}=7.5\text{V}.$$ This agrees with the ideal boost relation $V_o=\frac{V_{dc}}{1-D}$ at $D=0.5$.

6. Check by power balance: the average inductor (input) current is $\frac{6}{2}=3\text{A}$ (the triangle's mean), so $P_{in}=7.5\times 3=22.5\text{W}$. The output power is $P_{out}=\frac{V_o^2}{R}=\frac{15^2}{10}=22.5\text{W}$. They agree, so $V_{dc}=7.5\text{V}$, option (C).