GATE 2025 EE Q.34: solution and answer

Official answer: 90Source: Official answer key, GATE 2025 EE (IIT Roorkee)

1. The MOSFET is on for the duty ratio $D=0.4$ and off for $1-D=0.6$ of each switching cycle. The source is an ideal current source, so its current is $I=10\text{A}$ at every instant. Let $v_s$ be the voltage across it, which is the voltage of the top node.

2. MOSFET ON: the switch connects the top node to the bottom rail, so $v_s=0$. The series diode has its cathode at $V=15\text{V}$ and its anode at $0\text{V}$, so it is reverse biased. No current reaches $V$, and the source delivers $p=v_{s}I=0$.

3. MOSFET OFF: the current $I$ has no path through the switch, so it flows through the series diode into $V$. The diode conducts and, being ideal, drops no voltage, so $v_s=V=15\text{V}$. The source delivers $p=15\times 10=150\text{W}$ during this interval.

4. Average voltage across the current source over one cycle: $$V_{s,\text{avg}}=D\cdot 0+(1-D)V=0.6\times 15=9\text{V}.$$

5. Since $I$ is constant, the average power delivered by the source is $$P=I{V}_{s,\text{avg}}=10\times 9=90\text{W}.$$ This agrees with $(1-D)\times 150=0.6\times 150=90\text{W}$.

6. Check: the average current into $V$ is $(1-D)I=0.6\times 10=6\text{A}$, so the power absorbed by $V$ is $15\times 6=90\text{W}$. With ideal devices there is no loss, so this equals the power from the source.